INSTRUCTOR’S SOLUTIONS
MANUAL
D IFFERENTIAL E QUATIONS AND
L INEAR ALGEBRA
FOURTH EDITION
Stephen W. Goode
and
Scott A. Annin
California State University, Fullerton
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The author and publisher of this book have used their best efforts in preparing this book. These efforts include the
development, research, and testing of the theories and programs to determine their effectiveness. The author and
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damages in connection with, or arising out of, the furnishing, performance, or use of these programs.
Reproduced by Pearson from electronic files supplied by the author.
Copyright © 2017, 2011, 2005 Pearson Education, Inc.
Publishing as Pearson, 501 Boylston Street, Boston, MA 02116.
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any
form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written
permission of the publisher. Printed in the United States of America.
ISBN-13: 978-0-321-98551-4
ISBN-10: 0-321-98551-6
www.pearsonhighered.com
Contents
Chapter 1
1
Chapter 2
123
Chapter 3
207
Chapter 4
256
Chapter 5
348
Chapter 6
390
Chapter 7
442
Chapter 8
534
Chapter 9
631
Chapter 10
762
Chapter 11
861
Appendix A
979
Appendix B
982
Appendix C
987
1
Chapter 1 Solutions
Solutions to Section 1.1
True-False Review:
(a): FALSE. A derivative must involve some derivative of the function y = f (x), not necessarily the first
derivative.
(b): FALSE. The order of a differential equation is the order of the highest, not the lowest, derivative
appearing in the differential equation.
(c): FALSE. This differential equation has order two, since the highest order derivative that appears in the
equation is the second order expression y .
(d): FALSE. The carrying capacity refers to the maximum population size that the environment can
support in the long run; it is not related to the initial population in any way.
(e): TRUE. The value y(0) is called an initial condition to the differential equation for y(t).
(f ): TRUE. According to Newton’s Law of Cooling, the rate of cooling is proportional to the difference
between the object’s temperature and the medium’s temperature. Since that difference is greater for the
object at 100◦ F than the object at 90◦ F , the object whose temperature is 100◦ F has a greater rate of
cooling.
(g): FALSE. The temperature of the object is given by T (t) = Tm + ce−kt , where Tm is the temperature
of the medium, and c and k are constants. Since e−kt = 0, we see that T (t) = Tm for all times t. The
temperature of the object approaches the temperature of the surrounding medium, but never equals it.
(h): TRUE. Since the temperature of the coffee is falling, the temperature difference between the coffee
and the room is higher initially, during the first hour, than it is later, when the temperature of the coffee
has already decreased.
(i): FALSE. The slopes of the two curves are negative reciprocals of each other.
(j): TRUE. If the original family of parallel lines have slopes k for k = 0, then the family of orthogonal trajectories are parallel lines with slope − k1 . If the original family of parallel lines are vertical (resp. horizontal),
then the family of orthogonal trajectories are horizontal (resp. vertical) parallel lines.
(k): FALSE. The family of orthogonal trajectories for a family of circles centered at the origin is the family
of lines passing through the origin.
(l): TRUE. If v(t) denotes the velocity of the object at time t and a(t) denotes the velocity of the object
at time t, then we have a(t) = v (t), which is a differential equation for the unknown function v(t).
(m): FALSE. The restoring force is directed in the direction opposite to the displacement from the equilibrium position.
(n): TRUE. The allometric relationship B = B0 m3/4 , where B0 is a constant, relates the metabolic rate
and total body mass for any species.
Problems:
1. The order is 2.
2. The order is 1.
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3. The order is 3.
4. The order is 2.
5. We compute the first three derivatives of y(t) = ln t:
1
d2 y
= − 2,
dt2
t
dy
1
= ,
dt
t
Therefore,
2
dy
dt
3
=
d3 y
2
= 3.
dt3
t
2
d3 y
= 3,
3
t
dt
as required.
6. We compute the first two derivatives of y(x) = x/(x + 1):
dy
1
=
dx
(x + 1)2
d2 y
2
=−
.
2
dx
(x + 1)3
and
Then
y+
d2 y
x
x3 + 2x2 + x − 2
(x + 1) + (x3 + 2x2 − 3)
1
x3 + 2x2 − 3
2
=
=
=
=
+
,
−
2
3
3
3
2
dx
x + 1 (x + 1)
(x + 1)
(x + 1)
(x + 1)
(1 + x)3
as required.
7. We compute the first two derivatives of y(x) = ex sin x:
dy
= ex (sin x + cos x)
dx
Then
2y cot x −
and
d2 y
= 2ex cos x.
dx2
d2 y
= 2(ex sin x) cot x − 2ex cos x = 0,
dx2
as required.
dT
d
= −k =⇒ (ln |T − Tm |) = −k. The preceding equation can be integrated directly to
dt
dt
yield ln |T − Tm | = −kt + c1 . Exponentiating both sides of this equation gives |T − Tm | = e−kt+c1 , which
can be written as
T − Tm = ce−kt ,
8. (T − Tm )−1
where c = ±ec1 . Rearranging yields T (t) = Tm + ce−kt .
9. After 4 p.m. In the first two hours after noon, the water temperature increased from 50◦ F to 55◦
F, an increase of five degrees. Because the temperature of the water has grown closer to the ambient air
temperature, the temperature difference |T − Tm | is smaller, and thus, the rate of change of the temperature
of the water grows smaller, according to Newton’s Law of Cooling. Thus, it will take longer for the water
temperature to increase another five degrees. Therefore, the water temperature will reach 60◦ F more than
two hours later than 2 p.m., or after 4 p.m.
10. The object temperature cools a total of 40◦ F during the 40 minutes, but according to Newton’s Law of
Cooling, it cools faster in the beginning (since |T − Tm | is greater at first). Thus, the object cooled half-way
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from 70◦ F to 30◦ F in less than half the total cooling time. Therefore, it took less than 20 minutes for the
object to reach 50◦ F.
11. The given family of curves satisfies: x2 + 9y 2 = c =⇒ 2x + 18y
dy
dy
x
= 0 =⇒
=− .
dx
dx
9y
Orthogonal trajectories satisfy:
dy
9y
1 dy
9
d
9
=
=⇒
= =⇒
(ln |y|) = =⇒ ln |y| = 9 ln |x| + c1 =⇒ y = kx9 , where k = ±ec1
dx
x
y dx
x
dx
x
.
y(x)
0.8
0.4
x
-1.5
-1.0
0.5
-0.5
1.0
1.5
-0.4
-0.8
Figure 0.0.1: Figure for Problem 11
y
. Hence,
x2
y 2y
dy
= 2cx = c 2 x =
.
dx
x
x
12. Given family of curves satisfies: y = cx2 =⇒ c =
Orthogonal trajectories satisfy:
1
x
dy
d 2
dy
=−
=⇒ 2y
= −x =⇒
(y ) = −x =⇒ y 2 = − x2 + c1 =⇒ 2y 2 + x2 = c2 ,
dx
2y
dx
dx
2
where c2 = 2c1 .
c
dy
dy
y
13. Given a family of curves satisfies: y = =⇒ x
+ y = 0 =⇒
=− .
x
dx
dx
x
Orthogonal trajectories satisfy:
1
1
dy
x
dy
d 1 2
= =⇒ y
= x =⇒
y = x =⇒ y 2 = x2 + c1 =⇒ y 2 − x2 = c2 , where c2 = 2c1 .
dx
y
dx
dx 2
2
2
y
. Hence,
x5
y 5y
dy
= 5cx4 = 5 5 x4 =
.
dx
x
x
14. The given family of curves satisfies: y = cx5 =⇒ c =
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y(x)
2.0
1.6
1.2
0.8
0.4
x
-2
-1
1
-0.4
2
-0.8
-1.2
-1.6
-2.0
Figure 0.0.2: Figure for Problem 12
y(x)
4
2
x
-2
-4
4
2
-2
-4
Figure 0.0.3: Figure for Problem 13
Orthogonal trajectories satisfy:
dy
x
dy
d
=−
=⇒ 5y
= −x =⇒
dx
5y
dx
dx
5 2
y
2
= −x =⇒
5 2
1
y = − x2 + c1 =⇒ 5y 2 + x2 = c2 ,
2
2
where c2 = 2c1 .
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y(x)
0.8
0.4
x
-1.5
1.0
-1.0
1.5
-0.4
-0.8
Figure 0.0.4: Figure for Problem 14
dy
= cex = y. Orthogonal trajectories satisfy:
dx
dy
1
1
dy
d 1 2
= − =⇒ y
= −1 =⇒
y = −1 =⇒ y 2 = −x + c1 =⇒ y 2 = −2x + c2 .
dx
y
dx
dx 2
2
15. Given family of curves satisfies: y = cex =⇒
y(x)
2
1
x
-1
1
-1
-2
Figure 0.0.5: Figure for Problem 15
16. Given family of curves satisfies: y 2 = 2x + c =⇒
dy
1
= .Orthogonal trajectories satisfy:
dx
y
dy
d
dy
= −y =⇒ y −1
= −1 =⇒
(ln |y|) = −1 =⇒ ln |y| = −x + c1 =⇒ y = c2 e−x .
dx
dx
dx
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y(x)
4
3
2
1
x
1
-1
2
3
4
-1
-2
-3
-4
Figure 0.0.6: Figure for Problem 16
dy
y
dy
my
= cmxm−1 , but c = m so
=
. Orthogonal trajectories satisfy:
dx
x
dx
x
x
1 2
1
x
dy
x
d 1 2
1
dy
=−
=⇒ y
= − =⇒
y = − =⇒ y 2 = −
x + c1 =⇒ y 2 = − x2 + c2 .
dx
my
dx
m
dx 2
m
2
2m
m
17. y = cxm =⇒
dy
18. y = mx + c =⇒
= m.
dx
Orthogonal trajectories satisfy:
1
1
dy
= − =⇒ y = − x + c1 .
dx
m
m
dy
dy
m
= m =⇒
=
.
dx
dx
2y
Orthogonal trajectories satisfy:
19. y 2 = mx + c =⇒ 2y
2x
dy
2y
2
d
2
2
dy
=−
=⇒ y −1
= − =⇒
(ln |y|) = − =⇒ ln |y| = − x + c1 =⇒ y = c2 e− m .
dx
m
dx
m
dx
m
m
dy
dy
mx
+ 2mx = 0 =⇒
=−
.
dx
dx
y
Orthogonal trajectories satisfy:
20. y 2 + mx2 = c =⇒ 2y
dy
y
1
d
1
dy
=
=⇒ y −1
=
=⇒
(ln |y|) =
=⇒ m ln |y| = ln |x| + c1 =⇒ y m = c2 x.
dx
mx
dx
mx
dx
mx
21. The given family of curves satisfies: x2 + y 2 = 2cx =⇒ c =
2x + 2y
x2 + y 2
. Hence,
2x
dy
x2 + y 2
= 2c =
.
dx
x
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Therefore,
2y
x2 + y 2
y 2 − x2
dy
=
− 2x =
,
dx
x
x
so that
dy
y 2 − x2
=
.
dx
2xy
Orthogonal trajectories satisfy:
2xy
dy
2xy
= 2
.
=− 2
dx
y − x2
x − y2
22. u = x2 + 2y 2 =⇒ 0 = 2x + 4y
dy
dy
x
=⇒
=− .
dx
dx
2y
Orthogonal trajectories satisfy:
dy
dy
2y
2
d
2
=
=⇒ y −1
= =⇒
(ln |y|) = =⇒ ln |y| = 2 ln |x| + c1 =⇒ y = c2 x2 .
dx
x
dx
x
dx
x
y(x)
2.0
1.6
1.2
0.8
0.4
x
-2
-1
-0.4
1
2
-0.8
-1.2
-1.6
-2.0
Figure 0.0.7: Figure for Problem 22
23. m1 = tan (a1 ) = tan (a2 − a) =
2
tan (a2 ) − tan (a)
m2 − tan (a)
=
.
1 + tan (a2 ) tan (a)
1 + m2 tan (a)
2
gt
24. ddt2y = g =⇒ dy
dt = gt + c1 =⇒ y(t) = 2 + c1 t + c2 . Now impose the initial conditions. y(0) = 0 =⇒
2
c2 = 0. dy
is: y(t) = gt2 . The object hits the
dt (0) =⇒ c1 = 0. Hence, the solution to the initial-value problem
2
gt
ground at time, t0 , when y(t0 ) = 100. Hence 100 = 20 , so that t0 = 200
g ≈ 4.52 s, where we have taken
g = 9.8 ms−2 .
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d2 y
= g, we integrate twice to obtain the general equations for the velocity and the position of
dt2
dy
1
the ball, respectively:
= gt + c and y(t) = gt2 + ct + d, where c, d are constants of integration. Setting
dt
2
y = 0 to be at the top of the boy’s head (and positive direction downward), we know that y(0) = 0. Since the
object hits the ground 8 seconds later, we have that y(8) = 5 (since the ground lies at the position y = 5).
5 − 32g
From the values of y(0) and y(8), we find that d = 0 and 5 = 32g + 8c. Therefore, c =
.
8
(a). The ball reaches its maximum height at the moment when y (t) = 0. That is, gt + c = 0. Therefore,
25. From
t=−
c
32g − 5
=
≈ 3.98 s.
g
8g
(b). To find the maximum height of the tennis ball, we compute
y(3.98) ≈ −253.51 feet.
So the ball is 253.51 feet above the top of the boy’s head, which is 258.51 feet above the ground.
d2 y
= g, we integrate twice to obtain the general equations for the velocity and the position of
dt2
dy
1
the rocket, respectively:
= gt + c and y(t) = gt2 + ct + d, where c, d are constants of integration. Setting
dt
2
y = 0 to be at ground level, we know that y(0) = 0. Thus, d = 0.
26. From
(a). The rocket reaches maximum height at the moment when y (t) = 0. That is, gt + c = 0. Therefore, the
c
time that the rocket achieves its maximum height is t = − . At this time, y(t) = −90 (the negative sign
g
accounts for the fact that the positive direction is chosen to be downward). Hence,
2
1
c2
c
c
c
c2
c2
= g −
=
−90 = y −
+c −
−
=− .
g
2
g
g
2g
g
2g
√
Solving this for c, we find that c = ± 180g. However, since c represents the initial velocity of the rocket,
and the
√ initial velocity is negative (relative to the fact that the positive direction is downward), we choose
c = − 180g ≈ −42.02 ms−1 , and thus the initial speed at which the rocket must be launched for optimal
viewing is approximately 42.02 ms−1 .
c
−42.02
(b). The time that the rocket reaches its maximum height is t = − ≈ −
= 4.28 s.
g
9.81
d2 y
= g, we integrate twice to obtain the general equations for the velocity and the position of
dt2
dy
1
the rocket, respectively:
= gt + c and y(t) = gt2 + ct + d, where c, d are constants of integration. Setting
dt
2
y = 0 to be at the level of the platform (with positive direction downward), we know that y(0) = 0. Thus,
d = 0.
27. From
(a). The rocket reaches maximum height at the moment when y (t) = 0. That is, gt + c = 0. Therefore, the
c
time that the rocket achieves its maximum height is t = − . At this time, y(t) = −85 (this is 85 m above
g
the platform, or 90 m above the ground). Hence,
2
1
c2
c
c
c
c2
c2
= g −
=
−85 = y −
+c −
−
=− .
g
2
g
g
2g
g
2g
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√
Solving this for c, we find that c = ± 170g. However, since c represents the initial velocity of the rocket,
and the
√ initial velocity is negative (relative to the fact that the positive direction is downward), we choose
c = − 170g ≈ −40.84 ms−1 , and thus the initial speed at which the rocket must be launched for optimal
viewing is approximately 40.84 ms−1 .
c
−40.84
(b). The time that the rocket reaches its maximum height is t = − ≈ −
= 4.16 s.
g
9.81
28. If y(t) denotes the displacement of the object from its initial position at time t, the motion of the object
can be described by the initial-value problem
dy
d2 y
= g, y(0) = 0,
(0) = −2.
2
dt
dt
dy
gt2
d2 y
= g =⇒
= gt + c1 =⇒ y(t) =
+ c1 t + c2 . Now
2
dt
dt
2
dy
impose the initial conditions. y(0) = 0 =⇒ c2 = 0.
(0) = −2 =⇒ c1 = −2. Hence the solution to the
dt
2
gt
g(10)2
initial-value problem is y(t) =
− 2t. We are given that y(10) = h. Consequently, h =
− 2 · 10 =⇒
2
2
−2
h = 10(5g − 2) ≈ 470 m where we have taken g = 9.8 ms .
We first integrate this differential equation:
29. If y(t) denotes the displacement of the object from its initial position at time t, the motion of the object
can be described by the initial-value problem
d2 y
dy
= g, y(0) = 0,
(0) = v0 .
2
dt
dt
d2 y
dy
gt2
=
g
=⇒
=⇒
y(t)
=
=
gt
+
c
+ c1 t + c2 . Now impose
1
dt2
dt
2
dy
the initial conditions. y(0) = 0 =⇒ c2 = 0.
(0) = v0 =⇒ c1 = v0 . Hence the solution to the initial-value
dt
2
gt
problem is y(t) =
+ v0 t. We are given that y(t0 ) = h. Consequently, h = gt20 + v0 t0 . Solving for v0 yields
2
2h − gt20
v0 =
.
2t0
We first integrate the differential equation:
30. From y(t) = A cos (ωt − φ), we obtain
dy
= −Aω sin (ωt − φ)
dt
Hence,
and
d2 y
= −Aω 2 cos (ωt − φ).
dt2
d2 y
+ ω 2 y = −Aω 2 cos (ωt − φ) + Aω 2 cos (ωt − φ) = 0.
dt2
Substituting y(0) = a, we obtain a = A cos(−φ) = A cos(φ). Also, from dy
dt (0) = 0, we obtain 0 =
−Aω sin(−φ) = Aω sin(φ). Since A = 0 and ω = 0 and |φ| < π, we have φ = 0. It follows that a = A.
dy
d2 y
= −c1 ω 2 cos (ωt) −
= −c1 ω sin (ωt) + c2 ω cos (ωt) =⇒
dt
dt2
d2 y
c2 ω 2 sin (ωt) = −ω 2 [c1 cos (ωt) + c2 cos (ωt)] = −ω 2 y. Consequently,
+ ω 2 y = 0. To determine the
dt2
31. y(t) = c1 cos (ωt) + c2 sin (ωt) =⇒
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amplitude of the motion we write the solution to the differential equation in the equivalent form:
c1
c2
y(t) = c21 + c22 2
cos (ωt) + 2
sin (ωt) .
2
c1 + c2
c1 + c22
We can now define an angle φ by
c1
cos φ = 2
c1 + c22
and
c2
sin φ = 2
.
c1 + c22
Then the expression for the solution to the differential equation is
y(t) = c21 + c22 [cos (ωt) cos φ + sin (ωt) sin φ] = c21 + c22 cos (ωt + φ).
Consequently the motion corresponds to an oscillation with amplitude A =
c21 + c22 .
32. In this problem we have m0 = 3g, M = 2700g, a = 1.5. Substituting these values into Equation (1.1.26)
yields
4
1/4
1
−1.5t/(4(2700)1/4 )
m(t) = 2700 1 − 1 −
e
.
900
Therefore the mass of the heron after 30 days is
m(30) = 2700 1 − 1 −
1
900
1/4
4
1/4
e−45/(4(2700)
)
≈ 1271.18 g.
33. In this problem we have m0 = 8g, M = 280g, a = 0.25. Substituting these values into Equation (1.1.26)
yields
4
1/4
1
−t/(16(280)1/4 )
m(t) = 280 1 − 1 −
e
.
35
We need to find the time, t when the mass of the rat reaches 75% of its fully grown size. Therefore we need
to find t such that
4
1/4
75
1
−t/(16(280)1/4 )
e
.
· 280 = 280 1 − 1 −
100
35
Solving algebraically for t yields
t = 16 · (280)
1/4
· ln
1 − (1/35)
1/4
1 − (75/100))
1/4
Solutions to Section 1.2
True-False Review:
(a): TRUE. This is condition 1 in Definition 1.2.8.
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11
(b): TRUE. This is the content of Theorem 1.2.12.
(c): FALSE. There are solutions to y + y = 0 that do not have the form c1 cos x + 5c2 cos x, such as
y(x) = sin x. Therefore, c1 cos x + 5c2 cos x does not meet the second requirement set forth in Definition
1.2.8 for the general solution.
(d): FALSE. There are solutions to y + y = 0 that do not have the form c1 cos x + 5c1 sin x, such
as y(x) = cos x + sin x. Therefore, c1 cos x + 5c1 sin x does not meet the second requirement set form in
Definition 1.2.8 for the general solution.
(e): TRUE. Since the right-hand side of the differential equation is a function of x only, we can integrate
both sides n times to obtain the formula for the solution y(x).
Problems:
1. Linear.
2. Non-linear, because of the y 2 expression on the right side of the equation.
3. Non-linear, because of the term yy on tthe left side of the equation.
4. Non-linear, because of the expression tan y appearing on the left side of the equation.
5. Linear.
6. Non-linear, because of the expression
1
on the left side of the equation.
y
7. y(x) = c1 e−5x + c2 e5x =⇒ y = −5c1 e−5x + 5c2 e5x =⇒ y = 25c1 e−5x + 25c2 e5x =⇒ y − 25y =
(25c1 e−5x + 25c2 e5x ) − 25(c1 e−5x + c2 e5x ) = 0. Thus y(x) = c1 e−5x + c2 e5x is a solution of the given
differential equation for all x ∈ R.
8. y(x) = c1 cos 2x + c2 sin 2x =⇒ y = −2c1 sin 2x + 2c2 cos 2x =⇒ y = −4c1 cos 2x − 4c2 sin 2x =⇒
y + 4y = (−4c1 cos 2x − 4c2 sin 2x) + 4(c1 cos 2x + c2 sin 2x) = 0. Thus y(x) = c1 cos 2x + c2 sin 2x is a
solution of the given differential equation for all x ∈ R.
9. y(x) = c1 ex + c2 e−2x =⇒ y = c1 ex − 2c2 e−2x =⇒ y = c1 ex + 4c2 e−2x =⇒ y + y − 2y = (c1 ex +
4c2 e−2x ) + (c1 ex − 2c2 e−2x ) − 2(c1 ex + c2 e−2x ) = 0. Thus y(x) = c1 ex + c2 e−2x is a solution of the given
differential equation for all x ∈ R.
1
1
1
= −y 2 . Thus y(x) =
=⇒ y = −
is a solution of the given differential
x+4
(x + 4)2
x+4
equation for x ∈ (−∞, −4) or x ∈ (−4, ∞).
10. y(x) =
√
√
y
c1
11. y(x) = c1 x =⇒ y = √ =
. Thus y(x) = c1 x is a solution of the given differential equation for
2x
2 x
all x ∈ {x : x > 0}.
12. y(x) = c1 e−x sin (2x) =⇒ y = 2c1 e−x cos (2x)−c1 e−x sin (2x) =⇒ y = −3c1 e−x sin (2x)−4c1 e−x cos (2x) =⇒
y + 2y + 5y = −3c1 e−x sin (2x) − 4c1 e−x cos (2x) + 2[2c1 e−x cos (2x) − c1 e−x sin (2x)] + 5[c1 e−x sin (2x)] = 0.
Thus y(x) = c1 e−x sin (2x) is a solution to the given differential equation for all x ∈ R.
13. y(x) = c1 cosh (3x) + c2 sinh (3x) =⇒ y = 3c1 sinh (3x) + 3c2 cosh (3x) =⇒ y = 9c1 cosh (3x) +
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9c2 sinh (3x) =⇒ y − 9y = [9c1 cosh (3x) + 9c2 sinh (3x)] − 9[c1 cosh (3x) + c2 sinh (3x)] = 0. Thus y(x) =
c1 cosh (3x) + c2 sinh (3x) is a solution to the given differential equation for all x ∈ R.
c1
12c1
c2
3c1
c2
12c1
2c2
2c2
+
+
=⇒ y = − 4 − 2 =⇒ y = 5 + 3 =⇒ x2 y + 5xy + 3y = x2
14. y(x) = 3 +
x x
x
x
x
x
x5
x3
3c1
c1
c2
c2
c2
c1
5x − 4 − 2 + 3 3 +
= 0. Thus y(x) = 3 +
is a solution to the given differential equation
x
x
x
x
x
x
for all x ∈ (−∞, 0) or x ∈ (0, ∞).
15. y(x) = c1 x2 ln x =⇒ y = c1 (2x ln x + x) =⇒ y = c1 (2 ln x + 3) =⇒ x2 y − 3xy + 4y = x2 · c1 (2 ln x +
3) − 3x · c1 (2x ln x + x) + 4c1 x2 ln x = c1 x2 [(2 ln x + 3) − 3(2x ln x + 1) + 4 ln x] = 0. Thus y(x) = c1 x2 ln x
is a solution of the given differential equation for all x > 0.
16. y(x) = c1 x2 cos(3 ln x) =⇒ y = c1 [2x cos(3 ln x)−3x sin(3 ln x)] =⇒ y = c1 [−7 cos(3 ln x)−6 sin(3 ln x)] =⇒
x2 y −3xy +13y = x2 ·c1 [−7 cos(3 ln x)−9 sin(3 ln x)]−3x·c1 [2x cos(3 ln x)−3x sin(3 ln x)]+13c1 x2 cos(3 ln x) =
c1 x2 {[−7 cos(3 ln x) − 9 sin(3 ln x)] − 3[2 cos(3 ln x) − 3 sin(3 ln x)] + 13 cos(3 ln x)} = 0. Thus y(x) = c1 x2 cos(3 ln x)
is a solution of the given differential equation for all x > 0.
√
c1
c1
c1
2
2
2
17. y(x) = c1 x + 3x =⇒ y = √ + 6x =⇒ y = − √ + 6 =⇒ 2x y − xy + y = 2x − √ + 6 −
2 x
4 x3
4 x3
√
√
c1
√ + 6x +(c1 x+3x2 ) = 9x2 . Thus y(x) = c1 x+3x2 is a solution to the given differential equation
x
2 x
for all x ∈ {x : x > 0}.
18. y(x) = c1 x2 + c2 x3 − x2 sin x =⇒ y = 2c1 x + 3c2 x2 − x2 cos x − 2x sin x =⇒ y = 2c1 + 6c2 x + x2 sin x −
2x cos x − 2x cos −2 sin x. Substituting these results into the given differential equation yields
x2 y − 4xy + 6y = x2 (2c1 + 6c2 x + x2 sin x − 4x cos x − 2 sin x) − 4x(2c1 x + 3c2 x2 − x2 cos x − 2x sin x)
+ 6(c1 x2 + c2 x3 − x2 sin x)
= 2c1 x2 + 6c2 x3 + x4 sin x − 4x3 cos x − 2x2 sin x − 8c1 x2 − 12c2 x3 + 4x3 cos x + 8x2 sin x
+ 6c1 x2 + 6c2 x3 − 6x2 sin x
= x4 sin x.
Hence, y(x) = c1 x2 + c2 x3 − x2 sin x is a solution to the differential equation for all x ∈ R.
19. y(x) = c1 eax + c2 ebx =⇒ y = ac1 eax + bc2 ebx =⇒ y = a2 c1 eax + b2 c2 ebx . Substituting these results
into the differential equation yields
y − (a + b)y + aby = a2 c1 eax + b2 c2 ebx − (a + b)(ac1 eax + bc2 ebx ) + ab(c1 eax + c2 ebx )
= (a2 c1 − a2 c1 − abc1 + abc1 )eax + (b2 c2 − abc2 − b2 c2 + abc2 )ebx
= 0.
Hence, y(x) = c1 eax + c2 ebx is a solution to the given differential equation for all x ∈ R.
20. y(x) = eax (c1 + c2 x) =⇒ y = eax (c2 ) + aeax (c1 + c2 x) = eax (c2 + ac1 + ac2 x) =⇒ y = eaax (ac2 ) +
aeax (c2 + ac1 + ac2 x) = aeax (2c2 + ac1 + ac2 x). Substituting these into the differential equation yields
y − 2ay + a2 y = aeax (2c2 + ac1 + ac2 x) − 2aeax (c2 + ac1 + ac2 x) + a2 eax (c1 + c2 x)
= aeax (2c2 + ac1 + ac2 x − 2c2 − 2ac1 − 2ac2 x + ac1 + ac2 x)
= 0.
Thus, y(x) = eax (c1 + c2 x) is a solution to the given differential eqaution for all x ∈ R.
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21. y(x) = eax (c1 cos bx + c2 sin bx) so,
y = eax (−bc1 sin bx + bc2 cos bx) + aeax (c1 cos bx + c2 sin bx)
= eax [(bc2 + ac1 ) cos bx + (ac2 − bc1 ) sin bx] so,
y = eax [−b(bc2 + ac1 ) sin bx + b(ac2 + bc1 ) cos bx] + aeax [(bc2 + ac1 ) cos bx + (ac2 + bc1 ) sin bx]
= eax [(a2 c1 − b2 c1 + 2abc2 ) cos bx + (a2 c2 − b2 c2 − abc1 ) sin bx].
Substituting these results into the differential equation yields
y − 2ay + (a2 + b2 )y = (eax [(a2 c1 − b2 c1 + 2abc2 ) cos bx + (a2 c2 − b2 c2 − abc1 ) sin bx])
− 2a(eax [(bc2 + ac1 ) cos bx + (ac2 − bc1 ) sin bx]) + (a2 + b2 )(eax (c1 cos bx + c2 sin bx))
= eax [(a2 c1 − b2 c1 + 2abc2 − 2abc2 − 2a2 c1 + a2 c1 + b2 c1 ) cos bx
2
2
2
2
2
+ (a c2 − b c2 − 2abc1 + 2abc1 − 2a c2 + a c2 + b c2 ) sin bx]
=0
Thus, y(x) = eax (c1 cos bx + c2 sin bx) is a solution to the given differential equation for all x ∈ R.
22. y(x) = erx =⇒ y = rerx =⇒ y = r2 erx . Substituting these results into the given differential equation
yields erx (r2 − r − 6) = 0, so that r must satisfy r2 − r − 6 = 0, or (r − 3)(r + 2) = 0. Consequently r = 3
and r = −2 are the only values of r for which y(x) = erx is a solution to the given differential equation. The
corresponding solutions are y(x) = e3x and y(x) = e−2x .
23. y(x) = erx =⇒ y = rerx =⇒ y = r2 erx . Substituting these results into the given differential equation
yields erx (r2 + 6r + 9) = 0, so that r must satisfy r2 + 6r + 9 = 0, or (r + 3)2 = 0. Consequently r = −3 is
the only value of r for which y(x) = erx is a solution to the given differential equation. The corresponding
solution are y(x) = e−3x .
24. y(x) = xr =⇒ y = rxr−1 =⇒ y = r(r − 1)xr−2 . Substitution into the given differential equation yields
xr [r(r − 1) + r − 1] = 0, so that r must satisfy r2 − 1 = 0. Consequently r = −1 and r = 1 are the only
values of r for which y(x) = xr is a solution to the given differential equation. The corresponding solutions
are y(x) = x−1 and y(x) = x.
25. y(x) = xr =⇒ y = rxr−1 =⇒ y = r(r − 1)xr−2 . Substitution into the given differential equation yields
xr [r(r − 1) + 5r + 4] = 0, so that r must satisfy r2 + 4r + 4 = 0, or equivalently (r + 2)2 = 0. Consequently
r = −2 is the only value of r for which y(x) = xr is a solution to the given differential equation. The
corresponding solution is y(x) = x−2 .
26. y(x) = 21 x(5x2 − 3) = 12 (5x3 − 3x) =⇒ y = 12 (15x2 − 3) =⇒ y = 15x. Substitution into the Legendre
equation with N = 3 yields (1 − x2 )y − 2xy + 12y = (1 − x2 )(15x) + x(15x2 − 3) + 6x(5x2 − 3) = 0.
Consequently the given function is a solution to the Legendre equation with N = 3.
27. y(x) = a0 +a1 x+a2 x2 =⇒ y = a1 +2a2 x =⇒ y = 4a2 . Substitution into the given differential equation
yields (1−x2 )(2a2 )−x(a1 +2a2 x)+4(a0 +a1 x+a2 x2 ) = 0 =⇒ 3a1 x+2a2 +4a0 = 0. For this equation to hold
for all x we require 3a1 = 0, and 2a2 + 4a0 = 0. Consequently a1 = 0, and a2 = −2a0 . The corresponding
solution to the differential equation is y(x) = a0 (1 − 2x2 ). Imposing the normalization condition y(1) = 1
requires that a0 = −1. Hence, the required solution to the differential equation is y(x) = 2x2 − 1.
28. x sin y − ex = c =⇒ x cos y
dy
ex − sin y
dy
+ sin y − ex = 0 =⇒
=
.
dx
dx
x cos y
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29. xy 2 + 2y − x = c =⇒ 2xy
dy
dy
dy
1 − y2
+ y2 + 2
− 1 = 0 =⇒
=
.
dx
dx
dx
2(xy + 1)
dy
dy
1 − yexy
. Given y(1) = 0 =⇒
+ y] − 1 = 0 =⇒ xexy
+ yexy = 1 =⇒
dx
dx
xexy
ln
x
e0(1) − 1 = c =⇒ c = 0. Therefore, exy − x = 0, so that y =
.
x
30. exy + x = c =⇒ exy [x
31. ey/x + xy 2 − x = c =⇒ ey/x
x
dy
−y
dy
dy
x2 (1 − y 2 ) + yey/x
dx
2
+
2xy
−
1
=
0
=⇒
.
+
y
=
x2
dx
dx
x(ey/x + 2x2 y)
dy
dy
cos x − 2xy 2
1
32. x2 y 2 − sin x = c =⇒ 2x2 y
+ 2xy 2 − cos x = 0 =⇒
=
. Since y(π) = , then
dx
dx
2x2 y
π
2
1
1 + sin x
1
− sin π = c =⇒ c = 1. Hence, x2 y 2 − sin x = 1 so that y 2 =
. Since y(π) = , take the
π2
2
π
x
π
√
1 + sin x
branch of y where x < 0 so y(x) =
.
x
33.
dy
= sin x =⇒ y(x) = − cos x + c for all x ∈ R.
dx
34.
dy
= x−2/3 =⇒ y(x) = 3x1/3 + c for all x = 0.
dx
35.
dy
d2 y
= xex =⇒
= xex − ex + c1 =⇒ y(x) = xex − 2ex + c1 x + c2 for all x ∈ R.
dx2
dx
d2 y
= xn , where n is an integer.
dx2
dy
If n = −1 then
= ln |x| + c1 =⇒ y(x) = x ln |x| + c1 x + c2 for all x ∈ (−∞, 0) or x ∈ (0, ∞).
dx
dy
If n = −2 then
= −x−1 + c1 =⇒ y(x) = c1 x + c2 − ln |x| for all x ∈ (−∞, 0) or x ∈ (0, ∞).
dx
dy
xn+2
xn+1
If n = −1 and n = −2 then
=
+ c1 =⇒ y =
+ c1 x + c2 for all x ∈ R.
dx
n+1
(n + 1)(n + 2)
36.
1
1
1
1
dy
= x2 ln x =⇒ y(x) = x3 ln x − x3 + c1 = x3 (3 ln x − 1) + c1 . y(1) = 2 =⇒ 2 = (0 − 1) + c1 =⇒
dx
3
9
9
9
19
19
1 3
1 3
c1 =
. Therefore, y(x) = x (3 ln x − 1) +
=
x (3 ln x − 1) + 19 .
9
9
9
9
37.
dy
d2 y
= cos x =⇒
= sin x + c1 =⇒ y(x) = − cos x + c1 x + c2 .
2
dx dx
Thus, y (0) = 1 =⇒ c1 = 1, and y(0) = 2 =⇒ c2 = 3. Thus, y(x) = 3 + x − cos x.
38.
d2 y
dy
d3 y
= 6x =⇒ 2 = 3x2 + c1 =⇒
= x3 + c1 x + c2 =⇒ y = 14 x4 + 12 c1 x2 + c2 x + c3 .
3
dx dx
dx
Thus, y (0) = 4 =⇒ c1 = 4, and y (0) = −1 =⇒ c2 = −1, and y(0) = 1 =⇒ c3 = 1. Thus, y(x) =
1 4
2
4 x + 2x − x + 1.
39.
40. y = xex =⇒ y = xex − ex + c1 =⇒ y = xex − 2ex + c1 x + c2 .
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Thus, y (0) = 4 =⇒ c1 = 5, and y(0) = 3 =⇒ c2 = 5. Thus, y(x) = xex − 2ex + 5x + 5.
41. Starting with y(x) = c1 ex + c2 e−x , we find that y (x) = c1 ex − c2 e−x and y (x) = c1 ex + c2 e−x . Thus,
y − y = 0, so y(x) = c1 ex + c2 e−x is a solution to the differential equation on (−∞, ∞). Next we establish
that every solution to the differential equation has the form c1 ex + c2 e−x . Suppose that y = f (x) is any
solution to the differential equation. Then according to Theorem 1.2.12, y = f (x) is the unique solution to
the initial-value problem
y − y = 0,
y(0) = f (0),
y (0) = f (0).
However, consider the function
y(x) =
f (0) + f (0) x f (0) − f (0) −x
e +
e .
2
2
(0)
(0)
This is of the form y(x) = c1 ex + c2 e−x , where c1 = f (0)+f
and c2 = f (0)−f
, and therefore solves the
2
2
differential equation y − y = 0. Furthermore, evaluation this function at x = 0 yields
y(0) = f (0)
and
y (0) = f (0).
Consequently, this function solves the initial-value problem above. However, by assumption, y(x) = f (x)
solves the same initial-value problem. Owing to the uniqueness of the solution to this initial-value problem,
it follows that these two solutions are the same:
f (x) = c1 ex + c2 e−x .
Consequently, every solution to the differential equation has the form y(x) = c1 ex + c2 e−x , and therefore
this is the general solution on any interval I.
d2 y
dy
= e−x =⇒
= −e−x + c1 =⇒ y(x) = e−x + c1 x + c2 . Thus, y(0) = 1 =⇒ c2 = 0, and
2
dx
dx
y(1) = 0 =⇒ c1 = − 1e . Hence, y(x) = e−x − 1e x.
42.
dy
d2 y
= −6 − 4 ln x =⇒
= −2x − 4x ln x + c1 =⇒ y(x) = −2x2 ln x + c1 x + c2 . Since, y(1) = 0 =⇒
dx2
dx
2e2
2e2
c1 + c2 = 0, and since, y(e) = 0 =⇒ ec1 + c2 = 2e2 . Solving this system yields c1 =
, c2 = −
.
e−1
e−1
2
2e
Thus, y(x) =
(x − 1) − 2x2 ln x.
e−1
43.
44. y(x) = c1 cos x + c2 sin x
(a). y(0) = 0 =⇒ 0 = c1 (1) + c2 (0) =⇒ c1 = 0. y(π) = 1 =⇒ 1 = c2 (0), which is impossible. No solutions.
(b). y(0) = 0 =⇒ 0 = c1 (1) + c2 (0) =⇒ c1 = 0. y(π) = 0 =⇒ 0 = c2 (0), so c2 can be anything. Infinitely
many solutions.
45-50. Use some kind of technology to define each of the given functions. Then use the technology to
simplify the expression given on the left-hand side of each differential equation and verify that the result
corresponds to the expression on the right-hand side.
51. (a). Use some form of technology to substitute y(x) = a + bx + cx2 + dx3 + ex4 + f x5 where a, b, c, d, e, f
are constants, into the given Legendre equation and set the coefficients of each power of x in the resulting
equation to zero. The result is:
e = 0, 20f + 18d = 0, e + 2c = 0, 3d + 14b = 0, c + 15a = 0.
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9
21
Now solve for the constants to find: a = c = e = 0, d = − 14
3 b, f = − 10 d = 5 b. Consequently the
corresponding solution to the Legendre equation is:
14
21
y(x) = bx 1 − x2 + x4 .
3
5
21
15
Imposing the normalization condition y(1) = 1 requires 1 = b(1 − 14
3 + 5 ) =⇒ b = 8 . Consequently the
required solution is y(x) = 18 x(15 − 70x2 + 63x4 ).
52. (a). J0 (x) =
∞ (−1)k x 2k
1 4
= 1 − 14 x2 + 64
x + ...
2
2
k=0 (k!)
(b). A Maple plot of J(0, x, 4) is given in the accompanying figure.
J(0, x, 4)
1
0.8
0.6
Approximation to the first
positive zero of J0(x)
0.4
0.2
0
1
2
3
4
x
–0.2
Figure 0.0.8: Figure for Problem 52(b)
(c). From this graph, an approximation to the first positive zero of J0 (x) is 2.4. Using the Maple internal
function BesselJZeros gives the approximation 2.404825558.
(c) A Maple plot of the functions J0 (x) and J(0, x, 4) on the interval [0,2] is given in the accompanying
figure. We see that to the printer resolution, these graphs are indistinguishable. On a larger interval, for
example, [0,3], the two graphs would begin to differ dramatically from one another.
(d). By trial and error, we find the smallest value of m to be m = 11. A plot of the functions J(0, x) and
J(0, x, 11) is given in the accompanying figure.
Solutions to Section 1.3
True-False Review:
(a): TRUE. This is precisely the remark after Theorem 1.3.2.
(b): FALSE. For instance, the differential equation in Example 1.3.7 has no equilibrium solutions.
(c): FALSE. This differential equation has equilibrium solutions y(x) = 2 and y(x) = −2.
(d): TRUE. For this differential equation, we have f (x, y) = x2 + y 2 . Therefore, any equation of the form
x2 + y 2 = k is an isocline, by definition.
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J0(x), J(0, x, 4)
1
0.8
0.6
0.4
0 0.2 0.4 0.6 0.8
1
1.2 1.4 1.6 1.8
x
2
Figure 0.0.9: Figure for Problem 52(c)
J(0, x), J(0, x, 11)
1
0.8
0.6
0.4
0.2
0
2
4
6
8
10
x
J(0, x)
–0.2
–0.4
J(0, x, 11)
Figure 0.0.10: Figure for Problem 52(d)
(e): TRUE. Equilibrium solutions are always horizontal lines. These are always parallel to each other.
2
2
+y
(f ): TRUE. The isoclines have the form x 2y
= k, or x2 +y 2 = 2ky, or x2 +(y −k)2 = k 2 , so the statement
is valid.
(g): TRUE. An equilibrium solution is a solution, and two solution curves to the differential equation
dy
dx = f (x, y) do not intersect.
Problems:
1. y = ce2x =⇒ c = ye−2x . Hence,
2. y = ecx =⇒ ln y = cx =⇒ c =
dy
= 2ce2x = 2y.
dx
ln y
y
dy
, x = 0. Hence,
= cecx = ln y, x = 0.
x
dx
x
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3. y = cx2 =⇒ c =
y
dy
2y
y
. Hence,
= 2cx = 2 2 x =
.
2
x
dx
x
x
4. y = cx−1 =⇒ c = xy. Hence,
5. y 2 = cx =⇒ c =
dy
y
= −cx−2 = −(xy)x−2 = − .
dx
x
y2
dy
dy
c
y
. Hence, 2y
= c, so that,
=
=
.
x
dx
dx
2y
2x
x2 + y 2
dy
x2 + y 2
dy
x2 + y 2
= c. Hence, 2x + 2y
= 2c =
, so that, y
=
− x.
2x
dx
x
dx
2x
2
2
dy
y −x
Consequently,
=
.
dx
2xy
6. x2 + y 2 = 2cx =⇒
x2 + y 2
. Differentiating the given
7. (x − c)2 + (y − c)2 = 2c2 =⇒ x2 − 2cx + y 2 − 2cy = 0 =⇒ c =
2(x + y) 2
2
2
x + y 2 dy
dy
x +y
+2 y−
equation yields 2(x − c) + 2(y − c)
= 0, so that 2 x −
= 0, that is
dx
2(x + y)
2(x + y) dx
dy
x2 + 2xy − y 2
.
=− 2
dx
y + 2xy − x2
−2y ± 4y 2 + 4x2
dy
2
2
2
2
8. 2cy = x − c =⇒ c + 2cy − x = 0 =⇒ c =
= 2x,
= −y ± x2 + y 2 . Hence, 2c dx
2
dy
x
x
√ 2 2.
so that dx = c =
−y±
x +y
9. x2 + y 2 = c =⇒ 2x + 2y
dy
dy
x
= 0 =⇒
=− .
dx
dx
y
y(x)
2.0
1.6
1.2
0.8
0.4
x
-2
-1
-0.4
1
2
-0.8
-1.2
-1.6
-2.0
Figure 0.0.11: Figure for Problem 9
dy
y
3y
= 3cx2 = 3 3 x2 =
. The initial condition y(2) = 8 =⇒ 8 = c(2)3 =⇒ c = 1. Thus
dx
x
x
the unique solution to the initial value problem is y = x3 .
10. y = cx3 =⇒
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y(x)
8
(2, 8)
4
x
2
-2
-4
-8
Figure 0.0.12: Figure for Problem 10
dy
dy
y2
dy
= c =⇒ 2y
=
=⇒
= y2x =⇒ 2x · dy − y · dx = 0. The initial condition
dx
dx
x
dx
y(1) = 2 =⇒ c = 4, so that the unique solution to the initial value problem is y 2 = 4x.
11. y 2 = cx =⇒ 2y
y(x)
3
(1, 2)
1
x
-3
-2
1
-1
2
3
-1
-3
Figure 0.0.13: Figure for Problem 11
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12. (x − c)2 + y 2 = c2 =⇒ x2 − 2cx + c2 + y 2 = c2 , so that
x2 − 2cx + y 2 = 0.
(0.0.1)
dy
= 0.
dx
(0.0.2)
Differentiating with respect to x yields
2x − 2c + 2y
But from (0.0.1), c =
x2 + y 2
which, when substituted into (0.0.2), yields 2x −
2x
2
dy
x + y2
+ 2y
= 0,
x
dx
dy
y 2 − x2
=
. Imposing the initial condition y(2) = 2 =⇒ from (0.0.1) c = 2, so that the unique
dx
2xy
solution to the initial value problem is y = + x(4 − x).
that is,
y(x)
3
(2, 2)
2
1
x
1
2
3
4
5
6
-1
-2
-3
Figure 0.0.14: Figure for Problem 12
13. Let f (x, y) = x sin (x + y), which is continuous for all x, y ∈ R.
∂f
= x cos (x + y), which is continuous for all x, y ∈ R.
∂y
dy
= x sin (x + y), y(x0 ) = y0 has a unique solution for some interval I ∈ R.
By Theorem 1.3.2,
dx
dy
x
= 2
(y 2 − 9), y(0) = 3.
dx
x +1
x
f (x, y) = 2
(y 2 − 9), which is continuous for all x, y ∈ R.
x +1
∂f
2xy
= 2
, which is continuous for all x, y ∈ R.
∂y
x +1
So the initial value problem stated above has a unique solution on any interval containing (0, 3). By inspection
we see that y(x) = 3 is the unique solution.
14.
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15. The initial-value problem does not necessarily have a unique solution since the hypothesis of the existence
∂f
= 12 xy −1/2
and uniqueness theorem are not satisfied at (0,0). This follows since f (x, y) = xy 1/2 , so that
∂y
which is not continuous at (0, 0).
∂f
= −4xy. Both of these functions are continuous for all (x, y), and
∂y
therefore the hypothesis of the uniqueness and existence theorem are satisfied for any (x0 , y0 ).
16. (a). f (x, y) = −2xy 2 =⇒
(b). y(x) =
1
2x
= −2xy 2 .
=⇒ y = − 2
x2 + c
(x + c)2
(c). y(x) =
1
.
x2 + c
(i). y(0) = 1 =⇒ 1 =
1
1
=⇒ c = 1. Hence, y(x) = 2
. The solution is valid on the interval (−∞, ∞).
c
x +1
y(x)
1.2
0.8
0.4
x
-2
2
Figure 0.0.15: Figure for Problem 16c(i)
1
1
=⇒ c = 0. Hence, y(x) = 2 . This solution is valid on the interval (0, ∞).
1+c
x
1
1
(iii). y(0) = −1 =⇒ −1 = =⇒ c = −1. Hence, y(x) = 2
. This solution is valid on the interval
c
x −1
(−1, 1).
(ii). y(1) = 1 =⇒ 1 =
(d). Since, by inspection, y(x) = 0 satisfies the given initial-value problem, it must be the unique solution
to the initial-value problem.
∂f
= 2y − 1 are continuous at all points (x, y). Consequently, the
∂y
hypothesis of the existence and uniqueness theorem are satisfied by the given initial-value problem for any
x0 , y0 .
17. (a). Both f (x, y) = y(y − 1) and
(b). Equilibrium solutions: y(x) = 0, y(x) = 1.
dy
d2 y
= (2y − 1)
= (2y − 1)y(y − 1). Hence the
dx2
dx
1
solution curves are concave up for 0 < y < 2 , and y > 1, and concave down for y < 0, and 12 < y < 1.
(c). Differentiating the given differential equation yields
(d). The solutions will be bounded provided 0 ≤ y0 ≤ 1.
18. (a). Equilibrium solutions: y(x) = −2, y(x) = 1.
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y(x)
8
6
4
2
x
1
2
3
4
5
6
Figure 0.0.16: Figure for Problem 16c(ii)
y(x)
-1.0
-0.5
0.5
1.0
x
-1
-2
-3
Figure 0.0.17: Figure for Problem 16c(iii)
(b).
dy
= (y + 2)(y − 1) =⇒ the solutions are increasing when y < −2 and y > 1, and the solutions are
dx
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y(x)
2
1
–2
–1
0
1
2
x
–1
–2
Figure 0.0.18: Figure for Problem 17(d)
decreasing when −2 < y < 1.
dy
d2 y
= (2y + 1)
= (2y + 1)(y + 2)(y − 1). Hence
2
dx
dx
1
the solution curves are concave up for −2 < y < − 2 , and y > 1, and concave down for y < −2, and
− 12 < y < 1.
(c). Differentiating the given differential equation yields
19. (a). Equilibrium solution: y(x) = 2.
dy
= (y − 2)2 =⇒ the solutions are increasing when y < 2 and y > 2.
dx
d2 y
dy
(c). Differentiating the given differential equation yields
= 2(y − 2)
= 2(y − 2)3 . Hence the solution
2
dx
dx
curves are concave up for y > 2, and concave down for y < 2.
(b).
20. (a). Equilibrium solutions: y(x) = 0, y(x) = 1.
dy
(b).
= y 2 (y − 1) =⇒ the solutions are increasing when y < 1, and the solutions are decreasing when
dx
y < 1.
d2 y
dy
= (3y 2 − 2y)
= y 3 (3y − 2)(y − 1). Hence the
2
dx
dx
solution curves are concave up for 0 < y < − 23 , and y > 1, and concave down for y < 0, and 2/3 < y < 1.
(c). Differentiating the given differential equation yields
21. (a). Equilibrium solutions: y(x) = 0, y(x) = 1, y(x) = −1.
dy
(b).
= (y + 2)(y − 1) =⇒ the solutions are increasing when −1 < y < 0 and y > 1, and the solutions are
dx
decreasing when y < −1, and 0 < y < 1.
dy
d2 y
= (3y 2 − 1)
= (3y 2 − 1)y(y − 1)(y + 1).
dx2
dx
Hence the solution curves are concave up for −1 < y < − √13 , and 0 < y < √13 , and y > 1, and concave down
for y < −1, and − √13 < y < 0, and √13 < y < 1.
(c). Differentiating the given differential equation yields
22. y = 4x. There are no equilibrium solutions. The slope of the solution curves is positive for x > 0 and
is negative for x < 0. The isoclines are the lines x = k4 .
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Slope of Solution Curve
-4
-2
0
2
4
Equation of Isocline
x = −1
x = −1/2
x=0
x = 1/2
x=1
y(x)
1.5
1
0.5
–1.5
–1
–0.5
0
0.5
1
1.5
x
–0.5
–1
–1.5
Figure 0.0.19: Figure for Problem 22
23. y = x1 . There are no equilibrium solutions. The slope of the solution curves is positive for x > 0 and
increases without bound as x → 0+ . The slope of the curve is negative for x < 0 and decreases without
bound as x → 0− . The isoclines are the lines x1 = k.
Slope of Solution Curve
±4
±2
±1/2
±1/4
±1/10
Equation of Isocline
x = ±1/4
x = ±1/2
x = ±2
x = ±4
x = ±10
24. y = x + y. There are no equilibrium solutions. The slope of the solution curves is positive for y > −x,
and negative for y < −x. The isoclines are the lines y + x = k.
Slope of Solution Curve
−2
−1
0
1
2
Equation of Isocline
y = −x − 2
y = −x − 1
y = −x
y = −x + 1
y = −x + 2
Since the slope of the solution curve along the isocline y = −x − 1 coincides with the slope of the isocline,
it follows that y = −x − 1 is a solution to the differential equation. Differentiating the given differential
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y(x)
2
1
–2
0
–1
1
2
x
–1
–2
Figure 0.0.20: Figure for Problem 23
equation yields: y = 1 + y = 1 + x + y. Hence the solution curves are concave up for y > −x − 1, and
concave down for y < −x − 1. Putting this information together leads to the slope field in the accompanying
figure.
y(x)
3
2
1
-3
-2
1
-1
2
3
x
-1
-2
-3
Figure 0.0.21: Figure for Problem 24
25. y = xy . There are no equilibrium solutions. The slope of the solution curves is zero when x = 0. The
solution has a vertical tangent line at all points along the x-axis (except the origin). Differentiating the
1
1 x2
1
x
differential equation yields: y = − 2 y = − 3 = 3 (y 2 − x2 ). Hence the solution curves are concave
y
y
y
y
y
up for y > 0 and y 2 > x2 ; y < 0 and y 2 < x2 and concave down for y > 0 and y 2 < x2 ; y < 0 and y 2 > x2 .
The isoclines are the lines xy = k.
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Slope of Solution Curve
±2
±1
±1/2
±1/4
±1/10
Equation of Isocline
y = ±x/2
y = ±x
y = ±2x
y = ±4x
y = ±10x
Note that y = ±x are solutions to the differential equation.
y(x)
2
1
–2
–1
0
1
2
x
–1
–2
Figure 0.0.22: Figure for Problem 25
26. y = − 4x
y . Slope is zero when x = 0 (y = 0). The solutions have a vertical tangent line at all points
along the x-axis(except the origin). The isoclines are the lines − 4x
y = k. Some values are given in the table
below.
Slope of Solution Curve
±1
±2
±3
Equation of Isocline
y = ±4x
y = ±2x
y = ±4x/3
4
4
4(y 2 + 4x2 )
4xy 16x2
+ 2 = − − 3 = −
.
y
y
y
y
y
Consequently the solution curves are concave up for y < 0, and concave down for y > 0. Putting this
information together leads to the slope field in the accompanying figure.
Differentiating the given differential equation yields: y = −
27. y = x2 y. Equilibrium solution: y(x) = 0 =⇒ no solution curve can cross the x-axis. Slope: zero
when x = 0 or y = 0. Positive when y > 0 (x = 0), negative when y < 0 (x = 0). Differentiating the given
d2 y
dy
differential equation yields:
= 2xy +x2
= 2xy +x4 y = xy(2+x3 ). So, when y > 0, the solution curves
dx2
dx
are concave up for x ∈ (−∞, (−2)1/3 ), and for x > 0, and are concave down for x ∈ ((−2)1/3 , 0). When
y < 0, the solution curves are concave up for x ∈ ((−2)1/3 , 0), and concave down for x ∈ (−∞, (−2)1/3 ) and
for x > 0. The isoclines are the hyperbolas x2 y = k.
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y(x)
4
3
2
1
x
-2
-1
1
2
Figure 0.0.23: Figure for Problem 26
Slope of Solution Curve
±2
±1
±1/2
±1/4
±1/10
0
Equation of Isocline
y = ±2/x2
y = ±1/x2
y = ±1/(2x)2
y = ±1/(4x)2
y = ±1/(10x)2
y=0
28. y = x2 cos y. The slope is zero when x = 0. There are equilibrium solutions when y = (2k + 1) π2 . The
slope field is best sketched using technology. The accompanying figure gives the slope field for − π2 < y < 3π
2 .
29. y = x2 + y 2 . The slope of the solution curves is zero at the origin, and positive at all the other points.
There are no equilibrium solutions. The isoclines are the circles x2 + y 2 = k.
Slope of Solution Curve
1
2
3
4
5
30.
dT
dt
Equation of Isocline
x = ±1/4
x = ±1/2
x = ±2
x = ±4
x = ±10
1
= − 80
(T − 70). Equilibrium solution: T (t) = 70. The slope of the solution curves is positive for
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y(x)
2
1
–2
0
–1
1
x
2
–1
–2
Figure 0.0.24: Figure for Problem 27
y(x)
5
4
3
2
1
–3
–2
–1
0
1
2
3
x
–1
Figure 0.0.25: Figure for Problem 28
d2 T
1 dT
1
=−
=
(T − 70). Hence the solution curves are concave
dt2
80 dt
6400
1
(T − 70) = k.
up for T > 70, and concave down for T < 70. The isoclines are the horizontal lines − 80
T > 70, and negative for T < 70.
Slope of Solution Curve
−1/4
1/5
0
1/5
1/4
Equation of Isocline
T = 90
T = 86
T = 70
T = 54
T = 50
31. y = −2xy.
32. y =
x sin x
.
1 + y2
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y(x)
2
1
–2
0
–1
1
2
x
–1
–2
Figure 0.0.26: Figure for Problem 29
T(t)
100
80
60
40
20
0
20
40
60
80
100 120 140
t
Figure 0.0.27: Figure for Problem 30
33. y = 3x − y.
34. y = 2x2 sin y.
35. y =
2 + y2
.
3 + 0.5x2
36. y =
1 − y2
.
2 + 0.5x2
37. (a). Slope field for the differential equation y = x−1 (3 sin x − y).
(b). Slope field with solution curves included.
The figure suggests that the solution to the differential equation are unbounded as x → 0+ .
(c). Slope field with solution curve corresponding to the initial condition y( π2 ) = π6 .
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y(x)
2
1
–2
–1
0
1
2
x
–1
–2
Figure 0.0.28: Figure for Problem 31
y(x)
2
1
–2
–1
0
1
2
x
–1
–2
Figure 0.0.29: Figure for Problem 32
y(x)
2
1
–2
–1
0
1
2
x
–1
–2
Figure 0.0.30: Figure for Problem 33
This solution curve is bounded as x → 0+ .
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y(x)
3
2
1
–2
–1
0
1
2
x
–1
–2
–3
Figure 0.0.31: Figure for Problem 34
y(x)
2
1
–2
–1
0
1
2
x
–1
–2
Figure 0.0.32: Figure for Problem 35
y(x)
2
1
–2
–1
0
1
2
x
–1
–2
Figure 0.0.33: Figure for Problem 36
(d). In the accompanying figure we have sketched several solution curves on the interval (0,15].
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y(x)
4
2
0
2
4
6
8
10
x
–2
–4
Figure 0.0.34: Figure for Problem 37(a)
y(x)
4
2
0
2
4
6
8
10
x
–2
–4
Figure 0.0.35: Figure for Problem 37(b)
y(x)
4
2
0
2
4
6
8
10
x
–2
–4
Figure 0.0.36: Figure for Problem 37(c)
The figure suggests that the solution curves approach the x-axis as x → ∞.
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y(x)
4
2
0
2
4
6
8
10
12
14
x
–2
–4
Figure 0.0.37: Figure for Problem 37(d)
dy
y
38. (a). Differentiating the given equation gives
= 2kx = 2 . Hence the differential equation of the
dx
x
x
dy
=− .
orthogonal trajectories is
dx
2y
y(x)
4
2
–4
–2
0
2
4
x
–2
–4
Figure 0.0.38: Figure for Problem 38(a)
(b). The orthogonal trajectories appear to be ellipses. This can be verified by integrating the differential
equation derived in (a).
39. If a > 0, then as illustrated in the following slope field (a = 0.5, b = 1), it appears that limt→∞ i(t) = ab .
If a < 0, then as illustrated in the following slope field (a = −0.5, b = 1) it appears that i(t) diverges as
t → ∞.
If a = 0 and b = 0, then once more i(t) diverges as t → ∞. The accompanying figure shows a representative case when b > 0. Here we see that limt→∞ i(t) = +∞. If b < 0, then limt→∞ i(t) = −∞.
If a = b = 0, then the general solution to the differential equation is i(t) = i0 where i0 is a constant.
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y(x)
4
2
–3
–2
–1
0
1
2
3
x
–2
–4
Figure 0.0.39: Figure for Problem 39 when a > 0
i(t)
4
2
–3
–2
–1
0
1
2
3
t
–2
–4
Figure 0.0.40: Figure for Problem 39 when a < 0
y(x)
4
2
–3
–2
–1
0
1
2
3
x
–2
–4
Figure 0.0.41: Figure for Problem 39 when a = 0
Solutions to Section 1.4
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True-False Review:
dy
1 dy
(a): TRUE. The differential equation dx
= f (x)g(y) can be written g(y)
dx = f (x), which is the proper
form, according to Definition 1.4.1, for a separable differential equation.
(b): TRUE. A separable differential equation is a first-order differential equation, so the general solution
contains one constant. The value of that constant can be determined from an initial condition, as usual.
(c): TRUE. Newton’s Law of Cooling is usually expressed as dT
dt = −k(T − Tm ), and this can be rewritten
as
1
dT
= −k,
T − Tm dt
and this form shows that the equation is separable.
(d): FALSE. The expression x2 + y 2 cannot be separated in the form f (x)g(y), so the equation is not
separable.
(e): FALSE. The expression x sin(xy) cannot be separated in the form f (x)g(y), so the equation is not
separable.
dy
(f ): TRUE. We can write the given equation as e−y dx
= ex , which is the proper form for a separable
equation.
dy
(g): TRUE. We can write the given equation as (1 + y 2 ) dx
= x12 , which is the proper form for a separable
equation.
(h): FALSE. The expression x+4y
4x+y cannot be separated in the form f (x)g(y), so the equation is not
separable.
3
2 2
y
dy
(i): TRUE. We can write x xy+x
= xy, so we can write the given differential equation as y1 dx
= x, which
2 +xy
is the proper form for a separable equation.
Problems:
1. Separating the variables and integrating yields
2
dy
= 2 xdx =⇒ ln |y| = x2 + c1 =⇒ y(x) = cex .
y
2. Separating the variables and integrating yields
dx
1
−2
y dy =
=⇒ y(x) = −
.
x2 + 1
tan−1 x + c
3. Separating the variables and integrating yields
ey dy = e−x dx = 0 =⇒ ey + e−x = c =⇒ y(x) = ln (c − e−x ).
4. Separating the variables and integrating yields
dy
(ln x)−1
=
dx =⇒ y(x) = c ln x.
y
x
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5. Separating the variables and integrating yields
dx
dy
=
=⇒ ln |x − 2| − ln |y| = c1 =⇒ y(x) = c(x − 2).
x−2
y
6. Separating the variables and integrating yields
dy
2x
=
dx =⇒ ln |y − 1| = ln |x2 + 3| + c1 =⇒ y(x) = c(x2 + 3) + 1.
2
y−1
x +3
7. y − x
dy
dy
dy
= 3 − 2x2
=⇒ x(2x − 1)
= (3 − y). Separating the variables and integrating yields
dx
dx
dx
dx
dx
2
dy
=
=⇒ − ln |y − 3| = −
+
dx
−
y−3
x(2x − 1)
x
2x − 1
=⇒ − ln |y − 3| = − ln |x| + ln |2x − 1| + c1
x
cx − 3
=⇒
= c2 =⇒ y(x) =
.
(y − 3)(2x − 1)
2x − 1
sin y
cos x
dy
cos (x − y)
dy
cos x cos y
8.
=
−1 =⇒
=
=⇒
dy =
dx =⇒ − ln | cos y| = ln | sin x|+c1 =⇒
dx
sin x sin y
dx
sin x sin y
cos y
cos y
cos y = c csc x.
dy
x(y 2 − 1)
dy
1
xdx
=
=⇒
=
, y = ±1. Thus,
dx
2(x − 2)(x − 1)
(y + 1)(y − 1)
2 (x − 2)(x − 1)
dy
dy
dx
1
1
dx
1
2
=⇒ − ln |y + 1|+ln |y − 1| = 2 ln |x − 2|−ln |x − 1|+c1
+
=
−
−
2
y+1 2
y−1
2
x−2
x−1
9.
y−1
(x − 2)2
(x − 1) + c(x − 2)2
. By inspection we see that y(x) = 1, and y(x) = −1
=c
=⇒ y(x) =
y+1
x−1
(x − 1) − c(x − 2)2
are solutions of the given differential equation. The former is included in the above solution when c = 0.
dy
dy
16
x2 y − 32
x2
dx =⇒ ln |y − 2| =
10.
+ 2 =⇒
dx =⇒ ln |y − 2| = −
1+ 2
=
=
dx
16 − x2
y−2
16 − x2
x − 16
dx
dx
dx
=⇒ ln |y − 2| = −x + 2 ln |x + 4| −
−x − 16
=⇒ ln |y − 2| = −x − 16 − 18
+ 18
2
x − 16
x+4
x−4
2
x+4
2 ln |x − 4| + c1 =⇒ y(x) = 2 + c
e−x .
x−4
dy
dy
dy
dx
1 1
1
dx =⇒
11. (x−a)(x−b) −(y−c) = 0 =⇒
=
=⇒
=
−
y − c
(x − a)(x − b) y−c
a−b
x−a x−b
dx
1/(a−b)
1/(a−b) 1/(a−b)
x − a
x−b
x−a
(y
−
c)
=
c
=⇒
ln |y − c| = ln c1 =⇒
y
−
c
=
c
=⇒
1
2
x − b
x−a
x−b
1/(a−b)
x−a
.
y(x) = c + c2
x−b
=⇒
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dy
dx
dy
π
=−
=⇒ tan−1 y = tan−1 x + c, but y(0) = 1 so c = .
+ y 2 = −1 =⇒
2
2
dx
1+y
1+x
4
π
1−x
−1
−1
Thus, tan y = tan x + or y(x) =
.
4
1+x
12. (x2 + 1)
dy
dy
2x
dx =⇒ − ln |a − y| = − 12 ln |1 − x2 | + c1 =⇒ y(x) =
13. (1 − x2 )
+ xy = ax =⇒
= − 12 −
dx
a
−
y
1
−
x2
√
√
a + c 1 − x2 , but y(0) = 2a so c = a and therefore, y(x) = a(1 + 1 − x2 ).
sin y
sin x
dy
sin (x + y)
dy
14.
= 1−
=⇒
= − tan x cot y =⇒ −
dy =
dx =⇒ − ln | cos x cos y| = c, but
dx
sin x sin y
dx
cos y
cos x
π
π
y( ) = so c = ln (2). Hence, − ln | cos x cos y| = ln (2) =⇒ y(x) = cos−1 12 sec x .
4
4
dy dy
1
= sin xdx for y = 0. Thus − 2 = − cos x+c. However, we cannot impose the
= y 3 sin x =⇒
dx
y3
2y
initial condition y(0) = 0 on the last equation since it is not defined at y = 0. But, by inspection, y(x) = 0
is a solution to the given differential equation and further, y(0) = 0; thus, the unique solution to the initial
value problem is y(x) = 0.
15.
dy
dy
= 23 dx if y = 1 =⇒ 2(y − 1)1/2 = 23 x + c but y(1) = 1 so
= 23 (y − 1)1/2 =⇒
1/2
dx
(y
−
1)
√
√
c = − 23 =⇒ 2 y − 1 = 23 x − 23 =⇒ y − 1 = 13 (x − 1). This does not contradict the Existence-Uniqueness
theorem because the hypothesis of the theorem is not satisfied when x = 1.
16.
dv
m
= mg − kv 2 =⇒
dv = dt. If we let a = mg
k then the preceding equation can
2
dt
k [(mg/k) − v ]
m
1
be written as
dv = dt which can be integrated directly to obtain
k a2 − v 2
m
a+v
= t + c,
ln
2ak
a−v
17. (a). m
that is, upon exponentiating both sides,
2ak
a+v
= c1 e m t .
a−v
Imposing the initial condition v(0) = 0, yields c = 0 so that
2ak
a+v
= e m t.
a−v
Therefore,
v(t) = a
2akt
e m −1
2akt
e m +1
which can be written in the equivalent form
v(t) = a tanh
gt
.
a
(b). No. As t → ∞, v → a and as t → 0+ , v → 0.
(c)2017 Pearson Education. Inc.
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(c). v(t) = a tanh
gt y(0) = 0 then y(t) =
a
=⇒
dy
a2
=⇒ a tanh gt
dt =⇒ y(t) =
= a tanh gt
ln(cosh ( gt
a
a
a )) + c1 and if
dt
g
a2
ln cosh ( gt
a) .
g
dy
x
= − , y(0) = 12 . Separating
dx
4y
the variables in the differential equation yields 4y −1 dy = −1dx, which can be integrated directly to obtain
x2
2y 2 = − + c. Imposing the initial condition we obtain c = 12 , so that the solution curve has the equation
2
2y 2 = −x2 + 12 , or equivalently, 4y 2 + 2x2 = 1.
18. The required curve is the solution curve to the initial-value problem
dy
19. The required curve is the solution curve to the initial-value problem
= ex−y , y(3) = 1. Separating
dx
the variables in the differential equation yields ey dy = ex dx, which can be integrated directly to obtain
ey = ex + c. Imposing the initial condition we obtain c = e − e3 , so that the solution curve has the equation
ey = ex + e − e3 , or equivalently, y = ln(ex + e − e3 ).
dy
= x2 y 2 , y(−1) = 1. Separating
dx
the variables in the differential equation yields y12 dy = x2 dx, which can be integrated directly to obtain
− y1 = 13 x3 + c. Imposing the initial condition we obtain c = − 23 , so that the solution curve has the equation
3
y = − 1 x31− 2 , or equivalently, y = 2−x
3.
20. The required curve is the solution curve to the initial-value problem
3
3
1
dv = −dt. Integrating we
1 + v2
−1
−1
obtain tan (v) = −t + c. The initial condition v(0) = v0 implies that c = tan (v0 ), so that tan−1 (v) =
−t + tan−1 (v0 ). The object will come to rest if there is time t, at which the velocity is zero. To determine tr ,
we set v = 0 in the previous equation which yields tan−1 (0) = tr +tan−1 (v0 ). Consequently, tr = tan−1 (v0 ).
dv
The object does not remain at rest since we see from the given differential equation that
< 0 at t = tr ,
dt
and so v is decreasing with time. Consequently v passes through zero and becomes negative for t < tr .
21. (a). Separating the variables in the given differential equation yields
dv
dx
dv
dv
(b). From the chain rule we have
=
. Then
= v . Substituting this result into the differential
dt
dt
dx
dx
v
dv
dv = −dx. Integrating we
= −(1 + v 2 ). We now separate the variables:
equation (1.4.22) yields v
dx
1 + v2
2
obtain ln (1 + v ) = −2x + c. Imposing the initial condition v(0) = v0 , x(0) = 0 implies that c = ln (1 + v02 ),
so that ln (1 + v 2 ) = −2x + ln (1 + v02 ). When the object comes to rest the distance travelled by the object
is x = 12 ln (1 + v02 ).
dv
= −kv n =⇒ v −n dv = −kdt.
dt
1
1
n = 1 :=⇒
v 1−n = −kt + c. Imposing the initial condition v(0) + v0 yields c =
v 1−n , so that
1−n
1−n 0
v = [v01−n + (n − 1)kt]1/(1−n) . The object comes to rest in a finite time if there is a positive value of t for
which v = 0.
n = 1 :=⇒ Integratingv −n dv = −kdt and imposing the initial conditions yields v = v0 e−kt , and the object
does not come to rest in a finite amount of time.
dx
(b). If n = 1, 2, then
= [v01−n + (n − 1)kt]1/(1−n) , where x(t) denotes the distanced travelled by the
dt
22. (a).
(c)2017 Pearson Education. Inc.
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1
[v 1−n + (n − 1)kt](2−n)/(1−n) + c. Imposing the initial condition
k(2 − n) 0
1
1
1
x(0) = 0 yields c =
v 2−n , so that x(t) = −
[v 1−n + n(n − 1)kt](2−n)/(1−n) +
v 2−n .
k(2 − n) 0
k(2 − n) o
k(2 − n) 0
2−n
1
For 1 < n < 2, we have
< 0, so that limt→∞ x(t) =
. Hence the maximum distance that the
1−n
k(2 − n)
1
object can travel in a finite time is less than
.
k(2 − n)
v0
If n = 1, then we can integrate to obtain x(t) = (1 − e−kt ), where we have imposed the initial condition
k
v0
x(0) = 0. Consequently, limt→∞ x(t) =
. Thus in this case the maximum distance that the object can
k
v0
travel in a finite time is less than
.
k
object. Consequently, x(t) = −
(c). If n > 2, then x(t) = −
1
1
[v 1−n + n(n − 1)kt](2−n)/(1−n) +
v 2−n is still valid. However,
k(2 − n) o
k(2 − n) 0
2−n
> 0, and so limt→∞ x(t) = +∞. Consequently, there is no limit to the distance that the
1−n
object can travel.
dx
= (v0−1 + kt)−1 , which can
If n = 2, then we return to v = [v01−n + (n − 1)kt]1/(1−n) . In this case
dt
1
be integrated directly to obtain x(t) = ln (1 + v0 kt), where we have imposed the initial condition that
k
x(0) = 0. Once more we see that limt→∞ x(t) = +∞, so that there is no limit to the distance that the object
can travel.
in this case
ρ 1/γ
p
) . Consequently the given differential equation can be written as dp = −gρ0 ( )1/γ dy,
ρ0
p0
gρ0
γp(γ−1)/γ
gρ0 y
or equivalently, p−1/γ dp = − 1/γ dy. This can be integrated directly to obtain
= − 1/γ + c. At
γ−1
p0
p0
the center of the Earth we have p = p0 . Imposing this initial condition on the preceding solution gives
(γ−1)/γ
γp0
. Substituting this value of c into the general solution to the differential equation we find,
c =
γ−1
(γ−1)/γ
(γ − 1)ρ0 gy
(γ − 1)ρ0 gy
(γ−1)/γ
(γ−1)/γ
1−
, so that p = p0 1 −
= p0
.
after some simplification, p
γp0
γp0
23. Solving p = p0 (
dT
dT
dT
= −k(T − Tm ) =⇒
= −k(T − 75) =⇒
= −kdt =⇒ ln |T − 75| = −kt + c1 =⇒ T (t) =
dt
dt
T − 75
−kt
−kt
75 + ce . T (0) = 135 =⇒ c = 60 so T = 75 + 60e . T (1) = 95 =⇒ 95 = 75 + 60e−k =⇒ k = ln 3 =⇒
T (t) = 75 + 60e−t ln 3 . Now if T (t) = 615 then 615 = 75 + 60−t ln 3 =⇒ t = −2h. Thus the object was placed
in the room at 2p.m.
24.
dT
= −k(T − 450) =⇒ T (t) = 450 + Ce−kt .T (0) = 50 =⇒ C = −400 so T (t) = 450 − 400e−kt and
dt
1
T (20) = 150 =⇒ k =
ln 4 ; hence, T (t) = 450 − 400( 34 )t/20 .
20 3
(i) T (40) = 450 − 400( 34 )2 = 225◦ F.
20 ln 4
(ii) T (t) = 350 = 450 − 400( 34 )t/20 =⇒ ( 34 )t/20 = 14 =⇒ t =
≈ 96.4 minutes.
ln(4/3)
25.
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dT
dT
= −k(T − 34) =⇒
= −kdt =⇒ T (t) = 34 + ce−kt . T (0) = 38 =⇒ c = 4 so that
dt
T − 34
T (t) = 34 + 4e−kt .T (1) = 36 =⇒ k = ln 2; hence, T (t) = 34 + 4e−t ln 2 . Now T (t) = 98 =⇒ T (t) =
34 + 4e−kt = 98 =⇒ 2−t = 16 =⇒ t = −4h. Thus T (−4) = 98 and Holmes was right, the time of death was
10 a.m.
26.
27. T (t) = 75 + ce−kt . T (10) = 415 =⇒ 75 + ce−10k = 415 =⇒ 340 = ce−10k and T (20) = 347 =⇒
1
75 + ce−20k = 347 =⇒ 272 = ce−20k . Solving these two equations yields k = 10
ln 54 and c = 425; hence,
4 t/10
T = 75 + 425( 5 )
(a) Furnace temperature: T (0) = 500◦ F.
10 ln 17
≈ 126.96 minutes. Thus the temperature of
(b) If T (t) = 100 then 100 = 75 + 425( 45 )t/10 =⇒ t =
ln 54
the coal was 100◦ F at 6:07 p.m.
dT
dT
dT
= −k(T − 72) =⇒
= −kdt =⇒ T (t) = 72 + ce−kt . Since
= −20, −k(T − 72) = −20 or
dt
T − 72
dt
10
−10/39
10/39
=⇒ c = 78e
; consequently, T (t) = 72 + 78e10(1−t)/39 .
k = 39 . Since T (1) = 150 =⇒ 150 = 72 + ce
28.
(i). Initial temperature of the object: t = 0 =⇒ T (t) = 72 + 78e10/30 ≈ 173◦ F
(ii). Rate of change of the temperature after 10 minutes: T (10) = 72 + 78e−30/13 so after 10 minutes,
dT
dT
10
260 −30/13
≈ 2◦ F per minute.
= − (72 + 78e−30/13 − 72) =⇒
=−
e
dt
39
dt
13
29. Substituting a = 0.5, M = 2000 g, and m0 = 4 g into the initial-value problem (1.4.17) yields
m 1/4 dm
3/4
1−
, m(0) = 4.
= 0.5m
dt
2000
Separating the variables in the preceding differential equation gives
m3/4 1 −
so that
m3/4 1 −
1
m 1/4 dm = 0.5 dt
2000
1
m 1/4 dm = 0.5t + c.
2000
To evaluate the integral on the left-hand-side of the preceding equation, we make the change of variable
m 1/4
1
1 m −3/4
w=
, dw = ·
dm
2000
4 2000 2000
and simplify to obtain
4 · (2000)
1/4
1
dw = 0.5t + c
1−w
which can be integrated directly to obtain
−4 · (2000)
1/4
ln(1 − w) = 0.5t + c.
Exponentiating both sides of the preceding equation, and solving for w yields
1/4
w = 1 − c1 e−0.125t/(2000)
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or equivalently,
m 1/4
2000
Consequently,
1/4
= 1 − c1 e−0.125t/(2000)
.
1/4 4
m(t) = 2000 1 − c1 e−0.125t/(2000)
.
(0.0.3)
Imposing the initial condition m(0) = 4 yields
4 = 2000 (1 − c1 )
so that
c1 = 1 −
1
500
4
1/4
≈ 0.7885.
Inserting this expression for c1 into Equation (0.0.3) gives
1/4 4
m(t) = 2000 1 − 0.7885e−0.125t/(2000)
.
Consequently,
1/4 4
m(100) = 2000 1 − 0.7885e−12.5/(2000)
≈ 1190.5 g.
30. Substituting a = 0.10, M = 0.15 g, and m0 = 0.008 g into the initial-value problem (1.4.17) yields
m 1/4 dm
3/4
1−
, m(0) = 0.008.
= 0.1m
dt
0.15
Separating the variables in the preceding differential equation gives
m3/4 1 −
so that
m3/4 1 −
1
m 1/4 dm = 0.1 dt
0.15
1
m 1/4 dm = 0.1t + c.
0.15
To evaluate the integral on the left-hand-side of the preceding equation, we make the change of variable
w=
m 1/4
0.15
,
dw =
and simplify to obtain
4 · (0.15)
1/4
1
1 m −3/4
dm
·
4 0.15 0.15
1
dw = 0.1t + c
1−w
which can be integrated directly to obtain
−4 · (0.15)
1/4
ln(1 − w) = 0.1t + c.
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Exponentiating both sides of the preceding equation, and solving for w yields
w = 1 − c1 e−0.025t/(0.15)
or equivalently,
m 1/4
0.15
Consequently,
1/4
= 1 − c1 e−0.025t/(0.15)
1/4
.
1/4 4
m(t) = 0.15 1 − c1 e−0.025t/(0.15)
.
(0.0.4)
Imposing the initial condition m(0) = 0.008 yields
0.008 = 0.15 (1 − c1 )
so that
c1 = 1 −
4
75
4
1/4
≈ 0.5194.
Inserting this expression for c1 into Equation (0.0.4) gives
1/4 4
.
m(t) = 0.15 1 − 0.5194e−0.025t/(0.15)
Consequently,
1/4 4
m(30) = 0.15 1 − 0.5194e−0.75/(0.15)
≈ 0.076 g.
The guppy will have reached 90% of its fully grown mass at time t where
1/4 4
.
0.9 · 0.15 = 0.15 1 − 0.5194e−0.025t/(0.15)
Solving algebraically for t yields
t=−
(0.15)1/4
1 − (0.9)1/4
≈ 74.5 days.
ln
0.025
0.5194
31. Since the chemicals A and B combine in the ratio 2:1, the amounts of A and B that are unconverted at
time t are (20 − 23 Q) grams and (20 − 13 Q) grams, respectively. Thus, according to the law of mass action,
the differential equation governing the behavior of Q(t) is
dQ
2
1
dQ
1
= k1 (20 − Q)(20 − Q) =⇒
= k(30 − Q)(60 − Q) =⇒
dQ = k dt
dt
3
3
dt
(30 − Q)(60 − Q)
1
dQ
60 − Q
1
60 − Q
1
= kt + c =⇒
−
= kt + c =⇒ ln
= c1 e30kt .
=⇒
30 (30 − Q) 60 − Q) dt
30 − Q
30 − Q
Imposing the initial condition Q(0) = 0 yields c1 = 2. Further, Q(10) = 15 =⇒ 45/15 = 2e300k , so that
1
k=
ln(3/2). Therefore,
300
t/10
60 (3/2)t/10 − 1
t
3
60 − Q
=⇒ Q(t) =
.
= 2e 10 ln(3/2) = 2
30 − Q
2
2(3/2)t/10 − 1
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Therefore, Q(20) =
60 (3/2)2 − 1
150
150
=
. Hence,
grams of C are produced in 20 minutes.
2(3/2)2 − 1
7
7
32. Since the chemicals A and B combine in the ratio 2:3, the amounts of A and B that are unconverted at
time t are (10 − 25 Q) grams and (15 − 35 Q) grams, respectively. Thus, according to the law of mass action,
the differential equation governing the behavior of Q(t) is
dQ
2
1
1
3
dQ
dQ
=
k dt =⇒
= k1 (10 − Q)(15 − Q) =⇒
= k(25 − Q)2 =⇒
= kt + c.
2
dt
5
5
dt
(25 − Q)
25 − Q
Hence, Q(t) = 25 −
1
. Imposing the initial condition Q(0) = 0 yields c = 1/25, so that
kt + c
625kt
1
=
Q(t) = 25 1 −
.
25kt + 1
25kt + 1
50t
. Therefore, Q(30) = 1500/75 = 20 grams. Hence, 20
2t + 15
grams of C are produced in 30 minutes. The reaction will be 50% complete when Q(t) = 12.5 this will occur
50t
=⇒ t ≈ 7.5 minutes.
after t minutes where 12.5 =
2t + 15
Q(5) = 10 =⇒ k = 2/375, so that Q(t) =
33. Since A and B combine in the ratio 3:5 to produce C. Therefore, producing 30 g of C will require
150
5
· 30 =
g of A.
8
8
34. (a).Since the chemicals A and B combine in the ratio a : b to produce chemical C, when Q grams of C
b
a
Q grams of A and
grams of B. Consequently, the amounts of A and
are produced , the consist of
a+b
a+b
a
b
Q grams and B0 −
Q grams, respectively. Therefore,
B that are unconverted at time t are A0 −
a+b
a+b
according to the law of mass action, the chemical reaction is governed by the differential equation
a
b
dQ
= k A0 −
Q
B0 −
Q .
dt
a+b
a+b
a+b
a
b
a+b
dQ
abk
−
Q
−
Q
, which can be
= k A0 −
Q
B0 −
Q =
A
B
0
0
dt
a+b
a+b
(a + b)2
a
b
a+b
a+b
dQ
abk
,α =
written as
= r(α − Q)(β − Q), where r =
A0 , β =
B0 .
dt
(a + b)2
a
b
1
dQ
1
1
1
dQ =
35.
= r(α − Q)(β − Q) =⇒
dQ = r dt + c =⇒
−
dt
(α − Q)(β − Q)
α−β
β−Q α−Q
α−Q
1
α − βcer(α−β)t
α−Q
= rt + c =⇒
rt + c. Hence,
. Imposing the
ln
= cer(α−β)t =⇒ Q(t) =
α−β
β−Q
β−Q
1 − cer(α−β)t
r(α−β)t
αβ 1 − e
initial condition Q(0) = 0 yields c = α/β, so that Q(t) =
. When α > β, limt→∞ Q(t) =
β − αer(α−β)t
αβ
= β.
α
(b).
dQ
dQ
= r(Q − α)2 =⇒ (α − Q)−2
= r =⇒ −(α − Q)−1 =
36. When α = β, Equation (1.4.24) reduces to
dt
dt
rt + c.
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1
α2 rt
. Imposing the initial condition Q(0) = 0 yields c = 1/α so that Q(t) =
.
rt + c
αrt + 1
Therefore, limt→∞ Q(t) = α.
Hence, Q(t) = α −
37. Separating the variables in the given differential equation yields
1
dQ = kdt
(α − Q)(β − Q)(γ − Q)
so that
1
1
1
dQ = kt + c,
+
+
(β − α)(γ − α)(α − Q) (α − β)(γ − β)(β − Q) (α − γ)(β − γ)(γ − Q)
so that
1
1
1
ln(α − Q) +
ln(β − Q) +
ln(γ − Q) = −kt − c,
(β − α)(γ − α)
(α − β)(γ − β)
(α − γ)(β − γ)
which can be written as
(β − γ) ln(α − Q) + (γ − α) ln(α − Q) + (α − β) ln(γ − Q) = (α − β)(β − γ)(γ − α)kt + c1
or equivalently,
ln(α − Q)β−γ + ln(α − Q)γ−α + ln(γ − Q)α−β = (α − β)(β − γ)(γ − α)kt + c1 .
Exponentiating both sides and simplifying yields
(α − Q)β−γ (β − Q)γ−α (γ − Q)α−β = c2 e(α−β)(β−γ)(γ−α)kt .
Q(0) = 0 =⇒ c2 = αβ−γ β γ−α γ α−β , so that
(1 − Q/α)β−γ (1 − Q/β)γ−α (1 − Q/γ)α−β = e(α−β)(β−γ)(γ−α)kt .
Solutions to Section 1.5
True-False Review:
(a): TRUE. The differential equation for such a population growth is dP
dt = kP , where P (t) is the population
as a function of time, and this is the Malthusian growth model described at the beginning of this section.
(b): FALSE. The initial population could be greater than the carrying capacity, although in this case the
population will asymptotically decrease towards the value of the carrying capacity.
(c): TRUE. The differential equation governing the logistic model is (1.5.2), which is certainly separable
as
D
dP
= r.
P (C − P ) dt
Likewise, the differential equation governing the Malthusian growth model is dP
dt = kP , and this is separable
as P1 dP
=
k.
dt
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(d): TRUE. As (1.5.3) shows, as t → ∞, the population does indeed tend to the carrying capacity C
independently of the initial population P0 . As it does so, its rate of change dP
dt slows to zero (this is best
seen from (1.5.2) with P ≈ C).
(e): TRUE. Every five minutes, the population doubles (increase 2-fold). Over 30 minutes, this population
will double a total of 6 times, for an overall 26 = 64-fold increase.
(f ): TRUE. An 8-fold increase would take 30 years, and a 16-fold increase would take 40 years. Therefore,
a 10-fold increase would take between 30 and 40 years.
dP
(g): FALSE. The growth rate is dP
dt = kP , and so as P changes, dt changes. Therefore, it is not always
constant.
(h): TRUE. From (1.5.2), the equilibrium solutions are P (t) = 0 and P (t) = C, where C is the carrying
capacity of the population.
(i): FALSE. If the initial population is in the interval ( C2 , C), then although it is less than the carrying
capacity, its concavity does not change. To get a true statement, it should be stated instead that the initial
population is less than half of the carrying capacity.
(j): TRUE. Since P (t) = kP , then P (t) = kP (t) = k 2 P > 0 for all t. Therefore, the concavity is always
positive, and does not change, regardless of the initial population.
Problems:
dP
1.
= kP =⇒ P (t) = P0 ekt . Since P (0) = 10, then P = 10ekt . Since P (3) = 20, then 2 = e3k =⇒ k =
dt
ln 2
. Thus P (t) = 10e(t/3) ln 3 . Therefore, P (24) = 10e(24/3) ln 3 = 10 · 28 = 2560 bacteria.
3
2. Using P (t) = P0 ekt we obtain P (10) = 5000 =⇒ 5000 = p0 e10k and P (12) = 6000 =⇒ 6000 = P0 e12k
which implies that e2k = 65 =⇒ k = 12 ln 65 . Hence, P (0) = 5000( 56 )5 = 2009.4. Also, P = 2P0 when
2 ln 2
t = 12 ln 2 =
≈ 7.6h.
ln 65
3. From P (t) = P0 ekt and P (0) = 2000 it follows that P (t) = 2000ekt . Since td = 4, k = 14 ln 2 so
P = 2000et ln 2/4 . Therefore, P (t) = 106 =⇒ 106 = 2000et ln 2/4 =⇒ t ≈ 35.86 hours.
dP
= kP =⇒ P (t) = P0 ekt . Since, P (0) = 10000 then P (t) = 10000ekt . Since P (5) = 20000 then
dt
20000 = 10000ekt =⇒ k = 15 ln 2. Hence P (t) = 10000e(t ln 2)/5 .
4.
(a). P (20) = 10000e4 ln 2 = 160000.
(b). 1000000 = 10000e(t ln 2)/5 =⇒ 100 = e(t ln 2)/5 =⇒ t =
5 ln 100
≈ 33.22 years.
ln 2
50C
. In formulas (1.5.5) and (1.5.6) we have P0 = 500, P1 = 800, P2 = 1000, t1 = 5,
50 + (C − 50)e−rt 1
(1000)(300)
1
800[(800)(1500) − 2(500)(1000)]
and t2 = 10. Hence, r = ln
= ln 3, C =
≈ 1142.86, so
5
(500)(200)
5
8002 − (500)(1000)
571430
1142.86)(500)
≈
. Inserting t = 15 into the preceding formula
that P (t) =
500 + 642.86e−0.2t ln 3
500 + 642.86e−0.2t ln 3
yields P (15) = 1091.
5. P (t) =
(c)2017 Pearson Education. Inc.
46
50C
In formulas (1.5.5) and (1.5.6) we have P0 = 50, P1 = 62, P2 = 76, t1 = 2,
50 + (C − 50)e−rt
62[(62)(126) − 2(50)(76)]
(76)(12)
1
≈ 0.132, C =
and t2 = 2t1 = 4. Hence, r = ln
≈ 298.727, so that
2
(50)(14)
622 − (50)(76)
14936.35
. Inserting t = 20 into the preceding formula yields P (20) ≈ 221.
P (t) =
50 + 248.727e−0.132t
6. P (t) =
P2 (P1 − P0 )
> 1. Rearranging the terms in this inequality and
P0 (P2 − P1 )
2P0 P2
P1 (P0 + P2 ) − 2P0 P2
. Further, C > 0 requires that
> 0.
using the fact that P2 > P1 yields P1 >
P0 + P2
P12 − P0 P2
2P0 P2
From P1 >
we see that the numerator in the preceding inequality is positive, and therefore the
P0 + P2
2P0 P2
denominator must also be positive. Hence in addition to P1 >
, we must also have P12 > P0 P2 .
P0 + P2
7. From equation (1.5.5) r > 0 requires
dy
=
dt
ky(1500 − y), y(0) = 5, y(1) = 10, where k is a positive constant. Separating the differential equation
1
dy = k dt. Using a partial fraction decomposition on the left-hand
and integrating yields
y(1500
−
y)
1
1
y
1
dy = kt + c, so that
= kt + c, which upon
+
ln
side gives
1500y 1500(1500 − y)
1500
1500 − y
y
1
exponentiation yields
= c1 e1500kt . Imposing the initial condition y(0) = 5, we find that c1 =
.
1500 − y
299
10
y
1 1500kt
1 1500k
. The further condition y(1) = 10 requires
. solving
Hence,
=
e
=
e
1500 − y
299
1490
299
1
299
y
1 t ln (299/149)
for k gives k =
. Solving algebraically for y we find
ln
. Therefore,
=
e
1500 149
1500 − y
299
1500et ln (299/149)
1500
1500
y(t) =
=
. Hence, y(14) =
= 1474.
299 + et ln (299/149)
1 + 299e−t ln (299/149)
1 + 299e−14 ln (299/149)
8. Let y(t) denote the number of passengers who have the flu at time t. Then we must solve
9.(a). Equilibrium solutions: P (t) = 0, P (t) = T .
dP
dP
> 0, 0 < P < T =⇒
< 0.
Slope: P > T =⇒
dt
dt
k
rT 2 + 4k
1
2
T±
. We see that slope of the
Isoclines: r(P − T ) = k =⇒ P − T P − = 0 =⇒ P =
r
2
r
−rT 2
solution curves satisfies k ≥
.
4
2
dP
d P
= r(2P − T )
Concavity:
= r2 (2P − T )(P − T )P . Hence, the solution curves are concave up for
dt2
dt
t
T
P > , and are concave down for 0 < P < .
2
2
(b). See accompanying figure.
(c). For 0 < P0 < T , the population dies out with time. For P0 > T , there is a population growth. The
term threshold level is appropriate since T gives the minimum value of P0 above which there is a population
growth.
(c)2017 Pearson Education. Inc.
47
P(t)
T
t
0
Figure 0.0.42: Figure for Problem 9(b)
1
dP
10. (a). Separating the variables in differential equation (1.5.7) gives
= r, which can be
P
(P
−
T
)
dt 1
1 dP
1
P −T
= rt+c, so that
written in the equivalent form
−
= r. Integrating yields ln
T (P − T ) T P dt
T
P
P −T
P0 − T
P −T
P0 − T
T rt
erT t .
= c1 , so that
= c1 e . The initial condition P (0) = P0 requires
=
P
P0
P
P0
T P0
Solving algebraically for P yields P (t) =
.
P0 − (P0 − T )erT t
T P0
is positive, and increases without bound as
P0 − (P0 − T )erT t
t → ∞. Consequently limt→∞ P (t) = 0. In this case the population dies out as t increases.
(b). If P0 < T , then the denominator in
T P0
(c). If P0 > T , then the denominator of
vanishes when (P0 − T )erT t = P0 , that is when
P0 − (P0 − T )erT t
1
P0
t=
ln
. This means that within a finite time the population grows without bound. We can
rT
P0 − T
interpret this as a mathematical model of a population explosion.
dP
= r(C − P )(P − T )P, P (0) = P0 , r > 0, 0 < T < C.
dt
Equilibrium solutions: P (t) = 0, P (t) = T, P (t) = C. The slope of the solution curves is negative for
0 < P < T , and for P > C. It is positive for T < P < C.
d2 P
Concavity:
= r2 [(C − P )(P − T ) − (P − T )P + (C − P )P ](C − P )(P − T )P , which simplifies to
dt2
d2 P
= r2 (−3P 2 + 2P T + 2CP − CT )(C − P )(P − T ). Hence changes in concavity occur when P = 13 (C +
dt2 √
T ± C 2 − CT + T 2 ). A representative slope field with some solution curves is shown in the accompanying
figure. We see that for 0 < P0 < T the population dies out, whereas for T < P0 < C the population grows
and asymptotes to the equilibrium solution P (t) = C. If P0 > C, then the solution decays towards the
equilibrium solution P (t) = C.
11.
dP
= rP (ln C − ln P ), P (0) = P0 , and r, C, and P0 are positive constants.
dt
Equilibrium solutions: P (t) = C. The slope of the solution curves is positive for 0 < P < C, and negative
12.
(c)2017 Pearson Education. Inc.
48
P(t)
C
T
t
0
Figure 0.0.43: Figure for Problem 11
dP
C
d2 P
C
C
2
−
1
−
1
P ln . Hence, the solution curves are concave
ln
=
r
ln
=
r
2
dt
P
dt
P
P
C
C
up for 0 < P <
and P > C. They are concave down for
< P < C. A representative slope field with
e
e
some solution curves are shown in the accompanying figure.
for P > C.
Concavity:
25
20
15
P(t)
10
5
0
1
2
3
t
4
5
Figure 0.0.44: Figure for Problem 12
1
dP
= r which can be integrated directly to
P (ln C − ln P ) dt
−rt
. The initial condition P (0) = P0 requires that
obtain − ln (ln C − ln P ) = rt + c so that ln ( C
P ) = c1 e
−rt
C
C
C
−rt
ln ( P0 ) = c1 . Hence, ln ( P ) = e
ln ( P0 ) so that P (t) = Celn (P0 /k)e . Since limt→∞ e−rt = 0, it follows
that limt→∞ P (t) = C.
13. Separating the variables in (1.5.8) yields
dP
= kP , which is easily integrated to obtain P (t) = P0 ekt .
dt
kt
The initial condition P (0) = 400 requires that P0 = 400, so that
P (t) = 400e . We also know that
1
17
. Consequently,
ln
P (30) = 340. This requires that 340 = 400e30k so that k =
30
20
14. Using the exponential decay model we have
17
P (t) = 400e 30 ln( 20 ) .
t
(c)2017 Pearson Education. Inc.
(0.0.5)
49
17
(a). From (0.0.5), P (60)400e2 ln( 20 ) = 289.
10
17
(b). From (0.0.5), P (100) = 400e 3 ln( 20 ) ≈ 233
(c). From (0.0.5), the half-life, tH , is determine from
tH
17
200 = 400e 30 ln( 20 ) =⇒ tH = 30
ln 2
≈ 128 days.
ln 20
17
15. (a). More.
dP
= kP , which is easily integrated to obtain P (t) = P0 ekt .
dt
100,000ekt . We also know
The initial condition P (0) = 100, 000 requires that P0 = 100, 000, so that P (t) = 1
4
. Consequently,
that P (10) = 80, 000. This requires that 100, 000 = 80, 000e10k so that k =
ln
10
5
(b). Using the exponential decay model we have
4
P (t) = 100, 000e 10 ln( 5 ) .
t
(0.0.6)
Using (0.0.6), the half-life is determined from
tH
4
50, 000 = 100, 000e 10 ln( 5 ) =⇒ tH = 10
ln 2
≈ 31.06 min.
ln 54
(c). Using (0.0.6) there will be 15,000 fans left in the stadium at time t0 , where
15, 000 = 100, 000e
t0
10 ln
3
( 45 ) =⇒ t = 10 ln 20 ≈ 85.02 min.
0
4
ln 15
dP
16. Using the exponential decay model we have
= kP , which is easily integrated to obtain P (t) = P0 ekt .
dt
Since the half-life is 5.2 years, we have
ln 2
1
P0 = P0 e5.2k =⇒ k = −
.
2
5.2
Therefore,
ln 2
P (t) = P0 e−t 5.2 .
Consequently, only 4% of the original amount will remain at time t0 where
ln 2
ln 25
4
P0 = P0 e−t0 5.2 =⇒ t0 = 5.2
≈ 24.15 years.
100
ln 2
17. Maple, or even a TI 92 plus, has no problem in solving these equations.
18. (a). Malthusian model is P (t) = 151.3ekt . Since P (1) = 179.4, then 179.4 = 151.3e10k =⇒ k =
t ln (179.4/151.1)
1
179.4
10
.
10 ln 151.3 . Hence, P (t) = 151.3e
(c)2017 Pearson Education. Inc.
50
P(t)
320
300
280
260
240
220
200
180
160
0
10
20
30
40
t
Figure 0.0.45: Figure for Problem 18(c)
151.3C
. Imposing the initial conditions P (10) = 179.4 and P (20) = 203.3
151.3 + (C − 151.3)e−rt
10 ln (179.4/151.1)
20 ln (179.4/151.1)
10
10
and 203.3 = 151.3e
whose solution is
gives the pair of equations 179.4 = 151.3e
39935.6
.
C ≈ 263.95, r ≈ 0.046. Using these values for C and r gives P (t) =
151.3 + 112.65e−0.046t
(b). P (t) =
(c). Malthusian model: P (30) ≈ 253 million; P (40) ≈ 300 million.
Logistics model: P (30) ≈ 222 million; P (40) ≈ 236 million.
The logistics model fits the data better than the Malthusian model, but still gives a significant underestimate
of the 1990 population.
50C
. Imposing the conditions P (5) = 100, P (15) = 250 gives the pair of equations
50 + (C − 50)e−rt
50C
50C
and 250 =
whose positive solutions are C ≈ 370.32, r ≈ 0.17.
100 =
50 + (C − 50)e−5r
50 + (C − 50)e−15r
18500
Using these values for C and r gives P (t) =
. From the figure we see that it will take
50 + 18450e−0.17t
approximately 52 years to reach 95% of the carrying capacity.
19. P (t) =
Solutions to Section 1.6
True-False Review:
(a): FALSE. Any solution to the differential equation (1.6.7) serves as an integrating factor
for the differ
ential equation. There are infinitely many solutions to (1.6.7), taking the form I(x) = c1 e p(x)dx , where c1
is an arbitrary constant.
(b): TRUE. Any solution to the differential equation (1.6.7) serves as an integrating factor
for the differ
ential equation. There are infinitely many solutions to (1.6.7), taking the form I(x) = c1 e p(x)dx , where c1
is an arbitrary constant. The most natural choice is c1 = 1, giving the integrating factor I(x) = e p(x)dx .
(c): TRUE. Multiplying y + p(x)y = q(x) by I(x) yields y I + pIy = qI. Assuming that I = pI, the
requirement on the integrating factor, we have y I + I y = qI, or by the product rule, (I · y) = qI, as
requested.
(c)2017 Pearson Education. Inc.
51
P(t)
Carrying Capacity
350
300
250
200
150
100
50
0
10
20
30
40
50
60
70
t
Figure 0.0.46: Figure for Problem 19
(d): FALSE. Rewriting the differential equation as
dy
− x2 y = sin x,
dx
2
3
we have p(x) = −x2 , and so an integrating factor must have the form I(x) = e p(x)dx = e (−x )dx = e−x /3 ,
3
2
or any constant multiple of e−x /3 . Since ex is not of this form, then it is not an integrating factor.
(e): FALSE. Rewriting the differential equation as
dy
1
+ y = x,
dx x
1
we have p(x) = , and so an integrating factor must have the form I(x) = e p(x)dx = e (1/x)dx = x, or any
x
constant multiple of x. Since x + 5 is not of this form, then it is not an integrating factor.
Problems:
In this section the function I(x) = e p(x)dx will represent the integrating factor for a differential equation
of the form y + p(x)y = q(x).
1. y + y = 4ex . I(x) = e dx = ex =⇒
2. y +
d(ex y)
= 4e2x =⇒ ex y = 2e2x + c =⇒ y(x) = e−x (2e2x + c).
dx
2
d(x2 y)
y = 5x2 . I(x) = e (2/x) dx = x2 =⇒
= 5x4 =⇒ x2 y = x5 + c =⇒ y(x) = x−2 (x5 + c).
x
dx
3. x2 y − 4xy = x7 sin x, x > 0 =⇒ y − x4 y = x5 sin x. I(x) = x−4 =⇒
sin x − x cos x + c =⇒ y(x) = x4 (sin x − x cos x + c).
4. y + 2xy = 2x3 . I(x) = e2
2
xdx
2
= ex =⇒
2
ex (x2 − 1) + c =⇒ y(x) = x2 − 1 + ce−x .
d(x−4 y)
= x sin x =⇒ x−4 y =
dx
2
d x2
2
2
2
(e y) = 2ex x3 =⇒ ex y = 2 ex x3 dx =⇒ ex y =
dx
(c)2017 Pearson Education. Inc.
52
2x
1
d
y = 4x, −1 < x < 1. I(x) =
=⇒
2
2
1−x
1−x
dx
c =⇒ y(x) = (1 − x2 )[− ln (1 − x2 )2 + c].
5. y +
y
1 − x2
=
4x
y
=⇒
= − ln(1 − x2 )2 +
2
1−x
1 − x2
2x
2x
4
d
4
dx
1+x2
y
=
.
I(x)
=
e
= 1 + x2 =⇒
=⇒ (1 + x2 )y =
[(1 + x2 )y] =
2
2
2
1+x
(1 + x )
dx
(1 + x2 )2
dx
1
4
=⇒ (1 + x2 )y = 4 tan−1 x + c =⇒ y(x) =
(4 tan−1 x + c).
2
1+x
1 + x2
6. y +
dy
sin 2x
1
d
+ y sin 2x = 4 cos4 x, 0 ≤ x ≤ π2 =⇒ y +
y = 2 cos2 x. I(x) =
=⇒
(y sec x) =
dx
2 cos2 x
cos x
dx
cos x =⇒ y(x) = cos x(2 sin x + c) =⇒ y(x) = sin 2x + c cos x.
7. 2 cos2 x
dx
1
d
y = 9x2 . I(x) = e x ln x = ln x =⇒
(y ln x) = 9 x2 ln xdx =⇒ y ln x = 3x3 ln x − x3 + c =⇒
x ln x
dx
3x3 ln x − x3 + c
y(x) =
.
ln x
8. y +
d
9. y − y tan x = 8 sin3 x. I(x) = cos x =⇒
(y cos x) = 8 cos x sin3 x =⇒ y cos x = 8 cos x sin3 xdx + c =⇒
dx
1
y cos x = 2 sin4 x + c =⇒ y(x) =
(2 sin4 x + c).
cos x
dt
2
d
dx
4et
+ 2x = 4et =⇒ x + x =
. I(x) = e2 t = t2 =⇒ (t2 x) = 4tet =⇒ t2 x = 4 tet dt + c =⇒
dt
t
t
dt
4et (t − 1) + c
2
t
t x = 4e (t − 1) + c =⇒ x(t) =
.
t2
10. t
d
y = (sin x sec x)y − 2 sin x =⇒ y − (sin x sec x)y = −2 sin x. I(x) = cos x =⇒
(y cos x) =
dx
1
1
1
cos 2x + c .
−2 sin x cos x =⇒ y cos x = −2 sin x cos xdx + c = cos 2x + c =⇒ y(x) =
2
cos x 2
11.
d
(y sec x) =
12. (1−y sin x)dx−cos xdy = 0 =⇒ y +(sin x sec x)y = sec x. I(x) = e sin x sec xdx = sec x =⇒
dx
2
2
sec x =⇒ y sec x = sec xdx+c =⇒ y sec x = tan x+c =⇒ y(x) = cos x(tan x+c) =⇒ y(x) = sin x+c cos x.
1
d −1
13. y − x−1 y = 2x2 ln x. I(x) = e− x dx = x−1 =⇒
(x y) = 2x ln x =⇒ x−1 y = 2 x ln xdx + c =⇒
dx
1 3
1 2
−1
x y = x (2 ln x − 1) + c. Hence, y(x) = x (2 ln x − 1) + cx.
2
2
d αx
(e y) = e(α+β)x =⇒ eαx y = e(α+β)x dx + c. If α + β = 0,
dx
e(α+β)x
eβx
αx
−αx
(x + c). If α + β = 0, then eαx y =
then e y = x + c =⇒ y(x) = e
+ c =⇒ y(x) =
+ ce−αx .
α+β
α+β
14. y + αy = eβx . I(x) = eα
dx
= eαx =⇒
m
d m
y = ln x. I(x) = xm =⇒
(x y) = xm ln x =⇒ xm y = xm ln xdx + c. If m = −1, then
x
dx
(ln x)2
xm+1
(ln x)2
xm+1
xm y =
+ c =⇒ y(x) =
+ c =⇒ y(x) = x
+ c . If m = −1, then xm y =
ln x −
2
2
m+1
(m + 1)2
15. y +
(c)2017 Pearson Education. Inc.
53
c
x
x
+ m.
ln x −
m+1
(m + 1)2
x
dx
2
d 2
16. y + y = 4x. I(x) = e2 x = e2 ln x = x2 =⇒
(x y) = 4x3 =⇒ x2 y = 4 x3 dx + c =⇒ x2 y = x4 + c,
x
dx
x4 + 1
but y(1) = 2 so c = 1; thus, y(x) =
.
x2
d
(y csc x) = 2 csc x cos x =⇒
dx
π
y csc x = 2 ln (sin x) + c, but y( 2 ) = 2 so c = 2; thus, y(x) = 2 sin x[ln (sin x) + 1].
17. y sin x − y cos x = sin 2x =⇒ y − y cot x = 2 cos x. I(x) = csc x =⇒
dt
2
d
18. x +
x = 5. I(t) = e2 4−t = e−2 ln(4−t) = (4−t)t−2 =⇒ ((4−t)−2 x) = 5(4−t)−2 =⇒ (4−t)−2 x =
4−t
dt
5 (4 − t)−2 dt + c =⇒ (4 − t)−2 x = 5(4 − t)−1 + c, but x(0) = 4 so c = −1; thus, x(t) = (4 − t)2 [5(4 − t)−1 − 1]
or x(t) = (4 − t)(1 + t).
d x
e2x
(e y) = e2x =⇒ ex y =
+ c, but y(0) = 1 so
19. (y − e−x )dx + dy = 0 =⇒ y + y = ex . I(x) = ex =⇒
dx
2
1
1 x
c = ; thus, y(x) = (e + e−x ) = cosh x.
2
2
20. y + y = f (x), y(0) = 3,
f (x) =
1,
0,
if x ≤ 1,
if x > 1.
x
x
d x
I(x) = e dx = ex =⇒
(e y) = ex f (x) =⇒ [ex y]x0 = 0 ex f (x)dx =⇒ ex y − y(0) = 0 ex f (x)dx =⇒
dx
x
x
ex y − 3 = 0 ex dx =⇒ y(x) = e−x 3 + 0 ex f (x)dx .
x x
x x
If x ≤ 1, 0 e f (x)dx = 0 e dx = ex − 1 =⇒ y(x) = e−x (2 + ex )
x
x
If x > 1, 0 ex f (x)dx = 0 ex dx = e − 1 =⇒ y(x) = e−x (2 + e).
21. y − 2y = f (x), y(0) = 1,
f (x) =
1 − x,
0,
if x < 1,
if x ≥ 1.
x
d −2x
y) = e−2x f (x) =⇒ [e−2x y]x0 = 0 e−2x f (x)dx =⇒ e−2x y − y(0) =
I(x) = e− 2dx = e−2x =⇒
(e
dx x
x
x −2x
e
f (x)dx =⇒ e−2x y − 1 = 0 e−2x f (x) =⇒ y(x) = e2x 1 + 0 e−2x f (x)dx .
0
x −2x
x −2x
1 −2x
1 −2x
2x
2x
2x
1+ e
f (x)dx = 0 e
(1 − x)dx = e
(2x−1+e ) =⇒ y(x) = e
(2x − 1 + e ) =
If x < 1, 0 e
4
4
1 2x
(5e + 2x − 1).
4
x
x
1
1
1
If x ≥ 1, 0 e−2x f (x)dx = 0 e−2x (1 − x)dx = (1 + e−2 ) =⇒ y(x) = e2x 1 + (1 + e−2 ) = e2x (5 + e−2 ).
4
4
4
22. On (−∞, 1), y − y = 1 =⇒ I(x) = e−x =⇒ y(x) = c1 ex − 1. Imposing the initial condition y(0) = 0
requires c1 = 1, so that y(x) = ex − 1, for x < 1.
d −x
On [1, ∞), y − y = 2 − x =⇒ I(x) = e−x =⇒
(e y) = (2 − x)e−x =⇒ y(x) = x − 1 + c2 e−x .
dx
Continuity at x = 1 requires that limx→1 y(x) = y(1). Consequently we must choose c2 to satisfy c2 e = e − 1,
so that c2 = 1 − e−1 . Hence, for x ≥ 1, y(x) = x − 1 + (1 − e−1 )ex .
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d2 y
1 dy
du 1
dy
du
d2 y
+
= 9x, x > 0. Let u =
so
= 2 . The first equation becomes
+ u = 9x which is
2
dx
x dx
dx
dx
dx
dx x
d
(xu) = 9x2 =⇒ xu = x2 dx + c =⇒ xu =
first-order linear. An integrating factor for this is I(x) = x so
dx
dy
dy
3x3 + c1 =⇒ u = 3x2 + c1 x−1 , but u =
so
= 3x2 + c1 x−1 =⇒ y = (3x2 + c1 x−1 )dx + c2 =⇒ y(x) =
dx
dx
x3 + c1 ln x + c2 .
23.
24. The differential equation for Newton’s Law of Cooling is dT
dt = −k(T − Tm ). We can re-write this
equation in the form of a first-order linear differential equation: dT
dt + kT = kTm . An integrating factor
d
for this differential equation is I = e k dt = ekt . Thus, (T ekt ) = kTm ekt . Integrating both sides, we get
dt
T ekt = Tm ekt + c, and hence, T = Tm + ce−kt , which is the solution to Newton’s Law of Cooling.
dT
dT
dTm
= α =⇒ Tm = αt + c1 so
= −k(T − αt − c1 ) =⇒
+ kT = k(αt + c1 ). An integrating
dt
dt
dt
d kt
factor for this differential equation is I = ek dt = ekt . Thus,
(e T ) = kekt (αt + c1 ) =⇒ ekt T =
dt
α
α
1
ekt (αt − + c1 ) + c2 =⇒ T = αt − + c1 + c2 e−kt =⇒ T (t) = α(t − ) + β + T0 e−kt where β = c1 and
k
k
k
T0 = c 2 .
25.
dTm
dT
= 10 =⇒ Tm = 10t + c1 but Tm = 65 when t = 0 so c1 = 65 and Tm = 10t + 65.
=
dt
dt
dT
dT
1
= −k(T − 10t − 65), but
(1) = 5, so k = . The last differential equation can be
−k(T − Tm ) =⇒
dt
dt
8
2
13
d kt
2
dT
kt
+ c =⇒
+ kT = k(10t + 65) =⇒ (e T ) = 5ke (2t + 13) =⇒ ekt T = 5kekt
t− 2 +
written
dt
dt
k
k
k
2
1
t
1
T = 5(2t − + 13) + ce−kt , but k =
so T (t) = 5(2t − 3) + ce− 8 . Since T (1) = 35, c = 40e 8 . Thus,
k
8
1
T (t) = 10t − 15 + 40e 8 (1−t) .
26.
dT
1
27. (a). In this case, Newton’s law of cooling is
= − (t − 80e−t/20 ). This linear differential equation
dt
40
dT
1
has standard form
+ T = 2e−t/20 , with integrating factor I(t) = et/40 . Consequently the differential
dt
40
d t/40
T ) = 2e−t/40 , so that T (t) = −80e−t/20 + ce−t/40 .
equation can be written in the integrable form
(e
dt
Then T (0) = 0 =⇒ c = 80, so that T (t) = 80(e−t/40 − e−t/20 ).
(b). We see that limt→∞ = 0. This is a reasonable result since the temperature of the surrounding
medium also approaches zero as t → ∞. We would expect the temperature of the object to approach to the
temperature of the surrounding medium at late times.
dT
1 −t/20
1 −t/40
−t/40
−t/20
(c). T (t) = 80(e
. So T (t) has only one critical point
−e
) =⇒
+ e
= 80 − e
40
20
dt
1
1
when 80 − e−t/40 + e−t/20 = 0 =⇒ t = 40 ln 2. Since T (0) = 0, and limt→∞ T (t) = 0 the function
40
20
assumes a maximum value at tmax = 40 ln 2. T (tmax ) = 80(e− ln 2 − e−2 ln 2 ) = 20◦ F, Tm (tmax ) = 80e−2 ln 2 =
20◦ F.
(d). The behavior of T (t) and Tm (t) is given in the accompanying figure.
(c)2017 Pearson Education. Inc.
55
T(t), Tm(t)
80
Tm(t)
60
40
T(t)
20
t
0
0
20
40
60
80
100
120
Figure 0.0.47: Figure for Problem 27(d)
28. (a). The temperature varies from a minimum of A − B at t = 0 to a maximum of A + B when t = 12.
T
A+B
A-B
t
5
10
15
20
Figure 0.0.48: Figure for Problem 28(a)
dT
+ k1 T = k1 (A − B cos ωt) + T0 . Multiplying
dt
k1 t
by the integrating factor I = e
reduces this differential equation to the integrable form
(b). First write the differential equation in the linear form
d k1 t
(e T ) = k1 ek1 t (A − B cos ωt) + T0 ek1 t .
dt
Consequently,
ek1 t T (t) =
Aek1 t − Bk1
ek1 t cos ωtdt +
(c)2017 Pearson Education. Inc.
T0 k 1 t
e +c
k1
56
so that
T (t) = A +
T0
Bk1
− 2
(k1 cos ωt + ω sin ωt) + ce−k1 t .
k1
k1 + ω 2
This can be written in the equivalent form
T (t) = A +
T0
Bk1
− 2
cos (ωt − α) + ce−k1 t
k1
k1 + ω 2
for an approximate phase constant α.
dy
dy
dy
+ p(x)y = 0 =⇒
= −p(x)dx =⇒
= − p(x)dx =⇒ ln |y| = − p(x)dx + c =⇒ yH =
dx
y
y
c1 e− p(x)dx .
dy
dv
du
=u
+ v . Substituting this
(b). Replace c1 in part (a) by u(x) and let v = e− p(x)dx . y = uv =⇒
dx
dx
dx
dv
du
dy
+ p(x) = q(x), we obtain u + v + p(x)y = q(x), but
last result into the original differential equation,
dx
dx
dx
du
dv
= −vp, the last equation reduces to v
= q(x) =⇒ du = v −1 (x)q(x)dx =⇒ u = v −1 (x)q(x)dx +
since
dx
dx
e p(x)dx q(x)dx + c .
c. Substituting the values for u and v into y = uv, we obtain y = e− p(x)dx
29. (a).
dy
+x−1 y = 0, with solution yH = cx−1 . According to Problem
dx
29, we determine the function u(x) such that y(x) = x−1 u(x) is a solution to the given differential equation.
du
du 1
dy
dy
We have
= x−1 −x−2 u. Substituting into
+x−1 y = cos x yields x−1 − 2 u+x−1 (x−1 u) = cos x,
dx
dx
dx
dx x
du
so that
= x cos x. Integrating we obtain u = x sin x + cos x + c, so that y(x) = x−1 (x sin x + cos x + c).
dx
30. The associated homogeneous equation is
dy
+ y = 0, with solution yH = ce−x . According to Problem
dx
29, we determine the function u(x) such that y(x) = e−x u(x) is a solution to the given differential equation.
dy
du −x
dy
du −x
We have
=
e − e−x u. Substituting into
+ y = e−2x yields
e − e−x u + e−x u(x) = e−2x , so
dx
dx
dx
dx
du
that
= e−x . Integrating we obtain u = −e−x + c, so that y(x) = e−x (−e−x + c).
dx
31. The associated homogeneous equation is
dy
+ cot x · y = 0, with solution yH = c · csc x. According
dx
to Problem 29, we determine the function u(x) such that y(x) = csc x · u(x) is a solution to the given
du
dy
dy
= csc x ·
− csc x · cot x · u. Substituting into
+ cot x · y = 2 cos x
differential equation. We have
dx
dx
dx
du
du
− csc x · cot x · u + csc x · cot x · u = cos x, so that
= 2 cos x sin x. Integrating we obtain
yields csc x ·
dx
dx
2
2
u = sin x + c, so that y(x) = csc x(sin x + c).
32. The associated homogeneous equation is
dy
1
− y = 0, with solution yH = cx. We determine the
dx
x
dy
du
function u(x) such that y(x) = xu(x) is a solution of the given differential equation. We have
= x + u.
dx
dx
du
dy 1
− y = x ln x and simplifying yields
= ln x, so that u = x ln x − x + c. Consequently,
Substituting into
dx x
dx
33. The associated homogeneous equation is
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57
y(x) = x(x ln x − x + c).
Problems 34 - 39 are easily solved using a differential equation solver such as the dsolve package in Maple.
Solutions to Section 1.7
True-False Review:
(a): TRUE. Concentration of chemical is defined as the ratio of mass to volume; that is, c(t) = VA(t)
(t) .
Therefore, A(t) = c(t)V (t).
(b): FALSE. The rate of change of volume is “rate in” − “rate out”, which is r1 − r2 , not r2 − r1 .
(c): TRUE. This is reflected in the fact that c1 is always assumed to be a constant.
(d): FALSE. The concentration of chemical leaving the tank is c2 (t) = VA(t)
(t) , and since both A(t) and V (t)
can be nonconstant, c2 (t) can also be nonconstant.
(e): FALSE. Kirchhoff’s second law states that the sum of the voltage drops around a closed circuit is zero,
not that it is independent of time.
(f ): TRUE. This is essentially Ohm’s law, (1.7.10).
(g): TRUE. Due to the negative exponential in the formula for the transient current, iT (t), it decays to
zero as t → ∞. Meanwhile, the steady-state current, iS (t), oscillates with the same frequency ω as the
alternating current, albeit with a phase shift.
0
(h): TRUE. The amplitude is given in (1.7.19) as A = √R2E
, and so as ω gets larger, the amplitude
+ω 2 L2
A gets smaller.
Problems:
A(60)
. ΔV = r1 Δt −
V (60)
dV
dA
r2 Δt =⇒
= 3 =⇒ V (t) = 3(t + 200) since V (0) = 600. ΔA ≈ c1 r1 Δt − c2 r2 Δt =⇒
= 30 − 3c2 =
dt
dt
A
A
30 − 3 = 30 −
=⇒ (t + 200)A = 15(t + 200)2 + c. Since A(0) = 1500, c = −300000 and therefore
V
t + 200
15
A(60)
596
A(t) =
[(t + 200)2 − 20000]. Thus
=
g/L.
t + 200
V (60)
169
1. Given V (0) = 600, A(0) = 1500, c1 = 5, r1 = 6, and r2 = 3. We need to find
dV
2. Given V (0) = 10, A(0) = 20, c1 = 4, r1 = 2, and r2 = 1. Then ΔV = r1 Δt−r2 Δt =⇒
= 1 =⇒ V (t) =
dt
dA
A
A
dA
1
= 8 − c2 = 8 − = 8 −
=⇒
+
A=
t + 10 since V (0) = 10. ΔA ≈ c1 r1 Δt − c2 r2 Δt =⇒
dt
V
t + 10
dt
t + 10
4
8 =⇒ (t + 10)A = 4(t + 10)2 + c1 . Since A(0) = 20 =⇒ c1 = −200 so A(t) =
[(t + 10)2 − 50]. Therefore,
t + 10
A(40) = 196 g.
dV
3. Given V (0) = 100, A(0) = 100, c1 = 0.5, r1 = 6, and r2 = 4. Then ΔV = r1 Δt − r2 Δt =⇒
= 2 =⇒
dt
4A
dA
2A
d
dA
+
= 3 =⇒
+
= 3 =⇒ [(t + 50)2 A] =
V (t) = 2(t + 50) since V (0) = 100. Then
dt
2(t + 50)
dt
t + 50
dt
(c)2017 Pearson Education. Inc.
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125000
.
(t + 50)2
The tank is full when V (t) = 200, that is when 2(t+50) = 200 so that t = 50 min. Therefore the concentration
A(50)
9
just before the tank overflows is:
=
g/L.
V (50)
16
3(t+50)2 =⇒ (t+50)2 A = (t+50)3 +c but A(0) = 100 so c = 125000 and therefore A(t) = t+50+
dV
= 2 =⇒ V =
dt
2(t + 10) since V (0) = 20. Thus V (t) = 40 for t = 10, so we must find A(10).ΔA ≈ c1 r1 Δt − c2 r2 Δt =⇒
2A
dA
A
dA
1
d
= 40 − 2c2 = 40 −
= 40 −
=⇒
+
A = 40 =⇒ [(t + 10)A] = 40(t + 10)dt =⇒
dt
V
t + 10
dt
t + 10
dt
20
(t + 10)A = 20(t + 10)2 + c. Since A(0) = 0 =⇒ c = −2000 so A(t) =
[(t + 10)2 − 100] and A(10) = 300
t + 10
g.
4. Given V (0) = 20, A(0) = 0, c1 = 10, r1 = 4, and r2 = 2. Then ΔV = r1 Δt − r2 Δt =⇒
dV
= 1 =⇒
dt
(20 + t)3 + c
dA
2
d
V = t+20 since V (0) = 20. Then
+
A = 3 =⇒ [(t+20)2 A] = 3(t+20)2 =⇒ A(t) =
dt t + 20
dt
(t + 20)2
3
3
(t + 20) − 20
and since A(0) = 0, c = −203 which means that A(t) =
.
(t + 20)2
5. (a). We are given that V (0) = 20, c1 = 1, r1 = 3, and r2 = 2. Then ΔV = r1 Δt − r2 Δt =⇒
A(t)
A(t)
or c2 =
so from part (a),
V (t)
t + 20
√
1
(t + 20)3 − 203
1
(t + 20)3 − 203
. Therefore c2 = g/l when =
=⇒ t = 20( 3 2 − 1)minutes.
c2 =
3
3
(t + 20)
2
2
(t + 20)
(b). The concentration of chemical in the tank, c2 , is given by c2 =
6. We are given that V (0) = 10, A(0) = 0, c1 = 0.5, r1 = 3, r2 =, and
A(5)
= 0.2.
V (5)
dV
= 1 =⇒ V (t) = t + 10 since V (0) = 10. Then ΔA ≈ c1 r1 Δt − c2 r2 Δt =⇒
(a). ΔV = r1 Δt − r2 Δt =⇒
dt
A
dA
2A
dA
2dt
= −2c2 = −2 = −
=⇒
=−
=⇒ ln |A| = −2 ln |t + 10| + c =⇒ A = k(t+ 10)−2 . Then
dt
V
t + 10
A
t + 10
A(5)
675
A(5) = 3 since V (5) = 15 and
. In particular, A(0) = 6.75
= 0.2. Thus, k = 675 and A(t) =
V (5)
(t + 10)2
g.
A(t)
A(t)
675
675
and V (t) = t + 10 =⇒
.
= 0.1. From part (a) A(t) =
=
2
V (t)
(t + 10)
V (t)
(t + 10)3
√
√
A(t)
Since
= 0.1 =⇒ (t + 10)3 = 6750 =⇒ t + 10 = 15 3 2 so V (t) = t + 10 = 15 3 2 L.
V (t)
(b). Find V (t) when
7. (a). We are given that V (0) = w, c1 = k, r1 = r, r2 = r, and A(0) = A0 . Then ΔV = r1 Δr − r2 Δt =⇒
dA
dV
A
A
= 0 =⇒ V (t) = V (0) = w for all t. Then ΔA = c1 r1 Δt − c2 r2 Δt =⇒
= kr − r = kr − r =
dt
dt
V
V
dA
r
d
r
+ A = kr =⇒ (e−rt/w A) = kre−rt/w =⇒ A(t) = kw + ce−rt/w . Since A(0) = A0 so
kr − A =⇒
w
dt
w
dt
c = A0 − kw =⇒ A(t) = e−rt/w [kw(ert/w − 1) + A0 ].
A(t)
e−rt/w
A0
(b). limt→∞
= limt→∞
[kw(ert/w − 1) + A0 ] = limt→∞ [k +
− k e−rt/w ] = k. This is
V (t)
w
w
reasonable since the volume remains constant, and the solution in the tank is gradually mixed with and
(c)2017 Pearson Education. Inc.
59
replaced by the solution of concentration k flowing in.
A1 (t)
dA1
dA1
dA1
= c1 r1 − c2 r2 =⇒
= c 1 r1 − r 2
=⇒
= c 1 r1 −
dt
dt
V1 (t)
dt
r2
dA1
r2
A1 (t) =⇒
A1 = c1 r1 .
+
(r1 − r2 )t + V1
dt
(r1 − r2 )t + v1
A1
A2 (t)
dA2
dA2
dA2
For the bottom tank we have:
− r3
= c2 r2 − c3 r3 =⇒
= r2
=⇒
=
dt
dt
(r1 − r2 )t + V1
V2 (t)
dt
A1
A2 (t)
dA2
r2 A1
r3
− r3
=⇒
A2 =
.
r2
+
(r1 − r2 )t + V1
(r2 − r3 )t + V2
dt
(r1 − r2 )t + V2
(r1 − r2 )t + V1
8. (a). For the top tank we have:
dA1
dA1
dA1
r2
4
2
A1 = c1 r1 =⇒
+
+
A1 = 3 =⇒
+
A1 =
dt
(r1 − r2 )t + v1
dt
2t + 40
dt
t + 20
d
c
3 =⇒ [(t + 20)2 A] = 3(t + 20)2 =⇒ A1 = t + 20 +
but A1 (0) = 4 so c = −6400. Consequently
dt
(t + 20)2
6400
dA2
6400
dA2
2
3
3
t + 20 −
=⇒
. Then
A1 (t) = t + 20 −
+
A2 =
+
A2 =
2
2
(t + 20)
dt
t + 20 t + 20
(t + 20)
dt
t + 20
2[(t + 20)3 − 6400]
d
t + 20
2[(t + 20)3 − 6400]
12800t
=⇒ A2 (t) =
=⇒ [(t + 20)3 A2 ] = (t + 20)3
+
−
3
(t + 20)
dt
(t + 20)3
2
(t + 20)3
t + 20
80000
k
12800t
but A2 (0) = 20 so k = 80000. Thus A2 (t) =
+
and in particular
−
(t + 20)3
2
(t + 20)3
(t + 20)3
119
A2 (10) =
≈ 13.2 g.
9
(b). From part (a)
di
R
1
di
d 40t
+ i = E(t) =⇒
+ 40i = 200 =⇒
(e i) =
dt
L
L
dt
dt
40t
−40t
−40t
200e =⇒ i(t) = 5 + ce
. But i(0) = 0 =⇒ c = −5. Consequently i(t) = 5(1 − e
).
1
. Then
9. Let E(t) = 20, R = 4 and L = 10
dq
1
E
dq
d
+
q=
=⇒
+10q = 20 =⇒ (qe10t ) = 20e10t =⇒
dt RC
R
dt
dt
q(t) = 2 + ce−10t . But q(0) = 0 =⇒ c = −2 so q(t) = 2(1 − e−40t ).
1
and E(t) = 100. Then
10. Let R = 5, C = 50
di
R
1
di
+ i = E(t) =⇒
+ 3i = 15 sin 4t =⇒
dt
L
L dt
3t
3e
d 3t
3
4
(e i) = 15e3t sin 4t =⇒ e3t i =
(3 sin 4t − 4 cos 4t) + c =⇒ i = 3
sin 4t − cos 4t + ce−3t , but
dt
5
5
5
12
3
−3t
i(0) = 0 =⇒ c =
so i(t) = (3 sin 4t − 4 cos 4t + 4e ).
5
5
11. Let R = 2, L = 23 and E(t) = 10 sin 4t. Then
12. Let R = 2, C = 18 and E(t) = 10 cos 3t. Then
dq
1
E
dq
d
+
q=
=⇒
+ 4q = 5 cos 3t =⇒ (e4t q) =
dt
RC
R
dt
dt
e4t
5e4t cos 3t =⇒ e4t q =
(4 cos 3t + 3 sin 3t) + c =⇒ q(t) = 15 (4 cos 3t + 3 sin 3t) + ce4t , but q(0) = 1 =⇒ c =
5
dq
1
1
1
(4 cos 3t + 3 sin 3t) + e−4t and i(t) =
= (9 cos 3t − 12 sin 3t − 4e−4t ).
5
5
dt
5
dq
1
E
13. In an RC circuit for t > 0 the differential equation is given by
+
q = . If E(t) = 0 then
dt
RC
R
1
d t/RC
dq
t/RC
−t/RC
q) = 0 =⇒ q = ce
and if q(0) = 5 then q(t) = 5e
. Then limt→∞ q(t) =
+
q = 0 =⇒ (e
dt RC
dt
0. Yes, this is reasonable. As the time increases and E(t) = 0, the charge will dissipate to zero.
(c)2017 Pearson Education. Inc.
60
1
E0
14. In an RC circuit the differential equation is given by i +
q=
. Differentiating this equation with
RC
R
1 dq
dq
di
1
d
di
+
= 0, but
= i so
+
i = 0 =⇒ (et/RC i) = 0 =⇒ i(t) = ce−t/RC .
respect to t we obtain
dt RC dt
dt
dt RC
dt
E0
E0 −t/RC
Since q(0) = 0, i(0) =
=⇒ d = E0 k so q(t) = E0 k(1 − e−t/RC ). Then
and so i(t) =
e
R
R
limt→∞ q(t) = E0 k, and lim t → ∞i(t) = 0.
q(t)
E0k
t
Figure 0.0.49: Figure for Problem 14
di
di
R
E(t)
R
E0
15. In an RL circuit,
+ i =
and since E(t) = E0 sin ωt, then
+ i =
sin ωt =⇒
dt
L
L
dt
L
L
E0 Rt/L
E0
d Rt/L
i) =
sin ωt =⇒ i(t) = 2
[R sin ωt − ωL cos ωt] + Ae−Rt/L . We can write this
(e
e
dt
L
R + L2 ω 2
R
ωL
E0
√
sin ωt − √
cos ωt + Ae−Rt/L . Defining the phase φ by
as i(t) = √
R 2 + L2 ω 2
R 2 + L2 ω 2
R 2 + L2 ω 2
R
ωL
E0
cos φ = √
, sin φ = √
, we have i(t) = √
[cos φ sin ωt−sin φ cos ωt]+Ae−Rt/L .
2
2
2
2
2
2
2
R +L ω
R +L ω
R + L2 ω 2
E0
That is, i(t) = √
sin (ωt − φ) + Ae−Rt/L .
2
R + L2 ω 2
Transient part of the solution: iT (t) = Ae−Rt/L .
E0
Steady state part of the solution: iS (t) = √
sin (ωt − φ).
2
R + L2 ω 2
di
E0
R
+ ai =
, i(0) = 0, where a = , and E0 denotes the
dt
L
L
constant EMF. An integrating factor for the differential equation is I = eat , so that the differential equation
E0 at
E0
d at
(e i) =
e . Integrating yields i(t) =
+ c1 e−at . The given initial
can be written in the form
dt
L
aL
E0
E0
E0
E0
condition requires c1 +
= 0, so that c1 = − . Hence i(t) =
(1 − e−at ) =
(1 − e−at ).
aL
aL
aL
R
16. We must solve the initial value problem
17.
d
E0 (1/RC−a)t
1
E(t)
dq
1
E0 −at
dq
=⇒ (et/RC q) =
=⇒
+
q=
=⇒
+
q=
e
e
dt
RC
R
dt
RC
R
dt
R
(c)2017 Pearson Education. Inc.
61
E0 C
E0 C
e(1/RC−a)t + k =⇒ q(t) =
e−at + ke−t/RC . Imposing the initial condition
1 − aRC
1 − aRC
E0 C
E0 C
q(0) = 0 (capacitor initially uncharged) requires k = −
, so that q(t) =
(e−at − e−t/RC ).
1 − aRC
1 − aRC
1 −t/RC
dq
E0 C
− ae−at .
Thus i(t) =
=
e
dt
1 − aRC RC
q(t) = e−t/RC
d2 q
1
1 2
di
1
dq
1
+
q = 0 =⇒ i
+
q = 0, since i =
. Then idi = −
qdq =⇒ i2 = −
q + k but
2
dt
LC
dq
LC
dt
LC
LC
q2 − q2
q2 − q2
q2
1 2
q2
dq
=⇒
=⇒
q(0) = q0 and i(0) = 0 so k = 0 =⇒ i2 = −
q + 0 =⇒ i = ± √0
= ± √0
LC
LC
LC
LC
LC
dt
dt
t
t
dq
= ±√
=⇒ sin−1 ( qq0 ) = ± √
+ k1 =⇒ q = q0 sin ± √
+ k1 but q(0) = q0 so q0 =
2
2
LC
LC
LC qo − q
1
t
π
=⇒ q(t) = q0 cos √
q0 sin k1 =⇒ k1 = π2 + 2nπ where n is an integer =⇒ q = q0 sin ± √
+
2
LC
LC
t
q0
dq
sin √
.
= −√
and i(t) =
dt
LC
LC
18.
1
di dq
d2 q
E0
dq
d2
di
+
=
q
=
.
Since
i
=
then
= i . Hence the original equation can be written
2
2
dt
LC
L
dt
dt
dq dt
dq
di
1
E0
1
E0
i2
q2
E0 q
as i
+
q =
=⇒ idi +
qdq =
dq or
+
=
+ A. Since i(0) = 0 and q(0) = q0
dq
LC
L
LC
L
2
2LC
L
1/2
q02
E0 q 0
i2
q2
E0 q
2E0 q
q2
then A =
=⇒ i =
−
. From
+
=
+ A we get that i = 2A +
−
2LC
L
2
2LC
L
L
LC
1/2
2 1/2
(E0 C)2
(2E0 C)2
(q − E0 C)2
dq
q − E0 C
2
√
and we let D = 2A +
=⇒
2A +
−
then i
=D 1−
LC
LC
LC
dt
D LC
√
√
q − E0 C
q − E0 C
√
√
LC sin−1
= t + B. Then since q(0) = 0 so B = LC sin−1
and therefore
D LC D LC √
q − E0 C
dq
t+B
t+B
t+B
√
= sin √
=⇒ q(t) = D LC sin √
+ E0 c =⇒ i =
. Since
= D cos √
dt
D LC
LC
LC
LC
2
2
2A + (E0 C)
q
E0 q 0
|q0 − E0 C|
.
D2 =
and A = 0 −
we can substitute to eliminate A and obtain D = ± √
LC
LC
2LC L
t+B
+ E0 c.
Thus q(t) = ±|q0 − E0 C| sin √
LC
19.
Solutions to Section 1.8
True-False Review:
(a): TRUE. We have
f (tx, ty) =
3(yt)2 − 5(xt)(yt)
3y 2 t2 − 5xyt2
3y 2 − 5xy
=
=
= f (x, y),
2(xt)(yt) + (yt)2
2xyt2 + y 2 t2
2xy + y 2
so f is homogeneous of degree zero.
(c)2017 Pearson Education. Inc.
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(b): FALSE. We have
f (tx, ty) =
y 2 t2 + xt
y2 t + x
(yt)2 + xt
y2 + x
= 2 2
= 2
,
= 2
2
2
2
2
2
(xt) + 2(yt)
x t + 2y t
x t + 2y t
x + 2y 2
so f is not homogeneous of degree zero.
3
2
(c): FALSE. Setting f (x, y) = xy+xy
3 +1 , we have
f (tx, ty) =
(xt)3 + (xt)(yt)2
x3 t3 + xy 2 t3
=
= f (x, y),
(yt)3 + 1
y 3 t3 + 1
so f is not homogeneous of degree zero. Therefore, the differential equation is not homogeneous.
4 −2
y
(d): TRUE. Setting f (x, y) = xx2 +y
2 , we have
f (tx, ty) =
x4 y −2 t2
x4 y −2
(xt)4 (yt)−2
=
=
= f (x, y).
(xt)2 + (yt)2
x 2 t 2 + y 2 t2
x2 + y 2
Therefore, f is homogeneous of degree zero, and therefore, the differential equation is homogeneous.
(e): TRUE. This is verified in the calculation leading to Theorem 1.8.5.
(f ): TRUE. This is verified in the calculation leading to (1.8.12).
(g): TRUE. We can rewrite the equation as
y −
√
xy =
√
xy 1/2 ,
√
√
which is the proper form for a Bernoulli equation, with p(x) = − x, q(x) = x, and n = 1/2.
(h): FALSE. The presence of an exponential exy involving y prohibits this equation from having the proper
form for a Bernoulli equation.
(i): TRUE. After dividing the differential equation through by y, it becomes
Bernoulli equation with p(x) = x, q(x) = x2 , and n = 2/3.
Unless otherwise indicated in this section v =
dy
+ xy = x2 y 2/3 , which is a
dx
y dy
dv
,
=v+x
and t > 0.
x dx
dx
Problems:
5(xt) + 2(yt)
t(5x + 2y)
5x + 2y
=
=
= f (x, y). Thus, f is homogeneous of degree zero.
9(xt) − 4(yt)
t(9x − 4y)
9x − 4y
y
5 + 2x
5 + 2v
=
f (x, y) =
= F (v).
9 − 4 xy
9 − 4v
1. f (tx, ty) =
2. f (tx, ty) = 2(xt) − 5(yt) = t(2x − 5y) = f (x, y). Thus, f is not homogeneous of degree zero.
3. f (tx, ty) =
zero. f (x, y) =
ty
(tx) sin ( tx
ty ) − (ty) cos ( tx )
y
x
sin xy − xy cos xy
y
x
=
=
x sin xy − y cos xy
y
= f (x, y). Thus f is homogeneous of degree
sin v1 − v cos v
= F (v).
v
(c)2017 Pearson Education. Inc.
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3(tx)2 + 5(ty)2
3x2 + 5y 2
4. f (tx, ty) =
=
= f (x, y). Thus f is homogeneous of degree zero. f (x, y) =
2(tx) + 5(ty) √
2x + 5y
y
3 + 5( x )2
3x2 + 5y 2
3 + 5V 2
=
=
= F (v).
y
2x + 5y
2 + 5( x )
2 + 5V
5. f (tx, ty) =
tx + 7
x+7
=
. Thus f is not homogeneous of degree zero.
2ty
2y
tx − 2
5(ty) + 3
(3tx − 6) + (10ty + 6)
t(3x + 10y)
3x + 10y
x−2
5y + 3
+
=
=
=
=
+
.
2(ty)
3(ty)
6ty
6ty
6y
2y
3y
3x + 10y
x
5
1
5
Thus, f is homogeneous of degree zero. We have f (x, y) =
=
+ =
+ = F (v).
6y
2y 3
2v 3
6. f (tx, ty) =
(tx)2 + (ty)2
x2 + y 2
7. f (tx, ty) =
=
= f (x, y). Thus f is homogeneous of degree zero. f (x, y) =
tx
x
y 2
2
2
2
√
|x| 1 + ( x )
x +y
y
= − 1 + v 2 = F (v).
=
=− 1+
x
x
x
(tx)2 + 4(ty)2 − (tx) + (ty)
x2 + 4y 2 − x + y
8. f (tx, ty) =
=
= f (x, y). Thus f is homogeneous of
x + 3y
(tx) + 3(ty)
√
1 + 4( xy )2 − 1 + xy
x2 + 4y 2 − x + y
1 + 4v 2 − 1 + v
=
=
= F (v).
degree zero. f (x, y) =
y
x + 3y
1 + 3x
1 + 3v
9. By inspection the differential equation is first-order homogeneous. We therefore let y = xV in which case
y = xV + V . Substituting these results into the given differential equation yields xV + V = V 2 + V + 1,
or equivalently, xV = V 2 + 1. Separating the variables and integrating yields
1
dV =
V2+1
y
1
dx =⇒ arctan
= ln |x| + c1 =⇒ y(x) = tan(x ln cx).
x
x
dy
y dy
y
dv
dv
3v
= 3y =⇒ (3 − 2 )
= 3 =⇒ (3 − 2v) v + x
= 3v =⇒ x
=
− v =⇒
dx
x dx
x
dx
dx
3 − 2v
dx
3 − 2v
3x
3x
3
y
dv =
=⇒ −
− ln |v| = ln |x| + c1 =⇒ −
− ln | | = ln |x| + c1 =⇒ ln y = −
+ c2 =⇒
2
2v
x
2v
2y
x
2y
y 2 = ce−3x/y .
10. (3x − 2y)
dv
dx
dy
dy
dv
(x + y)2
1
y 2
1
=⇒
=⇒ v + x
=
=
1+
= (1 + v)2 =⇒
=
=⇒ tan−1 v =
2
2
dx
2x
dx
2
x
dx
2
v
+
1
x
y 1
1
ln |x| + c =⇒ tan−1
= ln |x| + c.
2
x
2
y
y dy
y
dy
x
y
dv
= cos
12. sin
x
− y = x cos
=⇒ sin
−
=⇒ sin v v + x
−v =
x dx x
y
dx x
x
dx
y sin v
dx
dv
= cos v =⇒
cos v =⇒ sin v x
dv =
=⇒ − ln | cos v| = ln |x| + c1 =⇒ x cos
= c2 =⇒
dx
cos
v
x
x
c
y(x) = x cos−1
.
x
11.
(c)2017 Pearson Education. Inc.
64
√
dy
16x2 − y 2 + y
dv
dv
dy
13.
=
=
=⇒
= 16 − ( xy )2 + xy =⇒ v + x
= 16 − v 2 + v =⇒ √
dx
x
dx
dx
16 − v 2
dx
y
) = ln |x| + c.
=⇒ sin−1 ( v4 ) = ln |x| + c =⇒ sin−1 ( 4x
x
(9x2 + y 2 ) + y
14. We first rewrite the given differential equation in the equivalent form y =
. Factoring
x
y 2
|x| 9 + ( x ) + y
out an x2 from the square root yields y =
. Since we are told to solve the differential
x
equation on the interval x > 0 we have |x| = x, so that y = 9 + ( xy )2 + xy , which we recognize as being
homogeneous. We therefore√let y = xV , so that y = xV√ + V . Substitution into the preceding differential
equation yields xV + V = 9 + V 2 + V , that is xV = 9 + V 2 . Separating the variables in this equation
√
1
1
dV = dx. Integrating we obtain ln (V + 9 + V 2 ) = ln c1 x. Exponentiating both sides
we obtain √
2
x
9+V
√
y
yields V + 9 + V 2 = c1 x. Substituting
= V and multiplying through by x yields the general solution
x
y + 9x2 + y 2 = c1 x2 .
15. The given differential equation can be written in the equivalent form
dy
y(x2 − y 2 )
=
,
dx
x(x2 + y 2 )
which we recognize as being first order homogeneous. The substitution y = xv yields
v+x
so that
dv
v(1 − v 2 )
2v 3
dv
=⇒ x
,
=
=−
2
dx
1+v
dx
1 + v2
1 + v2
dv = −2
v3
dx
v −2
=⇒ −
+ ln |v| = −2 ln |x| + c1 .
x
2
Consequently,
−
x2
+ ln |xy| = c1 .
2y 2
dx
dy
dy
y y
dv
dv
+ y ln x = y ln y =⇒
= ln =⇒ v + x
= v ln v =⇒
=⇒ ln | ln v − 1| =
dx
dx
x x
dx
v(ln v − 1)
x
ln xy − 1
ln |x| + c1 =⇒
= c =⇒ y(x) = xe1+cx .
x
16. x
dv
dv
dy
y 2 + 2xy − 2x2
v 2 + 2v − 2
−v 3 + 2v 2 + v − 2
v2 − v + 1
=⇒
v+x
=⇒
x
= 2
=
=
=⇒
dv =
3
2
dx
x − xy + y 2
dx
1 − v + v2
v2 − v + 1
v − 2v − v + 2
dx
2
dx
dx
dx
v −v+1
1
1
1
dv = −
−
=⇒
dv = −
=⇒
−
+
=⇒
x
(v − 1)(v + 2)(v + 1)
x
v − 2 2(v − 1) 2(v + 1)
x
2
(v − 2)2 (v + 1) 1
1
= −2 ln x+c2 =⇒ (y − 2x) (y + x) =
ln |v − 2|− ln |v − 1|+ ln |v + 1| = − ln |x|+c1 =⇒ ln 2
2
v−1
y−x
c.
y 2 dv
y dy
2
2
2
2
2
− (e−v +
= 0 =⇒ 2v v + x
− e−y /x + 2
18. 2xydy − (x2 e−y /x + 2y 2 )dx = 0 =⇒ 2
x dx
x
dx
17.
(c)2017 Pearson Education. Inc.
65
2v 2 ) = 0 =⇒ 2vx
x2 ln (ln (cx)).
2
dx
dv
2
2
2
2
= e−v =⇒ ev (2vdv) =
=⇒ ev = ln |x| + c1 =⇒ ey /x = ln (cx) =⇒ y 2 =
dx
x
dy
dy
dv
dv
= y 2 + 3xy + x2 =⇒
= ( xy )2 + 3 xy + 1 =⇒ v + x
= v 2 + 3v + 1 =⇒ x
= (v + 1)2 =⇒
19. x2
dx
dx
dx
dx
dx
1
y
dv
1
1
=
=⇒ −
= ln |x| + c1 =⇒ − y
= ln |x| + c1 =⇒
= −
=⇒ y(x) =
(v + 1)2
x
v
+
1
+
1
x
ln
(cx)
x
1
.
−x 1 +
ln (cx)
√
√
1 + ( xy )2 − 1
dy
x2 + y 2 − x
1 + v2 −
1 + v2 − x
dv
dy
dv
20.
=⇒ v + x
=
=⇒
=
=
=⇒ x
=
=⇒
y
dx
y
dx
dx
v
dx
v
x
dx
c
v
√
dv =
=⇒ ln |1 − u| = ln |x| + c1 =⇒ |x(1 − u)| = c2 =⇒ 1 − u =
=⇒ u2 =
x
x
1 + v2 − 1 − v2
c2
c
c2
c
2
−
2
=
− 2 =⇒ y 2 = c2 − 2cx.
+
1
=⇒
v
x2
x
x2
x
y
dy
dv
dv
dy
= v(4 − v) =⇒ 2x
21. 2x(y + 2x)
= y(4x − y) =⇒ 2
+2
= xy (4 − xy ) =⇒ 2(v + 2) v + x
=
dx
x
dx
dx
dx
dx
v+2
3v 2
dv = −3
=⇒ 2
=⇒ 2 ln |v| − v4 = −3 ln |x| + c1 =⇒ y 2 = cxe4x/y .
−
v+2
v2
x
dx
dv
dy
dv
= x tan ( xy ) + y =⇒ v + x
= tan v + v =⇒ x
= tan v =⇒ cot vdv =
=⇒ ln | sin v| =
dx
dx
dx
x
−1
−1
ln |x| + c1 =⇒ sin v = cx =⇒ v = sin (cx) =⇒ y(x) = x sin (cx).
22. x
2
x
x2 + y 2 + y 2
y
dv
dy
dv
+1 +
( v1 )2 + 1 + v =⇒
23.
=⇒
=
=⇒ v + x
=
=
xy
dx
y
x
dx
dx
√
√
dv
dx
( v1 )2 + 1 =⇒ =
=⇒ ln |v + 1 + v 2 | = ln |x|+c =⇒ v + 1 + v 2 = cx =⇒ xy + 1 + ( xy )2 =
x
( 1 )2
dy
x
=
dx
v
[(cx)2 − 1]
y
xc =⇒ 2( x ) + 1 = (cx)2 =⇒ y 2 = x2
.
2
24. The given differential equation can be written as (x−4y)dy = (4x+y)dx. Converting to polar coordinates
we have x = r cos θ =⇒ dx = cos θdr − r sin θdθ, and y = r sin θdr + r cos θdθ. Substituting these results into
the preceding differential equation and simplifying yields the separable equation 4r−1 dr = dθ which can be
integrated directly to yield 4 ln r = θ + c, so that r = c1 eθ/4 .
x
dy
2(2y − x)
dy
y +1
=
. Since x = 0, divide the numerator and denominator by y yields
=
.
dx
x+y
dx
2(2 − xy )
2(2 − v)
dy
dx
dv
dv
v+1
= v+y
=⇒ v + y
=
=⇒
dv =
=⇒
Now let v = xy so that
2
dy
dy
dy
2(2 − v)
2v − 3v + 1
y
dv
dv
(v − 1)2
−6
= ln (c2 |y|) =⇒ (x − y)2 = c(y − 2x)3 . Since y(0) = 2
+2
= ln |y| + c1 =⇒ ln
2v − 1
v−1
|2v − 1|3
1
1
then c = . Thus, (x − y)2 = (y − 2x)3 .
2
2
25.
(c)2017 Pearson Education. Inc.
66
2 − xy
dy
dv
2x − y
dy
2−v
dv
2 − 2v − 4v 2
1
1 + 4v
=
=⇒
=
=
=⇒ x
=
=⇒
dv =
y =⇒ v+x
2
dx
x + 4y
dx
1 + 4x
dx
1 + 4v
dx
1 + 4v
2 2v + v − 1
dx
1
1
1
=⇒ ln |2v 2 + v − 1| = − ln |x| + c =⇒ ln |x2 (2v 2 + v − 1)| = c =⇒ ln |2y 2 + yx − x2 | = c, but
−
x
2
2
2
1
1
1
2
2
y(1) = 1 so c = ln 2. Thus ln |2y + yx − x | = ln 2 and since y(1) = 1 it must be the case that
2
2
2
2y 2 + yx − x2 = 2.
26.
√
dx
y
dv
dv
dy
y − x2 + y 2
dy
y
=−
27.
=
=⇒
= − 1 + ( )2 =⇒ x
= − 1 + v 2 =⇒ √
=⇒
2
dx
x
dx
x
x
dx
x
1+v
√
|x| 2
|x|
ln (v + 1 + v 2 ) = − ln |x| + c1 =⇒ y
+ x + y 2 = c2 . Since y(3) = 4 then c2 = 9. Then take
=1
x
x
since we must have y(3) = 4; thus y + x2 + y 2 = 9.
y 2
√
dx
dy
v
dv
dv
y
28.
=⇒ v + x
=
− = 4−
= v + 4 − v 2 =⇒ √
=⇒ sin−1 = ln |x| + c =⇒
dx x
x
dx
x
2
4 − v2
−1 y
sin
= ln x + c since x > 0.
2x
x + ay
dv
1 + v2
dy
=
. Substituting y = xv and simplifying yields x
=
. Separating the variables
dx
ax − y
dx
a−v
1
and integrating we obtain a tan−1 v − ln (1 + v 2 ) = ln x + ln c or equivalently, a tan−1 xy − 12 ln (x2 + y 2 ) =
2
ln c. Substituting for x = r cos θ, y = r sin θ yields aθ − ln r = ln c. Exponentiating then gives r = keaθ .
√
(b). The initial condition y(1) = 1 corresponds to r( π4 ) = 2. Imposing this condition on the polar form
√ −π/8
of the solution obtained in (a) yields k = 2e
. Hence, the solution to the initial value problem is
√ (θ−π/4)/2
1
dy
2x + y
. When a = , the differential equation is
r = 2e
=
. Consequently every solution
2
dx
x − 2y
x
curve has a vertical tangent line at points of intersection with the line y = . The maximum interval of
2
x
existence for the solution of the initial value problem can be obtained by determining where y = intersects
2
√ (θ−π/4)/2
x
1
. The line y = has a polar equation tan θ = . The corresponding values of θ
the curve r = 2e
2
2
1
are θ = θ1 = tan−1 ≈ 0.464, θ = θ2 = θ1 + π ≈ 3.61. Consequently, the x-coordinates of the intersection
2
√
√
points are x1 = r cos θ1 = 2e(θ1 −π/4)/2 cos θ1 ≈ 1.08, x2 = r cos θ2 = 2e(θ2 −π/4)/2 cos θ2 ≈ −5.18. Hence
the maximum interval of existence for the solution is approximately (−5.18, 1.08).
29. (a).
(c). See the accompanying figure.
x2 + y 2
dy
dy
dy
. Hence 2x + 2y
= 2c
=⇒
=
2y
dx
dx
dx
x
dy
2xy
y 2 − x2
dy
dv
. Orthogonal trajectories satisfies:
= 2
=
. Let y = vx so that
= v+x .
2
c−y
x −y
dx
2xy
dx
dx
dv
v2 + 1
1
Substituting these results into the last equation yields x
=−
=⇒ ln |v + 1| = − ln |x| + c1 =⇒
dx
2v
2
y
c2
+1=
=⇒ x2 + y 2 = 2kx.
2
x
x
30. Given family of curves satisfies: x2 + y 2 = 2cy =⇒ c
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y(x)
2
1
-6
-5
-4
-3
-2
-1
1
0
x
-1
-2
-3
-4
Figure 0.0.50: Figure for Problem 29(c)
y(x)
x
Figure 0.0.51: Figure for Problem 30
x2 + y 2
dy
. Hence 2(x−c)+2(y −c)
=
2(x + y)
dx
c−x
dy
y 2 − 2xy − x2
y 2 + 2xy − x2
0 =⇒
.
Orthogonal
trajectories
satisfies:
. Let y = vx so
= 2
=
y−c
y + 2xy − x2
dx
x2 + 2xy − y 2
dy
dv
dv
v 2 + 2v − 1
that
=⇒
= v + x . Substituting these results into the last equation yields v + x
=
dx
dx
dx
1 + 2v − v 2
31. Given family of curves satisfies: (x−c)2 +(y −c)2 = 2c2 =⇒ c =
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1 + 2v − v 2
1
dv = dx =⇒
3
2
v −v +v−1
x
2k 2 .
1
2v
− 2
v−1 v +1
dv =
1
dx =⇒ x2 +y 2 = 2k(x−y) =⇒ (x−k)2 +(y +k)2 =
x
y(x)
x
Figure 0.0.52: Figure for Problem 31
32. (a). Let r represent the radius
√ of one of the circles with center at (a.ma) and passing through (0, 0).
r =√ (a − 0)2 + (ma − 0)2 = |a| 1 + m2 . Thus, the circle’s equation can be written as (x−a)2 +(y−ma)2 =
(|a| 1 + m2 )2 or (x − a)2 + (y − ma)2 = a2 (1 + m2 ).
x2 + y 2
. Differentiating the first equation with respect
2(x + my)
dy
a−x
dy
y 2 − x2 − 2mxy
x and solving we obtain
=
. Substituting for a and simplifying yields
=
.
dx
y − ma
dx
my 2 − mx2 + 2xy
y
y
2
m − m( ) − 2 x
mx2 − my 2 − 2xy
dy
dy
. Let y = vx so that
= 2
=⇒
= y 2 x
Orthogonal trajectories satisfies:
2
dx
y − x − 2mxy
dx
( x ) − 1 − 2m xy
dy
dv
dv
m − mv 2 − 2v
xdv
= v + x . Substituting these results into the last equation yields v + x
= 2
=⇒
=
dx
dx
dx
v − 1 − 2mv
dx
2
2
dx
v − 2mv − 1
dx
dv
2v
(m − v)(1 + v )
dv =
=⇒
dv =
=⇒
−
=⇒ ln |v − m| −
v 2 − 2mv − 1
(m − v)(1 + v 2 )
x
v−m
1 + v2
x
2
2
2
2
2
2
ln (1 + v ) = ln |x| + c1 =⇒ v − m = c2 x(1 + v ) =⇒ y − mx = c2 x + c2 y =⇒ x + y + cmx − cy = 0.
Completing the square we obtain (x + cm/2)2 + (y − c/2)2 = c2 /4(m2 + 1). Now letting b = c/2, the last
equation becomes (x + bm)2 + (y − b)2 = b2 (m2 + 1) which is a family or circles lying on the line y = −my
and passing through the origin.
(b). (x − a)2 + (y − ma)2 = a2 (1 + m2 ) =⇒ a =
(c). See the accompanying figure.
33. x2 + y 2 = c =⇒
m2 − tan ( π4 )
dy
x
−x/y − 1
x+y
= − = m2 . m1 =
=
. Let y = vx so that
π =
dx
y
1 + m2 tan ( 4 )
1 − x/y
x−y
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y(x)
x
Figure 0.0.53: Figure for Problem 32(c)
dy
dv
dv
1+v
1−v
dv =
= v + x . Substituting these results into the last equation yields v + x
=
=⇒
dx
dx
1−v
1 + v2
dx
dx
1
dx
v
1
dv =
+ 2
=⇒
−
=⇒ − ln (1 + v 2 )+tan−1 v = ln |x|+c1 =⇒ Oblique trajectories:
2
x
1+v
v +1
x
2
ln (x2 + y 2 ) − 2 tan−1 (y/x) = c2 .
m2 − tan ( π4 )
dy
6y/x − 1
6y − x
= 6y/x = m2 . m1 =
=
=
. Let y = vx so that
dx
1 + m2 tan ( π4 )
1 + 6y/x
6y + x
dv
dv
6v − 1
dv
dy
= v + x . Substitute these results into the last equation yields v + x
=
=⇒ x
=
dx
dx
dx
6v
+
1
dx
dx
9
8
(3v − 1)(1 − 2v)
dv =
=⇒
−
=⇒ 3 ln |3v − 1| − 4 ln |2v − 1| = ln |x| + c1 =⇒
6v + 1
3v − 1 2v − 1
x
Oblique trajectories (3y − x)3 = k(2y − x)4 .
34. y = cx6 =⇒
35. x2 + y 2 = 2cx =⇒ c =
y 2 − x2 − 2xy
.
y 2 − x2 + 2xy
2
x +y
2x
2
2
2
m2 − tan ( π4 )
dy
y −x
and
=
= m2 . m1 =
=
dx
2xy
1 + m2 tan ( π4 )
y 2 − x2
−1
2xy
=
y 2 − x2
1+
2xy
dy
dv
= v + x . Substituting these results into the last equadx
dx
dv
v 2 − 2v − 1
dv
−v 3 − v 2 − v − 1
−v 2 − 2v + 1
dx
tion yields v + x
= 2
=⇒ x
=
=⇒
dv =
=⇒
2
3
2
dx
v + 2v − 1
dx
v + 2v − 1
v +v +v+1
x
dx
2v
1
dv =
− 2
=⇒ ln |v + 1| − ln (v 2 + 1) = ln |x| + c1 =⇒ ln |y + x| = ln |y 2 + x2 | + c1 =⇒
v+1 v +1
x
Oblique trajectories: x2 + y 2 = 2k(x + y) or, equivalently, (x − k)2 + (y − k)2 = 2k 2 .
Let y = vx so that
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dy
y
m2 − tan α0
−y/x − tan α0
=
. Let y = vx so
= −cx−2 = − . m1 =
dx
x
1 + m2 tan α0
1 − y/x tan α0
dy
dv
dv
tan α0 + v
that
= v + x . Substituting these results into the last equation yields v + x
=
=⇒
dx
dx
dx
v tan α0 − 1
2v tan α0 − 2
2dx
dv = −
=⇒ ln |v 2 tan α0 − 2v − tan α0 | = −2 ln |x| + c1 =⇒ (y 2 − x2 ) tan α0 −
v 2 tan α0 − 2v − tan α0
x
2xy = k.
36. (a). y = cx−1 =⇒
(b). See the accompanying figure.
y(x)
x
Figure 0.0.54: Figure for Problem 36(b)
dy
m2 − tan α0
−x/y − m
x
x + my
dy
=
= − . m1 =
=
. Let y = vx so that
=
dx
y
1 + m2 tan α0
1 − (x/y)m
mx − y
dx
dv
dv
1 + mv
dv
1 + v2
v + x . Substituting these results into the last equation yields v + x
=
=⇒ x
=
=⇒
dx
dx
m−v
dx
m−v
dx
dx
m
v
1
v−m
dv = −
dv = −
−
=⇒
=⇒ ln (1 + v 2 ) − m tan v = − ln |x| + c1 . In polar
1 + v2
x
1 + v2
1 + v2
x
2
coordinates, r = x2 + y 2 and θ = tan−1 y/x, so this result becomes ln r − mθ = c1 =⇒ r = emθ where k is
an arbitrary constant.
37. (a). x2 +y 2 = c =⇒
(b). See the accompanying figure.
dy 1
dy 1 2
− y = 4x2 y −1 cos x. This is a Bernoulli equation. Multiplying both sides y results in y
− y =
dx x
dx x
du
dy
dy
1 du
dy
1
4x2 cos x. Let u = y 2 so
= 2y
or y
=
. Substituting these results into y
− y 2 = 4x2 cos x
dx
dx
dx
2 dx
dx x
d −2
du 2
yields
− u = 8x2 cos x which has an integrating factor I(x) = x−2 =⇒
(x u) = 8 cos x =⇒ x−2 u =
dx x
dx
8 cos xdx + c =⇒ x−2 u = 8 sin x + c =⇒ u = x2 (8 sin x + c) =⇒ y 2 = x2 (8 sin x + c).
38.
39. y −3
dy
dy
du
1
+ y −2 tan x = 2 sin x. This is a Bernoulli equation. Let u = y −2 so
= −2y −3
or
dx
2
dx
dx
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y(x)
x
Figure 0.0.55: Figure for Problem 37(b)
dy
1 du
du
=
. Substituting these results into the last equation yields
−u tan x = −4 sin x. An integrating
dx
2 dx
dx
d
(u cos x) = −4 cos x sin x =⇒ u cos x = 4 cos x sin xdx =⇒
factor for this equation is I(x) = cos x. Thus,
dx
c
1
(cos2 x + c) =⇒ y −2 = 2 cos x +
.
u(x) =
cos x
cos x
y −3
dy
1 dy 3 2/3
dy
dy 3
2
= 6x2 ln x. Let u = y 2/3 =⇒
− y = 6y 1/3 x2 ln x or 1/3
− y
= y −1/3 . Substituting
dx 2x
dx 2x
dx
3
dx
y
du
3 2/3
1
1 dy
= 6x2 ln x yields
−
y
− u = 4x2 ln x. An integrating factor for this
these results into 1/3
dx
2x
dx
x
y
1
d −1
−1
equation is I(x) = so
(x u) = 4x ln x =⇒ x u = 4 x ln xdx + c =⇒ x−1 u = 2x2 ln x − x2 + c =⇒
x
dx
u(x) = x(2x2 ln x − x2 + c) =⇒ y 2/3 = x(2x2 ln x − x2 + c).
40.
√
√
dy
dy
dy
du
2
2
+ y = 6 1 + x2 y 1/2 or y −1/2
+ y 1/2 = 6 1 + x2 . Let u = y 1/2 =⇒ 2
= y −1/2 .
dx
x
dx
x
dx
dx
√
√
dy
du
2
1
+ y 1/2 = 6 1 + x2 yields
+ u = 3 1 + x2 . An integrating
Substituting these results into y −1/2
dx
x
dx
x
√
√
d
2
factor for this equation is I(x) = x so
(xu) = 3x 1 + x =⇒ xu = x 1 + x2 dx + c =⇒ xu =
dx
1
c
1
c
(1 + x2 )3/2 + c =⇒ u = (1 + x2 )3/2 + =⇒ y 1/2 = (1 + x2 )3/2 + .
x
x
x
x
41.
dy
dy
du
dy
2
2
+ y = 6y 2 x4 or y −2
+ y −1 = 6x4 . Let u = y −1 =⇒ −
= y −2 . Substituting these results
dx x
dx x
dx
dx
du 2
2
dy
+ y −1 = 6x4 yields
− u = −6x4 . An integrating factor for this equation is I(x) = x−2 so
into y −2
dx x
dx x
1
d −2
(x u) = −6x2 =⇒ x−2 u = −2x3 + c =⇒ u = −2x5 + cx2 =⇒ y −1 = −2x5 + cx2 =⇒ y(x) = 2
.
dx
x (c − 2x3 )
42.
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dy
+ y 3 x2
dx
dy
dy
1 du
1 −2
+
y = −x−2 . Let u = y −2 =⇒ −
= y −3 . Substituting
dx 2x
2 dx
dx
du 1
1 −2
−3 dy
−2
2
+
y = −x yields
− u = 2x . An integrating factor for this equation is
these results into y
dx 2x
dx x
1
d −1
I(x) = so
(x u) = 2x =⇒ x−1 u = x2 + c =⇒ u = x3 + cx =⇒ y −2 = x3 + cx.
x
dx
dy
dy
du
2(b − a)
44. (x − a)(x − b)
− y 1/2 = 2(b − a)y or y −1/2
−
y 1/2 = 1. Let u = y 1/2 =⇒ 2
=
dx
dx (x − a)(x − b)
dx
dy
dy
du
2(b − a)
(b − a)
1
y 1/2 = 1 yields
−
u= .
y −1/2 . Substituting these results into y −1/2 −
dx
dx (x − a)(x− b)
dx
(x
−
1)(x
−
b)
2
d x−a
x−a
x−a
1
x−a
so
u =
=⇒
u = [x+(b−
An integrating factor for this equation is I(x) =
x−b
dx x − b
2(x − b)
x−b
2
2
x−b
1 x−b
[x+(b−a) ln |x − b|+c]2 .
a) ln |x − b|+c] =⇒ y 1/2 =
[x+(b−a) ln |x − b|+c] =⇒ y(x) =
2(x − a)
4 x−a
43. 2x
+ y = 0 or y −3
dy
cos x
dy
dy
cos x
du
6
6
+ y = 3y 2/3
or y −2/3
+ y 1/3 = 3
. Let u = y 1/3 =⇒ 3
= y −2/3 . Substituting
dx x
x
dx x
x
dx
dx
dy 6 1/3
cos x
du 2
cos x
=3
+ y
yields
+ u=
. An integrating factor for this equation
these results into y −2/3
dx x
x
dx x
x
d 2
cos x + x sin x + c
.
(x u) = x cos x =⇒ x2 u = cos x + x sin x + c =⇒ y 1/3 =
is I(x) = x2 so
dx
x2
45.
dy
dy
dy
du
+ 4xy = 4x3 y 1/2 or y −1/2
+ 4xy 1/2 = 4x3 . Let u = y 1/2 =⇒ 2
= y −1/2 . Substituting these
dx
dx
dx
dx
dy
du
+ 4xy 1/2 = 4x3 yields
+ 2xu = 2x3 . An integrating factor for this equation is I(x) =
results into y −1/2
dx
dx
d x2
2
2
2
2
2
2
ex so
(e u) = 2ex x3 =⇒ ex u = ex (x2 −1)+c =⇒ y 1/2 = x2 −1+ce−x =⇒ y(x) = [(x2 −1)+ce−x ]2 .
dx
46.
dy
dy
dy
1 du
1
1
−
= 2xy 3 or y −3
−
y −2 = 2x. Let u = y −2 =⇒
= y −3 . Substituting these
dx 2x ln x
dx 2x ln x
2 dx
dx
dy
du
1
1
−
y −2 = 2x yields
+
u = −4x. An integrating factor for this equation is
results into y −3
dx 2x ln x
dx x ln x
ln x
d
.
(u ln x) = −4x ln x =⇒ u ln x = x2 − 2x2 ln x + c =⇒ y 2 = 2
I(x) = ln x so
dx
x (1 − 2 ln x) + c
47.
dy
dy
3x
1 du
1
3
1
dy
−
y=
xy π or y −π −
y 1−π =
. Let u = y 1−π =⇒
= y −π .
dx (π − 1)x
(1 − π)
dx (π − 1)x
1−π
1 − π dx
dx
3x
1
du 1
−π dy
1−π
=
Substituting these results into y
−
y
yields
+ u = 3x. An integrating factor for
dx (π − 1)x
1−π
dx x
1/(1−π)
3
x3 + c
x +c
d
2
3
1−π
=
.
(xu) = 3x =⇒ xu = x + c =⇒ y
=⇒ y(x) =
this equation is I(x) = x so
dx
x
x
48.
dy
dy
du
dy
+y cot x = 8y −1 cos3 x or 2y +y 2 cot x = 8 cos2 x. Let u = y 2 =⇒
= 2y . Substituting these
dx
dx
dx
dx
du
dy
+ y 2 cot x = 8 cos2 x yields
+ u sec x = sec x. An integrating factor for this equation is
results into 2y
dx
dx
−2 cos4 x + c
d
3
I(x) = sin x so
(u sin x) = 8 cos x sin x =⇒ u sin x = −2 cos4 x + c =⇒ y 2 =
.
dx
sin x
49. 2
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√
√ dy
√
√
√
√ dy
du
50. (1− 3) +y sec x = y 3 sec x or (1− 3)y − 3 +y 1− 3 sec x = sec x. Let u = y 1− 3 =⇒
= (1−
dx
dx
dx
√
√ −√3 dy
√ −√3 dy
du
3)y
. Substituting these results into (1− 3)y
+y 1− 3 sec x = sec x yields
+u sec x = sec x.
dx
dx
dx
d
[(sec x+tan x)u] = sec x(sec x+tan x) =⇒
An integrating factor for this equation is I(x) = sec x+tan x so
dx
1/(1−√3)
√
1
c
1− 3
=1+
.
=⇒ y(x) = 1 +
(sec x + tan x)u = tan x + sec x + c =⇒ y
sec x + tan x
sec x + tan x
dy
dy
1 dy
du
2x
2x 1
y = xy 2 or 2
+
+
= x. Let u = y −1 so
= −y −2 . Substituting these
dx
1 + x2
y dx
1 + x2 y
dx
dx
2x 1
du
2x
1 dy
u = −x. An integrating factor for this equation is
+
= x yields
−
results into 2
y dx 1 + x2 y
dx
1 + x2
xdx
u
x
d
u
u
1
1
=−
so
=⇒
=−
+c =⇒
= − ln (1 + x2 )+c =⇒
I(x) =
2
2
2
2
1 + x2 dx 1 + x2
1
+
x
1
+
x
1
+
x
1
+
x
2
1
1
u = (1 + x2 ) − ln (1 + x2 ) + c =⇒ y −1 = (1 + x2 ) − ln (1 + x2 ) + c . Since y(0) = 1 =⇒ c = 1 so
2
2
1
1
= (1 + x2 ) − ln (1 + x2 ) + 1 .
y
2
51.
dy
dy
1 du
dy
+ y cot x = y 3 sin3 x or y −3
+ y −2 cot x = sin3 x. Let u = y −2 =⇒ −
= y −3 . Substituting
dx
dx
2 dx
dx
dy
du
+ y −2 cot x = sin3 x yields
− 2u cot x = −2 sin3 x. An integrating factor for this
these results into y −3
dx
dx
d
equation is I(x) = csc2 x so
(u csc2 x) = − sin x =⇒ u csc2 x = 2 cos x + c. Since y(π/2) = 1 =⇒ c = 1.
dx
1
Thus y 2 =
.
sin2 x(2 cos x + 1)
52.
dy
dv
dy
dy
dv
dy
53.
= F (ax + by + c). Let v = ax + by + c so that
= a+b
=⇒ b
=
− a =⇒
=
dx
dx
dx
dx
dx
dx dv
dv
dv
1 dv
− a = F (v) =⇒
− a = bF (v) =⇒
= bF (v) + a =⇒
= dx.
b dx
dx
dx
bf (v) + a
54.
dv
dy
dy
dv
dv
= dx =⇒
= (9x − y)2 . Let v = 9x − y so that
= 9−
=⇒
= 9 − v 2 =⇒
2
dx
dx
dx
dx
9−v
1
tanh−1 (v/3) = x + c1 but y(0) = 0 so c = 0. Thus, tanh−1 (3x − y/3) = 3x or y(x) = 3(3x − tanh 3x).
3
55.
dv
dy
dv
dy
dv
= (4x + y + 2)2 . Let v = 4x + y + 2 so that
= 4+
=⇒ 2
= dx =⇒
= dx =⇒
2
dx
dx
dx
v +4
v +4
1
tan−1 v/2 = x + c1 =⇒ tan−1 (2x + y/2 + 1) = 2x + c =⇒ y(x) = 2[tan (2x + c) − 2x − 1].
2
dy
dy
1 dv
1 dv
56.
= sin2 (3x − 3y + 1). Let v = 3x − 3y + 1 so that
1−
=⇒ 1 −
= sin2 v =⇒
dx
dx
3 dx
3 dx
dv
= 3 cos2 v =⇒ sec2 vdv = 3 dx =⇒ tan v = 3x + c =⇒ tan (3x − 3y + 1) = 3x + c =⇒ y(x) =
dx
1
[3x − tan−1 (3x + c) + 1].
3
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57. V = xy =⇒ V = xy + y =⇒ y = (V − y)/x. Substitution into the differential equation yields
1
dV
1
(V − y)/x = yF (V )/x =⇒ V = y[F (V ) + 1] =⇒ V = V [F (V ) + 1]/x, so that
= .
V [F (V ) + 1] dx
x
1
dV
1
1
1
=
for F (V ) = ln V − 1 yields
dV = dx =⇒ ln ln V =
V [F (V ) + 1] dx
x
V ln V
x
1
ln cx =⇒ V = ecx =⇒ y(x) = ecx .
x
58. Substituting into
59. (a). y(x) = w(x)−x =⇒ y = w −1. Substituting these results into (1.8.18) yields w −1 = 2xw2 −1 =⇒
w = 2xw2 .
(b). Separating the variables in the preceding differential equation and integrating yields
1
1
dw
=
2
x dx + c =⇒ −w−1 = x2 + c =⇒ w(x) =
,
w2
c 1 − x2
where c1 = −c. Hence, the general solution to (1.8.18) is y(x) =
y(0) = 1 requires that 1 =
1
− x. Imposing the initial condition
c 1 − x2
1
1
=⇒ c1 = 1. Therefore, y(x) =
− x.
c1
1 − x2
dy
dv
60. (a). x = u − 1, y = v + 1 =⇒
=
. Substitution into the given differential equation yields
dx
du
u + 2v
dv
=
.
du
2u − v
(b). The differential equation obtained in (a) is first order homogeneous. We therefore let W = v/u, and
1 + 2W
1 + W2
substitute into the differential equation to obtain W u + W =
=⇒ W u =
. Separating the
2−W
2−W
1
W
2
dW = du. This can be integrated directly to obtain 2 tan−1 W −
−
variables yields
1 + W2
1 + W2
u
1
−1
−1
ln (1 + W 2 ) = ln u + ln c. Simplifying we obtain cu2 (1 + W 2 ) = e4 tan W =⇒ c(u2 + v 2 ) = etan (v/u) .
2
−1
Substituting back in for x and y yields c[(x + 1)2 + (y − 1)2 ] = etan [(y−1)/(x+1)] .
61. (a). y = Y (x) + v −1 (x) =⇒ y = Y (x) − v −2 (x)v (x). Now substitute into the given differential
equation and simplify algebraically to obtain Y (x) + p(x)Y (x) + q(x)Y 2 (x) − v −2 (x)v (x) + v −1 (x)p(x) +
q(x)[2Y (x)v −1 (x) + v −2 (x)] = r(x). We are told that Y (x) is a particular solution to the given differential
equation, and therefore Y (x) + p(x)Y (x) + q(x)Y 2 (x) = r(x). Consequently the transformed differential
equation reduces to −v −2 (x)v (x) + v −1 p(x) + q(x)[2Y (x)v −1 (x) + v −2 (x)] = 0, or equivalently v − [p(x) +
2Y (x)q(x)]v = q(x).
(b). The given differential equation can be written as y − x−1 y − y 2 = x−2 , which is a Riccati differential
equation with p(x) = −x−1 , q(x) = −1, and r(x) = x−2 . Since y(x) = −x−1 is a solution to the given
differential equation, we make a substitution y(x) = −x−1 + v −1 (x). According to the result from part (a),
the given differential equation then reduces to v −(−x−1 +2x−1 )v = −1, or equivalently v −x−1 v = −1. This
d −1
linear differential equation has an integrating factor I(x) = x−1 , so that v must satisfy
(x v) = −x−1 =⇒
dx
1
1
1
1
v(x) = x(c−ln x). Hence the solution to the original equation is y(x) = − +
=
−1 .
x x(c − ln x)
x c − ln x
62. (a). If y = axr , then y = arxr−1 . Substituting these expressions into the given differential equation
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yields arxr−1 + 2axr−1 − a2 x2r = −2x−2 . For this to hold for all x > 0, the powers of x must match up on
either side of the equation. Hence, r = −1. Then a is determined from the quadratic −a + 2a − a2 = −1 ⇐⇒
a2 − a − 2 = 0 ⇐⇒ (a − 2)(a + 1) = 0. Consequently, a = 2, −1 in order for us to have a solution to the given
differential equation. Therefore, two solutions to the differential equation are y1 (x) = 2x−1 , y2 (x) = −x−1 .
(b). Taking Y (x) = 2x−1 and using the result from Problem 61(a), we now substitute y(x) = 2x−1 + v −1
into the given Riccati equation. The result is (−2x−2 −v −2 v )+2x−1 (2x−1 +v −1 )−(4x−2 +4x−1 v −1 +v −2 ) =
−2x−2 . Simplifying this equation yields the linear equation v + 2x−1 v = −1. Multiplying by the integrating
d 2
−1
(x v) = −x2 . Integrating this
factor I(x) = e 2x dx = x2 results in the integrable differential equation
dx
1
1
differential equation we obtain v(x) = x2 − x3 + c = x2 (c1 − x3 ). Consequently, the general solution
3
3
2
3
to the Riccati equation is y(x) = + 2
.
x x (c1 − x3 )
63. (a). y = x−1 + w(x) =⇒ y = −x−2 + w . Substituting into the given differential equation yields
(−x−2 + w ) + 7x−1 (x−1 + w) − 3(x−2 + 2x−1 w + w2 ) = 3x−2 which simplifies to w + x−1 w − 3w2 = 0.
(b). The preceding equation can be written in the equivalent form w−2 w + x−1 w−1 = 3. We let u = w−1 ,
so that u = −w−2 w . Substitution into the differential equation gives, after simplification, u − x−1 u = −3.
An integrating factor for this linear differential equation is I(x) = x−1 , so that the differential equation
d −1
can be written in the integrable form
(x u) = −3x−1 . Integrating we obtain u(x) = x(−3 ln x +
dx
1
c), so that w(x) =
. Consequently the solution to the original Riccati equation is y(x) =
x(c − 3 ln x)
1
1
1+
.
x
c − 3 ln x
dy
du
1 dy
du
+ p(x) ln y = q(x). If we let u = ln y, then
=
and the given equation becomes
+
dx
dx
y dx
dx e p(x)dx q(x)dx + c .
p(x)u = q(x) which is a first order linear and has a solution of the form u = e− p(x)dx
p(x)dx
−1
q(x)dx + c into u = ln y we obtain y(x) = eI [ I(t)q(t)dt+c] where
e
Substituting ln y = e− p(x)dx
64. y −1
I(x) = e p(t)dt and c is an arbitrary constant.
dy
2
1 − 2 ln x
du 2
− ln y =
. Let u = ln y so using the technique of the preceding problem:
− u=
dx x
x⎡
dx x
⎤
dx
dx 1 − 2 ln x
−2
2
1 − 2 ln x
1 − 2 ln x
2
x ⎣ e
x
dx + c1 ⎦ = x2
dx
+
c
and u = e
1 = ln x + cx ,
x
x
x3
65. y −1
2
and since u = ln y, ln y = ln x + cx2 . Now y(1) = e so c = 1 =⇒ y(x) = xex .
du
dy
dy
du
= f (y)
and the given equation f (y)
+ p(x)f (y) = q(x) becomes
+
dx
dx
dx
dx
e p(x)dx q(x)dx + c . Substituting
p(x)u = q(x) which has a solution of the form u(x) = e− p(x)dx
f (y) = e− p(x)dx
e p(x)dx q(x)dx + c into u = f (y) and using the fact that f is invertible, we obtain
y(x) = f −1 I −1
I(t)q(t)dt + c where I(x) = e p(t)dt and c is and arbitrary constant.
66. If u = f (y), then
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dy
dy
1
du
1
tan y = √
. Let u = tan y so that
+ √
= sec2 y
and the given equation
dx
dx
dx
2 1+x
2 1+x
1
1
du
u= √
which is first order linear. An integrating factor for this equation is
+ √
becomes
dx
2 1+x
2 1+x√
√
√
√
√
√
e 1+x
d √1+x
e 1+x
1+x
1+x
√
√
I(x) = e
=⇒
u) =
u =
=⇒ e
=⇒ e 1+x u = e 1+x + c =⇒ u =
(e
dx
2 1+x √
2 1+x
√
√
− 1+x
− 1+x
1 + ce
. But u = tan y so tan y = 1 + ce
or y(x) = tan−1 (1 + ce− 1+x ).
67. sec2 y
Solutions to Section 1.9
True-False Review:
(a): FALSE. The requirement, as stated in Theorem 1.9.4, is that My = Nx , not Mx = Ny , as stated.
(b): FALSE. A potential function φ(x, y) is not an equation. The general solution to an exact differential
equation takes the form φ(x, y, ) = c, where φ(x, y) is a potential function.
(c): FALSE. According to Definition 1.9.2, M (x)dx + N (y)dy = 0 is only exact if there exists a function
φ(x, y) such that φx = M and φy = N for all (x, y) in a region R of the xy-plane.
(d): TRUE. This is the content of part 1 of Theorem 1.9.11.
(e): FALSE. If φ(x, y) is a potential function for M (x, y)dx + N (x, y)dy = 0, then so is φ(x, y) + c for any
constant c.
(f ): TRUE. We have
My = 2e2x − cos y
and
Nx = 2e2x − cos y,
and so since My = Nx , this equation is exact.
(g): FALSE. We have
My =
and
Nx =
(x2 + y)2 (−2x) + 4xy(x2 + y)
(x2 + y)4
(x2 + y)2 (2x) − 2x2 (x2 + y)(2x)
.
(x2 + y)4
Thus, My = Nx , and so this equation is not exact.
(h): FALSE. We have
My = 2y
and
Nx = 2y 2 ,
and since My = Nx , we conclude that this equation is not exact.
(i): FALSE. We have
My = ex sin y cos y + xex sin y cos y
and
Nx = cos y sin yex sin y ,
and since My = Nx , we conclude that this equation is not exact.
Problems:
1. yexy dx + (2y − xe−xy )dy = 0. M = yexy and N = 2y − xe−xy =⇒ My = yxexy + exy and Nx =
xye−xy − e−xy =⇒ My = Nx =⇒ the differential equation is not exact.
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2. [cos (xy) − xy sin (xy)]dx − x2 sin (xy)dy = 0 =⇒ M = cos (xy) − xy sin (xy) and N = −x2 sin (xy) =⇒
My = −2x sin (xy) − x2 y cos (xy) and Nx = −2x sin (xy) − x2 y cos (xy) =⇒ My = Nx =⇒ the differential
equation is exact.
3. (y + 3x2 )dx + xdy = 0. M = y + 3x2 and N = x =⇒ My = 1 and Nx = 1 =⇒ My = Nx =⇒ the
differential equation is exact.
4. 2xey dx + (3y 2 + x2 ey )dy = 0. M = 2xey and N = 3y 2 + x2 ey =⇒ My = 2xey and Nx = 2xey =⇒ My =
Nx =⇒ the differential equation is exact.
5. 2xydx + (x2 + 1)dy = 0. M = 2xy and N = x2 + 1 =⇒ My = 2x and Nx = 2x =⇒ My = Nx =⇒
∂φ
∂φ
the differential equation is exact so there exists a potential function φ such that (a)
= 2xy and (b)
=
∂x
∂x
dh(x)
dh(x)
dh(x)
so from (a), 2xy = 2xy +
=⇒
= 0 =⇒ h(x) is a constant. Since we need just one
2xy +
dx
dx
dx
2
potential function, let h(x) = 0. Thus, φ(x, y) = (x + 1)y; hence, (x2 + 1)y = c.
6. Given (y 2 − 2x)dx + 2xydy = 0 then My = Nx = 2xy so the differential equation is exact and there exists
∂φ
∂φ
a potential function φ such that (a)
= y 2 − 2x and (b)
= 2xy. From (b) φ(x, y) = xy 2 + h(x) =⇒
∂x
∂y
∂φ
dh(x)
dh(x)
dh(x)
= y2 +
so from (a) y 2 +
= y 2 − 2x =⇒
= −2x =⇒ h(x) = −2x where the constant of
∂x
dx
dx
dx
integration has been set to zero since we just need one potential function. φ(x, y) = xy 2 −x2 =⇒ xy 2 −x2 = c.
7. Given (4e2x + 2xy − y 2 )dx + (x − y)2 dy = 0 then My = Nx = 2y so the differential equation is exact
∂φ
∂φ
and there exists a potential function φ such that (a)
= 4e2x + 2xy − y 2 and (b)
= (x − y)2 .
∂x
∂y
y3
dh(x)
dh(x)
∂φ
From (b) φ(x, y) = x2 y − xy 2 +
+ h(x) =⇒
= 2xy − y 2 +
so from (a) 2xy − y 2 +
=
3
∂x
dx
dx
dh(x)
= 4e2x =⇒ h(x) = 2e2x where the constant of integration has been set to zero
4e2x + 2xy − y 2 =⇒
dx
y3
y3
since we need just one potential function. φ(x, y) = x2 y − xy 2 +
+ 2e2x =⇒ x2 y − xy 2 +
+ 2e2x =
3
3
c1 =⇒ 6e2x + 3x2 y − 3xy 2 + y 3 = c.
x
y 2 − x2
dy
=
0
then
M
=
N
=
so the differential equation
y
x
x2 + y 2
(x2 + y 2 )2
∂φ
∂φ
1
y
x
and (b)
.
is exact and there exists a potential function φ such that (a)
= − 2
= 2
2
∂x
x
x +y
∂y
x + y2
∂φ
dh(x)
dh(x)
y
y
From (b) φ(x, y) = tan−1 (y/x) + h(x) =⇒
+
+
= − 2
so from (a) − 2
=
2
2
∂x
x +y
dx
x +y
dx
dh
1
y
=⇒
−
= x−1 =⇒ h(x) = ln |x| where the constant of integration is set to zero since we only
x x2 + y 2
dx
need one potential function. φ(x, y) = tan−1 (y/x) + ln |x| =⇒ tan−1 (y/x) + ln |x| = c.
8. Given
1
y
− 2
x x + y2
dx +
9. Given [y cos (xy) − sin x]dx + x cos (xy)dy = 0 then My = Nx = −xy sin (xy) + cos (xy) so the differential
∂φ
equation is exact so there exists a potential function φ such that (a)
= y cos (xy) − sin x and (b)
∂x
∂φ
dh(x)
∂φ
= x cos (xy). From (b) φ(x, y) = sin (xy) + h(x) =⇒
= y cos (xy) +
so from (a) y cos (xy) +
∂x
∂x
dx
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dh(x)
dh
= y cos (xy) − sin x =⇒
= − sin x =⇒ h(x) = cos x where the constant of integration is set to zero
dx
dx
since we only need one potential function. φ(x, y) = sin (xy) + cos x =⇒ sin (xy) + cos x = c.
10. (2y 2 e2x + 3x2 )dx + 2ye2x dy = 0. M = 2y 2 e2x + 3x2 and N = 2ye2x =⇒ My = 4ye2x and Nx =
4ye2x =⇒ My = Nx =⇒ the differential equation is exact so there exists a potential function φ such that
∂φ
∂φ
∂φ
dh(y)
(a)
= 2y 2 e2x + 3x2 and (b)
= 2ye2x . From (a) φ(x, y) = y 2 e2x + x3 + h(y) =⇒
= 2ye2x +
∂x
∂y
∂y
dy
dh(y)
dh
2x
2x
so from (b) 2ye +
= 2ye =⇒
= 0 =⇒ h(y) = c1 . Since we only need one potential function we
dy
dy
2 2x
3
can set c1 = 0. Then φ(x, y) = y e + x =⇒ y 2 e2x + x3 = c.
11. (y 2 + cos x)dx + (2xy + sin y)dy = 0. M = y 2 + cos x and N = 2xy + sin y =⇒ My = 2y and
Nx = 2y =⇒ My = Nx =⇒ the differential equation is exact so there exists a potential function φ such that
∂φ
∂φ
∂φ
dh(y)
(a)
= y 2 + cos x and (b)
= 2xy + sin y. From (a) φ(x, y) = xy 2 + sin x + h(y) =⇒
= 2xy +
so
∂x
∂y
∂y
dy
dh(y)
dh
from (b) 2xy +
= 2xy + sin y =⇒
= sin y =⇒ h(y) = − cos y where the constant of integration has
dy
dy
been set to zero since we just need one potential function. φ(x, y) = xy 2 +sin x−cos y =⇒ xy 2 +sin x−cos y =
c.
12. (sin y + y cos x)dx + (x cos y + sin x)dy = 0. M = sin y + y cos x and N = x cos y + sin x =⇒ My =
cos y + cos x and Nx = cos y + cos x =⇒ My = Nx =⇒ the differential equation is exact so there exists
∂φ
∂φ
a potential function φ such that (a)
= sin y + y cos x and (b)
= x cos y + sin x. From (a) φ(x, y) =
∂x
∂y
∂φ
dh(y)
dh(y)
x sin y + y sin x + h(y) =⇒
= x cos y + sin x +
so from (b) x cos y + sin x +
= x cos y +
∂y
dy
dy
dh
sin x =⇒
= 0 =⇒ h(y) = c1 . Since we only need one potential function we can set c1 = 0. φ(x, y) =
dy
x sin y + y sin x =⇒ x sin y + y sin x = c.
x
13. Given [1+ln (xy)]dx+ dy = 0 then My = Nx = y −1 so the differential equation is exact and there exists
y
∂φ
∂φ
dh(x)
a potential function φ such that (a)
= 1+ln (xy) and (b) φ(x, y) = x ln y +h(x) =⇒
= ln y +
so
∂x
∂x
dx
dh
dh(x)
= 1 ln (xy) =⇒
= 1+ln x =⇒ h(x) = c ln x where the constant of integration is set to
from (a) ln y+
dx
dx
zero since we only need one potential function. φ(x, y) = x ln y+x ln x =⇒ x ln y+x ln x = c =⇒ x ln (xy) = c.
xy + 1
xy − 1
dx +
dy = 0 then My = Nx = 1 =⇒ then the differential equation is exact so
x
y
∂φ
xy − 1
∂φ
xy + 1
there exists a potential function φ such that (a)
=
and (b)
=
. From (a) φ(x, y) =
∂x
x
∂y
y
∂φ
dh(y)
dh(y)
xy + 1
dh(y)
xy − ln |x| + h(y) =⇒
=x+
so from (b), x +
=
=⇒
= y −1 =⇒ h(y) = ln |y|
∂y
dy
dy
y
dy
where the constant of integration has been set to zero since we need just one potential function. φ(x, y) =
xy + ln |y/x| =⇒ xy + ln |x/y| = c.
14. Given
15. Given (2xy + cos y)dx + (x2 − x sin y − 2y)dy = 0 then My = Nx = 2x − sin y so the differential equation
∂φ
∂φ
is exact so there is a potential function φ such that (a)
= 2xy + cos y and (b)
= x2 − x sin y − 2y.
∂x
∂y
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From (a) φ(x, y) = x2 y + x cos y + h(y) =⇒
∂φ
dh(y)
dh(y)
= x2 − x sin y +
so from (b) x2 − x sin y +
=
∂y
dy
dy
dh
= −2y =⇒ h(y) = −y 2 where the constant of integration has been set to zero since
dy
we only need one potential function. φ(x, y) = x2 y + x cos y − y 2 =⇒ x2 y + x cos y − y 2 = c.
x2 − x sin y − 2y =⇒
dx
+ 4xy = 3 sin x =⇒ (4xy − 3 sin x)dx + 2x2 dy = 0 then My = Nx = 4x so the differential
dy
∂φ
∂φ
equation is exact so there exists a potential function φ such that (a)
= 4xy − 3 sin x and (b)
= 2x2 .
∂x
∂y
∂φ
dh(x)
dh(x)
dh(x)
From (b) φ(x, y) = 2x2 y + h(x) =⇒
= 4xy +
so from (a) 4xy +
= 4xy − 3 sin x =⇒
=
∂x
dx
dx
dx
−3 sin x =⇒ h(x) = 3 cos x where the constant of integration has been set to zero since we only need one
potential function. φ(x, y) = 2x2 y + 3 cos x =⇒ 2x2 y + 3 cos x = c. Now since y(2π) = 0, c = 3; thus,
3 − 3 cos x
.
2x2 y + 3 cos x = 3 or y(x) =
2x2
16. Given 2x2
17. Given (3x2 ln x + x2 − y)dx − xdy = 0 then My = Nx = −1 so the differential equation is exact so
∂φ
∂φ
there exists a potential function φ such that (a)
= 3x2 ln x + x2 − y and (b)
= −x. From (b)
∂x
∂y
dh(x)
∂φ
dh(x)
dh(x)
φ(x, y) = −xy + h(x) =⇒
= −y +
so from (a) −y +
= 3x2 ln x + x2 − y =⇒
=
∂x
dx
dx
dx
2
2
3
3x ln x + x =⇒ h(x) = x ln x where the constant of integration has been set to zero since we only need
one potential function. φ(x, y) = −xy + x3 ln x =⇒ −xy + x3 ln x = c. Now since y(1) = 5, c = −5; thus,
x3 ln x + 5
x3 ln x − xy = −5 or y(x) =
.
x
18. Given (yexy + cos x)dx + xexy dy = 0 then My = Nx = xyexy + exy so the differential equation is
∂φ
∂φ
exact so there exists a potential function φ such that (a)
= yexy + cos x and (b)
= xexy . From (b)
∂x
∂y
∂φ
dh(x)
dh(x)
φ(x, y) = exy +h(x) =⇒
= yexy +
so from (a) yexy +cos x =⇒
= cos x =⇒ h(x) = sin x where
∂x
dx
dx
the constant of integration is set to zero since we only need one potential function. φ(x, y) = exy + sin x =⇒
ln (2 − sin x)
exy + sin x = c. Now since y(π/2) = 0, c = 2; thus, exy + sin x = 2 or y(x) =
.
x
19. If φ(x, y) is a potential function for M dx + N dy = 0 =⇒ d(φ(x, y)) = 0 so d(φ(x, y) + c) = d(φ(x, y)) +
d(c) = 0 + 0 = 0 =⇒ φ(x, y) + c is also a potential function.
20. M = cos (xy)[tan (xy) + xy] and N = x2 cos (xy) =⇒ My = 2x cos (xy) − x2 y sin (xy) = Nx =⇒ My =
Nx =⇒ M dx = N dy = 0 is exact so I(x, y) = cos (xy) is an integrating factor for [tan (xy)+xy]dx+x2 dy = 0.
21. M = e−x/y (x2 y −1 − 2x) and N = −e−x/y x3 y −2 =⇒ My = e−x/y (x3 y −3 − 3x2 y −2 ) = Nx =⇒ M dx +
N dy = 0 is exact so I(x, y) = y −2 e−x/y is an integrating factor for y[x2 − 2xy]dx − x3 dy = 0.
22. M = sec x[2x − (x2 + y 2 ) tan x] and N = 2y sec x =⇒ My = −2y sec x tan x and Nx = 2y sec x tan x =⇒
My = Nx =⇒ M dx + N dy = 0 is not exact so I(x) = sec x is not an integrating factor for [2x − (x2 +
y 2 ) tan x]dx + 2ydy = 0.
23. Given (y − x2 )dx + 2xdy = 0 then M = y − x2 and N = 2x. Thus My = 1 and Nx = 2 so
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My − N x
=
N
80
1
1
= f (x) is a function of x alone so I(x) = e f (x)dx = √ is an integrating factor for the given equation.
2x
x
Multiplying the given equation by I(x) results in the exact equation (x−1/2 y−x3/2 )dx+2x1/2 dy = 0. We find
2x5/2
2x5/2
that φ(x, y) = 2x1/2 y −
and hence the general solution of our differential equation is 2x1/2 y −
=c
5
5
c + 2x5/2
√ .
or y(x) =
10 x
−
24. Given (3xy − 2y −1 )dx + x(x + y −2 )dy = 0 then M = 3xy − 2y −1 and N = x(x + y −2 ). Thus
My − N x
1
My = 3x + 2y −2 and Nx = 2x + y −2 so
= = f (x) is a function of x alone so I(x) = e f (x)dx = x
N
x
is an integrating factor for the given equation. Multiplying the given equation by I(x) results in the exact
equation (3x2 y − 2xy −1 )dx + x2 (x + y −2 )dy = 0. We find that φ(x, y) = x3 y − x2 y −1 and hence the general
solution of our differential equation is x3 y − x2 y −1 = c.
25. Given x2 ydx + y(x3 + e−3y sin y)dy = 0 then M = x2 y and N = y(x3 + e−3y sin y). Thus My = x2
My − N x
and Nx = 3x2 y so
= y −1 − 3 = g(y) is a function of y alone so I(y) = e g(y)dy = e3y /y is an
M
integrating factor for the given equation. Multiplying the equation by I(y) results in the exact equation
x3 e3y
x2 e3y dx + e3y (x3 + e−3y sin y)dy = 0. We find that φ(x, y) =
− cos y and hence the general solution
3
3 3y
x e
of our differential equation is
− cos y = c.
3
26. Given (xy − 1)dx + x2 dy = 0 then M = xy − 1 and N = x2 . Thus My = x and Nx = 2x so
My − Nx
= −x−1 = f (x) is a function of x alone so I(x) = e f (x)dx = x−1 is an integrating factor for the
N
given equation. Multiplying the given equation by I(x) results in the exact equation (y − x−1 )dx + xdy = 0.
We find that φ(x, y) = xy −ln |x| and hence, the general solution of our differential equation is xy −ln |x| = c.
dy
1
2xy
=
=⇒ (2xy + 2x3 y − 1)dx + (1 + x2 )2 dy = 0 then M = 2xy + 2x3 y − 1 and
+
2
dx 1 + x
(1 + x2 )2
My − N x
2x
= f (x) is a function of
N = (1 + x2 )2 . Thus My = 2x + 2x3 and Nx = 4x(1 + x2 ) so
=−
N
1 + x2
1
x alone so I(x) = e f (x)dx =
is an integrating factor for the given equation. Multiplying the given
1 + x2
1
dx + (1 + x2 )dy = 0. We find that φ(x, y) =
equation by I(x) yields the exact equation 2xy −
1 + x2
(1 + x2 )y − tan−1 x and hence the general solution of our differential equation is (1 + x2 )y − tan−1 x = c or
tan−1 x + c
y(x) =
.
1 + x2
27. Given
28. Given xy[2 ln (xy) + 1]dx + x2 dy = 0 then M = xy[2 ln (xy) + 1] and N + x2 . Thus My = 3x + 2x ln (xy)
MY − N x
1
= y −1 = g(y) is a function of y only so I(y) = e g(y)dy = is an integrating factor
and Nx = 2x so
M
y
for the given equation. Multiplying the given equation by I(y) results in the exact equation x[2 ln (xy) +
1]dx + x2 y −1 dy = 0. We find that φ(x, y) = x2 ln y + x2 ln x and hence the general solution of our differential
2
equation is x2 ln y + x2 ln x = c or y(x) = xec/x .
29. Given ydx − (2x + y 4 )dy = 0 then M = y and N = −(2x + y 4 ). Thus My = 1 and Nx = −2 so
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My − N x
= 3y −1 = g(y) is a function of y alone so I(y) = e− g(y)dy = 1/y 3 is an integrating factor
M
for the given differential equation. Multiplying the given equation by I(y) results in the exact equation
y −2 dx − (2xy −3 + y)dy = 0. We find that φ(x, y) = xy −2 − y 2 /2 and hence, the general solution of our
differential equation is xy −2 − y 2 /2 = c1 =⇒ 2x − y 4 = cy 2 .
30. Given (y −1 − x−1 )dx + (xy −2 − 2y −1 )dy = 0 =⇒ xr y s (y −1 − x−1 )dx + xr y s (xy −2 − 2y −1 )dy = 0 =⇒
(xr y s−1 −xr−1 y s )dx+(xr+1 y s−2 −2xr y s−1 )dy = 0. Then M = xr y s−1 −xr−1 y s and N = xr+1 y s−2 −2xr y s−1
so My = xr (s − 1)y s−2 − xr−1 sy s−1 and Nx = (r + 1)xr y s−2 − 2rxr−1 y s−1 . The equation is exact if and
s−1
s
2r
r+1
only if My = Nx =⇒ xr y s−1 − xr−1 y s = (r + 1)xr y s−2 − 2rxr−1 y s−1 =⇒
−
−
=
=⇒
2
2
y
xy
y
xy
s − 2r
s−r−2
=
. From the last equation we require that s − r − 2 = 0 and s − 2r = 0. Solving this system
y2
xy
yields r = 2 and s = 4.
31. Given 2y(y + 2x2 )dx + x(4y + 3x2 )dy = 0 =⇒ xr y s 2y(y + 2x2 )dx + xr y s x(4y + 3x2 )dy = 0. Then
M = 2xr y s+2 + 4xr+2 y s+1 and N = 4xr+1 y s+1 + 3xr+3 y s so My = 2xr (s + 2)y s+1 + 4xr+2 (s + 1)y s and
Nx = 4(r + 1)xr y s+1 + 3(r + 3)xr+2 y s . The equation is exact if and only if My = Nx =⇒ 2xr (s + 2)y s+1 +
4xr+2 (s + 1)y s = 4(r + 1)xr y s+1 + 3(r + 3)xr+2 y s =⇒ 2(s + 2)y + 4x2 (s + 1) = 4(r + 1)y + 3(r + 3)x2 . From
this last equation we require that 2(s + 2) = 4(r + 1) and 4(s + 1) = 3(r + 3). Solving this system yields
r = 1 and s = 2.
32. Given y(5xy 2 + 4)dx + x(xy 2 − 1)dy = 0 =⇒ xr y s y(5xy 2 + 4)dx + xr y s x(xy 2 − 1)dy = 0. Then
M = xr y s+1 (5xy 2 + 4) and N = xr+1 y s (xy 2 − 1) so My = 5(s + 3)xr+1 y s+2 + 4(s + 1)xr y s and Nx =
(r + 2)xr+1 y s−2 − (r + 1)xr y s . The equation is exact if and only if My = Nx =⇒ 5(s + 3)xr+1 y s+2 + 4(s +
1)xr y s = (r + 2)xr+1 y s+2 − (r + 1)xr y s =⇒ 5(s + 3)xy 2 + 4(s + 1) = (r + 2)xy 2 − (r + 1). From the last
equation we require that 5(s + 3) = r + 2 and 4(s + 1) = −(r + 1). Solving this system yields r = 3 and
s = −2.
My − N x
= g(y) is a function of y only. Then dividing the equation (1.9.21) by M ,
M
it follows that I is an integrating factor for M (x, y)dx + N (x, y)dy = 0 if and only if it is a solution of
∂I
N ∂I
−
= Ig(y) (30.1). We must show that this differential equation has a solution I = I(y). However,
M ∂x ∂y
dI
= −Ig(y), which is a separable equation with solution I(y) = e− g(t)dt .
if I = I(y), then (30.1) reduces to
dy
33. Suppose that
dy
+ py = q can be written in the differential form as (py − q)dx + dy = 0 (34.1). This
dx
My − N x
has M = py − q and N = 1 so that
= p(x). Consequently, an integrating factor for (34.1) is
N
x
I(x) = e p(t)dt .
34. (a). Note
x
x
x
yields the exact equation e p(t)dt (py − q)dx + e p(t)dt dy = 0.
x
x
∂φ
∂φ
Hence, there exists a potential function φ such that (i)
= e p(t)dt (py − q) and (ii)
= e p(t)dt . From
∂x
∂y
x
x
x
x
dh(x)
dh(x)
p(t)dt
p(t)dt
p(t)dt
(i), p(x)ye
+
(py − q) =⇒
=⇒ h(x) = − q(x)e p(t)dt dx,
=e
= −q(x)e
dx
dx
where the constant
of integration
just need one potential function.
Consequently,
x
x has been set to zero since
x
we
x
φ(x, y) = ye p(t)dt − q(x)e p(t)dt dx =⇒ y(x) = I −1
Iq(t)dt + c , where I(x) = e p(t)dt .
(b). Multiplying (34.1) by I(x) = e
p(t)dt
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Solutions to Section 1.10
True-False Review:
(a): TRUE. This is well-illustrated by the calculations shown in Example 1.10.1.
(b): TRUE. The equation
y1 = y0 + f (x0 , y0 )(x1 − x0 )
dy
= f (x, y) at the point (x0 , y0 ). Once the point (x1 , y1 ) is determined, the
is the tangent line to the curve dx
procedure can be iterated over and over at the new points obtained to carry out Euler’s method.
(c): FALSE. It is possible, depending on the circumstances, for the errors associated with Euler’s method
to decrease from one step to the next.
(d): TRUE. This is illustrated in Figure 1.10.3.
Problems:
1. Applying Euler’s method with y = 4y −1, x0 = 0, y0 = 1, and h = 0.05 we have yn+1 = yn +0.05(4yn −1).
This generates the sequence of approximants given in the table below.
n
1
2
3
4
5
6
7
8
9
10
xn
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
yn
1.15
1.33
1.546
1.805
2.116
2.489
2.937
3.475
4.120
4.894
Consequently the Euler approximation to y(0.5) is y10 = 4.894. (Actual value: y(.05) = 5.792 rounded
to 3 decimal places).
2xy
x n yn
, x0 = 0, y0 = 1, and h = 0.1 we have yn+1 = yn − 0.2
.
1 + x2
1 + x2n
This generates the sequence of approximants given in the table below.
2. Applying Euler’s method with y = −
n
1
2
3
4
5
6
7
8
9
10
xn
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
yn
1
0.980
0.942
0.891
0.829
0.763
0.696
0.610
0.569
0.512
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Consequently the Euler approximation to y(1) is y10 = 0.512. (Actual value: y(1) = 0.5).
3. Applying Euler’s method with y = x−y 2 , x0 = 0, y0 = 2, and h = 0.05 we have yn+1 = yn +0.05(xn −yn2 ).
This generates the sequence of approximants given in the table below.
n
1
2
3
4
5
6
7
8
9
10
xn
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
yn
1.80
1.641
1.511
1.404
1.316
1.242
1.180
1.127
1.084
1.048
Consequently the Euler approximation to y(0.5) is y10 = 1.048. (Actual value: y(.05) = 1.0477 rounded
to four decimal places).
4. Applying Euler’s method with y = −x2 y, x0 = 0, y0 = 1, and h = 0.2 we have yn+1 = yn − 0.2x2n yn . This
generates the sequence of approximants given in the table below.
n
1
2
3
4
5
xn
0.2
0.4
0.6
0.8
1.0
yn
1
0.992
0.960
0.891
0.777
Consequently the Euler approximation to y(1) is y5 = 0.777. (Actual value: y(1) = 0.717 rounded to 3
decimal places).
5. Applying Euler’s method with y = 2xy 2 , x0 = 0, y0 = 1, and h = 0.1 we have yn+1 = yn + 0.1xn yn2 . This
generates the sequence of approximants given in the table below.
n
1
2
3
4
5
6
7
8
9
10
xn
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
yn
0.5
0.505
0.515
0.531
0.554
0.584
0.625
0.680
0.754
0.858
Consequently the Euler approximation to y(1) is y10 = 0.856. (Actual value: y(1) = 1).
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∗
=
6. Applying the modified Euler method with y = 4y − 1, x0 = 0, y0 = 1, and h = 0.05 we have yn+1
yn + 0.05(4yn − 1)
∗
yn+1 = yn + 0.025(4yn − 1 + 4yn+1
− 1). This generates the sequence of approximants given in the table
below.
n
1
2
3
4
5
6
7
8
9
10
xn
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
yn
1.165
1.3663
1.6119
1.9115
2.2770
2.7230
3.2670
3.9308
4.7406
5.7285
Consequently the modified Euler approximation to y(0.5) is y10 = 5.7285. (Actual value: y(.05) = 5.7918
rounded to 4 decimal places).
2xy
∗
7. Applying the modified Euler method with y = −
, x0 = 0, y0 = 1, and h = 0.1 we have yn+1
=
1 + x2
x n yn
yn − 0.2
1 + x2n ∗
xn+1 yn+1
x n yn
. This generates the sequence of approximants given in the table
yn+1 = yn + 0.05 −
−
2
1 + x2n
1 + x2n+1
below.
n
1
2
3
4
5
6
7
8
9
10
xn
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
yn
0.9900
0.9616
0.9177
0.8625
0.8007
0.7163
0.6721
0.6108
0.5536
0.5012
Consequently the modified Euler approximation to y(1) is y10 = 0.5012. (Actual value: y(1) = 0.5).
∗
8. Applying the modified Euler method with y = x − y 2 , x0 = 0, y0 = 2, and h = 0.05 we have yn+1
=
yn − 0.05(xn − yn2 )
∗
yn+1 = yn + 0.025(xn − yn2 + xn+1 − (yn+1
)2 ). This generates the sequence of approximants given in the
table below.
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n
1
2
3
4
5
6
7
8
9
10
xn
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
yn
1.8203
1.6725
1.5497
1.4468
1.3600
1.2866
1.2243
1.1715
1.1269
1.0895
Consequently the modified Euler approximation to y(0.5) is y10 = 1.0895. (Actual value: y(.05) = 1.0878
rounded to 4 decimal places).
∗
9. Applying the modified Euler method with y = −x2 y, x0 = 0, y0 = 1, and h = 0.2 we have yn+1
=
2
yn − 0.2xn yn
∗
yn+1 = yn − 0.1[x2n yn + x2n+1 yn+1
]. This generates the sequence of approximants given in the table below.
n
1
2
3
4
5
xn
0.2
0.4
0.6
0.8
1.0
yn
0.9960
0.9762
0.9266
0.8382
0.7114
Consequently the modified Euler approximation to y(1) is y5 = 0.7114. (Actual value: y(1) = 0.7165
rounded to 4 decimal places).
∗
10. Applying the modified Euler method with y = 2xy 2 , x0 = 0, y0 = 1, and h = 0.1 we have yn+1
=
2
yn + 0.1xn yn
∗
yn+1 = yn +0.05[xn yn2 +xn+1 (yn+1
)2 ]. This generates the sequence of approximants given in the table below.
n
1
2
3
4
5
6
7
8
9
10
xn
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
yn
0.5025
0.5102
0.5235
0.5434
0.5713
0.6095
0.6617
0.7342
0.8379
0.9941
Consequently the modified Euler approximation to y(1) is y10 = 0.9941. (Actual value: y(1) = 1).
11. We have y = 4y −1, x0 = 0, y0 = 1, and h = 0.05. So, k1 = 0.05(4yn −1), k2 = 0.05[4(yn + 12 k1 )−1], k3 =
0.05[4(yn + 12 k2 ) − 1], k4 = 0.05[4(yn + 12 k3 ) − 1],
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yn+1 = yn + 16 (k1 + k2 + k3 + k4 ). This generates the sequence of approximants given in the table below
(computations rounded to five decimal places).
n
1
2
3
4
5
6
7
8
9
10
xn
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
yn
1.16605
1.36886
1.61658
1.91914
2.28868
2.74005
3.29135
3.96471
4.78714
5.79167
Consequently the Runge-Kutta approximation to y(0.5) is y10 = 5.79167. (Actual value: y(.05) = 5.79179
rounded to 5 decimal places).
(xn + 0.05)(yn + k21 )
xy
x n yn
,
x
=
0,
y
=
1,
and
h
=
0.1.
So,
k
=
−0.2
,
k
=
−0.2
), k3 =
0
0
1
2
1 + x2
1 + x2n
[1 + (xn + 0.05)2 ]
(xn + 0.05)(yn + k22 )
xn+1 (yn + k3 )
−0.2
, k4 = −0.2
,
[1 + (xn + 0.05)2 ]
[1 + (xn+1 )2 ]
1
yn+1 = yn + 6 (k1 + k2 + k3 + k4 ). This generates the sequence of approximants given in the table below
(computations rounded to seven decimal places).
12. We have y = −2
n
1
2
3
4
5
6
7
8
9
10
xn
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
yn
0.9900990
0.9615383
0.9174309
0.8620686
0.7999996
0.7352937
0.6711406
0.6097558
0.5524860
0.4999999
Consequently the Runge-Kutta approximation to y(1) is y10 = 0.4999999. (Actual value: y(.05) = 0.5).
13. We have y = x − y 2 , x0 = 0, y0 = 2, and h = 0.05. So, k1 = 0.05(xn − yn2 ), k2 = 0.05[xn + 0.025 − (yn +
k1 2
k2 2
2
2 ) ], k3 = 0.05[xn + 0.025 − (yn + 2 ) ], k4 = 0.05[xn+1 − (yn + k3 ) ]],
1
yn+1 = yn + 6 (k1 + k2 + k3 + k4 ). This generates the sequence of approximants given in the table below
(computations rounded to six decimal places).
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n
1
2
3
4
5
6
7
8
9
10
xn
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
yn
1.1.81936
1.671135
1.548079
1.445025
1.358189
1.284738
1.222501
1.169789
1.125263
1.087845
Consequently the Runge-Kutta approximation to y(0.5) is y10 = 1.087845. (Actual value: y(0.5) = 1.087845
rounded to 6 decimal places).
14. We have y = −x2 y, x0 = 0, y0 = 1, and h = 0.2. So, k1 = −0.2x2n yn , k2 = −0.2(xn + 0.1)2 (yn + k21 ), k3 =
−0.2(xn + 0.1)2 (yn + k22 ), k4 = −0.2(xn+1 )2 (yn + k3 ),
yn+1 = yn + 16 (k1 + k2 + k3 + k4 ). This generates the sequence of approximants given in the table below
(computations rounded to six decimal places).
n
1
2
3
4
5
xn
0.2
0.4
0.6
0.8
1.0
yn
0.997337
0.978892
0.930530
0.843102
0.716530
Consequently the Runge-Kutta approximation to y(1) is y10 = 0.716530. (Actual value: y(1) = 0.716531
rounded to 6 decimal places).
15. We have y = 2xy 2 , x0 = 0, y0 = 1, and h = 0.1. So, k1 = 0.2xn − yn2 , k2 = 0.2(xn + 0.05)(yn + k21 )2 , k3 =
0.2(xn + 0.05)(yn + k22 )2 , k4 = 0.2xn+1 (yn + k3 )2 ]],
yn+1 = yn + 16 (k1 + k2 + k3 + k4 ). This generates the sequence of approximants given in the table below
(computations rounded to six decimal places).
n
1
2
3
4
5
6
7
8
9
10
xn
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
yn
0.502513
0.510204
0.523560
0.543478
0.571429
0.609756
0.662252
0.735295
0.840336
0.999996
Consequently the Runge-Kutta approximation to y(1) is y10 = 0.999996. (Actual value: y(1) = 1).
(c)2017 Pearson Education. Inc.
88
1
1
16. We have y + y = e−x/10 cos x, x0 = 0, y0 = 0, and h = 0.5 Hence, k1 = 0.5 − yn + exn /10 cos xn , k2 =
10 10
1
1
1
1
(−xn +0.25)/10
yn + k 1 + e
yn + k2 + e(−xn +0.25)/10 cos (xn + 0.25) ,
cos (xn + 0.25) , k3 = 0.5 −
0.5 −
10
2
10
2
1
k4 = 0.5 − (yn + k3 ) + e−xn+1 /10 cos xn+1 , yn+1 = yn + 16 (k1 + 2k2 + 2k3 + k4 ). This generates the se10
quence of approximants plotted in the accompanying figure. We see that the solution appears to be oscillating
with a diminishing amplitude. Indeed, the exact solution to the initial value problem is y(x) = e−x/10 sin x.
The corresponding solution curve is also given in the figure.
y(x)
0.75
0.5
0.25
5
10
x
15
20
25
-0.25
-0.5
Figure 0.0.56: Figure for Problem 16
Solutions to Section 1.11
Problems:
dy
dy
d2 y
du
d2 y
3x
−
2
.
Let
u
=
. Substituting these results into the first equation
=
6e
so
that
=
dx2
dx
dx
dx
dx2
du
yields
− 2u = 6e3x . An appropriate integrating factor for this equation is I(x) = e−2 dx = e−2x =⇒
dx
dy
d −2x
u) = 6ex =⇒ e−2x u = 6ex + c1 =⇒ u = 6e3x + c1 e2x =⇒
(e
= 6e3x + c1 e2x =⇒ y(x) =
dx
dx
2e3x + c1 e2x + c2 , where we have redefined the constant c1 in the last step.
1.
d2 y
2 dy
dy
du
d2 y
2
=
.
Let
u
=
. Substituting these results into the first equation
+
4x
so
that
=
dx2
x dx
dx
dx
dx2
du
du
2
2
yields
= u + 4x2 =⇒
− u = 4x2 . An appropriate integrating factor for this equation is I(x) =
dx
x
dx
x
2.
(c)2017 Pearson Education. Inc.
89
dx
dy
x = x−2 =⇒ d(x−2 u) = 4 =⇒ x−2 u = 4 dx =⇒ x−2 u = 4x + c1 =⇒ u = 4x3 + c1 x2 =⇒
=
dx
3
2
3
4
4x + c1 x =⇒ y(x) = c1 x + x + c2 .
e
−2
dy
d2 y
1
dy
du
d2 y
=
. Substituting these results into
−
1
. Let u =
so that
=
2
dx
(x − 1)(x − 2) dx
dx
dx
dx2
1
du
1
1
du
=
(u − 1) =⇒
−
u = −
. An
the first equation yields
dx
(x − 1)(x − 2)
dx
(x − 1)(x − 2)
(x − 1)(x − 2)
1
−
dx
x−1
d x−1
(x
−
1)(x
−
2)
=
=⇒
u =
appropriate integrating factor for this equation is I(x) = e
x−2
dx x − 2
x−1
1
dy
1
1
=⇒
u = (x − 2)−2 dx =⇒ u = −
+ c1 =⇒
=−
+ c1 =⇒ y(x) = − ln |x − 1| +
2
(x − 2)
x−2
x−1
dx
x−1
c1 x + c2 .
3.
d2 y
2
4.
+
dx2
y
2
dy
dy
du
du
d2 y
. Substituting these results into the first
. Let u =
so that
=u
=
dx
dx
dx
dy
dx2
du
2
2
du
+ u2 = u =⇒ u = 0 or
− u = 1. An appropriate integrating factor for the last
equation yields u
dy
y
dy
y
2
dy
d 2
y3
dy
y c1
(y u) = y 2 =⇒ y 2 u = y 2 dy =⇒ y 2 u =
+c1 =⇒
= + 2 =⇒
equation is I(y) = e y = y 2 =⇒
3
dx
3 y
√ dy
ln |y 3 + c2 | = x + c3 =⇒ y(x) = 3 c4 ex + c5 .
=
2
dy
du
du
d2 y
. Substituting these results into the first
so that
=u
=
dx
dx
dy
dx2
du
du
equation yields u
= u2 tan y. If u = 0 then
= 0 =⇒ y equals a constant and this is a solution to the
dy
dx
du
dy
du
= u tan y =⇒
= tan ydy =⇒ u = c1 sec y =⇒
=
equation. Now suppose that u = 0. Then
dy
u
dx
c1 sec y =⇒ y(x) = sin−1 (c1 x + c2 ).
5.
d2 y
=
dx2
dy
dx
dy
dx
tan y. Let u =
2
d2 y
dy
dy
dy
du
d2 y
6.
+
tan
x
.
Let
u
=
. Substituting these results into the first
=
so
that
=
dx2
dx
dx
dx
dx
dx2
1
du
du
dz
equation yields
+ tan xu = u2 which is a Bernoulli equation. Letting z = u−1 gives 2 =
=− .
dx
u
dx
dx
dz
− tan xz = −1. Then an integrating factor for
Substituting these results into the last equation yields
dx
d
− tan xdx
= cos x =⇒
(z cos x) = − cos x =⇒ z cos x = − cos xdx =⇒ z =
this equation is I(x) = e
dx
cos x
dy
cos x
− sin x + c1
=⇒ u =
=⇒
=
=⇒ y(x) = c2 − ln |c1 − sin x|.
cos x
c1 − sin x
dx
c1 − sin x
2
dx
dx
d2 x
dx
du
d2 x
=
+
2
. Substituting these results into the first equation
7.
.
Let
u
=
so
that
=
dt2
dt
dt
dt
dt
dt2
du
du
= u2 + 2u =⇒
− 2u = u2 which is a Bernoulli equation. If u = 0 then x is a constant which
yields
dt
dt
dz
1 du
satisfies the equation. Now suppose that u = 0. Let z = u−1 so that
=− 2
. Substituting these
dt
u dt
(c)2017 Pearson Education. Inc.
90
dz
+ 2z = −1. An integrating factor for this equation is I(x) = e2t =⇒
dt
2e2t
d 2t
1
2e2t
=⇒
x
=
dt =⇒ x(t) = c2 − ln |c1 − e2t |.
(e z) = −e2t =⇒ z = ce−2t − =⇒ u =
dt
2
2c − e2t
2c − e2t
results into the last equation yields
d2 y
dy
dy
du
d2 y
4
−
2
.
Let
u
=
. Substituting these results into the first equation
=
6x
so
that
=
dx2
dx
dx
dx
dx2
dx
−2
du
2
x = x−2 =⇒
yields
− u = 6x4 . An appropriate integrating factor for this equation is I(x) = e
dx
x
dy
1
d −2
(x u) = 6x2 =⇒ x−2 u = 6 x2 dx =⇒ u = 2x5 + cx2 =⇒
= 2x5 + cx2 =⇒ y(x) = x6 + c1 x3 + c2 .
dx
dx
3
8.
dx
d2 x
dx
du
d2 x
. Let u =
9. t 2 = 2 t +
so that
= 2 . Substituting these results into the first equation
dt
dt
dt
dt
dt
d
2
du
− u = 2. An integrating factor for this equation is I(x) = t−2 =⇒ (t−2 u) = 2t−2 =⇒ u =
yields
dt
t
dt
dx
−2t + ct2 =⇒
= −2t + ct2 =⇒ x(t) = c1 t3 − t2 + c2 .
dt
d2 y
10.
−α
dx2
dy
dx
2
−β
dy
dy
du
d2 y
. Substituting these results into the
= 0. Let u =
so that
=
dx
dx
dx
dx2
du
− βu = αu2 which is a Bernoulli equation. If u = 0 then y is a constant and
dx
du
dz
satisfies the equation. Now suppose that u = 0. Let z = u−1 so that
= −u−2 . Substituting
dx
dx
dz
+ βz = −α. The an integrating factor for this equation is
these results into the last equation yields
dx
βx
α
βeβx
dy
βeβx
β dx
βx
βx
= e =⇒ e z = −α e dx =⇒ − + ce−βx =⇒ u =
=⇒
=⇒
=
I(x) = e
βx
β
cβ − αe
dx
cβ − αeβx
1
βeβx
dx =⇒ y(x) = − ln |c1 + c2 eβx |.
y=
cβ − αeβx
α
first equation yields
d2 y
2 dy
dy
du
d2 y
4
−
.
Let
u
=
. Substituting these results into the first equation
=
18x
so
that
=
dx2
x dx
dx
dx
dx2
d −2
du
2
yields
− u = 18x4 which has I(x) = x−2 as an integrating factor so
(x u) = 18x2 =⇒ u =
dx
x
dx
dy
6x5 + cx2 =⇒
= 6x5 + cx2 =⇒ y(x) = x6 + c1 x3 + c2 .
dx
11.
d2 y
2x dy
dy
du
= −
. Let u =
so that
= df racd2 ydx2 . If u = 0 then y is a constant and
dx2
1 + x2 dx
dx
dx
satisfies the equation. Now suppose that u = 0. Substituting these results into the first equation yields
c1
dy
2x
c1
du
u =⇒ ln |u| = − ln (1 + x2 ) + c =⇒ u =
=⇒
=⇒ y(x) = c1 tan−1 x + c2 .
=−
=
dx
1 + x2
1 + x2
dx
1 + x2
12.
3
dy
dy
du
du
d2 y
. Let u =
so that
=u
= 2 . Substituting these results into
dx
dx
dx
dy
dx
du 1 2
the first equation yields u
+ u = ye−y u3 . If u = 0 then y is a constant and satisfies the equation. Now
dx y
13.
d2 y
1
+
dx2
y
dy
dx
2
= ye−3
(c)2017 Pearson Education. Inc.
91
suppose that u = 0. Substituting these results into the first equation yields
du
u
+ = ye−y u2 which is a
dx
y
du
dv
= −u−2 . Substituting these results into the last equation
dy
dy
d −1
dv v
yields
− = −ye−y . Then I(y) = y −1 is an integrating factor for the equation thus
(y v) = −e−1 =⇒
dy y
dy
ey
dy
ey
=⇒
=⇒ (ye−y + cy)dy = dx =⇒ e−y (y + 1) + c1 y 2 − x.
v = y(e−1 + c) =⇒ u =
=
y
y + cye
dx
y + cyey
Bernoulli equation. Let v = u−1 so that
dy
d2 y
dy
du
d2 y
− tan x
. Substituting these results into the first equation
= 1. Let u =
so that u
=
2
dx
dx
dx
dy
dx2
du
yields
− u tan x = 1. An appropriate integrating factor for this equation is I(x) = e− tan xdx = cos x =⇒
dx
d
dy
(u cos x) = cos x =⇒ u cos x = sin x + c =⇒ u(x) = tan x + c sec x =⇒
= tan x + c sec x =⇒ y(x) =
dx
dx
ln sec x + c1 ln (sec x + tan x) + c2 .
14.
2
dy
dy
du
du
d2 y
+ y 2 . Let u =
. Substituting these results into the first
so that
=u
=
dx
dx
dx
dy
dx2
du
du
2
1 dz
equation yields u
− u2 = y, a Bernoulli equation. Let z = u2 so that u
=
. Substituting these
dy
y
dy
2 dy
dz 4
− z = 2y which has I(y) = y −4 as an integrating factor. Therefore,
results into the last equation yields
dy y
dy
d −4
(y z) = 2y −3 =⇒ z = c1 y 4 − y 2 =⇒ u2 = c1 y 4 − y 2 =⇒ u = ± c1 y 4 − y 2 =⇒
= ± c1 y 4 − y 2 =⇒
dy dx
1
−1
= ±x + c2 . Using the facts that f (0) = 1 and y (0) = 0 we find that c1 = 1 and c2 = 0; thus
cos
√
y c1
y(x) = sec x.
15. y
d2 y
=2
dx2
d2 y
dy
du
du
d2 y
2
=
ω
y
where
ω
>
0.
Let
u
=
. Substituting these results into the first
so
that
=
u
=
dx2
dx
dx
dy
dx2
du
equation yields u
= ω 2 y =⇒ u2 = ω 2 y 2 + c2 . Using the given that y(0) = a and y (0) = 0 we find that
dy
dy
1
c2 = a2 ω 2 . Then
= ±ω y 2 − a2 =⇒ cosh−1 (y/a) = ±x + c =⇒ y(x) = a cosh [ω(c ± x)] =⇒ y =
dx
ω
±aω sinh [ω(c ± x)] and since y (0) = 0, c = 0; hence, y(x) = a cosh (ωx).
16.
dy
du
d2 y
17. Let u =
. Substituting these results into the differential equation yields
so that u
=
dx
dx
dx2
√
1
du
1√
1 + u2 . Separating the variables and integrating we obtain 1 + u2 = y +c. Imposing the initial
u
=
dy
a
a
√
1
1
dy
2
conditions y(0) = a, (0) = 0 gives c = 0. Hence, 1 + u = y so that 1 + u2 = 2 y 2 or equivalently,
dx
a
a
dy
1
1
= ± dx which
u = ± y 2 /a2 − 1. Substituting u =
and separating the variables gives dx
|a|
y 2 − a2
can be integrated to obtain cosh−1 (y/a) = ±x/a + c1 so that y = a cosh (±x/a + c1 ). Imposing the initial
conditions y(0) = a gives c1 = 0 so that y(x) = a cosh (x/a).
18.
d2 y
dy
dy
du
d2 y
+
p(x)
. Substituting these results into the first equation
=
q(x).
Let
u
=
so
that
=
dx2
dx
dx
dx
dx2
(c)2017 Pearson Education. Inc.
92
du
e− p(x)dx q(x)dx + c1
gives us the equivalent system:
+p(x)u = q(x) which has a solution u = e− p(x)dx
dx
#
" − p(x)dx − p(x)dx
dy
e− p(x)dx q(x)dx + c1 . Thus y =
e
e
q(x)dx + c1 dx + c2 is
so
= e− p(x)dx
dx
a solution to the original equation.
du1
du2
du3
d3 y
d2 y
dy
d2 y
d3 y
since
=⇒
;
thus
=
F
x,
=
=⇒ u3 =
=
=
dx
dx
dx
dx2
dx
dx3
dx3
dx2
du3
the latter equation is equivalent to
= F (x, u3 ).
dx
d3 y
1 d2 y
du1
du2
(b).
=
− 1 . Replace this equation by the equivalent first order system:
= u2 ,
= u3 ,
dx3
x dx2
dx
dx
du3
dx
du2
K
1
du3
= (u3 − 1) =⇒
=
=⇒ u3 = Kx + 1 =⇒
= Kx + 1 =⇒ u2 = x2 + x + c2 =⇒
and
dx
x
u3 − 1
x
dx
2
du1
K 3 1 2
1 2
K 2
3
= x + x + c2 =⇒ u1 = x + x + c2 x + c3 =⇒ y(x) = u1 = c1 x + x + c2 x + c3 .
dx
2
6
2
2
19. (a). u1 = y =⇒ u2 =
19. Given
d2 θ
g
dθ
+ sin θ = 0, θ(0) = θ0 , and
(0) = 0.
2
dt
L
dt
g
du dθ
d2 θ
dθ
du
d2 θ
du
+
=
θ
=
0.
Let
u
=
so
that
=
= u . Substituting these results into
2
2
dt
L
dt
dt
dt
dθ dt
dθ
g
dθ
g
du
+ θ = 0 =⇒ u2 = − θ2 + c21 , but
(0) = 0 and θ(0) = θ0 so c21 =
the last equation yields u
dθ
L
L
dt
g 2
θ
g 2
g
g
= ±
θ0 − θ2 =⇒ sin−1
θ0 =⇒ u2 = (θ02 − θ2 ) =⇒ u = ±
t + c2 , but θ(0) = θ0 so
L
L L
θ0
L
θ
π
g
g
g
π
π
c2 = =⇒ sin−1
= ±
t =⇒ θ = θ0 sin
±
t =⇒ θ = θ0 cos
t . Yes, the predicted
2
θ0
2
L
2
L
L
motion is reasonable.
(a).
du dθ
dθ
du
d2 θ
du
d2 θ g
+ sin θ = 0. Let u =
so that
= 2 =
= u . Substituting these results into the last
2
dt
L
dt
dt
dt
dθ dt
dθ
2g
dθ
2g
du g
2
cos θ + c. Since θ(0) = θ0 and
(0) = 0, then c = − cos θ0
equation yields u + sin θ = 0 =⇒ u =
dθ L
L
dt L
2g
2g
2g
dθ
dθ
2g
2g
and so u2 =
cos θ −
cos θ0 =⇒
=±
cos θ −
cos θ0 =⇒
=±
[cos θ − cos θ0 ]1/2 .
L
L
dt
L
L
dt
L
L
dθ
= ±dt. When the pendulum goes from θ = θ0 to θ = 0
(c). From part (b),
2g [cos θ − cos θ0 ]1/2
dθ
(which corresponds to one quarter of a period)
is negative; hence, choose the negative sign. Thus,
dt
0
L
dθ
L θ0
dθ
=⇒
T
=
.
T =−
1/2
2g θ0 [cos θ − cos θ0 ]
2g 0 [cos θ − cos θ0 ]1/2
θ0
L
dθ
=⇒
(d). T =
2g 0 [cos θ − cos θ0 ]1/2
(b).
T =
L
2g
θ0
0
2 sin
2
θ0
2
dθ
− 2 sin
2
1
1/2 = 2
θ
L
2g
θ0
0
2
(c)2017 Pearson Education. Inc.
sin
2
θ0
2
dθ
1/2 .
θ
− sin
2
2
93
Let k = sin
θ0
2
so that
1
T =
2
L
2g
θ0
0
dθ
1/2 .
θ
2
k 2 − sin
2
(0.0.7)
2k cos (u)du
Now let sin θ/2 = k sin u. When θ = 0, u = 0 and when θ = θ0 , u = π/2; moreover, dθ =
=⇒
cos (θ/2)
2 k 2 − sin2 (θ/2)du
2k 1 − sin2 (u)du
2 k 2 − (k sin (u))2 du
=⇒ dθ =
=⇒ dθ =
. Making this
dθ = 1 − sin2 (θ/2)
1 − k 2 sin2 (u)
1 − k 2 sin2 (u)
change of variables in equation (0.0.7) yields
L π/2
du
T =
where k = sin θ0 /2.
g 0
1 − k 2 sin2 (u)
Solutions to Section 1.12
1. The acceleration of gravity is a = 9.8 meters/sec2 . Integrating, we find that the vertical component of
the velocity of the rocket is v(t) = 4.9t + c1 . We are given that v(0) = −10, so that c1 = −10. Thus,
v(t) = 4.9t − 10. Integrating again, we find the position s(t) = 2.495t2 − 10t + c2 . Setting s = 0 at two
meters above the ground, we have s(0) = 0 so that s(t) = 2.495t2 − 10t.
10
≈ 2.04 seconds. Thus,
(a). The highest point above the ground is obtained when v(t) = 0. That is, t = 4.9
2
the highest point is approximately s(2.04) = 2.495 · (2.04) − 10(2.04) ≈ −10.02, which is 12.02 meters above
the ground.
(b). The rocket hits the ground when s(t) = 2. That is 2.495t2 − 10t − 2 = 0. Solving for t with the
quadratic formula, we find that t = −0.19 or t = 4.27. Since we must report a positive answer, we conclude
that the rocket hits the ground 4.27 seconds after launch.
2. The acceleration of gravity is a = 32 ft/sec2 . Integrating, we find that the vertical component of the
velocity of the ball is v(t) = 16t + c1 . Since the ball is initially hit horizontally, we have v(0) = 0, so that
c1 = 0. Hence, v(t) = 16t. Integrating again, we find the position s(t) = 8t2 + c2 . Setting s = 0 at two feet
above the ground, we have s(0) = 0 so that c2 = 0. Thus, s(t) = 8t2 . The ball hits the ground when s(t) = 2,
so that t2 = 14 . Therefore, t = 12 . Since 80 miles per hour equates to over 117 ft/sec. In one-half second, the
horizontal change in position of the ball is therefore more than 117
2 = 58.5 feet, more than enough to span
the necessary 40 feet for the ball to reach the front wall. Therefore, the ball does reach the front wall before
hitting the ground.
3. We first determine the slope of the given family at the point (x, y). Differentiating
y = cx3
(0.0.8)
with respect to x yields
dy
= 3cx2 .
dx
From (0.0.8) we have c = xy3 which, when substituted into Equation (0.0.9) yields
3y
dy
=
.
dx
x
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94
Consequently, the differential equation for the orthogonal trajectories is
dy
x
=− .
dx
3y
Separating the variables and integrating gives
1
3 2
y = − x2 + C,
2
2
which can be written in the equivalent form
x2 + 3y 2 = k.
4. We first determine the slope of the given family at the point (x, y). Differentiating
y = ln(cx)
(0.0.10)
with respect to x yields
1
dy
= .
dx
x
Consequently, the differential equation for the orthogonal trajectories is
dy
= −x.
dx
which can be integrated directly to obtain
1
y = − x2 + k.
2
5. We first determine the slope of the given family at the point (x, y). Differentiating
y 2 = cx3
(0.0.11)
with respect to x yields
2y
so that
dy
= 3cx2
dx
dy
3cx2
=
.
dx
2y
2
From (0.0.11) we have c = xy 3 which, when substituted into Equation (0.0.12) yields
dy
3y
=
.
dx
2x
Consequently, the differential equation for the orthogonal trajectories is
2x
dy
=− .
dx
3y
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Separating the variables and integrating gives
3 2
y = −x2 + C,
2
which can be written in the equivalent form
2x2 + 3y 2 = k.
6. We first determine the slope of the given family at the point (x, y). Differentiating
x4 + y 4 = c
(0.0.13)
with respect to x yields
4x3 + 4y 3
so that
dy
=0
dx
dy
x3
= − 3.
dx
y
(0.0.14)
Consequently, the differential equation for the orthogonal trajectories is
dy
y3
= 3.
dx
x
Separating the variables and integrating gives
1
1
− y −2 = − x−2 + C,
2
2
which can be written in the equivalent form
y 2 − x2 = kx2 y 2 .
7. (a). We first determine the slope of the given family at the point (x, y). Differentiating
x2 + 3y 2 = 2cy
(0.0.15)
with respect to x yields
2x + 6y
so that
dy
dy
= 2c
dx
dx
dy
x
=
.
dx
c − 3y
2
(0.0.16)
2
From (0.0.15) we have c = x +3y
which, when substituted into Equation (0.0.16) yields
2y
dy
2xy
x
= 2
,
= x2 +3y2
dx
x
− 3y 2
− 3y
2y
as required.
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96
(b). It follows from Equation(0.0.17) that the differential equation for the orthogonal trajectories is
dy
3y 2 − x2
=−
.
dx
2xy
This differential equation is first-order homogeneous. Substituting y = xV into the preceding differential
equation gives
3V 2 − 1
dV
+V =
x
dx
2V
which simplifies to
dV
V2−1
=
.
dx
2V
Separating the variables and integrating we obtain
ln(V 2 − 1) = ln x + C,
or, upon exponentiation,
V 2 − 1 = kx.
Inserting V = y/x into the preceding equation yields
y2
− 1 = kx,
x2
that is,
y 2 − x2 = kx3 .
8. Slope field.
9. Slope field.
10. Slope field.
11. See accompanying figure.
12. Slope field.
13. (a). If v(t) = 25, then
1
dv
= 0 = (25 − v).
dt
2
(b). The accompanying figure suggests that
lim v(t) = 25.
t→∞
14. (a). The equilibrium solutions are any constant values of m that satisfy
m 1/4 = 0.
am3/4 1 −
M
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y(x)
x
Figure 0.0.57: Figure for Problem 11
v(t)
25
20
15
10
5
t
5
10
Figure 0.0.58: Figure for Problem 13
Hence, there are two equilibrium solutions, namely, m = 0 and m = M .
(b).This follows since a > 0, and 0 < m(t) < M .
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(c). The given differential equation can be written in the equivalent form
dm
1
3/4
= a m − 1/4 m ,
dt
M
dm
3 −1/4
1
− 1/4
m
4
dt
M
m 1/4 1
2 3/4
3/4
=a m
m − 1/4 m 1 −
M
M
1/4
m
m 1/4
1 2 1/2
3−4
1−
.
= a m
4
M
M
so that
d2 m
=a
dt2
Since 0 < m < M , the expression on the right-hand side of the preceding equation is positive when
3−4
m 1/4
M
> 0,
m 1/4
3
81
< , or equivalently, m <
M . Consequently, the solution curves are concave up for
M
4
256
81
81
0<m<
M , and concave down for
M < m < M.
256
256
81
(d). From the results of (c), there is a change in concavity when m =
M . Substituting this value of
256
m into the right-hand side of the given differential equation yields the following value for the slope of the
solutions curves at the point of inflection:
3/4 1/4
81
27
81
1−
=
a.
a
256
256
256
that is,
(e). Slope field.
15. (a). Separating the variables in Equation (1.12.6) yields
mv
dv
=1
mg − kv 2 dy
which can be integrated to obtain
m
ln(mg − kv 2 ) = y + c.
2k
Multiplying both sides of this equation by −1 and exponentiating gives
−
2k
mg − kv 2 = c1 e− m y .
The initial condition v(0) = 0 requires that c1 = mg, which, when inserted into the preceding equation yields
2k
mg − kv 2 = mge− m y ,
or equivalently,
v2 =
2k
mg 1 − e− m y ,
k
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v2(y)
mg/k
y
Figure 0.0.59: Figure for Problem 15
as required.
(b). See accompanying figure.
16. By inspection the differential equation is separable, but not first-order homogeneous. Further, the
differential equation can be re-written as
dy
+ x2 y = x2 y 2
dx
which reveals that it is also a Bernoulli equation, but is not linear. Finally, a different rearrangement of
terms yields
1
x2 dx −
dy = 0,
y(y − 1)
which is an exact differential equation. Separating the variables in the given differential equation yields
1
1
1
2
dy = x dx =⇒
dy = x3 + c
y(y − 1)
y(y − 1)
3
Using a partial fraction decomposition of the integrand on the left-hand side of the preceding equation we
obtain
3
1
y−1
y−1
1
1
1
−
dy = x3 + c =⇒ ln
= x3 + c =⇒
= c1 ex /3
y−1 y
3
y
3
y
so that
1
.
y(x) =
1 − c1 ex3 /3
41. Writing the differential equation in the form
e−y
dy
= xex ·
dx
1+y
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we see that it is separable, but not homogeneous. It is also neither a linear differential equation nor a
Bernoulli differential equation. Rearranging the differential equation we have
xex dx − ey (1 + y)dy = 0
which is exact. Separating the variables in the given differential equation and integrating yields
ey (1 + y)dy = xex dx + c =⇒ yey = ex (x − 1) + c.
17. The given differential equation is separable. Separating the variables gives
y
dy
ln x
=2
,
dx
x
which can be integrated directly to obtain
1 2
y = (ln x)2 + c,
2
or, equivalently,
y 2 = 2(ln x)2 + c1 .
18. The given differential equation is first-order linear. We first divide by x to put the differential equation
in standard form:
dy
2
− y = 2x ln x.
(0.0.18)
dx x
An integrating factor for this equation is I = e (−2/x)dx = x−2 . Multiplying Equation (0.0.18) by x−2 reduces
it to
d −2
(x y) = 2x−1 ln x,
dx
which can be integrated to obtain
x−2 y = (ln x)2 + c
so that
y(x) = x2 [(ln x)2 + c].
19. We first re-write the given differential equation in the differential form
2xy dx + (x2 + 2y)dy = 0.
(0.0.19)
Then
My = 2x = Nx
so that the differential equation is exact. Consequently, there exists a potential function φ satisfying
∂φ
= 2xy,
∂x
∂φ
= x2 + 2y.
∂y
Integrating these two equations in the usual manner yields
φ(x, y) = x2 y + y 2 .
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Therefore Equation (0.0.19) can be written in the equivalent form
d(x2 y + y 2 ) = 0
with general solution
x2 y + y 2 = c.
20. We first rewrite the given differential equation as
dy
y 2 + 3xy + x2
,
=
dx
x2
which is first order homogeneous. Substituting y = xV into the preceding equation yields
x
so that
x
dV
+ V = V 2 + 3V + 1
dx
dV
= V 2 + 2V + 1 = (V + 1)2 ,
dx
or, in separable form,
1
dV
1
= .
2
(V + 1) dx
x
This equation can be integrated to obtain
−(V + 1)−1 = ln x + c
so that
V +1=
1
.
c1 − ln x
Inserting V = y/x into the preceding equation yields
1
y
+1=
,
x
c1 − ln x
so that
y(x) =
x
− x.
c1 − ln x
21. We first rewrite the given differential equation in the equivalent form
dy
+ y · tan x = −y 2 sin x,
dx
which is a Bernoulli equation. Dividing this equation by y 2 yields
y −2
dy
+ y −1 tan x = − sin x.
dx
Now make the change of variables u = y −1 in which case
Equation (0.0.20) gives the linear differential equation
−
(0.0.20)
dy
du
= −y −2 . Substituting these results into
dx
dx
du
+ u · tan x = − sin x
dx
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or, in standard form,
du
− u · tan x = sin x.
dx
(0.0.21)
An integrating factor for this differential equation is I = e− tan x dx = cos x. Multiplying Equation (0.0.21)
by cos x reduces it to
d
(u · cos x) = sin x cos x
dx
which can be integrated directly to obtain
1
u · cos x = − cos2 x + c1 ,
2
so that
u=
− cos2 x + c2
.
cos x
Inserting u = y −1 into the preceding equation and rearranging yields
y(x) =
−2 cos x
2 cos x
=
.
− cos2 x + c2
cos2 x + c
22. The given differential equation is linear with integrating factor
I=e
2e2x
dx
1+e2x
2x
= eln(1+e
)
= 1 + e2x .
Multiplying the given differential equation by 1 + e2x yields
e2x + 1
d
2e2x
(1 + e2x )y = 2x
= −1 + 2x
dx
e −1
e −1
which can be integrated directly to obtain
(1 + e2x )y = −x + ln |e2x − 1| + c,
so that
y(x) =
−x + ln |e2x − 1| + c
.
1 + e2x
23. We first rewrite the given differential equation in the equivalent form
dy
y + x2 − y 2
=
,
dx
x
which we recognize as being first order homogeneous. Inserting y = xV into the preceding equation yields
x
|x| dV
1 − V 2,
+V =V +
dx
x
that is,
√
dV
1
1
=± .
x
1 − V 2 dx
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Integrating we obtain
sin−1 V = ± ln |x| + c,
so that
V = sin(c ± ln |x|).
Inserting V = y/x into the preceding equation yields
y(x) = x sin(c ± ln |x|).
24. We first rewrite the given differential equation in the equivalent form
(sin y + y cos x + 1)dx − (1 − x cos y − sin x)dy = 0.
Then
My = cos y + cos x = Nx
so that the differential equation is exact. Consequently, there is a potential function satisfying
∂φ
= sin y + y cos x + 1,
∂x
∂φ
= −(1 − x cos y − sin x).
∂y
Integrating these two equations in the usual manner yields
φ(x, y) = x − y + x sin y + y sin x,
so that the differential equation can be written as
d(x − y + x sin y + y sin x) = 0,
and therefore has general solution
x − y + x sin y + y sin x = c.
25. Writing the given differential equation as
dy
1
25 −1 2
+ y=
y x ln x,
dx x
2
we see that it is a Bernoulli equation with n = −1. We therefore divide the equation by y −1 to obtain
y
dy
25 2
1
+ y2 =
x ln x.
dx x
2
dy
We now make the change of variables u = y 2 , in which case, du
dx = 2y dx . Inserting these results into the
preceding differential equation yields
1 du 1
25 2
+ u=
x ln x,
2 dx x
2
or, in standard form,
du 2
+ u = 25x2 ln x.
dx x
An integrating factor for this linear differential equation is I = e (2/x)dx = x2 . Multiplying the previous
differential equation by x2 reduces it to
d 2
(x u) = 25x4 ln x
dx
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which can be integrated directly to obtain
2
x u = 25
so that
1 5
1
x ln x − x5
5
25
+c
u = x3 (5 ln x − 1) + cx−2 .
Making the replacement u = y 2 in this equation gives
y 2 = x3 (5 ln x − 1) + cx−2 .
26. The given differential equation can be written in the equivalent form
dy
ex−y
= 2x+y = e−x e−2y ,
dx
e
which we recognize as being separable. Separating the variables gives
e2y
dy
= e−x
dx
which can be integrated to obtain
1 2y
e = −e−x + c
2
so that
y(x) =
1
ln(c1 − 2e−x ).
2
27. The given differential equation is linear with integrating factor I = e cot x dx = sin x. Multiplying the
given differential equation by sin x reduces it to
d
sin x
(y sin x) =
dx
cos x
which can be integrated directly to obtain
y sin x = − ln(cos x) + c,
so that
y(x) =
c − ln(cos x)
.
sin x
28. Writing the given differential equation as
1
2ex
dy
y = 2y 2 e−x ,
+
x
dx 1 + e
1
we see that it is a Bernoulli equation with n = 1/2. We therefore divide the equation by y 2 to obtain
1
y− 2
dy
2ex 1
y 2 = 2e−x .
+
dx 1 + ex
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1
1
1 − 2 dy
We now make the change of variables u = y 2 , in which case, du
dx = 2 y
dx . Inserting these results into the
preceding differential equation yields
du
2ex
u = 2e−x ,
2
+
dx 1 + ex
or, in standard form,
ex
du
u = e−x .
+
dx 1 + ex
An integrating factor for this linear differential equation is
I=e
ex
1+ex dx
= eln(1+e ) = 1 + ex .
x
Multiplying the previous differential equation by 1 + ex reduces it to
d
[(1 + ex )u] = e−x (1 + ex ) = e−x + 1
dx
which can be integrated directly to obtain
(1 + ex )u = −e−x + x + c
so that
u=
x − e−x + c
.
1 + ex
1
Making the replacement u = y 2 in this equation gives
1
y2 =
x − e−x + c
.
1 + ex
29. We first rewrite the given differential equation in the equivalent form
dy
y y
=
ln
+1 .
dx
x
x
The function appearing on the right of this equation is homogeneous of degree zero, and therefore the
differential equation itself is first order homogeneous. We therefore insert y = xV into the differential
equation to obtain
dV
x
+ V = V (ln V + 1),
x
so that
dV
= V ln V.
x
dx
Separating the variables yields
1 dV
1
=
V ln V dx
x
which can be integrated to obtain
ln(ln V ) = ln x + c.
Exponentiating both side of this equation gives
ln V = c1 x,
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or equivalently,
V = e c1 x .
Inserting V = y/x in the preceding equation yields
y = xec1 x .
30. For the given differential equation we have
M (x, y) = 1 + 2xey ,
N (x, y) = −(ey + x),
so that
My − N x
1 + 2xey
= 1.
=
M
1 + 2xey
Consequently, an integrating factor for the given differential equation is
I = e−
dy
= e−y .
Multiplying the given differential equation by e−y yields the exact differential equation
(2x + e−y )dx − (1 + xe−y )dy = 0.
(0.0.22)
Therefore, there exists a potential function φ satisfying
∂φ
= 2x + e−y ,
∂x
∂φ
= −(1 + xe−y ).
∂y
Integrating these two equations in the usual manner yields
φ(x, y) = x2 − y + xe−y .
Therefore Equation (0.0.22) can be written in the equivalent form
d(x2 − y + xe−y ) = 0
with general solution
x2 − y + xe−y = c.
31. The given differential equation is first-order linear. However, it can also e written in the equivalent form
dy
= (1 − y) sin x
dx
which is separable. Separating the variables and integrating yields
− ln |1 − y| = − cos x + c,
so that
Hence,
1 − y = c1 ecos x .
y(x) = 1 − c1 ecos x .
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32. For the given differential equation we have
M (x, y) = 3y 2 + x2 ,
N (x, y) = −2xy,
so that
My − N x
4
=− .
N
x
Consequently, an integrating factor for the given differential equation is
I = e−
4
x dx
= x−4 .
Multiplying the given differential equation by x−4 yields the exact differential equation
(3y 2 x−4 + x−2 )dx − 2yx−3 dy = 0.
(0.0.23)
Therefore, there exists a potential function φ satisfying
∂φ
= 3y 2 x−4 + x−2 ,
∂x
∂φ
= −2yx−3 .
∂y
Integrating these two equations in the usual manner yields
φ(x, y) = −y 2 x−3 − x−1 .
Therefore Equation (0.0.23) can be written in the equivalent form
d(−y 2 x−3 − x−1 ) = 0
with general solution
−y 2 x−3 − x−1 = c,
or equivalently,
x2 + y 2 = c 1 x3 .
Notice that the given differential equation can be written in the equivalent form
dy
3y 2 + x2
=
,
dx
2xy
which is first-order homogeneous. Another equivalent way of writing the given differential equation is
dy
3
1
−
y = xy −1 ,
dx 2x
2
which is a Bernoulli equation.
33. The given differential equation can be written in the equivalent form
dy
1
9
−
y = − x2 y 3 ,
dx 2x ln x
2
which is a Bernoulli equation with n = 3. We therefore divide the equation by y 3 to obtain
y −3
dy
9
1
−
y −2 = − x2 .
dx 2x ln x
2
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−3 dy
We now make the change of variables u = y −2 , in which case, du
dx = −2y
dx . Inserting these results into
the preceding differential equation yields
1 du
1
9
−
u = − x2 ,
2 dx 2x ln x
2
−
or, in standard form,
du
1
+
u = 9x2 .
dx x ln x
An integrating factor for this linear differential equation is
1
x ln x dx
I=e
= eln(ln x) = ln x.
Multiplying the previous differential equation by ln x reduces it to
d
(ln x · u) = 9x2 ln x
dx
which can be integrated to obtain
ln x · u = x3 (3 ln x − 1) + c
so that
x3 (3 ln x − 1) + c
.
ln x
Making the replacement u = y 3 in this equation gives
u=
y3 =
x3 (3 ln x − 1) + c
.
ln x
34. Separating the variables in the given differential equation yields
1 dy
2+x
1
=
=1+
,
y dx
1+x
1+x
which can be integrated to obtain
ln |y| = x + ln |1 + x| + c.
Exponentiating both sides of this equation gives
y(x) = c1 (1 + x)ex .
35. The given differential equation can be written in the equivalent form
2
dy
+
y=1
dx x2 − 1
(0.0.24)
which is first-order linear. An integrating factor is
I=e
2
dx
x2 −1
1
1
= e ( x−1 − x+1 )dx = e[ln(x−1)−ln(x+1)] =
Multiplying (0.0.24) by (x − 1)/(x + 1) reduces it to the integrable form
x−1
2
d x−1
·y =
=1−
.
dx x + 1
x+1
x+1
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.
x+1
109
Integrating both sides of this differential equation yields
x−1
· y = x − 2 ln(x + 1) + c
x+1
so that
y(x) =
x+1
x−1
[x − 2 ln(x + 1) + c].
36. The given differential equation can be written in the equivalent form
[y sec2 (xy) + 2x]dx + x sec2 (xy)dy = 0
Then
My = sec2 (xy) + 2xy sec2 (x) tan(xy) = Nx
so that the differential equation is exact. Consequently, there is a potential function satisfying
∂φ
= y sec2 (xy) + 2x,
∂x
∂φ
= x sec2 (xy).
∂y
Integrating these two equations in the usual manner yields
φ(x, y) = x2 + tan(xy),
so that the differential equation can be written as
d(x2 + tan(xy)) = 0,
and therefore has general solution
x2 + tan(xy) = c,
or equivalently,
y(x) =
tan−1 (c − x2 )
.
x
37. The given differential equation is first-order homogeneous. Inserting y = xV into the given equation
yields
dV
1
x
+ V,
+V =
dx
1+V2
that is,
(1 + V 2 )
dV
1
= .
dx
x
Integrating we obtain
1
V + V 3 = ln |x| + c.
3
Inserting V = y/x into the preceding equation yields
y3
y
+ 3 = ln |x| + c,
x 3x
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or equivalently,
3x2 y + y 3 = 3x3 (ln |x| + c).
dy
38. The differential equation is first-order homogeneous. We therefore let y = xv, in which case
=
dx
dv
+ v. Substituting these results into the given differential equation yields
x
dx
x
so that
x
1 + v2
dv
,
+v =
dx
1 − 3v 2
dv
(v + 1)(3v 2 − 2v + 1)
1 + v 2 − v + 3v 3
=
.
=
dx
1 − 3v 2
1 − 3v 2
Separating the variables gives
1 − 3v 2
1
dv = dx.
(v + 1)(3v 2 − 2v + 1)
x
Decomposing the left-hand side into partial fractions yields
1
1
2(3v − 2)
dv = dx,
−
−
2
3(v + 1) 3(3v − 3v + 1)
x
or equivalently,
−
1
1
2(3v − 2)
−
2 2 dv = dx.
1
3(v + 1)
x
+
v−
3
9
This can be integrated to yield
1
3v − 1
3
2
1
3
√
− ln(v + 1) = ln x + c.
− ln (3v − 1)2 +
2 √ arctan
2
9
9
3
2
2
Therefore,
2
3
√ arctan
2
3y − x
√
2x
−
3
2
(3y − x)2
1
y+x
+
ln
−
ln
= ln x + c.
2
9x2
9
3
x
39. The given differential equation is a Bernoulli equation with n = −1. We therefore divide the equation
by y −1 to obtain
25 ln x
1
dy
.
+ y2 =
y
dx x
2x3
dy
We now make the change of variables u = y 2 , in which case, du
dx = 2y dx . Inserting these results into the
preceding differential equation yields
1 du 1
25 ln x
,
+ u=
2 dx x
2x3
or, in standard form,
du 2
+ u = 25x−3 ln x.
dx x
An integrating factor for this linear differential equation is
I=e
2
x dx
= x2 .
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Multiplying the previous differential equation by x2 reduces it to
d 2
(x u) = 25x−1 ln x,
dx
which can be integrated directly to obtain
x2 u =
so that
u=
25
(ln x)2 + c
2
25(ln x)2 + c
.
2x2
Making the replacement u = y 2 in this equation gives
y2 =
25(ln x)2 + c
.
2x2
dy
40. The differential equation is first-order homogeneous. We therefore let y = xv, in which case
=
dx
dv
+ v. Substituting these results into the given differential equation yields
x
dx
x
so that
x
dv
1 + v2
,
+v =
dx
1 − 3v 2
dv
(v + 1)(3v 2 − 2v + 1)
1 + v 2 − v + 3v 3
=
.
=
dx
1 − 3v 2
1 − 3v 2
Separating the variables gives
1 − 3v 2
1
dv = dx.
2
(v + 1)(3v − 2v + 1)
x
Decomposing the left-hand side into partial fractions yields
2(3v − 2)
1
1
−
dv = dx,
−
2
3(v + 1) 3(3v − 3v + 1)
x
or equivalently,
−
1
1
2(3v − 2)
dv = dx.
−
1 2
2
3(v + 1)
x
+
v−
3
9
This can be integrated to yield
1
3
2
1
3v − 1
3
2
√
− ln(v + 1) = ln x + c.
− ln (3v − 1) +
2 √ arctan
2
9
9
3
2
2
Therefore,
2
3
√ arctan
2
3y − x
√
2x
−
1
3
2
(3y − x)2
y+x
−
= ln x + c.
+
ln
ln
2
9x2
9
3
x
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41. The given differential equation can be written in the equivalent form
ey (1 + y)
dy
= xex
dx
which is separable. Integrating both sides of this equation gives
yey = ex (x − 1) + c.
42. The given differential equation can be written in the equivalent form
dy
cos x
−
y = − cos x
dx
sin x
which is first order linear with integrating factor
I = e−
cos x
sin x dx
= e− ln(sin x) =
1
.
sin x
Multiplying the preceding differential equation by sin1 x reduces it to
1
d
cos x
·y =−
dx sin x
sin x
which can be integrated directly to obtain
1
· y = − ln(sin x) + c
sin x
so that
y(x) = sin x[c − ln(sin x)].
43. The given differential equation is linear, and therefore can be solved using an appropriate integrating
factor. However, if we rearrange the terms in the given differential equation then it can be written in the
equivalent form
1 dy
= x2
1 + y dx
which is separable. Integrating both sides of the preceding differential equation yields
1 3
x +c
3
ln(1 + y) =
so that
1
3
y(x) = c1 e 3 x − 1.
Imposing the initial condition y(0) = 5 we find c1 = 6. Therefore the solution to the initial-value problem is
1
3
y(x) = 6e 3 x − 1.
44. The given differential equation can be written in the equivalent form
e−6y
dy
= −e−4x
dx
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which is separable. Integrating both sides of the preceding equation yields
1
1
− e−6y = e−4x + c
6
4
so that
3 −4x
1
.
y(x) = − ln c1 − e
6
2
Imposing the initial condition y(0) = 0 requires that
3
.
0 = ln c1 −
2
Hence, c1 = 52 , and so
1
y(x) = − ln
6
5 − 3e−4x
2
.
45. For the given differential equation we have
My = 4xy = Nx
so that the differential equation is exact. Consequently, there is a potential function satisfying
∂φ
= 3x2 + 2xy 2 ,
∂x
∂φ
= 2x2 y.
∂y
Integrating these two equations in the usual manner yields
φ(x, y) = x2 y 2 + x3 ,
so that the differential equation can be written as
d(x2 y 2 + x3 ) = 0,
and therefore has general solution
x2 y 2 + x3 = c.
Imposing the initial condition y(1) = 3 yields c = 10. Therefore,
x2 y 2 + x3 = 10
so that
10 − x3
.
x2
Note that the given differential equation can be written in the equivalent form
y2 =
1
3
dy
+ y = − y −1 ,
dx x
2
which is a Bernoulli equation with n = −1. Consequently, the Bernoulli technique could also have been used
to solve the differential equation.
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46. The given differential equation is linear with integrating factor
I = e−
sin x dx
= ecos x .
Multiplying the given differential equation by ecos x reduces it to the integrable form
d cos x
· y) = 1,
(e
dx
which can be integrated directly to obtain
ecos x · y = x + c..
Hence,
y(x) = e− cos x (x + c).
Imposing the given initial condition y(0) = 1e requires that c = 1. Consequently,
y(x) = e− cos x (x + 1).
47. (a). For the given differential equation we have
My = my m−1 ,
Nx = −nxn−1 y 3 .
We see that the only values for m and n for which My = Nx are m = n0. Consequently, these are the only
values of m and n for which the differential equation is exact.
(b). We rewrite the given differential equation in the equivalent form
dy
x5 + y m
,
=
dx
xn y 3
(0.0.25)
from which we see that the differential equation is separable provided m = 0. In this case there are no
restrictions on n.
(c). From Equation (0.0.25) we see that the only values of m and n for which the differential equation is
first-order homogeneous are m = 5 and n = 2.
(d). We now rewrite the given differential equation in the equivalent form
dy
− x−n y m−3 = x5−n y −3 .
dx
(0.0.26)
Due to the y −3 term on the right-hand side of the preceding differential equation, it follows that there are
no values of m and n for which the equation is linear.
(e). From Equation (0.0.26) we see that the differential equation is a Bernoulli equation whenever m = 4.
There are no constraints on n in this case.
48. In Newton’s Law of Cooling we have
Tm = 180◦ F,
T (0) = 80◦ F,
T (3) = 100◦ F.
We need to determine the time, t0 when T (t0 ) = 140◦ F. The temperature of the sandals at time t is governed
by the differential equation
dT
= −k(T − 180).
dt
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This separable differential equation is easily integrated to obtain
T (t) = 180 + ce−kt .
Since T (0) = 80 we have
80 = 180 + c =⇒ c = −100.
Hence,
T (t) = 180 − 100e−kt .
Imposing the condition T (3) = 100 requires
100 = 180 − 100e−3k .
Solving for k we find k = 13 ln
5
4
. Inserting this value for k into the preceding expression for T (t) yields
t
5
t0
5
T (t) = 180 − 100e− 3 ln( 4 ) .
We need to find t0 such that
140 = 180 − 100e− 3 ln( 4 ) .
Solving for t0 we find
ln 52
≈ 12.32 min.
t0 = 3
ln 54
49. In Newton’s Law of Cooling we have
Tm = 70◦ F,
T (0) = 150◦ F,
T (10) = 125◦ F.
We need to determine the time, t0 when T (t0 ) = 100◦ F. The temperature of the plate at time t is governed
by the differential equation
dT
= −k(T − 70).
dt
This separable differential equation is easily integrated to obtain
T (t) = 70 + ce−kt .
Since T (0) = 150 we have
150 = 70 + c =⇒ c = 80.
Hence,
T (t) = 70 + 80e−kt .
Imposing the condition T (10) = 125 requires
125 = 70 + 80e−10k .
1
ln
Solving for k we find k = 10
16 11
. Inserting this value for k into the preceding expression for T (t) yields
t
16
t0
16
T (t) = 70 + 80e− 10 ln( 11 ) .
We need to find t0 such that
100 = 70 + 80e− 10 ln( 11 ) .
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Solving for t0 we find
ln 83
≈ 26.18 min.
t0 = 10
ln 16
11
50. Let T (t) denote the temperature of the object at time t, and let Tm denote the temperature of the
surrounding medium. Then we must solve the initial-value problem
dT
= k(T − Tm )2 ,
dt
T (0) = T0 ,
where k is a constant. The differential equation can be written in separated form as
1
dT
= k.
(T − Tm )2 dt
Integrating both sides of this differential equation yields
−
1
= kt + c
T − Tm
so that
T (t) = Tm −
1
.
kt + c
Imposing the initial condition T (0) = T0 we find that
c=
1
Tm − T0
which, when substituted back into the preceding expression for T (t) yields
T (t) = Tm −
1
kt + Tm1−T0
= Tm −
Tm − T0
.
k(Tm − T0 )t + 1
As t → ∞, T (t) approaches Tm .
dv
(0) = 2, the velocity is increasing at the rate of 2 m/s2 at t = 0.
dt
(b). Evaluating the given differential equation at t = 0, using the given initial conditions yields
51.(a). Since
2 + 20k = 80k =⇒ k =
1
.
3
(c). An integrating factor for the given differential equation is I = e k dt = ekt . Multiplying the given
d
differential equation by this integrating factor reduces is to (ekt ·v) = 80k =⇒ v(t) = e−kt (80t+c). Imposing
dt
4
the initial condition v(0) = 20 yields c = 20, so that v(t) = 20e−kt (4kt + 1) =⇒ v(t) = e−t/30 (2t + 15).
3
4 −t/30
(d). v(t) = e
(2t + 15) =⇒ there is no finite t > 0 when v(t) = 0. Hence the object does not come to
3
rest in a finite time.
(e). limt→∞ v(t) = 0.
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52. We are given the differential equation
dT
= −k(T − 5 cos 2t)
dt
(0.0.27)
dT
(0) = 5.
dt
(0.0.28)
together with the initial conditions
T (0) = 0;
(a). Setting t = 0 in (0.0.27) and using (0.0.28) yields
5 = −k(0 − 5)
so that k = 1.
(b). Substituting k = 1 into the differential equation (0.0.27) and rearranging terms yields
dT
+ T = 5 cos t.
dt
An integrating factor for this linear differential equation is I = e dt = et . Multiplying the preceding
differential equation by et reduces it to
d t
(e · T ) = 5et cos 2t
dt
which upon integration yields
et · T = et (cos 2t + 2 sin 2t) + c,
so that
T (t) = ce−t + cos 2t + 2 sin 2t.
Imposing the initial condition T (0) = 0 we find that c = −1. Hence,
T (t) = cos 2t + 2 sin 2t − e−t .
(c). For large values of t we have
T (t) ≈ cos 2t + 2 sin 2t,
which can be written in phase-amplitude form as
T (t) ≈
√
5 cos(2t − φ),
where tan φ√= 2. consequently, for large t, the temperature is approximately oscillatory with period π and
amplitude 5.
53. If we let C(t) denote the number of sandhill cranes in the Platte River valley t days after April 1, then
C(t) is governed by the differential equation
dC
= −kC
dt
(0.0.29)
together with the auxiliary conditions
C(0) = 500, 000;
C(15) = 100, 000.
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118
Separating the variables in the differential equation (0.0.29) yields
1 dC
= −k,
C dt
which can be integrated directly to obtain
ln C = −kt + c.
Exponentiation yields
C(t) = c0 e−kt .
The initial condition C(0) = 500, 000 requires c0 = 500, 000, so that
C(t) = 500, 000e−kt .
(0.0.31)
Imposing the auxiliary condition C(15) = 100, 000 yields
100, 000 = 500, 000e−15k .
1
Taking the natural logarithm of both sides of the preceding equation and simplifying we find that k = 15
ln 5.
Substituting this value for k into (0.0.31) gives
C(t) = 500, 000e− 15 ln 5 .
t
(0.0.32)
1
= 20, 000 sandhile cranes.
(a). C(3) = 500, 000e−2 ln 5 = 500, 000 · 25
35
(b). C(35) = 500, 000e− 15 ln 5 ≈ 11696 sandhile cranes.
(c). We need to determine t0 such that
t0
1000 = 500, 000e− 15 ln 5
that is,
1
.
500
Taking the natural logarithm of both sides of this equation and simplifying yields
t0
e− 15 ln 5 =
t0 = 15 ·
ln 500
≈ 57.9 days after April 1.
ln 5
54. Substituting P0 = 200, 000 into Equation (1.5.3) in the text yields
P (t) =
200, 000C
.
200, 000 + (C − 200, 000)e−rt
(0.0.33)
We are given
P (3) = P (t1 ) = 230, 000,
P (6) = P (t2 ) = 250, 000.
Since t2 = 2t1 we can use the formulas (1.5.5) and (1.5.6) of the text to obtain r and C directly as follows:
1
1
25(23 − 20)
15
= ln
≈ 0.21.
r = ln
3
20(25 − 23)
3
8
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230, 000[(23)(45) − (40)(25)]
= 277586.
(23)2 − (20)(25)
C=
Substituting these values for r and C into (0.0.33) yields
P (t) =
55517200000
.
200, 000 + (77586)e−0.21t
Therefore,
P (10) =
55517200000
≈ 264, 997,
200, 000 + (77586)e−2.1
P (20) =
55517200000
≈ 275981.
200, 000 + (77586)e−4.2
and
55. The differential equation for determining q(t) is
dq 5
3
+ q = cos 2t,
dt
4
2
which has integrating factor I = e
it to the integrable form
5
4 dt
5
5
= e 4 t . Multiplying the preceding differential equation by e 4 t reduces
d 5t 3 5t
e 4 · q = e 4 cos 2t.
dt
2
Integrating and simplifying we find
q(t) =
5
6
(5 cos 2t + 8 sin 2t) + ce− 4 t .
89
(0.0.34)
The initial condition q(0) = 3 requires
3=
30
+ c,
89
so that c = 237
89 . Making this replacement in (0.0.34) yields
q(t) =
6
237 − 5 t
(5 cos 2t + 8 sin 2t) +
e 4 .
89
89
The current in the circuit is
i(t) =
dq
12
1185 − 5 t
=
(8 cos 2t − 5 sin 2t) −
e 4 .
dt
89
356
56. The current in the circuit is governed by the differential equation
di
100
+ 10i =
,
dt
3
which has integrating factor I = e 10 dt = e10t . Multiplying the preceding differential equation by e10t
reduces it to the integrable form
d 10t 100 10t
e ·i =
e .
dt
3
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Integrating and simplifying we find
i(t) =
10
+ ce−10t .
3
(0.0.35)
The initial condition i(0) = 3 requires
10
+ c,
3
so that c = − 13 . Making this replacement in (0.0.35) yields
3=
i(t) =
1
(10 − e−10t ).
3
57. We are given:
r1 = 6 L/min,
c1 = 3 g/L,
r2 = 4 L/min,
V (0) = 30 L,
A(0) = 0 g,
and we need to determine the amount of salt in the tank when V (t) = 60L. Consider a small time interval
Δt. Using the preceding information we have:
ΔV = 6Δt − 4Δt = 2Δt,
and
A
Δt.
V
Dividing both of these equations by Δt and letting Δt → 0 yields
ΔA ≈ 18Δt − 4
dV
= 2.
dt
(0.0.36)
A
dA
+ 4 = 18.
dt
V
Integrating (0.0.36) and imposing the initial condition V (0) = 30 yields
V (t) = 2(t + 15).
(0.0.37)
(0.0.38)
We now insert this expression for V (t) into (0.0.37) to obtain
dA
2
+
A = 18.
dt
t + 15
2
An integrating factor for this differential equation is I = e t+15 dt = (t + 15)2 . Multiplying the preceding
differential equation by (t + 15)2 reduces it to the integrable form
d
(t + 15)2 A = 18(t + 15)2 .
dt
Integrating and simplifying we find
A(t) =
6(t + 15)3 + c
.
(t + 15)2
Imposing the initial condition A(0) = 0 requires
0=
6(15)3 + c
,
(15)2
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so that c = −20250. Consequently,
A(t) =
6(t + 15)3 − 20250
.
(t + 15)2
We need to determine the time when the solution overflows. Since the tank can hold 60 L of solution, from
(0.0.38) overflow will occur when
60 = 2(t + 15) =⇒ t = 15.
The amount of chemical in the tank at this time is
A(15) =
6(30)3 − 20250
≈ 157.5 g.
(30)2
58. Applying Euler’s method with y = x2 + 2y 2 , x0 = 0, y0 = −3, and h = 0.1 we have yn+1 = yn + 0.1(x2n +
2yn2 ). This generates the sequence of approximants given in the table below.
n
1
2
3
4
5
6
7
8
9
10
xn
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
yn
−1.2
−0.911
−0.74102
−0.62219
−0.52877
−0.44785
−0.371736
−0.29510
−0.21368
−0.12355
Consequently the Euler approximation to y(1) is y10 = −0.12355.
3x
+ 2, x0 = 1, y0 = 2, and h = 0.05 we have
y
3xn
+2 .
yn+1 = yn + 0.05
yn
59. Applying Euler’s method with y =
This generates the sequence of approximants given in the table below.
n
1
2
3
4
5
6
7
8
9
10
xn
1.05
1.10
1.15
1.20
1.25
1.30
1.35
1.40
1.45
1.50
yn
2.1750
2.34741
2.51770
2.68622
2.85323
3.01894
3.18353
3.34714
3.50988
3.67185
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Consequently, the Euler approximation to y(1.5) is y10 = 3.67185.
60. Applying the modified Euler method with y = x2 + 2y 2 , x0 = 0, y0 = −3, and h = 0.1 generates the
sequence of approximants given in the table below.
n
1
2
3
4
5
6
7
8
9
10
xn
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
yn
−1.9555
−1.42906
−1.11499
−0.90466
−0.74976
−0.62555
−0.51778
−0.41723
−0.31719
−0.21196
Consequently, the modified Euler approximation to y(1) is y10 = −0.21196. Comparing this to the
corresponding Euler approximation from Problem 58 we have
|yME − yE | = |0.21196 − 0.12355| = 0.8841.
61. Applying the modified Euler method with y =
3x
+ 2, x0 = 1, y0 = 2, and h = 0.05 generates the
y
sequence of approximants given in the table below.
n
1
2
3
4
5
6
7
8
9
10
xn
1.05
1.10
1.15
1.20
1.25
1.30
1.35
1.40
1.45
1.50
yn
2.17371
2.34510
2.51457
2.68241
2.84886
3.01411
3.17831
3.34159
3.50404
3.66576
Consequently, the modified Euler approximation to y(1.5) is y10 = 3.66576. Comparing this to the corresponding Euler approximation from Problem 59 we have
|yME − yE | = |3.66576 − 3.67185| = 0.00609.
62. Applying the Runge-Kutta method with y = x2 + 2y 2 , x0 = 0, y0 = −3, and h = 0.1 generates the
sequence of approximants given in the table below.
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n
1
2
3
4
5
6
7
8
9
10
xn
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
yn
−1.87392
−1.36127
−1.06476
−0.86734
−0.72143
−0.60353
−0.50028
−0.40303
−0.30541
−0.20195
Consequently the Runge-Kutta approximation to y(1) is y10 = −0.20195. Comparing this to the corresponding Euler approximation from Problem 58 we have
|yRK − yE | = |0.20195 − 0.12355| = 0.07840.
63. Applying the Runge-Kutta method with y =
3x
+ 2, x0 = 1, y0 = 2, and h = 0.05 generates the
y
sequence of approximants given in the table below.
n
1
2
3
4
5
6
7
8
9
10
xn
1.05
1.10
1.15
1.20
1.25
1.30
1.35
1.40
1.45
1.50
yn
2.17369
2.34506
2.51452
2.68235
2.84880
3.01404
3.17823
3.34151
3.50396
3.66568
Consequently the Runge-Kutta approximation to y(1.5) is y10 = 3.66568. Comparing this to the corresponding Euler approximation from Problem 59 we have
|yRK − yE | = |3.66568 − 3.67185| = 0.00617.
Chapter 2 Solutions
Solutions to Section 2.1
True-False Review:
(a): TRUE. A diagonal matrix has no entries below the main diagonal, so it is upper triangular. Likewise,
it has no entries above the main diagonal, so it is also lower triangular.
(b): FALSE. An m × n matrix has m row vectors and n column vectors.
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(c): TRUE. This is a square matrix, and all entries off the main diagonal are zero, so it is a diagonal matrix
(the entries on the diagonal also happen to be zero, but this is not required).
(d): FALSE. The main diagonal entries of a skew-symmetric matrix must be zero. In this case, a11 = 4 = 0,
so this matrix is not skew-symmetric.
(e): FALSE. The form presented uses the same number along the entire main diagonal, but a symmetric
matrix need not have identical entries on the main diagonal.
(f ): TRUE. Since A is symmetric, A = AT . Thus, (AT )T = A = AT , so AT is symmetric.
(g): FALSE. The trace of a matrix is the sum of the entries along the main diagonal.
(h): TRUE. If A is skew-symmetric, then AT = −A. But A and AT contain the same entries along the
main diagonal, so for AT = −A, both A and −A must have the same main diagonal. This is only possible
if all entries along the main diagonal are 0.
(i): TRUE. If A is both symmetric and skew-symmetric, then A = AT = −A, and A = −A is only possible
if all entries of A are zero.
(j): TRUE. Both matrix functions are defined for values of t such that t > 0.
(k): FALSE. The (3, 2)-entry contains a function that is not defined for values of t with t ≤ 3. So for
example, this matrix functions is not defined for t = 2.
(l): TRUE. Each numerical entry of the matrix function is a constant function, which has domain R.
(m): FALSE. For instance, the matrix function A(t) = [t] and B(t) = [t2 ] satisfy A(0) = B(0), but A and
B are not the same matrix function.
Problems:
1(a). a31 = 0, a24 = −1, a14 = 2, a32 = 2, a21 = 7, a34 = 4.
1(b). (1, 4) and (3, 2).
2(a). b12 = −1, b33 = 4, b41 = 0, b43 = 8, b51 = −1, and b52 = 9.
2(b). (1, 2), (1, 3), (2, 1), (3, 2), and (5, 1).
1 5
3.
; 2 × 2 matrix.
−1 3
2 1 −1
4.
; 2 × 3 matrix.
0 4 −2
⎤
⎡
−1
⎢ 1 ⎥
⎥
5. ⎢
⎣ 1 ⎦; 4 × 1 matrix.
−5
⎤
⎡
1 −3 −2
⎢ 3
6
0 ⎥
⎥; 4 × 3 matrix.
6. ⎢
⎣ 2
7
4 ⎦
−4 −1
5
⎡
⎤
0 −1 2
0 3 ⎦; 3 × 3 matrix.
7. ⎣ 1
−2 −3 0
(c)2017 Pearson Education. Inc.
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⎡
0
⎢ 1
⎢
8. ⎣
2
3
⎡
2
⎢ 3
9. ⎢
⎣ 4
5
⎤
−1 −2 −3
0 −1 −2 ⎥
⎥; 4 × 4 matrix.
1
0 −1 ⎦
2
1 −0
⎤
3 4 5
4 5 6 ⎥
⎥; 4 × 4 matrix.
5 6 7 ⎦
6 7 8
10. tr(A) = 1 + 3 = 4.
11. tr(A) = 1 + 2 + (−3) = 0.
12. tr(A) = 2 + 2 + (−5) = −1.
1
−1
13. Column vectors:
,
.
3
5
Row vectors: [1 − 1], [3 5].
⎡
⎤ ⎡
⎤ ⎡
⎤
1
3
−4
14. Column vectors: ⎣ −1 ⎦ , ⎣ −2 ⎦ , ⎣ 5 ⎦.
2
6
7
Row vectors: [1 3 − 4], [−1 − 2 5], [2 6 7].
2
10
6
15. Column vectors:
,
,
. Row vectors: [2 10 6], [5 − 1 3].
5
−1
3
⎡
⎤
⎡
⎤ ⎡
⎤
1 2
1
2
16. A = ⎣ 3 4 ⎦. Column vectors: ⎣ 3 ⎦ , ⎣ 4 ⎦.
5 1
5
1
−1
−1
−2
0
4 −1 −1
−2
0
4
,
.
17. A =
; column vectors:
,
,
,
0
8
9 −4 −4
0
8
9
−4
−4
⎤
⎡
−2 −4
⎢ −6 −6 ⎥
⎥
⎢
0 ⎥
18. B = ⎢
⎥; row vectors: −2 −4 , −6 −6 , 3 0 , −1 0 , −2 1 .
⎢ 3
⎣ −1
0 ⎦
−2
1
⎡
⎤
2
5 0 1
7 0 2 ⎦. Row vectors: [2 5 0 1], [−1 7 0 2], [4 − 6 0 3].
19. B = ⎣ −1
4 −6 0 3
20. A = [a1 , a2 , . . . , ap ] has p columns and each column q-vector has q rows, so the resulting matrix has
dimensions q × p.
⎡
⎤
2 0
0
0 ⎦.
21. One example: ⎣ 0 3
0 0 −1
⎤
⎡
2 3 1 2
⎢ 0 5 6 2 ⎥
⎥
22. One example: ⎢
⎣ 0 0 3 5 ⎦.
0 0 0 1
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⎤
1
3 −1
2
⎢ −3
0
4 −3 ⎥
⎥.
23. One example: ⎢
⎣ 1 −4
0
1 ⎦
−2
3 −1
0
⎡
⎤
3 0 0
24. One example: ⎣ 0 2 0 ⎦.
0 0 5
⎡
⎡
0
25. The only possibility here is the zero matrix: ⎣ 0
0
⎡
⎤
0 0 0
26. ⎣ 0 0 0 ⎦.
0 0 0
⎤
⎡ 2
t −t 0
⎢ 0
0 ⎥
⎥.
27. One example: ⎢
⎣ 0
0 ⎦
0
0
√
1
√
t+2 0
3−t
28. One example:
.
0
0
0
1 t2 +1
29. One example:
.
0
30. One example:
t2 + 1
1
1
1
0
0
0
⎤
0
0 ⎦.
0
1 .
31. One example: Let A and B be 1 × 1 matrix functions given by
A(t) = [t]
and
B(t) = [t2 ].
32. Let A be a symmetric upper triangular matrix. Then all elements below the main diagonal are zeros.
Consequently, since A is symmetric, all elements above the main diagonal must also be zero. Hence, the
only nonzero entries can occur along the main diagonal. That is, A is a diagonal matrix.
33. Since A is skew-symmetric, we know that aij = −aji for all (i, j). But since A is symmetric, we know
that aij = aji for all (i, j). Thus, for all (i, j), we must have −aji = aji . That is, aji = 0 for all (i, j). That
is, every element of A is zero.
Solutions to Section 2.2
True-False Review:
(a): FALSE. The correct statement is (AB)C = A(BC), the associative law. A counterexample to the
particular statement given in this review item can be found in Problem 5.
(b): TRUE. Multiplying from left to right, we note that AB is an m × p matrix, and right multiplying AB
by the p × q matrix C, we see that ABC is an m × q matrix.
(c): TRUE. We have (A + B)T = AT + B T = A + B, so A + B is symmetric.
(c)2017 Pearson Education. Inc.
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⎡
⎤
⎡
0 1 0
0
(d): FALSE. For example, let A = ⎣ −1 0 0 ⎦ , B = ⎣ 0
0 0 0
−3
⎡
⎤
0 0
0
symmetric, but AB = ⎣ 0 0 −3 ⎦ is not symmetric.
0 0
0
0
0
0
⎤
3
0 ⎦. Then A and B are skew0
2
. The
is false since AB
(e): FALSE. The correct equation is (A+B)2 = A2+AB +BA+B
+BA
statement
1 1
1 0
0 1
2
does not necessarily equal 2AB. For instance, if A =
and B =
, then (A+B) =
0 0
0 0
0 0
1
2
= (A + B)2 .
and A2 + 2AB + B 2 =
0 0
0 1
1 0
(f ): FALSE. For example, let A =
and B =
. Then AB = 0 even though A = 0 and
0 0
0 0
B = 0.
0 0
0 0
(g): FALSE. For example, let A =
and let B =
. Then A is not upper triangular,
1 0
0 0
despite the fact that AB is the zero matrix, hence automatically upper triangular.
1 0
(h): FALSE. For instance, the matrix A =
is neither the zero matrix nor the identity matrix,
0 0
and yet A2 = A.
(i): TRUE. The derivative of each entry of the matrix is zero, since in each entry, we take the derivative
of a constant, thus obtaining zero for each entry of the derivative of the matrix.
(j): FALSE. The correct statement is given in Problem 45. The problem with the statement as given is
dA
that the second term should be dA
dt B, not B dt .
t
0
2e
satisfies A = dA
(k): FALSE. For instance, the matrix function A =
dt , but A does not have
0 3et
t
0
ce
.
the form
0 cet
(l): TRUE. This follows by exactly the same proof as given in the text for matrices of numbers (see part
3 of Theorem 2.2.23).
Problems:
−10 30
5
.
−5 0 −15
−6 −3 3
1(b). −3B =
.
0 −12 12
⎡
⎤
−1 + i −1 + 2i
1(c). iC = ⎣ −1 + 3i −1 + 4i ⎦.
−1 + 5i −1 + 6i
−6 11
3
1(d). 2A − B =
.
−2 −4 −2
1 + 3i 15 + 3i 16 + 3i
.
1(e). A + 3C T =
5 + 3i 12 + 3i 15 + 3i
1(a). 5A =
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⎡
⎤
8 10 7
1(f ). 3D − 2E = ⎣ 1 4 9 ⎦.
1 7 12
⎡
12
−3 − 3i
1(g). D + E + F = ⎣ 3 + i 3 − 2i
6
4 + 2i
⎤
−1 + i
⎦.
8
2
1(h). Solving for G and simplifying, we have that
3
G=− A−B =
2
1 −10 −1/2
3/2 −4 17/2
.
1(i). Solving for H and simplifying, we have that H = 4E − D − 2F =
⎡
⎤ ⎡
⎤ ⎡
⎤ ⎡
8 −20 −8
4 0 1
12
4 − 6i 2i
−8
⎣ 4
4
12 ⎦ − ⎣ 1 2 5 ⎦ − ⎣ 2 + 2i
−4i
0 ⎦ = ⎣ 1 − 2i
16 −8 −12
3 1 2
−2
10 + 4i 6
15
−24 + 6i
2 + 4i
−19 − 4i
⎤
−9 − 2i
⎦.
7
−20
⎤ ⎡
2+i
−13 + i
4 + i ⎦ = ⎣ 21 + 3i
6+i
5 + 5i
⎤
−5 + 2i
−9 + 4i ⎦ .
−5 + 6i
1(j). We have K T = 2B − 3A, so that K = (2B − 3A)T = 2B T − 3AT . Thus,
⎡
⎤
⎡
⎤ ⎡
⎤
2
0
−2 −1
10 3
4 ⎦ − 3⎣ 6
0 ⎦ = ⎣ −16 8 ⎦ .
K = 2⎣ 1
−1 −4
1 −3
−5 1
⎡
⎤
−4
0 −1
2(a). −D = ⎣ −1 −2 −5 ⎦.
−3 −1 −2
⎡
⎤ ⎡
⎤
2
0
8
0
4 ⎦=⎣ 4
16 ⎦.
2(b). 4B T = 4 ⎣ 1
−1 −4
−4 −16
⎡
⎤ ⎡
⎤ ⎡
⎤
−2 −1
1+i 2+i
5+i
4+i
0 ⎦ + ⎣ 3 + i 4 + i ⎦ = ⎣ −9 + i 4 + i ⎦.
2(c). −2AT + C = −2 ⎣ 6
1 −3
5+i 6+i
3 + i 12 + i
⎡
⎤ ⎡
⎤ ⎡
⎤
10 −25 −10
4 0 1
14 −25 −9
5
15 ⎦ + ⎣ 1 2 5 ⎦ = ⎣ 6
7
20 ⎦.
2(d). 5E + D = ⎣ 5
20 −10 −15
3 1 2
23
−9 −13
2(e). We have
⎡
⎤
⎡
⎤
⎡
−2 −1
2
0
1+i
0 ⎦ − 2⎣ 1
4 ⎦ + i⎣ 3 + i
4AT − 2B T + iC = 4 ⎣ 6
1 −3
−1 −4
5+i
2(f ). We have
⎡
⎤ ⎡
8 −20 −8
12
4
12 ⎦ − ⎣ 0
4E − 3DT = ⎣ 4
16 −8 −12
3
3
6
15
⎤ ⎡
⎤
9
−4 −23 −17
3 ⎦ = ⎣ 4 −2
9 ⎦.
6
13 −23 −18
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2(g). We have (1 − 6i)F + iD =
⎡
⎤ ⎡
6 − 36i −16 − 15i
6+i
4i
⎣ 7 − 5i
⎦+⎣ i
−12 − 2i
0
−1 + 6i 17 − 28i 3 − 18i
3i
0
2i
i
⎤ ⎡
i
6 − 32i
5i ⎦ = ⎣ 7 − 4i
2i
−1 + 9i
−16 − 15i
−12
17 − 27i
⎤
6 + 2i
⎦.
5i
3 − 16i
2(h). Solving for G, we have
G = A + (1 − i)C =
T
−2
−1
6
1
0 −3
1+i
2+i
+ (1 − i)
−2 6
1
2
4 − 2i
=
+
−1 0 −3
3 − i 5 − 3i
0
10 − 2i 7 − 4i
=
.
2 − i 5 − 3i 4 − 5i
3+i
4+i
5+i
6+i
6 − 4i
7 − 5i
2(i). Solve for H, we have
3
3
D − E + 3I3
2
2
⎡
⎤ ⎡
⎤ ⎡
6
0
3/2
3 −15/2 −3
1
⎣
⎦
⎣
⎦
⎣
3/2
3
15/2
3/2
3/2
9/2
=
−
+ 0
9/2 3/2
3
6
−3
−9/2
0
⎡
⎤
6
15/2 9/2
9/2
3 ⎦.
=⎣ 0
−3/2 9/2 21/2
H=
2(j). We have K T = DT + E T − F T = (D + E − F )T , so that
⎡
0
−7 + 3i
K = D + E − F = ⎣ 1 − i 3 + 2i
8
−6 − 2i
3(a).
AB =
3(b).
5 10 −3
27 22
3
0
1
0
⎤
−1 − i
⎦.
8
−4
⎡
⎤
9
BC = ⎣ 8 ⎦
−6
3(c). CA cannot be computed.
3(d).
⎡
1
AT E = ⎣ −1
2
⎤
3 2−i
1 ⎦
−i
4
1+i
2 + 4i
⎡
2 − 4i
= ⎣ −2
4 − 6i
(c)2017 Pearson Education. Inc.
⎤
7 + 13i
1 + 3i ⎦
10 + 18i
⎤
0
0 ⎦
1
130
3(e).
⎤
2 −2
3
2 −3 ⎦ .
CD = ⎣ −2
4 −4
6
3(f ).
⎡
⎡
1 −1
C T AT =
3(g).
2
F =
3(h).
i 1 − 3i
0 4+i
2
⎤
1 3
⎣ −1 1 ⎦ =
2 4
i 1 − 3i
0 4+i
=
6
10
−1 10 − 10i
0 15 + 8i
⎡
⎤⎡
⎤ ⎡
⎤
2 −1
3
2
15
1
2 ⎦ ⎣ −2 ⎦ = ⎣ 14 ⎦
BDT = ⎣ 5
4
6 −2
3
−10
3(i).
⎡
⎤
⎡
3 10
1
−1
2
1 ⎦
=⎣ 2
3
1 4
4
14
1
AT A = ⎣ −1
2
3(j).
FE =
i 1 − 3i
0 4+i
4(a).
AC =
2−i
−i
1 −1 2
3
1 4
1+i
2 + 4i
=
⎤
2 14
2 2 ⎦
2 20
−2 + i
1 − 4i
13 − i
4 + 18i
⎡
⎤
1
6
⎣ −1 ⎦ =
10
2
4(b).
DC = [10]
4(c).
DB = [6 14 − 4]
4(d). AD cannot be computed.
i
2−i 1+i
4(e). EF =
−i 2 + 4i
0
1 − 3i
4+i
=
1 + 2i
1
2 − 2i
1 + 17i
.
4(f ). Since AT is a 3 × 2 matrix and B is a 3 × 3 matrix, the product AT B cannot be constructed.
4(g). Since C is a 3 × 1 matrix, it is impossible to form the product C · C = C 2 .
2−i 1+i
2−i 1+i
4 − 5i
1 + 7i
4(h). E 2 =
=
.
−i 2 + 4i
−i 2 + 4i
3 − 4i −11 + 15i
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4(i). ADT =
T
4(j). E A =
5. We have
⎤
2
⎣ −2 ⎦ = 10 .
16
3
−i
1 −1 2
2 − 4i
=
2 + 4i
3
1 4
7 + 13i
1 −1 2
3
1 4
2−i
1+i
⎡
⎛
⎜ −3
ABC = (AB)C = ⎜
⎝
6
=
=
2
7 −1
0 −3 −5
CAB = C(AB) =
=
=
6.
Ac =
1 3
−5 4
15 −95
−9
65
−185 −460
119
316
and
−6
1
1
5
−6 1
1 5
−99
−30
−2
1 + 3i
⎛
⎜
⎜
⎝
−3
6
1
5
15 −95
−9
65
635
.
230
=6
1
−5
.
⎤⎞
−2
8
⎢ 8 −3 ⎥⎟ −6
⎥⎟
⎢
⎣ −1 −9 ⎦⎠
1
0
2
⎡
2
7 −1
0 −3 −5
6
−2
−6
1
4 − 6i
10 + 18i
+ (−2)
3
4
1
5
⎤⎞
−2
8
⎢ 8 −3 ⎥⎟
⎥⎟
⎢
⎣ −1 −9 ⎦⎠
0
2
⎡
=
0
−38
.
7.
⎡
⎤⎡
⎤
⎡
⎤
⎡
⎤
⎡
⎤ ⎡
⎤
3 −1 4
2
3
−1
4
−13
1 5 ⎦ ⎣ 3 ⎦ = 2 ⎣ 2 ⎦ + 3 ⎣ 1 ⎦ + (−4) ⎣ 5 ⎦ = ⎣ −13 ⎦ .
Ac = ⎣ 2
7 −6 3
−4
7
−6
3
−16
8.
⎤ ⎡
⎤
⎤
⎡
⎤
⎡
−7
−1
2 −1
2
5
7 ⎦
Ac = ⎣ 4
= 5 ⎣ 4 ⎦ + (−1) ⎣ 7 ⎦ = ⎣ 13 ⎦ .
−1
−4
29
5 −4
5
⎡
9. We have
Ac = x
a
e
+y
b
f
+z
c
g
+w
d
h
=
xa + yb + zc + wd
xe + yf + zg + wh
.
10(a). The dimensions of B should be n × r in order that ABC is defined.
10(b). The elements of the ith row of A are ai1 , ai2 , . . . , ain and the elements of the jth column of BC are
r
,
m=1
b1m cmj ,
r
,
m=1
b2m cmj , . . . ,
r
,
m=1
(c)2017 Pearson Education. Inc.
bnm cmj ,
132
so the element in the ith row and jth column of ABC = A(BC) is
ai1
r
,
b1m cmj + ai2
m=1
=
n
,
aik
r
,
bkm cmj
bnm cmj
m=1
m=1
r
,
=
n
,
r
,
k=1
m=1
aik bkm
cmj .
1 −1
−1 −4
=
.
2
3
8
7
−1 −4
1 −1
−9 −11
=
.
A3 = A2 A =
8
7
2
3
22
13
−9 −11
1 −1
−31 −24
=
.
A4 = A3 A =
22
13
2
3
48
17
A2 = AA =
11(b).
b2m cmj + · · · + ain
m=1
k=1
11(a).
r
,
1 −1
2
3
⎡
⎤⎡
⎤ ⎡
⎤
0
1 0
0
1 0
−2
0
1
0 1 ⎦ ⎣ −2
0 1 ⎦ = ⎣ 4 −3
0 ⎦.
A2 = AA = ⎣ −2
4 −1 0
4 −1 0
2
4 −1
⎡
⎤⎡
⎤ ⎡
⎤
−2
0
1
0
1 0
4 −3
0
0 ⎦ ⎣ −2
0 1 ⎦=⎣
6
4 −3 ⎦ .
A3 = A2 A = ⎣ 4 −3
2
4 −1
4 −1 0
−12
3
4
⎡
⎤⎡
⎤ ⎡
⎤
4 −3
0
0
1 0
6
4 −3
6
4 −3 ⎦ ⎣ −2
0 1 ⎦ = ⎣ −20
9
4 ⎦.
A4 = A3 A = ⎣
−12
3
4
4 −1 0
10 −16
3
12(a). We apply the distributive property of matrix multiplication as follows:
(A+2B)2 = (A+2B)(A+2B) = A(A+2B)+(2B)(A+2B) = (A2 +A(2B))+((2B)A+(2B)2 ) = A2 +2AB+2BA+4B 2 ,
where scalar factors of 2 are moved in front of the terms since they commute with matrix multiplication.
12(b). We apply the distributive property of matrix multiplication as follows:
(A + B + C)2 = (A + B + C)(A + B + C) = A(A + B + C) + B(A + B + C) + C(A + B + C)
= A2 + AB + AC + BA + B 2 + BC + CA + CB + C 2
= A2 + B 2 + C 2 + AB + BA + AC + CA + BC + CB,
as required.
12(c). We can use the formula for (A + B)3 found in Example 2.2.20 and substitute −B for B throughout
the expression:
(A − B)3 = A3 + A(−B)A + (−B)A2 + (−B)2 A + A2 (−B) + A(−B)2 + (−B)A(−B) + (−B)3
= A3 − ABA − BA2 + B 2 A − A2 B + AB 2 + BAB − B 3 ,
as needed.
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133
13. We have
2
A =
so that
2
A + 4A + 18I2 =
2 −5
6 −6
−26
−24
20
6
2 −5
6 −6
+
⎡
⎤⎡
−1 0 4
−1
A2 = ⎣ 1 1 2 ⎦ ⎣ 1
−2 3 0
−2
14. We have
=
8 −20
24 −24
0
1
3
−26
−24
+
20
6
18
0
⎤ ⎡
4
−7
2 ⎦ = ⎣ −4
0
5
,
0
18
=
0
0
0
0
.
⎤
12 −4
7
6 ⎦
3 −2
and
⎡
−1
A3 = ⎣ 1
−2
0
1
3
⎤⎡
4
−1
2 ⎦⎣ 1
0
−2
⎤⎡
⎤ ⎡
⎤⎡
4
−1 0 4
−7 12 −4
−1
2 ⎦ ⎣ 1 1 2 ⎦ = ⎣ −4 7
6 ⎦⎣ 1
0
−2 3 0
5 3 −2
−2
0
1
3
0
1
3
⎤ ⎡
⎤
4
27
0 −4
2 ⎦ = ⎣ −1 25 −2 ⎦ .
0
2 −3 26
Therefore, we have
⎡
⎤ ⎡
27
0 −4
−1
A3 + A − 26I3 = ⎣ −1 25 −2 ⎦ + ⎣ 1
2 −3 26
−2
0
1
3
⎤ ⎡
4
26
2 ⎦−⎣ 0
0
0
0
26
0
⎤ ⎡
0
0
0 ⎦=⎣ 0
26
0
1
1
0
⎤
0
1 ⎦.
1
⎡
15.
⎡
1
Substituting A = ⎣ 0
0
⎤ ⎡
⎤ ⎡
1 0 0
0 −1
0
1
0 −1 ⎦ = ⎣ 0
A2 = ⎣ 0 1 0 ⎦ − ⎣ 0
0 0 1
0
0
0
0
⎤
x z
1 y ⎦ for A, we have
0 1
⎡
1
⎣ 0
0
that is,
⎡
x
1
0
1
⎣ 0
0
⎤⎡
z
1 x
y ⎦⎣ 0 1
1
0 0
2x
1
0
⎤ ⎡
z
1
y ⎦=⎣ 0
1
0
⎤ ⎡
2z + xy
1
⎦=⎣ 0
2y
1
0
1
1
0
1
1
0
0
0
0
⎤
0
0 ⎦.
0
⎤
0
1 ⎦,
1
⎤
0
1 ⎦.
1
Since corresponding elements of equal matrices are equal, we obtain the following implications:
2y = 1 =⇒ y = 1/2,
2x = 1 =⇒ x = 1/2,
2z + xy = 0 =⇒ 2z + (1/2)(1/2) = 0 =⇒ z = −1/8.
⎡
1
Thus, A = ⎣ 0
0
1/2
1
0
⎤
−1/8
1/2 ⎦.
1
(c)2017 Pearson Education. Inc.
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x+y
x 1
x 1
x 1
x2 − 2
16. In order that A2 = A, we require
=
=
, that is,
−2x − 2y −2 + y 2
−2 y
−2 y −2 y
x 1
x2 − x − 2
x+y−1
, or equivalently,
= 02 . Since corresponding elements of equal ma−2 y
−2x − 2y + 2 y 2 − y − 2
trices are equal, it follows that
x2 − x − 2 = 0 =⇒ x = −1 or x = 2, and
y 2 − y − 2 = 0 =⇒ y = −1 or y = 2.
Two cases arise from x + y − 1 = 0:
(a): If x = −1, then y = 2.
(b): If x = 2, then y = −1. Thus,
−1
−2
1
2
0
i
−i
0
A=
17.
σ1 σ 2 =
σ2 σ3 =
σ3 σ1 =
0
1
1
0
0 −i
i
0
1
0
0 −1
1
0
0
1
or A =
2
1
−2 −1
.
i
0
1
0
=
=i
= iσ3 .
0 −i
0 −1
0
0 i
0 1
=
=i
= iσ1 .
−1
i 0
1 0
1
0 1
0 −i
=
=i
= iσ2 .
0
−1 0
i
0
18.
[A, B] = AB − BA
1 −1
3 1
3
=
−
2
1
4 2
4
−1 −1
5 −2
=
−
10
4
8 −2
−6 1
=
= 02 .
2 6
1
2
1 −1
2
1
19.
[A1 , A2 ] = A1 A2 − A2 A1
1 0
0 1
0 1
1 0
=
−
0 1
0 0
0 0
0 1
0 1
0 1
=
−
= 02 , thus A1 and A2 commute.
0 0
0 0
[A1 , A3 ] = A1 A3 − A3 A1
1 0
0 0
0 0
1 0
=
−
0 1
1 0
1 0
0 1
0 0
0 0
=
−
= 02 , thus A1 and A3 commute.
1 0
1 0
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[A2 , A3 ] = A2 A3 − A3 A2
0 1
0 0
0 0
0 1
=
−
0 0
1 0
1 0
0 0
1 0
0 0
1
0
=
−
=
= 02 .
0 0
0 1
0 −1
−1 0
Then [A3 , A2 ] = −[A2 , A3 ] =
= 02 . Thus, A2 and A3 do not commute.
0 1
20.
[A1 , A2 ] = A1 A2 − A2 A1
1 0 −1
1 0 i
0 −1
0 i
−
=
1
0
0
i 0
4 i 0
4 1
1 −i 0
1 i
0
−
=
0
−i
0 i
4
4
1 i
1 2i
0
0
=
= A3 .
=
4 0 −2i
2 0 −i
[A2 , A3 ] = A2 A3 − A3 A2
1 i
1 0 −1
i
0
0
0 −1
−
=
0
0 −i
1
0
4 1
4 0 −1
1
1 0 i
0 −i
−
=
0
4 i 0
4 −i
1 0 i
1 0 2i
=
= A1 .
=
4 2i 0
2 i 0
[A3 , A1 ] = A3 A1 − A1 A3
1 0 i
1 i
0
0 i
i
0
−
=
i 0
0 −i
4 0 −i
4 i 0
1 0 1
1
0 −1
−
=
0
4 −1
4 1 0
1 0 −1
1 0 −2
=
= A2 .
=
0
0
4 2
2 1
21.
[A, [B, C]] + [B, [C, A]] + [C, [A, B]]
= [A, BC − CB] + [B, CA − AC] + [C, AB − BA]
= A(BC − CB) − (BC − CB)A + B(CA − AC) − (CA − AC)B + C(AB − BA) − (AB − BA)C
= ABC − ACB − BCA + CBA + BCA − BAC − CAB + ACB + CAB − CBA − ABC + BAC
= 0.
22.
Proof that A(BC) = (AB)C: Let A = [aij ] be of size m × n, B = [bjk ] be of size n × p, and C = [ckl ] be
of size p × q. Consider the (i, j)-element of (AB)C:
n
p
p
n
,
,
,
,
[(AB)C]ij =
aih bhk ckj =
aih
bhk ckj = [A(BC)]ij .
k=1
h=1
h=1
k=1
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Proof that A(B + C) = AB + AC: We have
[A(B + C)]ij =
=
=
n
,
k=1
n
,
aik (bkj + ckj )
(aij bkj + aik ckj )
k=1
n
,
k=1
aik bkj +
n
,
aik ckj
k=1
= [AB + AC]ij .
23.
Proof that (AT )T = A: Let A = [aij ]. Then AT = [aji ], so (AT )T = [aji ]T = aij = A, as needed.
Proof that (A + C)T = AT + C T : Let A = [aij ] and C = [cij ]. Then [(A + C)T ]ij = [A + C]ji =
[A]ji + [C]ji = aji + cji = [AT ]ij + [C T ]ij = [AT + C T ]ij . Hence, (A + C)T = AT + C T .
24. We have
(IA)ij =
m
,
δik akj = δii aij = aij ,
k=1
for 1 ≤ i ≤ m and 1 ≤ j ≤ p. Thus, Im Am×p = Am×p .
25. Let A = [aij ] and B = [bij ] be n × n matrices. Then
n
n
n
n
n
n
,
,
,
,
,
,
aki bik =
bik aki =
bik aki = tr(BA).
tr(AB) =
k=1
26(a). B T AT =
⎡
⎢
26(b). C T B T = ⎢
⎣
i=1
i=1
i=1
k=1
⎤
−3
1
⎣ −1 ⎦ =
.
−7
6
⎤
⎡
⎤
−4
64
6
1 ⎢ −4
1 −3 ⎥
0 −7 −1
1 ⎥
⎥
⎥
=⎢
⎣ −20 −16 −18 ⎦.
1 −3
5 ⎦ −4
8
12
8
−2
0 −7 −1
−4
1 −3
−9
0
3
−2
k=1
⎡
26(c). Since DT is a 3 × 3 matrix and A is a 1 × 3 matrix, it is not possible to compute the expression DT A.
⎡
⎤
−2 0
1
27(a). ADT = −3 −1 6 ⎣ 1 0 −2 ⎦ = 35 42 −7 .
5 7 −1
⎤
⎡
⎤
⎡
82
1 −22
16
−9
1 ⎢
⎢ 0
1
1
5 −2 ⎥
1 ⎥
⎥. Therefore,
⎥ −9 0 3 −2 = ⎢
27(b). First note that C T C = ⎢
⎣
⎣ 3
⎦
−22
5
34 −16 ⎦
1 1 5 −2
5
16 −2 −16
8
−2 −2
⎤
⎤ ⎡
⎤⎡
⎡
7465 −59 −2803
1790
82
1 −22
16
82
1 −22
16
⎢
⎢
⎢
−59
31
185
−82 ⎥
1
1
5 −2 ⎥
1
1
5 −2 ⎥
⎥.
⎥=⎢
⎥⎢
(C T C)2 = ⎢
⎦
⎣
⎦
⎣
⎣ −22
−2803 185
1921 −1034 ⎦
−22
5
34 −16
5
34 −16
1790 −82 −1034
580
16 −2 −16
8
16 −2 −16
8
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⎡
⎤⎡
⎤ ⎡
⎤
−2 0
1
0 −4
−1
5
1 ⎦=⎣
2
2 ⎦.
27(c). DT B = ⎣ 1 0 −2 ⎦ ⎣ −7
5 7 −1
−1 −3
−48 −10
28(a). We have
⎡
⎤
z
2z ⎦ ,
z
−x −y
y
S = [s1 , s2 , s3 ] = ⎣ 0
x −y
⎡
so
2
AS = ⎣ 2
1
2
5
2
28(b).
⎤⎡
1
−x −y
2 ⎦⎣ 0
y
2
x −y
⎡
−x
S T AS = S T (AS) = ⎣ −y
z
0
y
2z
⎤ ⎡
z
−x −y
2z ⎦ = ⎣ 0
y
z
x −y
⎤⎡
x
−x −y
−y ⎦ ⎣ 0
y
z
x −y
⎤
7z
14z ⎦ = [s1 , s2 , 7s3 ].
7z
⎤ ⎡
7z
2x2
14z ⎦ = ⎣ 0
0
7z
0
3y 2
0
⎤
0
0 ⎦,
42z 2
but S T AS = diag(1, 1, 7), so we have the following
√
2
2
√
3
3y 2 = 1 =⇒ y = ±
3
√
6
2
6z = 1 =⇒ z = ±
.
6
2
2x = 1 =⇒ x = ±
29(a). We have
⎡
⎤⎡
1 −4 0
0 2x
7 0 ⎦⎣ 0 x
AS = ⎣ −4
0
0 5
z 0
⎡
⎤
0 −2x
9y
−x −18y ⎦
=⎣ 0
5z
0
0
⎤
y
−2y ⎦
0
= [5s1 , −s2 , 9s3 ].
29(b). We have
⎡
0
S T AS = ⎣ 2x
y
⎤⎡
0 z
0
x 0 ⎦⎣ 0
−2y 0
5z
−2x
−x
0
⎤ ⎡
9y
5z 2
−18y ⎦ = ⎣ 0
0
0
0
−5x2
0
⎤
0
0 ⎦,
45y 2
so in order for this to be equal to diag(5, −1, 9), we must have
5z 2 = 5,
−5x2 = −1,
45y 2 = 9.
Thus, we must have z 2 = 1, x2 = 15 , and y 2 = 15 . Therefore, the values of x, y, and z that we are looking for
are x = ± 15 , y = ± 15 , and z = ±1.
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⎡
2
⎢ 0
⎢
30(a). ⎣
0
0
⎡
7
30(b). ⎣ 0
0
⎤
0
0 ⎥
⎥.
0 ⎦
2
⎤
0
2
0
0
0
0
2
0
0
7
0
0
0 ⎦.
7
31. Suppose A is an n × n scalar matrix with trace k. If A = aIn , then tr(A) = na = k, so we conclude that
a = k/n. So A = nk In , a uniquely determined matrix.
32. We have
1
B =
(A + AT )
2
T
T
and
CT =
1
(A − AT )
2
=
T
=
1
1
(A + AT )T = (AT + A) = B
2
2
1 T
1
(A − A) = − (A − AT ) = −C.
2
2
Thus, B is symmetric and C is skew-symmetric.
33. We have
B+C =
1
1
1
1
1
1
(A + AT ) + (A − AT ) = A + AT + A − AT = A.
2
2
2
2
2
2
34. We have
⎛⎡
⎤ ⎡
⎡
⎤⎞
4 −1 0
8
8
4
9 2
1
1
1
B = (A + AT ) = ⎝⎣ 9 −2 3 ⎦ + ⎣ −1 −2 5 ⎦⎠ = ⎣ 8 −4
2
2
2
2
5 5
2
8
0
3 5
⎤ ⎡
⎤
2
4
4 1
8 ⎦ = ⎣ 4 −2 4 ⎦
10
1
4 5
and
⎛⎡
⎤ ⎡
⎡
⎤ ⎡
⎤⎞
⎤
4 −1 0
0 −10 −2
4
9 2
0 −5 −1
1
1
1
0 −2 ⎦ = ⎣ 5
0 −1 ⎦ .
C = (A − AT ) = ⎝⎣ 9 −2 3 ⎦ − ⎣ −1 −2 5 ⎦⎠ = ⎣ 10
2
2
2
2
5 5
2
2
0
0
3 5
1
1
0
35.
⎛⎡
1
1 ⎝⎣
3
B=
2
7
⎛⎡
1
1
C = ⎝⎣ 3
2
7
⎤ ⎡
−5 3
1
2 4 ⎦ + ⎣ −5
−2 6
3
⎤ ⎡
−5 3
1
2 4 ⎦ − ⎣ −5
−2 6
3
⎡
⎤ ⎡
⎤⎞
⎤
2 −2 10
3
7
1 −1 5
1
4
2 ⎦ = ⎣ −1
2 −2 ⎦⎠ = ⎣ −2
2 1 ⎦.
2
10
2 12
4
6
5
1 6
⎤ ⎡
⎤
⎤⎞
⎡
3
7
0 −8 −4
0 −4 −2
1
2 −2 ⎦⎠ = ⎣ 8
0
6 ⎦=⎣ 4
0
3 ⎦.
2
4 −6
0
2 −3
0
4
6
36(a). If A is symmetric, then AT = A, so that
B=
1
1
1
(A + AT ) = (A + A) = (2A) = A
2
2
2
C=
1
1
1
(A − AT ) = (A − A) = (0n ) = 0n .
2
2
2
and
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36(b). If A is skew-symmetric, then AT = −A, so that
B=
1
1
1
(A + AT ) = (A + (−A)) = (0n ) = 0n
2
2
2
C=
1
1
1
(A − AT ) = (A − (−A)) = (2A) = A.
2
2
2
and
37. If A = [aij ] and D = diag(d1 , d2 , . . . , dn ), then we must show that the (i, j)-entry of DA is di aij . In
index notation, we have
n
,
(DA)ij =
di δik akj = di δii aij = di aij .
k=1
Hence, DA is the matrix obtained by multiplying the ith row vector of A by di , where 1 ≤ i ≤ n.
38. If A = [aij ] and D = diag(d1 , d2 , . . . , dn ), then we must show that the (i, j)-entry of AD is dj aij . In
index notation, we have
n
,
(AD)ij =
aik dj δkj = aij dj δjj = aij dj .
k=1
Hence, AD is the matrix obtained by multiplying the jth column vector of A by dj , where 1 ≤ j ≤ n.
39. Since A and B are symmetric, we have that AT = A and B T = B. Using properties of the transpose
operation, we therefore have
(AB)T = B T AT = BA = AB,
and this shows that AB is symmetric.
40(a). We have (AAT )T = (AT )T AT = AAT , so that AAT is symmetric.
40(b). We have (ABC)T = [(AB)C]T = C T (AB)T = C T (B T AT ) = C T B T AT , as needed.
1
cos t
41. A (t) =
.
− sin t
4
−2e−2t
.
42. A (t) =
cos t
⎡
⎤
cos t − sin t 0
43. A (t) = ⎣ sin t cos t 1 ⎦.
0
3
0
t
2e2t 2t
e
.
44. A (t) =
2et 8e2t 10t
45. We
that the (i, j)-entry of both sides of the equation agree. First, recall that the (i, j)-entry of
show
n
d
AB is k=1 aik bkj , and therefore, the (i, j)-entry of dt
(AB) is (by the product rule)
n
,
k=1
aik bkj + aik bkj =
n
,
k=1
aik bkj +
n
,
aik bkj .
k=1
The former term is precise the (i, j)-entry of the matrix dA
dt B, while the latter term is precise the (i, j)-entry
d
of the matrix A dB
.
Thus,
the
(i,
j)-entry
of
(AB)
is
precisely
the sum of the (i, j)-entry of dA
dt
dt
dt B and the
dB
(i, j)-entry of A dt . Thus, the equation we are proving follows immediately.
(c)2017 Pearson Education. Inc.
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46. We have
1 t
e
t
2e
0
e−t
5e−t
47. We have
π/2 0
cos t
sin t
48. We have
1 t
e
t
2e
0
dt =
e2t
4e2t
t2
5t2
1
0 =
−e−t
−5e−t
sin t
− cos t
π/2
0 =
dt =
et
2et
dt = =
=
sin(π/2)
− cos(π/2)
1 2t
2e
2t
t3
3
5 3
3t
et
2et
2e
e
2e
e2 /2 1/3
2e2 5/3
49. We have
⎡
⎤
⎡ 1 2t
1
sin 2t
e2t
2e
⎣ t2 − 5
⎦ dt = ⎣ t3 − 5t
tet
3
0
sec2 t 3t − sin t
tan t
⎡
e2
2
−1/e
−5/e
e
2e
= ⎣ −14/3
tan 1
−
−
sin 0
− cos 0
1
2
1/2 0
2
0
=
=
1
0
e−1
2e − 2
1 − 1/e
5 − 5/e
0
−1
−
=
.
1
1
.
1
0
−
⎤
− 12 cos 2t
tet − et ⎦ 10
3 2
2 t + cos t
⎤ ⎡ 1
− cos2 2
2
⎦−⎣ 0
0
3
0
2 + cos 1
3t
1
e dt = −5t tan−1 (t)
−5dt
A(t)dt =
t2 +1 dt
2 t
2t
dt
=
.
51.
t3
3t2
⎡
⎤
⎤
⎡
sin t
cos t 0
− cos t
sin t
0
⎣ − cos t sin t t ⎦ dt = ⎣ − sin t − cos t t2 /2 ⎦.
52.
0
3t2 /2
0
3t
1
t
t
e−t
−e−t
et
e
dt =
.
53.
t
−t
t
2e 5e
2e −5e−t
⎤
⎤
⎡
⎡ 1 2t
e
− 12 cos 2t
sin 2t
e2t
2
⎦ dt = ⎣ t3 − 5t
⎣ t2 − 5
tet
54.
tet − et ⎦.
3
2
3 2
sec t 3t − sin t
tan t
2 t + cos t
50.
1 −1
2 −5
=
e−1
2e − 2
1/3
2e − 2 5/3
⎤ ⎡ e2 −1
− 12
2
−1 ⎦ = ⎣ −14/3
1
tan 1
1 3t
3e
e2 −1
2
2
1−cos 2
2
1
.
⎤
⎦.
1
2 + cos 1
.
Solutions to Section 2.3
True-False Review:
(a): FALSE. The last column of the augmented matrix corresponds to the constants on the right-hand
side of the linear system, so if the augmented matrix has n columns, there are only n − 1 unknowns under
consideration in the system.
(c)2017 Pearson Education. Inc.
141
(b): FALSE. Three distinct planes can intersect in a line (e.g. Figure 2.3.1, lower right picture). For
instance, the xy-plane, the xz-plane, and the plane y = z intersect in the x-axis.
(c): FALSE. The right-hand side vector must have m components, not n components.
(d): TRUE. If a linear system has two distinct solutions x1 and x2 , then any point on the line containing
x1 and x2 is also a solution, giving us infinitely many solutions, not exactly two solutions.
(e): TRUE. The augmented matrix for a linear system has one additional column (containing the constants
on the right-hand side of the equation) beyond the matrix of coefficients.
(f ): FALSE. Because the vector (x1 , x2 , x3 , 0, 0) has five entries, this vector belongs to R5 . Vectors in R3
can only have three slots.
(g): FALSE. The two column vectors given have different numbers of components, so they are not the
same vectors.
Problems:
1.
2 · 1 − 3(−1) + 4 · 2 = 13,
1 + (−1) − 2 = −2,
5 · 1 + 4(−1) + 2 = 3.
2.
2 + (−3) − 2 · 1 = −3,
3 · 2 − (−3) − 7 · 1 = 2,
2 + (−3) + 1 = 0,
2 · 2 + 2(−3) − 4 · 1 = −6.
3.
(1 − t) + (2 + 3t) + (3 − 2t) = 6,
(1 − t) − (2 + 3t) − 2(3 − 2t) = −7,
5(1 − t) + (2 + 3t) − (3 − 2t) = 4.
4.
s + (s − 2t) − (2s + 3t) + 5t = 0,
2(s − 2t) − (2s + 3t) + 7t = 0,
4s + 2(s − 2t) − 3(2s + 3t) + 13t = 0.
5. The two given lines are the same line. Therefore, since this line contains an infinite number of points,
there must be an infinite number of solutions to this linear system.
6. These two lines are parallel and distinct, and therefore, there are no common points on these lines. In
other words, there are no solutions to this linear system.
7. These two lines have different slopes, and therefore, they will intersect in exactly one point. Thus, this
system of equations has exactly one solution.
(c)2017 Pearson Education. Inc.
142
8. The first and third equations describe lines that are parallel and distinct, and therefore, there are no
common points on these lines. In other words, there are no solutions to this linear system.
⎡
⎤
⎡
⎤
⎡
⎤
1 2 −3 1
1 2 −3
1
9. A = ⎣ 2 4 −5 ⎦ , b = ⎣ 2 ⎦ , A# = ⎣ 2 4 −5 2 ⎦.
7 2 −1
7 2 −1 3
3
1 1
1 −1 3
1 1
1 −1
3
.
10. A =
,b =
, A# =
2 4 −3
7 2
2 4 −3
7
2
⎡
⎤
⎡
⎤
⎡
⎤
1 2 −1 0
1 2 −1
0
11. A = ⎣ 2 3 −2 ⎦ , b = ⎣ 0 ⎦ , A# = ⎣ 2 3 −2 0 ⎦.
5 6 −5 0
5 6 −5
0
12. It is acceptable to use any variable names. We will use x1 , x2 , x3 , x4 :
x1 − x2
x1 + x2
3x1 + x2
+2x3 + 3x4
−2x3 + 6x4
+4x3 + 2x4
= 1,
= −1,
= 2.
13. It is acceptable to use any variable names. We will use x1 , x2 , x3 :
2x1 + x2
4x1 − x2
7x1 + 6x2
+3x3 = 3,
+2x3 = 1,
+3x3 = −5.
14. The system of equations here only contains one equation: 4x1 − 2x2 − 2x3 − 3x5 = −9.
15. This system of equations has three equations: −3x2 = −1,
2x1 − 7x2 = 6,
16. Given Ax = 0 and Ay = 0, and an arbitrary constant c,
(a). we have
Az = A(x + y) = Ax + Ay = 0 + 0 = 0
and
Aw = A(cx) = c(Ax) = c0 = 0.
(b). No, because
A(x + y) = Ax + Ay = b + b = 2b = b,
and
A(cx) = c(Ax) = cb = b
in general.
−4
3
x1
4t
x1
=
+
.
17.
x2
6 −4
x2
t2
−t
x1
t2
x1
=
.
18.
− sin t
1
x2
x2
0
e2t
x1
0
x1
=
+
.
19.
x2
− sin t 0
x2
1
(c)2017 Pearson Education. Inc.
5x1 + 5x2 = 7.
143
⎤ ⎡
x1
0
20. ⎣ x2 ⎦ = ⎣ −et
x3
−t
⎡
− sin t
0
t2
⎤ ⎡
⎤⎡
⎤
x1
1
t
t2 ⎦ ⎣ x 2 ⎦ + ⎣ t3 ⎦ .
0
x3
1
21. We have
4e4t
−2(4e4t )
x (t) =
and
Ax + b =
2 −1
−2
3
22. We have
x (t) =
e4t
−2e4t
0
0
+
=
=
4(−2e−2t ) + 2 cos t
3(−2e−2t ) + sin t
4e4t
−8e4t
2e4t + (−1)(−2e4t ) + 0
−2e4t + 3(−2e4t ) + 0
=
−8e−2t + 2 cos t
−6e−2t + sin t
=
4e4t
−8e4t
.
and
Ax + b =
=
1 −4
−3
2
4e−2t + 2 sin t
3e−2t − cos t
+
−2(cos t + sin t)
7 sin t + 2 cos t
4e−2t + 2 sin t − 4(3e−2t − cos t) − 2(cos t + sin t)
−3(4e−2t + 2 sin t) + 2(3e−2t − cos t) + 7 sin t + 2 cos t
23. We compute
x =
3et + 2tet
et + 2tet
=
−8e−2t + 2 cos t
−6e−2t + sin t
.
and
Ax + b =
2 −1
−1
2
2tet + et
2tet − et
+
0
4et
=
2(2tet + et ) − (2tet − et ) + 0
−(2tet + et ) + 2(2tet − et ) + 4et
=
2tet + 3et
2tet + et
.
Therefore, we see from these calculations that x = Ax + b.
24. We compute
⎤
−tet − et
⎦
−9e−t
x = ⎣
t
te + et − 6e−t
⎡
and
⎤ ⎡
⎤ ⎡
⎤ ⎡
⎤ ⎡
⎤⎡
−et
−tet
−tet − et
−et
1
0 0
−tet
⎦+⎣ 6e−t ⎦ = ⎣ 2(−tet ) − 3(9e−t ) + 2(tet + 6e−t ) ⎦+⎣ 6e−t ⎦ = ⎣
9e−t
−9e−t
Ax+b = ⎣ 2 −3 2 ⎦ ⎣
−tet − 2(9e−t ) + 2(tet + 6e−t )
tet + 6e−t
et
et
tet + et − 6e−
1 −2 2
⎡
Therefore, we see from these calculations that x = Ax + b.
Solutions to Section 2.4
True-False Review:
(a): TRUE. The precise row-echelon form obtained for a matrix depends on the particular elementary row
operations (and their order). However, Theorem 2.4.15 states that there is a unique reduced row-echelon
form for a matrix.
(c)2017 Pearson Education. Inc.
144
(b): FALSE. Upper triangular matrices could have pivotentries that are not 1. For instance, the following
2 0
matrix is upper triangular, but not in row echelon form:
.
0 0
(c): TRUE. The pivots in a row-echelon form of an n × n matrix must move down and to the right as
we look from one row to the next beneath it. Thus, the pivots must occur on or to the right of the main
diagonal of the matrix, and thus all entries below the main diagonal of the matrix are zero.
(d): FALSE. This would not be true, for example, if A was a zero matrix with 5 rows and B was a nonzero
matrix with 4 rows.
(e): FALSE. If A is a nonzero matrix and B = −A, then A + B = 0, so rank(A + B) = 0, but rank(A),
rank(B) ≥ 1 so rank(A)+ rank(B) ≥ 2.
0 1
(f ): FALSE. For example, if A = B =
, then AB = 0, so rank(AB) = 0, but rank(A)+
0 0
rank(B) = 1 + 1 = 2.
(g): TRUE. A matrix of rank zero cannot have any pivots, hence no nonzero rows. It must be the zero
matrix.
(h): TRUE. The matrices A and 2A have the same reduced row-echelon form, since we can move between
the two matrices by multiplying the rows of one of them by 2 or 1/2, a matter of carrying out elementary
row operations. If the two matrices have the same reduced row-echelon form, then they have the same rank.
(i): TRUE. The matrices A and 2A have the same reduced row-echelon form, since we can move between
the two matrices by multiplying the rows of one of them by 2 or 1/2, a matter of carrying out elementary
row operations.
Problems:
1. Neither.
2. Reduced row-echelon form.
3. Neither.
4. Row-echelon form.
5. Row-echelon form.
6. Reduced row-echelon form.
7. Reduced row-echelon form.
8. Reduced row-echelon form.
9.
10.
2
1
2 −4
−4
8
1
−3
1
∼
1
∼
1 −3
2
1
1 −2
−4
8
2
∼
1 −2
0
0
1. M1 ( 12 )
2. A12 (4)
1. P12
2
∼
1 −3
0
7
, Rank (A) = 1.
∼
1 −3
0
1
2. A12 (−2)
3. M2 ( 17 )
3
(c)2017 Pearson Education. Inc.
, Rank (A) = 2.
145
⎡
11.
0
⎣ 0
0
1
1
3
⎤ ⎡
3
0 1
1
4 ⎦∼⎣ 0 0
5
0 0
⎤ ⎡
3
0
2
1 ⎦∼⎣ 0
4
0
1. A12 (−1), A13 (−3)
2. A21 (−1)
3. A12 (−2), A13 (−3)
4. P23
5. M2 (−1)
6. A32 (5)
⎡
⎤ ⎡
⎤ ⎡
⎤ ⎡
⎤ ⎡
⎤
2 −1
3
3
1 −2
1
2 −5
1
2 −5
1
2 −5
1
2
3
4
⎣ 3
1 −2 ⎦ ∼ ⎣ 2 −1
3 ⎦ ∼ ⎣ 2 −1
3 ⎦ ∼ ⎣ 0 −5 13 ⎦ ∼ ⎣ 0 −1 −2 ⎦
2 −2
1
2 −2
1
0 −1 −2
0 −1 −2
0 −5 13
⎡
⎤ ⎡
⎤ ⎡
⎤
1
2 −5
1 2 −5
1 2 −5
5
6
7
1
2 ⎦∼⎣ 0 1
2 ⎦∼⎣ 0 1
2 ⎦ , Rank (A) = 3.
∼⎣ 0
0 −5 13
0 0 23
0 0
1
1. P12
2. A21 (−1), A23 (−1)
3. A12 (−2)
4. P23
5. M2 (−1)
⎡
6. A23 (5)
⎤ ⎡
⎤ ⎡
⎤ ⎡
⎤ ⎡
⎤ ⎡
2 −1
3
2
1
3
1
3
1
3
1
1
2
3
4
5
⎣ 3
2 ⎦ ∼ ⎣ 2 −1 ⎦ ∼ ⎣ 2 −1 ⎦ ∼ ⎣ 0 −7 ⎦ ∼ ⎣ 0 −1 ⎦ ∼ ⎣ 0
2
5
2
5
0 −1
0 −7
2
5
0
1. P12
15.
2. A23 (−4)
⎤ ⎡
⎤ ⎡
⎤ ⎡
⎤ ⎡
⎤
2
1 4
3 −2 6
1
1 2
1
1 2
1
1 2
1
2
3
4
⎣ 2 −3 4 ⎦ ∼
⎣ 2 −3 4 ⎦ ∼
⎣ 2 −3 4 ⎦ ∼
⎣ 0 −5 0 ⎦ ∼
⎣ 0 −1 0 ⎦
3 −2 6
2
1 4
2
1 4
0 −1 0
0 −5 0
⎡
⎤ ⎡
⎤
1
1 2
1 1 2
5
6
1 0 ⎦ ∼ ⎣ 0 1 0 ⎦ , Rank (A) = 2.
∼⎣ 0
0 −5 0
0 0 0
1. P13
14.
⎤
3
1 ⎦ , Rank (A) = 2.
0
⎡
12.
13.
1
0
0
⎡
2
⎢ 3
⎢
⎣ 1
2
−2
−2
−1
−1
2. A21 (−1)
3. A12 (−2), A13 (−2)
4. P23
7. M3 (1/23).
⎤
3
1 ⎦ , Rank (A) = 2.
0
5. M2 (−1), A23 (7).
⎤ ⎡
⎤ ⎡
⎤
⎤ ⎡
1 −1
1 0
1 −1
1 0
1 −1
1 0
−1 3
3 1 ⎥
1
0 1 ⎥
1
0 1 ⎥
3 1 ⎥
1 ⎢
2 ⎢
3 ⎢
⎥∼
⎥∼
⎥
⎥∼
⎢ 3 −2
⎢ 0
⎢ 0
⎦
⎦
⎦
⎣
⎣
⎣
2 −2 −1 3
0
0 −3 3
0
0 −3 3 ⎦
1 0
2 −1
2 2
0
1
0 2
0
0
0 1
2 2
⎤
⎡
1 −1 1
0
0
1 0
1 ⎥
4 ⎢
⎥ , Rank (A) = 4.
⎢
∼⎣
0
0 1 −1 ⎦
0
0 0
1
(c)2017 Pearson Education. Inc.
146
1. P13
2. A12 (−3), A13 (−2), A14 (−2)
3. A24 (−1)
4. M3 (1/3)
16.
⎡
⎤ ⎡
⎤ ⎡
⎤ ⎡
1 −2
2 −1 3 4
1 −2 1 3
1 −2 1
3
1
2
3
⎣ 1 −2 1 3 ⎦ ∼ ⎣ 2 −1 3 4 ⎦ ∼ ⎣ 0
3 1 −2 ⎦ ∼ ⎣ 0
1
1 −5 0 5
1 −5 0 5
0
0 0
0
0
0
1. P12
17.
⎡
2
⎣ 1
2
1
0
3
3
2
1
4
1
5
⎤ ⎡
2
1
1
3 ⎦∼⎣ 2
7
2
0
1
3
2. A12 (−2), A13 (−1)
2
3
1
1
4
5
⎡
1
∼⎣ 0
0
18.
2. A12 (−2), A13 (−2),
4. M3 (− 13 )
3. A23 (−3)
3
2
1 −1
1
∼
2. A12 (−3), A13 (−2), A14 (−5)
−4
−6
2
3
1
∼
1 −1
3
2
2
∼
1. P12
− 23 ⎦ , Rank (A) = 2.
0
⎤ ⎡
⎤
0
2 1
3
1 0
2
1
3
3
1 −1 2 −4 ⎦ ∼ ⎣ 0 1 −1
2 −4 ⎦
3 −3 3
1
0 0
0 −3
1
⎤
⎤ ⎡
⎤ ⎡
⎤ ⎡
1
2 1
2
1
2 1
2
1
4
7 4
7
⎥ 2 ⎢ 0 −1 0 −1 ⎥ 3 ⎢ 0
⎥ 1 ⎢ 3
⎢ 3
5
3
5
5
3
5
⎥ ⎢
⎥ ⎢
⎥ ⎢
⎢
⎣ 2 −2 2 −2 ⎦ ∼ ⎣ 2 −2 2 −2 ⎦ ∼ ⎣ 0 −6 0 −6 ⎦ ∼ ⎣ 0
5 −2 5 −2
0 −12 0 −12
0
5 −2 5 −2
⎤
⎡
1 2 1 2
4 ⎢ 0 1 0 1 ⎥
⎥
∼⎢
⎣ 0 0 0 0 ⎦ , Rank (A) = 2.
0 0 0 0
19.
⎤
3. M2 (1/3)
⎡
1. A21 (−1)
20.
1
3
0
3
0
2 1
3
1 −1 2 −4 ⎦ , Rank (A) = 3.
0
0 1 − 13
4
1. P12
⎤ ⎡
3
1
2
2 ⎦∼⎣ 0
7
0
1
1 −1/2
−6
3
3. M2 (−1)
∼
1. M1 (− 14 )
2. A12 (6)
1 −1
0
5
3
∼
2. A12 (−3)
1 −1
0
1
4
∼
3. M2 ( 15 )
(c)2017 Pearson Education. Inc.
4. A23 (6), A24 (12)
1 −1/2
0
0
2
⎤
2 1
2
1 0
1 ⎥
⎥
−6 0 −6 ⎦
−12 0 −12
,
Rank(A) = 1.
1
0
0
1
= I2 , Rank (A) = 2.
4. A21 (1)
147
⎤ ⎡
⎤ ⎡
⎤ ⎡
⎤ ⎡
1 2
1
1
2
1
1 2 1
1
3 7 10
1
2
3
4
⎣ 2 3 −1 ⎦ ∼
⎣ 0 −1 −3 ⎦ ∼
⎣ 0 1 3 ⎦∼
⎣ 0
⎣ 2 3 −1 ⎦ ∼
3 7 10
0
1
7
0 1 7
0
1 2
1
⎡
⎤ ⎡
⎤
1 0 −5
1 0 0
5
6
3 ⎦ ∼ ⎣ 0 1 0 ⎦ = I3 , Rank (A) = 3.
∼⎣ 0 1
0 0
1
0 0 1
⎡
21.
1. P13
2. A12 (−2), A13 (−3)
⎡
22.
3 −3
⎣ 2 −2
6 −6
3. M2 (−1)
5. M3 ( 41 )
4. A21 (−2), A23 (−1)
⎤
0 −5
1
3 ⎦
0
4
6. A31 (5), A32 (−3)
⎤ ⎡
⎤
6
1 −1 2
1
4 ⎦∼⎣ 0
0 0 ⎦ , Rank (A) = 1.
12
0
0 0
1. M1 ( 13 ), A12 (−2), A13 (−6)
23.
⎡
⎤ ⎡
⎤ ⎡
3
5 −12
1
2 −5
1
1
2
⎣ 2
3 −7 ⎦ ∼ ⎣ 0 −1
3 ⎦∼⎣ 0
−2 −1
1
0
3 −9
0
1. A21 (−1), A12 (−2), A13 (2)
⎡
24.
1
⎢ 3
⎢
⎣ 2
4
3. A21 (−2), A23 (−3)
2. A21 (1), A23 (−1), A24 (−2)
4. M4 (−1)
1
2
3
2. M2 (−1)
⎤
0
1
1 −3 ⎦ , Rank (A) = 2.
0
0
⎤ ⎡
⎤ ⎡
⎤ ⎡
1 −1 −1 2
1 0 2
3
1 0
−1 −1 2
⎥
⎥
⎢
⎢
⎢
0
1
3
1
0
1
3
1
−2
0 7 ⎥
1 ⎢
2 ⎢
3 ⎢ 0 1
⎥∼
⎥∼
⎥∼
1
4 0 ⎦ ⎣ 0 0 1 −1 ⎦ ⎣ 0 0
−1
2 4 ⎦ ⎣ 0
0
2
7 0
0 0 1 −2
0 0
−2
3 8
⎤ ⎡
⎤
⎡
1 0 0
5
1 0 0 0
⎥ 5 ⎢ 0 1 0 0 ⎥
0
1
0
4
4 ⎢
⎥ ⎢
⎥
∼⎢
⎣ 0 0 1 −1 ⎦ ∼ ⎣ 0 0 1 0 ⎦ = I4 , Rank (A) = 4.
0 0 0
1
0 0 0 1
1. A12 (−3), A13 (−2), A14 (−4)
25.
⎡
1 −2
⎣ 3 −6
4 −8
⎤ ⎡
2 −5
1
3
1 −3 ⎦ ∼ ⎣ 0
3 −9
0
⎤
0
5
0
4 ⎥
⎥
1 −1 ⎦
0 −1
3. A31 (−2), A32 (−3), A34 (−1)
5. A41 (−5), A42 (−4), A43 (1)
⎤ ⎡
⎤ ⎡
⎤ ⎡
⎤
3
1 −2
1
3
1 −2
1
3
1 −2 0 1
1
2
3
7 ⎦∼⎣ 0
0 −1 −2 ⎦ ∼ ⎣ 0
0
1
2 ⎦∼⎣ 0
0 1 2 ⎦ , Rank (A) = 2.
10
0
0 −1 −2
0
0 −1 −2
0
0 0 0
1. A12 (−3), A13 (−4)
2. M2 (−1)
3. A21 (−1), A23 (1)
(c)2017 Pearson Education. Inc.
148
26.
⎡
0
⎣ 0
0
1
3
2
2
1
0
⎤ ⎡
⎤ ⎡
⎤ ⎡
1
0 1
2
1
0 1
2
1
0 1
1
2
3
2 ⎦ ∼ ⎣ 0 0 −6 −2 ⎦ ∼ ⎣ 0 0
1 1/3 ⎦ ∼ ⎣ 0 0
1
0 0 −4 −1
0 0 −4 −1
0 0
⎡
⎤ ⎡
⎤
0 1 0 1/3
0 1 0 0
4
5
∼ ⎣ 0 0 1 1/3 ⎦ ∼ ⎣ 0 0 1 0 ⎦ , Rank (A) = 3.
0 0 0
1
0 0 0 1
1. A12 (−3), A13 (−2)
2. M2 (− 16 )
3. A21 (−2), A23 (4)
4. M3 (3)
0
1
0
⎤
1/3
1/3 ⎦
1/3
5. A32 (− 13 ), A31 (− 13 )
Solutions to Section 2.5
True-False Review:
(a): FALSE. This process is known as Gaussian elimination. Gauss-Jordan elimination is the process by
which a matrix is brought to reduced row echelon form via elementary row operations.
(b): TRUE. A homogeneous linear system always has the trivial solution x = 0, hence it is consistent.
(c): TRUE. The columns of the row-echelon form that contain leading 1s correspond to leading variables,
while columns of the row-echelon form that do not contain leading 1s correspond to free variables.
(d): TRUE. If the last column of the row-reduced augmented matrix for the system does not contain a
pivot, then the system can be solved by back-substitution. On the other hand, if this column does contain
a pivot, then that row of the row-reduced matrix containing the pivot in the last column corresponds to the
impossible equation 0 = 1.
(e): FALSE. The linear system x = 0, y = 0, z = 0 has a solution in (0, 0, 0) even though none of the
variables here is free.
(f ): FALSE. The columns containing the leading 1s correspond to the leading variables, not the free
variables.
Problems:
For the problems of this section, A will denote the coefficient matrix of the given system, and
A# will denote the augmented matrix of the given system.
1. Converting the given system of equations to an augmented matrix and using Gaussian elimination we
obtain the following equivalent matrices:
1 −5 3
1 −5 3 2 1 −5 3
1
.
∼
∼
3 −9 15
0
6 6
0
1 1
1. A12 (−3)
2. M2 ( 16 )
By back substitution, we find that x2 = 1, and then x1 = 8. Therefore, the solution is (8, 1).
2. Converting the given system of equations to an augmented matrix and using Gaussian elimination we
obtain the following equivalent matrices:
2 3 1 − 14
4 −1 8 1 1 − 14 2 2 1 − 14
2
∼
∼
∼
.
3
2
1 1
2
1 1
0
1 −2
−3
0
2
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149
1. M1 ( 14 )
2. A12 (−2)
3. M2 ( 23 )
3
. Therefore, the solution is ( 32 , −2).
2
3. Converting the given system of equations to an augmented matrix and using Gaussian elimination we
obtain the following equivalent matrices:
7 −3 5
7 −3 5 2 1 − 37 57
1
.
∼
∼
14 −6 10
0
0 0
0
0 0
By back substitution, we find that x2 = −2, and then x1 =
1. A12 (−2)
2. M2 ( 17 )
Observe that x2 is a free variable, so we set x2 = t. Then by back substitution, we have x1 = 37 t + 57 .
Therefore, the solution set to this system is
3
5
t + ,t : t ∈ R .
7
7
4. Converting the given system of equations to an augmented matrix and using Gaussian elimination we
obtain the following equivalent matrices:
⎡
⎤ ⎡
⎤ ⎡
⎤ ⎡
⎤
1
1
1
1 2 1 1
1
2
1
1 2 1
1 2 1
1
2
3
⎣ 3 5 1 3 ⎦ ∼ ⎣ 0 −1 −2
0 ⎦∼⎣ 0 1 2
0 ⎦∼⎣ 0 1 2
0 ⎦.
2 6 7 1
0
2
5 −1
0 2 5 −1
0 0 1 −1
1. A12 (−3), A13 (−2)
2. M2 (−1)
3. A23 (−2)
The last augmented matrix results in the system:
x1 + 2x2 + x3 = 1,
x2 + 2x3 = 0,
x3 = −1.
By back substitution we obtain the solution (−2, 2, −1).
5. Converting the given system of equations to an augmented matrix and using Gaussian elimination, we
obtain the following equivalent matrices:
⎡
⎤ ⎡
⎤ ⎡
⎤
1
3 −1
0
1 −2 −5 −3
1 −2 −5 −3
1
2
⎣ 2
1
5
1
5
5 15 10 ⎦
4 ⎦∼⎣ 2
4 ⎦∼⎣ 0
7 −5 −8 −3
7 −5 −8 −3
0
9 27 18
⎡
⎤ ⎡
⎤
1 −2 −5 −3
1 0 1 1
3
4
1
3
2 ⎦ ∼ ⎣ 0 1 3 2 ⎦.
∼⎣ 0
0
9 27 18
0 0 0 0
1. A21 (−1)
2. A12 (−2), A13 (−7)
3. M2 ( 15 )
(c)2017 Pearson Education. Inc.
4. A21 (2), A23 (−9)
150
The last augmented matrix results in the system:
x1
+ x3 = 1,
x2 + 3x3 = 2.
Let the free variable x3 = t, a real number. By back substitution we find that the system has the solution
set {(1 − t, 2 − 3t, t) : for all real numbers t}.
6. Converting the given system of equations to an augmented matrix and using Gaussian elimination we
obtain the following equivalent matrices:
⎡
⎤ ⎡
⎤ ⎡
⎤ ⎡
⎤
3
3
3 5 −1 14
1 2
1 3
1
2
1
1 2 1
1
2
3
⎣ 1 2
1 3 ⎦ ∼ ⎣ 3 5 −1 4 ⎦ ∼ ⎣ 0 −1 −4 −5 ⎦ ∼ ⎣ 0 1 4
5 ⎦
2 5
6 2
2 5
6 2
0
1
4 −4
0 1 4 −4
⎡
⎤ ⎡
⎤
3
1 2 1
1 2 1 3
4
5
5 ⎦ ∼ ⎣ 0 1 4 5 ⎦.
∼⎣ 0 1 4
0 0 0 −9
0 0 0 1
1. P12
2. A12 (−3), A13 (−2)
3. M2 (−1)
4. A23 (−1)
5. M4 (− 19 )
This system of equations is inconsistent since 2 = rank(A) < rank(A# ) = 3.
7. Converting the given system of equations to an augmented matrix and using Gaussian elimination we
obtain the following equivalent matrices:
⎡
⎤ ⎡
⎤ ⎡
⎤
6 −3
3 12
2
1 − 12 − 12
1 − 12 12 2
1
1
⎣ 2 −1
1
4 ⎦ ∼ ⎣ 2 −1
1
0 0 0 ⎦.
4 ⎦∼⎣ 0
−4
2 −2 −8
−4
2 −2 −8
0
0 0 0
1. M1 ( 16 )
2. A12 (−2), A13 (4)
Since x2 and x3 are free variables, let x2 = s and x3 = t. The single equation obtained from the augmented
matrix is given by x1 − 12 x2 + 12 x3 = 2. Thus, the solution set of our system is given by
{(2 +
s
t
− , s, t) : s, t any real numbers }.
2 2
8. Converting the given system of equations to an augmented matrix and using Gaussian elimination we
obtain the following equivalent matrices:
⎤ ⎡
⎤ ⎡
⎤ ⎡
⎤
3
1 −2 −1
1
2 −5 −15
1
2 −5 −15
2 −1
3 14
⎢ 3
1 −2 −1 ⎥
3 14 ⎥
3 −14 ⎥
−5 13
44 ⎥
1 ⎢
2 ⎢
3 ⎢
⎢ 2 −1
⎢ 2 −1
⎢ 0
⎥∼
⎥∼
⎥∼
⎥
⎢
⎣ 7
2 −3
2 −3
2 −3
3 ⎦ ⎣ 7
3 ⎦ ⎣ 7
3 ⎦ ⎣ 0 −12 32 108 ⎦
5 −1 −2
5 −1 −2
5 −1 −2
0 −11 23
5
5
5
80
⎡
⎡
⎤ ⎡
⎤ ⎡
⎤
1
2 −5 −15
1
2 −5 −15
1
2 −5 −15
32 108 ⎥
9
1 −9 −28 ⎥
28 ⎥
4 ⎢ 0 −12
5 ⎢
6 ⎢
⎢ 0 −1
⎢ 0
⎥∼
⎥∼
⎥
∼⎢
⎣ 0 −5 13
⎣
⎣
⎦
⎦
0 −5 13
0 −5 13
44
44
44 ⎦
0 −11 23
0 −11 23
0 −11 23
80
80
80
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⎡
⎤ ⎡
⎤ ⎡
⎤
1 2 −5 −15
1 2 −5 −15
1 2 −5 −15
⎢
⎥ 9 ⎢ 0 1 −9 −28 ⎥
−9 −28 ⎥
7 ⎢ 0 1
⎢
⎥ 8 ⎢ 0 1 −9 −28 ⎥ ∼
⎥.
∼⎢
⎣ 0 0 −32 −96 ⎦ ∼ ⎣ 0 0
32
1
96 ⎦ ⎣ 0 0
3 ⎦
0 0 −76 −228
0 0 −76 −228
0 0
0
0
1. P12
5. A42 (−1)
2. A21 (−1)
6. M2 (−1)
3. A12 (−2), A13 (−7), A14 (−5)
7. A23 (5), A24 (11)
4. P23
8. M3 (−1)
1
9. M3 ( 32
), A34 (76).
The last augmented matrix results in the system of equations:
x1 − 2x2 − 5x3 = −15,
x2 − 9x3 = −28,
3.
x3 =
Thus, using back substitution, the solution set for our system is given by {(2, −1, 3)}.
9. Converting the given system of equations to an augmented matrix and using Gaussian elimination we
obtain the following equivalent matrices:
⎡
⎤ ⎡
⎤ ⎡
⎤ ⎡
⎤
2 −1 −4
5
1
1 −3 −3
1
1 −3 −3
1
1 −3 −3
⎢ 3
2 −5
2 −5
4 17 ⎥
1 −4 −17 ⎥
8 ⎥
8 ⎥
1 ⎢
2 ⎢
3 ⎢
⎢
⎢ 3
⎢ 0 −1
⎢ 0
⎥∼
⎥∼
⎥∼
⎥
⎣ 5
⎣
⎣
⎣
⎦
⎦
⎦
6 −6 20
5
6 −6 20
0
1
9 35
0
1
9
35 ⎦
1
1 −3 −3
2 −1 −4 −5
0 −3
2 11
0 −3
2
11
⎡
⎤ ⎡
⎤ ⎡
⎤
1 1 −3 −3
1 1 −3 −3
1 1 −3 −3
⎥ 5 ⎢ 0 1 −4 −17 ⎥ 6 ⎢ 0 1 −4 −17 ⎥
0
1
−4
−17
4 ⎢
⎥∼⎢
⎥∼⎢
⎥.
∼⎢
⎣ 0 0
13
1
1
52 ⎦ ⎣ 0 0
4 ⎦ ⎣ 0 0
4 ⎦
0 0 −10 −40
0 0 −10 −40
0 0
0
0
1. P14
2. A12 (−3), A13 (−5), A14 (−2)
3. M2 (−1)
4. A23 (−1), A24 (3)
1
5. M3 ( 13
)
6. A34 (10)
The last augmented matrix results in the system of equations:
x1 + x2 − 3x3 = − 3,
x2 − 4x3 = −17,
x3 =
4.
By back substitution, we obtain the solution set {(10, −1, 4)}.
10. Converting the given system of equations to an augmented matrix and using Gaussian elimination we
obtain the following equivalent matrices:
⎡
⎤ ⎡
⎤
1 2 −1 1 1
1 2 −1 1 1
1
⎣ 2 4 −2 2 2 ⎦ ∼ ⎣ 0 0
0 0 0 ⎦.
5 10 −5 5 5
0 0
0 0 0
1. A12 (−2), A13 (−5)
The last augmented matrix results in the equation x1 + 2x3 − x3 + x4 = 1. Now x2 , x3 , and x4 are free
variables, so we let x2 = r, x3 = s, and x4 = t. It follows that x1 = 1 − 2r + s − t. Consequently, the solution
set of the system is given by {(1 − 2r + s − t, r, s, t) : r, s, t and real numbers }.
(c)2017 Pearson Education. Inc.
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11. Converting the given system of equations to an augmented matrix and using Gaussian elimination we
obtain the following equivalent matrices:
⎡
⎤ ⎡
⎤ ⎡
⎤
1
1
2 −1
1
1
2 −1
1 1
1
2 −1
1
1
3
⎢ 2 −3
1 −1 2 ⎥
3 −3
0 ⎥
0 ⎥
1 − 37
1 ⎢
2 ⎢
7
⎢
⎢ 0 −7
⎢ 0
⎥∼
⎥∼
⎥
⎣ 1 −5
2 −2 1 ⎦ ⎣ 0 −7
3 −3
0 ⎦ ⎣ 0 −7
3 −3
0 ⎦
4
1 −1
1 3
0 −7
3 −3 −1
0 −7
3 −3 −1
⎡
1 2 −1
⎢
0 1 − 37
3
∼⎢
⎣ 0 0
0
0 0
0
1
3
7
0
0
⎤ ⎡
1
1 2 −1
3
⎢
0 ⎥
4 ⎢ 0 1 −7
⎥∼
0
0 ⎦ ⎣ 0 0
0 0
0
−1
1. A12 (−2), A13 (−1), A14 (−4)
2. M2 (− 17 )
1
3
7
0
0
⎤ ⎡
1
1 2 −1
3
⎢
0 ⎥
5 ⎢ 0 1 −7
⎥∼
0
−1 ⎦ ⎣ 0 0
0 0
0
0
3. A23 (7), A24 (7)
4. P34
1
3
7
0
0
⎤
1
0 ⎥
⎥.
1 ⎦
0
5. M3 (−1)
The given system of equations is inconsistent since 2 = rank(A) < rank(A# ) = 3.
12. Converting the given system of equations to an augmented matrix and using Gauss-Jordan elimination
we obtain the following equivalent matrices:
⎡
⎤ ⎡
⎤ ⎡
⎤
1 2
1
1 −2 3
1 2
1
1 −2 3
1 2 1 1 −2 3
1
2
⎣ 0 0
1
4 −3 2 ⎦ ∼ ⎣ 0 0
1
4 −3 2 ⎦ ∼ ⎣ 0 0 1 4 −3 2 ⎦ .
2 4 −1 −10
5 0
0 0 −3 −12
9 −6
0 0 0 0
0 0
1. A13 (−2)
2. A23 (3)
The last augmented matrix indicates that the first two equations of the initial system completely determine
its solution. We see that x4 and x5 are free variables, so let x4 = s and x5 = t. Then x3 = 2 − 4x4 + 3x5 =
2−4s+3t. Moreover, x2 is a free variable, say x2 = r, so then x1 = 3−2r−(2−4s+3t)−s+2t = 1−2r+3s−t.
Hence, the solution set for the system is
{(1 − 2r + 3s − t, r, 2 − 4s + 3t, s, t) : r, s, t any real numbers }.
13. Converting the given system of equations to an augmented matrix and using Gauss-Jordan elimination
we obtain the following equivalent matrices:
⎡
⎤ ⎡
⎤ ⎡
⎤ ⎡
⎤
−1 −2
2
4
4
4
1
4
1
1
4
1
1
4
1
1
2
3
3 −2 −1 ⎦ ∼ ⎣ 4
3 −2 −1 ⎦ ∼ ⎣ 0 −13 −6 −17 ⎦ ∼ ⎣ 0 −9 −3 −6 ⎦
4
1
2 −1 −1
0 −9 −3 −6
0 −13 −6 −17
4
2
2
⎣ 4
1
⎤ ⎡
⎤ ⎡
⎤ ⎡
⎤
4
4
4
1
4
1
1
4
1
1
4
1
1 4 1 4
5
6
7
12
4
4
8 ⎦ ∼ ⎣ 0 12
8 ⎦ ∼ ⎣ 0 −1 −2 −9 ⎦ ∼ ⎣ 0 1 2 9 ⎦
∼⎣ 0
0 −13 −6 −17
0 −1 −2 −9
0 12
4
0 12 4 8
8
⎡
⎤ ⎡
⎤ ⎡
⎤
3
1 0 −7 −32
1 0 −7 −32
1 0 0
8
9
10
2
2
9 ⎦∼⎣ 0 1
9 ⎦ ∼ ⎣ 0 1 0 −1 ⎦ .
∼⎣ 0 1
0 0 −20 −100
0 0
1
0 0 1
5
5
⎡
4
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1. P13
6. P23
2. A12 (−4), A13 (−2)
7. M2 (−1)
4. M2 (− 43 )
3. P23
8. A21 (−4), A23 (−12)
5. A23 (1)
1
9. M3 (− 20
)
10. A31 (7), A32 (−2)
The last augmented matrix results in the solution (3, −1, 5).
14. Converting the given system of equations to an augmented matrix and using Gauss-Jordan elimination
we obtain the following equivalent matrices:
⎡
⎤ ⎡
⎤ ⎡
⎤
1
3 1
5 2
1 1 −1 1
1
1 −1
1
2
⎣ 1 1 −1 1 ⎦ ∼
⎣ 3 1
5 2 ⎦ ∼ ⎣ 0 −2
8 −1 ⎦
2 1
2 3
2 1
2 3
0 −1
4
1
⎡
1
∼⎣ 0
0
3
1 −1
1 −4
−1
4
⎤
1 1 −1 1
1 ⎦
∼ ⎣ 0 1 −4 1/2 ⎦ .
2
0 0
0 3/2
1
1
⎡
⎤
4
We can stop here, since we see from this last augmented matrix that the system is inconsistent. In particular,
2 = rank(A) < rank(A# ) = 3.
1. P12
2. A12 (−3), A13 (−2)
3. M2 (− 12 )
4. A23 (1)
15. Converting the given system of equations to an augmented matrix and using Gauss-Jordan elimination
we obtain the following equivalent matrices:
⎡
⎤ ⎡
⎤ ⎡
⎤ ⎡
⎤
1
0 −2 −3
1
0 −2 −3
1
0 −2 −3
1 0 −2 −3
1
2
3
⎣ 3 −2
4 −9 ⎦ ∼ ⎣ 0 −2
2
1 −1
0 ⎦∼⎣ 0
0 ⎦ ∼ ⎣ 0 1 −1
0 ⎦
1 −4
2 −3
0 −4
4
0 −4
4
0 0
0
0
0
0
.
1. A12 (−3), A13 (−1)
2. M2 (− 12 )
3. A23 (4)
The last augmented matrix results in the following system of equations:
x1 − 2x3 = −3
and
x2 − x3 = 0.
Since x3 is free, let x3 = t. Thus, from the system we obtain the solutions {(2t − 3, t, t) : t any real number }.
16. Converting the given system of equations to an augmented matrix and using Gauss-Jordan elimination
we obtain the following equivalent matrices:
⎡
⎤ ⎡
⎤ ⎡
⎤
3
6
6
2 −1 3 −1
1 −2 3
1
1 −2
3
1
1
2
⎣ 3
2 1 −5 −6 ⎦ ∼ ⎣ 3
2 1 −5 −6 ⎦ ∼ ⎣ 0
8 −8 −8 −24 ⎦
1 −2 3
1
2 −1 3 −1
0
3 −3 −3 −9
6
3
⎡
⎤ ⎡
⎤
6
0
1 −2
3
1
1 0
1 −1
3
4
1 −1 −1 −3 ⎦ ∼ ⎣ 0 1 −1 −1 −3 ⎦ .
∼⎣ 0
0
3 −3 −3 −9
0 0
0
0
0
1. P13
2. A12 (−3), A13 (−2)
3. M2 ( 18 )
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4. A21 (2), A23 (−3)
154
The last augmented matrix results in the following system of equations:
x1 + x3 − x4 = 0
and
x2 − x3 − x4 = −3.
Since x3 and x4 are free variables, we can let x3 = s and x4 = t, where s and t are real numbers. It follows
that the solution set of the system is given by {(t − s, s + t − 3, s, t) : s, t any real numbers }.
17. Converting the given system of equations to an augmented matrix and using Gauss-Jordan elimination
we obtain the following equivalent matrices:
⎡
⎤ ⎡
⎤ ⎡
⎤
1
1
1 −1
1
1
1 −1
1 1 1 −1 4
4
4
⎢ 1 −1 −1 −1
0 −2 ⎥
0 1 ⎥
2 ⎥
1 ⎢
2 ⎢
⎢
⎢ 0 −2 −2
⎢ 0 1 1
⎥∼
⎥∼
⎥
⎣ 1
1 −1
1 −2 ⎦ ⎣ 0
0 −2
2 −6 ⎦ ⎣ 0 0 1 −1 3 ⎦
1 −1
1
1 −8
0 −2
0
2 −12
0 1 0 −1 6
⎡
⎤ ⎡
⎤ ⎡
⎤ ⎡
⎤
1 0
0 −1 3
1 0 0 −1
1 0 0 −1
1 0 0 0 −1
3
3
1
0 1 ⎥
1 −2 ⎥
1 −2 ⎥
2 ⎥
3 ⎢ 0 1
4 ⎢
5 ⎢
6 ⎢
⎢ 0 1 0
⎢ 0 1 0
⎢ 0 1 0 0
⎥∼
⎥∼
⎥∼
⎥.
∼⎢
⎣ 0 0
1 −1 3 ⎦ ⎣ 0 0 1 −1
3 ⎦ ⎣ 0 0 1 −1
3 ⎦ ⎣ 0 0 1 0 −1 ⎦
0 0 −1 −1 5
0 0 0 −2
0 0 0
1 −4
0 0 0 1 −4
8
1. A12 (−1), A13 (−1), A14 (−1)
4. A32 (−1), A34 (1)
2. M2 (− 12 ), M3 (− 12 ), M4 (− 12 )
5. M4 (− 12 )
3. A24 (−1)
6. A41 (1), A42 (−1), A43 (1)
It follows from the last augmented matrix that the solution to the system is given by (−1, 2, −1, −4).
18. Converting the given system of equations to an augmented matrix and using Gauss-Jordan elimination
we obtain the following equivalent matrices:
⎡
⎤ ⎡
⎤ ⎡
⎤
2 −1
3
1 −1 11
2
2
1 −3 −2 −1 −2
1 −3 −2 −1 −2
⎢ 1 −3 −2 −1 −2
⎢
⎢
3
1 −1 11 ⎥
5
7
3
3
2 ⎥
7 ⎥
⎢
⎥ 1 ⎢ 2 −1
⎥ 2 ⎢ 0
⎥
⎢ 3
⎥∼⎢ 3
⎥ ∼ ⎢ 0 10
⎥
1
−2
−1
1
1
−2
−1
1
4
2
7
−2
−2
−8
⎢
⎥ ⎢
⎥ ⎢
⎥
⎣ 1
2
1
2
3 −3 ⎦ ⎣ 1
2
1
2
3 −3 ⎦ ⎣ 0
5
3
3
5 −5 ⎦
5 −3 −3
1
2
5 −3 −3
1
2
0 12
7
6 12 −8
2
2
⎡
⎤ ⎡
⎤
⎤ ⎡
4
11
31
4
11
31
1 0
− 15
1 −3 −2 −1 −2
1 0
2
− 15
5
5
5
5
5
5
3
3
7
7 ⎥
7
7 ⎥
7
7 ⎥
3
3
3
3
⎢ 0
⎢
⎢
1
5
5
5
5 ⎥
5
5
5
5 ⎥ 4 ⎢ 0 1
5
5
5
5 ⎥ 5 ⎢ 0 1
3 ⎢
2
1
11
⎢
⎢
⎥
4
2
7 −8 ⎥
1 −22 ⎥
1
− 10
∼⎢
5
5 ⎥
⎢ 0 10
⎥∼⎢ 0 0
⎥ ∼ ⎢ 0 0 −10 −4
⎣ 0
5
3
3
5 −5 ⎦ ⎣ 0 0
−4
0
2 −12 ⎦ ⎣ 0 0
−4
0
2 −12 ⎦
24
24
0 12
7
6 12 −8
− 65
− 124
0 0 − 49
− 65
− 124
0 0 − 49
5
5
5
5
5
5
⎡
⎤ ⎡
⎤ ⎡
⎤
1
2
34
1
2
34
1
6
1 0 0 − 25
1 0 0 − 25
1 0 0 0 10
50
25
50
25
5
37
1
42 ⎥
37
1
42 ⎥
7
⎢ 0 1 0
⎢ 0 1 0
⎢ 0 1 0 0
−
−
− 85 ⎥
25
50
25 ⎥ 7 ⎢
25
50
25 ⎥ 8 ⎢
10
⎥
6 ⎢
2
1
11 ⎥
2
1
11 ⎥
1
⎢
⎢
− 10
− 10
3 ⎥
∼⎢
5
5 ⎥∼⎢ 0 0 1
5
5 ⎥ ∼ ⎢ 0 0 1 0 −2
⎢ 0 0 1
⎥
8
8
⎣ 0 0 0
⎦ ⎣ 0 0 0
1 −2 ⎦
1
1
−2 ⎦ ⎣ 0 0 0 1
− 16
5
5
5
191
68
11
191
68
− 81
0 0 0
0 0 0 0 11
0 0 0
− 81
25
50
25
10
5
25
50
25
⎤ ⎡
⎡
⎤
1
6
1 0 0 0 10
1 0 0 0 0
1
5
7
⎢ 0 1 0 0
⎢ 0 1 0 0 0 −3 ⎥
− 85 ⎥
10
⎥
⎢
⎢
⎥
9
10 ⎢
1
4 ⎥
3 ⎥
∼⎢
⎥∼⎢ 0 0 1 0 0
⎢ 0 0 1 0 −2
⎥
⎣ 0 0 0 1
1 −2 ⎦ ⎣ 0 0 0 1 0 −4 ⎦
0 0 0 0 1
2
0 0 0 0
1
2
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1. P12
2. A12 (−2), A13 (−3), A14 (−1), A15 (−5)
3. M2 ( 51 )
4. A21 (3), A23 (−10), A24 (−5), A25 (−12)
1
11
7
5. M3 (− 10 ) 6. A31 (− 5 ), A32 (− 5 ), A34 (4), A35 ( 49
7. M4 ( 58 )
5 )
2
1
2
68
10
1
7
8. A41 ( 25 ), A42 (− 25 ), A43 (− 5 ), A45 (− 25 ) 9. M5 ( 11 ) 10. A51 (− 10 ), A52 (− 10
), A53 ( 12 ), A54 (−1)
It follows from the last augmented matrix that the solution to the system is given by (1, −3, 4, −4, 2).
19. The equation Ax = b reads
⎤ ⎡
⎤⎡
⎤
1 −3
1
x1
8
⎣ 5 −4
1 ⎦ ⎣ x2 ⎦ = ⎣ 15 ⎦ .
2
4 −3
x3
−4
⎡
Converting the given system of equations to an augmented matrix and using Gauss-Jordan elimination we
obtain the following equivalent matrices:
⎤ ⎡
⎤ ⎡
⎤
⎡
1 −3
1
8
8
8
1 −3
1
1 −3
1
1
2
⎣ 5 −4
1 15 ⎦ ∼ ⎣ 0 11 −4 −25 ⎦ ∼ ⎣ 0
1
1 −5 ⎦
2
4 −3 −4
0 10 −5 −20
0 10 −5 −20
⎡
⎤ ⎡
⎤ ⎡
⎤
1
1 0
4 −7
1 0 4 −7
1 0 0
3
4
5
1 −5 ⎦ ∼ ⎣ 0 1 1 −5 ⎦ ∼ ⎣ 0 1 0 −3 ⎦ .
∼⎣ 0 1
0 0 −15 30
0 0 1 −2
0 0 1 −2
1. A12 (−5), A13 (−2)
2. A32 (−1)
3. A21 (3), A23 (−10)
1
4. M3 (− 15
)
5. A31 (−4), A32 (−1)
Thus, from the last augmented matrix, we see that x1 = 1, x2 = −3, and x3 = −2.
20. The equation Ax = b reads
⎤ ⎡
⎤⎡
⎤
1
0
5
x1
0
⎣ 3 −2 11 ⎦ ⎣ x2 ⎦ = ⎣ 2 ⎦ .
2 −2
6
x3
2
⎡
Converting the given system of equations to an augmented matrix and using Gauss-Jordan elimination we
obtain the following equivalent matrices:
⎤ ⎡
⎤ ⎡
⎤
⎡
0
1
0
5 0
1
0
5
1
0
5 0
1
2
⎣ 0 −2 −4 2 ⎦ ∼
⎣ 0
⎣ 3 −2 11 2 ⎦ ∼
1
2 −1 ⎦
2 −2 6 2
0 −2 −4 2
0 −2 −4
2
⎡
⎤
0
1 0 5
3
∼ ⎣ 0 1 2 −1 ⎦ .
0 0 0
0
1. A12 (−3), A13 (−2)
2. M2 (−1/2)
3. A23 (2)
Hence, we have x1 + 5x3 = 0 and x2 + 2x3 = −1. Since x3 is a free variable, we can let x3 = t, where t is
any real number. It follows that the solution set for the given system is given by {(−5t, −2t − 1, t) : t ∈ R}.
21. The equation Ax = b reads
⎡
0
⎣ 0
0
⎤⎡
⎤ ⎡
⎤
x1
1 −1
−2
5
1 ⎦ ⎣ x2 ⎦ = ⎣ 8 ⎦ .
2
1
5
x3
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Converting the given system of equations to an augmented matrix using Gauss-Jordan elimination we obtain
the following equivalent matrices:
⎡
⎤ ⎡
⎤ ⎡
⎤ ⎡
⎤
0 1 −1 −2
0 1 −1 −2
0 1 −1 −2
0 1 0 1
1
2
3
⎣ 0 5
1
6 18 ⎦ ∼ ⎣ 0 0
1
8 ⎦∼⎣ 0 0
3 ⎦ ∼ ⎣ 0 0 1 3 ⎦.
0 2
1
0 0
3
0 0
3
0 0 0 0
5
9
9
1. A12 (−5), A13 (−2)
2. M2 (1/6)
3. A21 (1), A23 (−3)
Consequently, from the last augmented matrix it follows that the solution set for the matrix equation is
given by {(t, 1, 3) : t ∈ R}.
22. The equation Ax = b reads
⎤⎡
⎤ ⎡
⎤
x1
1 −1 0 −1
2
⎣ 2
1 3
7 ⎦ ⎣ x2 ⎦ = ⎣ 2 ⎦ .
3 −2 1
0
4
x3
⎡
Converting the given system of equations to an augmented matrix and using Gauss-Jordan elimination we
obtain the following equivalent matrices:
⎡
⎤ ⎡
⎤ ⎡
⎤ ⎡
⎤
2
2
0
1 −1 0 −1 2
1 −1 0 −1
1 −1 0 −1
1 0 1 2
1
2
3
⎣ 2
1 3
7 2 ⎦∼⎣ 0
3 3
9 −2 ⎦ ∼ ⎣ 0
1 1
3 −2 ⎦ ∼ ⎣ 0 1 1 3 −2 ⎦ .
3 −2 1
0 4
0
1 1
3 −2
0
3 3
9 −2
0 0 0 0
4
1. A12 (−2), A13 (−3)
2. P23
3. A21 (1), A23 (−3)
From the last row of the last augmented matrix, it is clear that the given system is inconsistent.
23. The equation Ax = b reads
⎤⎡
⎤
⎤ ⎡
x1
1 1
0 −1
2
⎢
⎥
⎢ 3 1 −2
⎥ ⎢
3 ⎥
⎥ ⎢ x2 ⎥ = ⎢ 8 ⎥ .
⎢
⎣ 2 3
1
1 ⎦ ⎣ x3 ⎦ ⎣ 3 ⎦
−2 3
5 −2
−9
x4
⎡
Converting the given system of equations to an augmented matrix and using Gauss-Jordan elimination we
obtain the following equivalent matrices:
⎡
⎤ ⎡
⎤ ⎡
⎤ ⎡
⎤
1
1
0 1
1
1
0 1
1 0 −1 1
2
2
2
3
1 1
0
1
⎢ 3 1 −2
3
1
1 0 −1 ⎥
1 0 −1 ⎥
8 ⎥
2 ⎥
1 ⎢
2 ⎢
3 ⎢
⎢
⎢ 0 −2 −2 0
⎢ 0
⎢ 0 1
⎥∼
⎥∼
⎥∼
⎥.
⎣ 2 3
⎣
⎣
⎣
⎦
⎦
⎦
⎦
1
2
0
1
1 0 −1
0 −2 −2 0
0 0
0 0
3
2
−2 3
5 −2 −9
0
5
5 0 −5
0
5
5 0 −5
0 0
0 0
0
1. A12 (−3), A13 (−2), A14 (2)
2. P23
3. A21 (−1), A23 (2), A24 (−5)
From the last augmented matrix, we obtain the system of equations: x1 − x3 + x4 = 3, x2 + x3 = −1. Since
both x3 and x4 are free variables, we may let x3 = r and x4 = t, where r and t are real numbers. The
solution set for the system is given by {(3 + r − t, −r − 1, r, t) : r, t ∈ R}.
24. Converting the given system of equations to an augmented matrix and using Gauss-Jordan elimination
we obtain the following equivalent matrices:
⎡
⎤ ⎡
⎤ ⎡
⎤
3
3
3
1 2
−1
1
2
−1
1 2
−1
1
2
⎣ 2 5
⎦∼⎣ 0 1
⎦.
1
1
3
3
7 ⎦∼⎣ 0
1
1
2
2
2
1 1 −k −k
0 −1 1 − k −3 − k
0 0 4 − k −2 − k
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1. A12 (−2), A13 (−1)
2. A23 (1)
(a). If k = 2, then the last row of the last augmented matrix reveals an inconsistency; hence the system has
no solutions in this case.
(b). If k = −2, then the last row of the last augmented matrix consists entirely of zeros, and hence we have
only two pivots (first two columns) and a free variable x3 ; hence the system has infinitely many solutions.
(c). If k = ±2, then the last augmented matrix above contains a pivot for each variable x1 , x2 , and x3 , and
can be solved for a unique solution by back-substitution.
25. Converting the given system of equations to an augmented matrix and using Gauss-Jordan elimination
we obtain the following equivalent matrices:
⎤ ⎡
⎤ ⎡
⎤
⎡
1
1
1 −1 0
1
1
1
−1 0
2
1 −1
1 0
⎢ 1
1
1 −1 0 ⎥
1 −1
1 0 ⎥
3 0 ⎥
1 ⎢
2 ⎢
⎢ 2
⎢ 0 −1 −3
⎥∼
⎥∼
⎥
⎢
⎣
⎣
⎦
⎦
⎣ 4
2 −1
1 0
4
2 −1
1 0
0 −2 −5
5 0 ⎦
3 −1
1
k 0
3 −1
1
k 0
0 −4 −2 k + 3 0
⎡
⎤ ⎡
1 1
0
⎢ 0 1
0 ⎥
4
⎥∼⎢
0 ⎦ ⎣ 0 0
0 0
0
1
−1
3
−3
1
−1
10 k − 9
1. P12
2. A12 (−2), A13 (−4), A14 (−3)
3. M2 (−1)
1
1
1
−1
0
1
3
−3
3 ⎢
∼⎢
⎣ 0 −2 −5
5
0 −4 −2 k + 3
⎤ ⎡
1 1 1 −1
0
⎢ 0 1 3 −3
0 ⎥
5
⎥∼⎢
0 ⎦ ⎣ 0 0 1 −1
0 0 0 k+1
0
4. A23 (2), A24 (4)
⎤
0
0 ⎥
⎥.
0 ⎦
0
5. A34 (−10)
(a). Note that the trivial solution (0, 0, 0, 0) exists under all circumstances, so there are no values of k for
which there is no solution.
(b). From the last row of the last augmented matrix, we see that if k = −1, then the variable x4 corresponds
to an unpivoted column, and hence it is a free variable. In this case, therefore, we have infinitely solutions.
(c). Provided that k = −1, then each variable in the system corresponds to a pivoted column of the last
augmented matrix above. Therefore, we can solve the system by back-substitution. The conclusion from
this is that there is a unique solution, (0, 0, 0, 0).
26. Converting the given system of equations to an augmented matrix and using Gauss-Jordan elimination
we obtain the following equivalent matrices:
⎡
⎤ ⎡
⎤ ⎡
⎤ ⎡
⎤
1 1 −2
4
4
4
2
1 1 −2
1 1 −2
1 0 −3
1
2
3
⎣ 3 5 −4 16 ⎦ ∼
⎣ 0 2
⎦.
2
1
1
4 ⎦∼⎣ 0 1
2 ⎦∼⎣ 0 1
2
2 3 −a
0 1 4−a b−8
0 1 4−a b−8
0 0 3 − a b − 10
b
1. A12 (−3), A13 (−2)
2. M2 ( 12 )
3. A21 (−1), A23 (−1)
(a). From the last row of the last augmented matrix above, we see that there is no solution if a = 3 and
b = 10.
(b). From the last row of the augmented matrix above, we see that there are infinitely many solutions
if a = 3 and b = 10, because in that case, there is no pivot in the column of the last augmented matrix
corresponding to the third variable x3 .
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(c). From the last row of the augmented matrix above, we see that if a = 3, then regardless of the value
of b, there is a pivot corresponding to each variable x1 , x2 , and x3 . Therefore, we can uniquely solve the
corresponding system by back-substitution.
27. Converting the given system of equations to an augmented matrix and using Gauss-Jordan elimination
we obtain the following equivalent matrices:
⎡
⎤ ⎡
⎤
1 −a 3
3
1
−a
1
⎣ 2
1
6 ⎦ ∼ ⎣ 0 1 + 2a 0 ⎦ .
−3 a + b 1
0 b − 2a 10
From the middle row, we see that if a = − 12 , then we must have x2 = 0, but this leads to an inconsistency in
solving for x1 (the first equation would require x1 = 3 while the last equationwould require x1 = − 13 . Now
1 −1/2 3
. If b = −1,
suppose that a = − 12 . Then the augmented matrix on the right reduces to
0 b + 1 10
then once more we have an inconsistency in the last row. However, if b = −1, then the row-echelon form
obtained has full rank, and there is a unique solution. Therefore, we draw the following conclusions:
(a). There is no solution to the system if a = − 12 or if a = − 12 and b = −1.
(b). Under no circumstances are there an infinite number of solutions to the linear system.
(c). There is a unique solution if a = − 12 and b = −1.
28. The corresponding augmented matrix for this linear system can be reduced to row-echelon form via
⎡
⎤ ⎡
⎤ ⎡
⎤
y1
y1
1 1 1 y1
1 1
1
1 1
1
1
2
⎣ 2 3 1 y2 ⎦ ∼
⎦.
⎣ 0 1 −1 y2 − 2y1 ⎦ ∼
⎣ 0 1 −1
y2 − 2y1
3 5 1 y3
0 2 −2 y3 − 3y1
0 0
0 y1 − 2y2 + y3
1. A12 (−2), A13 (−3)
2. A23 (−2)
For consistency, we must have rank(A) = rank(A# ), which requires (y1 , y2 , y3 ) to satisfy y1 − 2y2 + y3 = 0.
If this holds, then the system has an infinite number of solutions, because the column of the augmented
matrix corresponding to y3 will be unpivoted, indicating that y3 is a free variable in the solution set.
29. Converting the given system of equations to an augmented matrix and using Gaussian elimination we
obtain the following row-equivalent matrices. Since a11 = 0:
a12
b1
b1
1
1 aa12
a11 a12 b1
1
2
a
a
a11
11
11
11
.
∼
∼
a11 b2 −a21 b1
21 a12
a21 a22 b2
0 aΔ11 aΔ112
0 a22 a11a−a
a11
11
1. M1 (1/a11 ), A12 (−a21 )
2. Definition of Δ and Δ2
(a). If Δ = 0, then rank(A) = rank(A# ) = 2, so the system
a unique solution (of course, we are assuming
has
Δ
a11 = 0 here). Using the last augmented matrix above, a11 x2 = aΔ112 , so that x2 = ΔΔ2 . Using this, we can
1
x2 = ab11
for x1 to obtain x1 = ΔΔ1 , where we have used the fact that Δ1 = a22 b1 − a12 b2 .
solve x1 + aa12
11
b1
1 aa12
a
11
11
, so it follows that
(b). If Δ = 0 and a11 = 0, then the augmented matrix of the system is
0 0
Δ2
the system has (i) no solution if Δ2 = 0, since rank(A) < rank(A# ) = 2, and (ii) an infinite number of
solutions if Δ2 = 0, since rank(A# ) < 2.
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(c). An infinite number of solutions would be represented as one line. No solution would be two parallel
lines. A unique solution would be the intersection of two distinct lines at one point.
30. We first use the partial pivoting algorithm to reduce the augmented matrix of the system:
⎡
⎤ ⎡
⎤ ⎡
⎤
3
1 2 1 1
3 5 1 3
3
5
1
1
2
⎣ 3 5 1 3 ⎦ ∼ ⎣ 1 2 1 1 ⎦ ∼ ⎣ 0 1/3 2/3
0 ⎦
2 6 7 1
2 6 7 1
0 8/3 19/3 −1
⎤ ⎡
⎤
⎡
3
3
3 5
1
3
5
1
3
4
∼ ⎣ 0 8/3 19/3 −1 ⎦ ∼ ⎣ 0 8/3 19/3 −1 ⎦ .
0 1/3 2/3
0
0 −1/8 1/8
0
1. P12
2. A12 (−1/3), A13 (−2/3)
3. P23
4. A23 (−1/8)
Using back substitution to solve the equivalent system yields the unique solution (−2, 2, −1).
31. We first use the partial pivoting algorithm to reduce the augmented matrix of the system:
⎤ ⎡
⎤ ⎡
⎤
⎡
3
3
7
2 −3
7
2
−3
2 −1
3 14
⎢ 3
1 −2 −1 ⎥
1 −2 −1 ⎥
1/7
−5/7 −16/7 ⎥
1 ⎢
2 ⎢
⎢ 3
⎢ 0
⎥∼
⎥∼
⎥
⎢
⎣ 7
2 −3
3 14 ⎦ ⎣ 0 −11/7 27/7
3 ⎦ ⎣ 2 −1
92/7 ⎦
5 −1 −2
5 −1 −2
0 −17/7 1/7
5
5
20/7
⎡
⎤ ⎡
⎤
3
3
7
2
−3
7
2
−3
1/7
1/7
20/7 ⎥
20/7 ⎥
3 ⎢ 0 −17/7
4 ⎢
⎢ 0 −17/7
⎥∼
⎥
∼⎢
⎣ 0 −11/7 27/7
⎣
⎦
0
0
64/17
92/7
192/17 ⎦
0
1/7
−5/7 −16/7
0
0
−12/17 −36/17
⎡
⎤
3
7
2
−3
⎥
0
−17/7
1/7
20/7
5 ⎢
⎥.
∼⎢
⎣ 0
0
64/17 192/17 ⎦
0
0
0
0
1. P13
2. A12 (−3/7), A13 (−2/7), A14 (−5/7)
4. A23 (−11/17), A24 (1/17)
3. P24
5. A34 (3/16)
Using back substitution to solve the equivalent system yields the unique solution (2, −1, 3).
32. We first use the partial pivoting algorithm to reduce the augmented matrix of the system:
⎤ ⎡
⎤ ⎡
⎤
⎡
5
6 −6 −20
5
6
−6
5
20
2 −1 −4
⎥
⎢ 3
2 −5
2 −5
8 ⎥
8 ⎥
1 ⎢
2 ⎢
⎢ 3
⎢ 0 −8/5 −7/5 −4 ⎥
⎥∼
⎥∼
⎢
⎣
⎣
⎦
⎦
⎣ 5
6 −6 20
2 −1 −4
0 −17/5 −8/5 −3 ⎦
5
1
1 −3 −3
1
1 −3 −3
0 −1/5 −9/5 −7
⎡
⎤ ⎡
⎤
20
20
5
6
−6
5
6
−6
⎢
⎥
−3
3 ⎢ 0 −17/5 −8/5 −3 ⎥
⎥ 4 ⎢ 0 −17/5 −8/5
⎥
∼⎢
⎣ 0 −8/5 −7/5 −4 ⎦ ∼ ⎣ 0
0
−11/17 −44/17 ⎦
0 −1/5 −9/5 −7
0
0
−29/17 −116/17
⎡
⎤ ⎡
⎤
5
6
−6
5
6
−6
20
20
⎥ 6 ⎢ 0 −17/5 −8/5
⎥
−8/5
−3
−3
5 ⎢ 0 −17/5
⎥∼⎢
⎥.
∼⎢
⎣
⎣ 0
⎦
0
−29/17 −116/17
0
0
−29/17 −116/17 ⎦
0
0
−11/17 −44/17
0
0
0
0
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1. P13
2. A12 (−3/5), A13 (−2/5), A14 (−1/5)
4. A23 (−8/17), A24 (−1/17)
5. P34
3. P23
6. A34 (−11/29)
Using back substitution to solve the equivalent system yields the unique solution (10, −1, 4).
33. We first use the partial pivoting algorithm to reduce the augmented matrix of the system:
⎡
⎤ ⎡
⎤ ⎡
⎤
2
2 −1 −1
4
3 −2 −1
4
3
−2 −1
1
2
⎣ 4
3 −2 −1 ⎦ ∼ ⎣ 2 −1 −1
0
2 ⎦ ∼ ⎣ 0 −5/2
5/2 ⎦
1
4
1
1
4
1
0 13/4 3/2 17/4
4
4
⎡
4
3
∼ ⎣ 0 13/4
0 −5/2
3
1. P12
⎤ ⎡
−2 −1
4
4
3/2 17/4 ⎦ ∼ ⎣ 0
0
0
5/2
⎤
−1
3
−2
13/4
3/2
17/4 ⎦ .
0
15/13 75/13
2. A12 (−1/2), A13 (−1/4)
3. P23
4. A23 (10/13)
Using back substitution to solve the equivalent system yields the unique solution (3, −1, 5).
34.
(a). Let
⎡
a11
⎢ a21
⎢
A# = ⎢
⎢ a31
⎣ ...
an1
0
a22
a32
...
an2
0
0
a33
...
an3
...
...
...
...
...
0
0
0
...
ann
⎤
b1
b2 ⎥
⎥
b3 ⎥
⎥
... ⎦
bn
represent the corresponding augmented matrix of the given system. Since a11 x1 = b1 , we can solve for x1
easily:
b1
x1 =
,
(a11 = 0).
a11
Now since a21 x1 + a22 x2 = b2 , by using the expression for x1 we just obtained, we can solve for x2 :
x2 =
a11 b2 − a21 b1
.
a11 a22
In a similar manner, we can solve for x3 , x4 , . . . , xn .
(b). We solve instantly for x1 from the first equation: x1 = 2. Substituting this into the middle equation,
we obtain 2 · 2 − 3 · x2 = 1, from which it quickly follows that x2 = 1. Substituting for x1 and x2 in the
bottom equation yields 3 · 2 + 1 − x3 = 8, from which it quickly follows that x3 = −1. Consequently, the
solution of the given system is (2, 1, −1).
35. This system of equations is not linear in x1 , x2 , and x3 ; however, the system is linear in x31 , x22 , and x3 ,
so we can first solve for x31 , x22 , and x3 . Converting the given system of equations to an augmented matrix
and using Gauss-Jordan elimination we obtain the following equivalent matrices:
⎡
⎤ ⎡
⎤ ⎡
⎤
4
2
3 12
2
1 −1
1 2
1 −1
1
1
2
⎣ 1 −1
1 2 ⎦∼⎣ 4
2
3 12 ⎦ ∼ ⎣ 0
6 −1
4 ⎦
3
1 −1 2
3
1 −1 2
0
4 −4 −4
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⎡
⎤ ⎡
⎤ ⎡
⎤
2
2
1
1 −1
1
1 −1
1
1 0
0
4
5
4 −4 −4 ⎦ ∼ ⎣ 0
1 −1 −1 ⎦ ∼ ⎣ 0 1 −1 −1 ⎦
∼⎣ 0
0
6 −1
0
6 −1
0 0
5 10
4
4
⎤ ⎡
⎤
⎡
1
1 0
0
1 0 0 1
6
7
∼ ⎣ 0 1 −1 −1 ⎦ ∼ ⎣ 0 1 0 1 ⎦ .
0 0
1
0 0 1 2
2
3
1. P12
2. A12 (−4), A13 (−3)
5. A21 (1), A23 (−6)
3. P23
6. M2 (1/5)
4. M2 (1/4)
7. A32 (1)
Thus, taking only real solutions, we have x31 = 1, x22 = 1, and x3 = 2. Therefore, x1 = 1, x2 = ±1, and
x3 = 2, leading to the two solutions (1, 1, 2) and (1, −1, 2) to the original system of equations. There is no
contradiction of Theorem 2.5.9 here since, as mentioned above, this system is not linear in x1 , x2 , and x3 .
36. Reduce the augmented matrix of the system:
⎤ ⎡
⎤ ⎡
⎡
1
1 −2 0
1
1
3
2 −1 0
1
2
⎣ 2
1
1 0 ⎦ ∼ ⎣ 0 −1
5 0 ⎦∼⎣ 0
1
5 −4
1 0
0 −9 11 0
0 −9
⎡
⎤ ⎡
1 0
3 0
1 0
4
5
∼ ⎣ 0 1 −5 0 ⎦ ∼ ⎣ 0 1
0 0
1 0
0 0
1. A21 (−1), A12 (−2), A13 (−5)
4. M3 (−1/34)
⎤ ⎡
⎤
−2 0
1 0
3 0
3
−5 0 ⎦ ∼ ⎣ 0 1 −5 0 ⎦
11 0
0 0 −34 0
⎤
0 0
0 0 ⎦.
1 0
2. M2 (−1)
3. A21 (−1), A23 (9)
5. A31 (−3), A32 (5)
Therefore, the unique solution to this system is x1 = x2 = x3 = 0: (0, 0, 0).
37. Reduce the augmented matrix of the system:
⎤ ⎡
⎤ ⎡
⎤ ⎡
⎤
⎡
2
1 −1 0
1 −1 −1 0
1 −1 −1 0
1 −1 −1 0
⎢ 3 −1
⎢
2 0 ⎥
2 0 ⎥
2
5 0 ⎥
3
1 0 ⎥
2 ⎢
3 ⎢
⎥ 1 ⎢ 3 −1
⎥∼
⎥∼
⎥
⎢
⎢ 0
⎢ 0
⎣ 1 −1 −1 0 ⎦ ∼ ⎣ 2
1 −1 0 ⎦ ⎣ 0
3
1 0 ⎦ ⎣ 0
2
5 0 ⎦
5
2 −2 0
5
2 −2 0
0
7
3 0
0
7
3 0
⎡
⎤ ⎡
⎤ ⎡
⎤ ⎡
⎤
1 −1 −1 0
1 0 −5 0
1 0 −5 0
1 0 0 0
⎥ 6 ⎢ 0 1 −4 0 ⎥ 7 ⎢ 0 1 0 0 ⎥
1 −4 0 ⎥
4 ⎢ 0
5 ⎢
⎢ 0 1 −4 0 ⎥ ∼
⎢
⎥∼
⎥∼⎢
⎥.
∼⎢
⎣ 0
⎣
⎦
2
5 0
0 0 13 0 ⎦ ⎣ 0 0
1 0 ⎦ ⎣ 0 0 1 0 ⎦
0
7
3 0
0 0 31 0
0 0 31 0
0 0 0 0
1. P13
2. A12 (−3), A13 (−2), A14 (−5)
5. A21 (1), A23 (−2), A24 (−7)
6. M3 (1/13)
3. P23
4. A32 (−1)
7. A31 (5), A32 (4), A34 (−31)
Therefore, the unique solution to this system is x1 = x2 = x3 = 0: (0, 0, 0).
38. Reduce the augmented matrix of the system:
⎤ ⎡
⎤ ⎡
⎤
⎡
1
1
4 0
1
1
4 0
2 −1 −1 0
1
2
⎣ 5 −1
2 0 ⎦ ∼ ⎣ 5 −1
2 0 ⎦ ∼ ⎣ 0 −6 −18 0 ⎦
1
1
4 0
2 −1 −1 0
0 −3 −9 0
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⎡
1
∼⎣ 0
0
3
1. P13
⎤ ⎡
⎤
1
4 0
1 0 1 0
4
1
3 0 ⎦ ∼ ⎣ 0 1 3 0 ⎦.
−3 −9 0
0 0 0 0
2. A12 (−5), A13 (−2)
3. M2 (−1/6)
4. A21 (−1), A23 (3)
It follows that x1 + x3 = 0 and x2 + 3x3 = 0. Setting x3 = t, where t is a free variable, we get x2 = −3t and
x1 = −t. Thus we have that the solution set of the system is {(−t, −3t, t) : t ∈ R}.
39. Reduce the augmented matrix of the system:
⎡
⎤ ⎡
⎤ ⎡
⎤
1 + 2i 1 − i
1
0
0
0
i
1+i
−i
1
1−i
−1
1
2
⎣
i
1+i
−i
1
1
0 ⎦ ∼ ⎣ 1 + 2i 1 − i
0 ⎦ ∼ ⎣ 1 + 2i 1 − i
0 ⎦
2i
1
1 + 3i 0
2i
1
1 + 3i 0
2i
1
1 + 3i 0
⎡
⎤ ⎡
⎤ ⎡
⎤
0
0
0
1
1−i
−1
1
1−i
−1
1 1−i
−1
3
4
5
0
−5 + 8i 0 ⎦
∼ ⎣ 0 −2 − 2i 1 + 2i 0 ⎦ ∼ ⎣ 0 −2 − 2i 1 + 2i 0 ⎦ ∼ ⎣ 0
0 −1 − 2i 1 + 5i 0
0
1
3i
0
1
3i
0
0
⎡
⎤ ⎡
⎤ ⎡
⎤
0
1 1−i
−1
1 1 − i −1 0
1 0 0 0
6
7
8
1
3i
1
3i 0 ⎦ ∼ ⎣ 0 1 0 0 ⎦ .
0 ⎦∼⎣ 0
∼⎣ 0
0
0
−5 + 8i 0
0
0
1 0
0 0 1 0
1. P12
2. M1 (−i)
6. P23
3. A12 (−1 − 2i), A13 (−2i)
1
7. M3 ( −5+8i
)
4. A23 (−1)
5. A32 (2 + 2i)
8. A21 (−1 + i), A31 (1), A32 (−3i)
Therefore, the unique solution to this system is x1 = x2 = x3 = 0: (0, 0, 0).
40. Reduce the augmented matrix of the system:
⎤ ⎡
⎤ ⎡
⎡
1
2
2
3
2 1 0
0
1
1
3
3
3
1
2
⎣ 0 −5
⎣ 6 −1 2 0 ⎦ ∼ ⎣ 6 −1 2 0 ⎦ ∼
12
6 4 0
12
6 4 0
0 −2
⎡
⎡
⎤
⎤
1
2
0
1
1 0 13 0
3
3
3
4
1 0 0 ⎦ ∼ ⎣ 0 1 0 0 ⎦.
∼⎣ 0
0 −2 0 0
0 0 0 0
1. M1 (1/3)
2. A12 (−6), A13 (−12)
3. M2 (−1/5)
1
3
0
0
⎤
0
0 ⎦
0
4. A21 (−2/3), A23 (2)
From the last augmented matrix, we have x1 + 13 x3 = 0 and x2 = 0. Since x3 is a free variable, we let x3 = t,
where t is a real number. It follows that the solution set for the given system is given by {(t, 0, −3t) : t ∈ R}.
41. Reduce the augmented matrix of the system:
⎡
⎤ ⎡
⎤ ⎡
⎤
2
1 −8 0
3 −2 −5 0
1 −3
3 0
⎢ 3 −2 −5 0 ⎥ 1 ⎢ 2
⎢
1 −8 0 ⎥
1 −8 0 ⎥
⎢
⎥ ⎢
⎥ 2 ⎢ 2
⎥
⎣ 5 −6 −3 0 ⎦ ∼ ⎣ 5 −6 −3 0 ⎦ ∼ ⎣ 5 −6 −3 0 ⎦
3 −5
1 0
3 −5
1 0
3 −5
1 0
⎡
⎤ ⎡
⎤ ⎡
⎤
1 −3
3 0
1 −3
3 0
1 0 −3 0
⎥
7 −14 0 ⎥
1 −2 0 ⎥
3 ⎢ 0
4 ⎢
5 ⎢
⎢ 0
⎢ 0 1 −2 0 ⎥ .
⎥∼
⎥∼
∼⎢
⎣ 0
⎣
⎣
⎦
⎦
9 −18 0
0
9 −18 0
0 0
0 0 ⎦
0
4 −8 0
0
4 −8 0
0 0
0 0
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1. P12
2. A21 (−1)
3. A12 (−2), A13 (−5), A14 (−3)
4. M2 (1/7)
5. A21 (3), A23 (−9), A24 (−4)
From the last augmented matrix we have: x1 − 3x3 = 0 and x2 − 2x3 = 0. Since x3 is a free variable, we let
x3 = t, where t is a real number. It follows that x2 = 2t and x1 = 3t. Thus, the solution set for the given
system is given by {(3t, 2t, t) : t ∈ R}.
42. Reduce the augmented matrix of the system:
⎡
⎤ ⎡
1
1+i
1−i 0
1
1+i
1
⎣
i
1
i
2−i
0 ⎦∼⎣ 0
1 − 2i −1 + i 1 − 3i 0
0 −4 + 2i
⎡
1 1+i
3
1
∼⎣ 0
0
0
1−i
−2−i
5
0
1. A12 (−i), A13 (−1 + 2i)
1−i
−1
2
⎤ ⎡
0
1 0
4
0 ⎦∼⎣ 0 1
0
0 0
⎤ ⎡
0
1 1+i
2
0 ⎦∼⎣ 0 2−i
0
0
0
6−2i
5
−2−i
5
0
⎤
0
0 ⎦
0
⎤
0
0 ⎦.
0
1
3. M2 ( 2−i
)
2. A23 (2)
1−i
−1
0
4. A21 (−1 − i)
From the last augmented matrix we see that x3 is a free variable. We set x3 = 5s, where s ∈ C. Then
x1 = 2(i − 3)s and x2 = (2 + i)s. Thus, the solution set of the system is {(2(i − 3)s, (2 + i)s, 5s) : s ∈ C}.
43. Reduce the augmented matrix of the system:
⎤ ⎡
⎤ ⎡
⎡
1 −1
1 0
1 −1
1 −1
1 0
⎥ 1 ⎢ 0
⎥ 2 ⎢ 0
⎢ 0
3
2
3
2
1
0
0
⎥∼⎢
⎥∼⎢
⎢
⎣ 3
0 −1 0 ⎦ ⎣ 0
3 −4 0 ⎦ ⎣ 0
3
5
1 −1 0
0
6 −6 0
0
6
⎡
1 0 5/3
0 1 2/3
3 ⎢
∼⎢
⎣ 0 0 −6
0 0 −10
⎤ ⎡
1 0 5/3
0
⎢ 0 1 2/3
0 ⎥
4
⎥∼⎢
1
0 ⎦ ⎣ 0 0
0 0 −10
0
1. A13 (−3), A14 (−5)
4. M3 (−1/6)
2. M2 (1/3)
1
2/3
−4
−6
⎤
0
0 ⎥
⎥
0 ⎦
0
⎤ ⎡
⎤
1 0 0 0
0
⎥
0 ⎥
5 ⎢
⎢ 0 1 0 0 ⎥.
⎥∼
⎣
⎦
0 0 1 0 ⎦
0
0 0 0 0
0
3. A21 (1), A23 (−3), A24 (−6)
5. A31 (−5/3), A32 (−2/3), A34 (10)
Therefore, the unique solution to this system is x1 = x2 = x3 = 0: (0, 0, 0).
44. Reduce the augmented matrix of the system:
⎤ ⎡
⎡
1
−2
3
2 −4
6 0
⎢ 3 −6 9 0 ⎥ 1 ⎢ 3
−6
9
⎥ ⎢
⎢
⎣ 1 −2 3 0 ⎦ ∼ ⎣ 2
−4
6
5 −10 15 0
5 −10 15
1. M1 (1/2)
⎤ ⎡
⎤
0
1 −2 3 0
0 0 0 ⎥
0 ⎥
2 ⎢
⎢ 0
⎥∼
⎥.
⎣
⎦
0
0 0 0 ⎦
0
0
0 0 0
0
2. A12 (−3), A13 (−2), A14 (−5)
From the last matrix we have that x1 − 2x3 + 3x3 = 0. Since x2 and x3 are free variables, let x2 = s and
let x3 = t, where s and t are real numbers. The solution set of the given system is therefore {(2s − 3t, s, t) :
s, t ∈ R}.
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45. Reduce the augmented matrix of the system:
⎡
⎤ ⎡
⎤ ⎡
⎤
4 −2 −1 −1 0
1 −3
1 −4 0
1 −3
1 −4 0
1
2
⎣ 3
1 −2
3 0 ⎦∼⎣ 3
1 −2
3 0 ⎦ ∼ ⎣ 0 10 −5 15 0 ⎦
5 −1 −2
1 0
5 −1 −2
1 0
0 14 −7 21 0
⎡
⎤ ⎡
⎤ ⎡
1 −3
1 −4 0
1 −3
1 −4 0
1 −3
1
4
5
2 −1
3 0 ⎦∼⎣ 0
2 −1
3 0 ⎦∼⎣ 0
1 −1/2
∼⎣ 0
0
2 −1
3 0
0
0
0
0 0
0
0
0
3
1. A21 (−1)
2. A12 (−3), A13 (−5)
4. A23 (−1)
⎤
−4 0
3/2 0 ⎦ .
0 0
3. M2 (1/5), M3 (1/7)
5. M2 (1/2)
From the last augmented matrix above we have that x2 − 12 x3 + 32 x4 = 0 and x1 −3x2 +x3 −4x4 = 0. Since x3
and x4 are free variables, we can set x3 = 2s and x4 = 2t, where s and t are real numbers. Then x2 = s − 3t
and x1 = s − t. It follows that the solution set of the given system is {(s − t, s − 3t, 2s, 2t) : s, t ∈ R}.
46. Reduce the augmented matrix of the system:
⎤ ⎡
⎤ ⎡
⎤
⎡
1
1
1 −1 0
1
1
1 −1 0
2
1 −1
1 0
⎢ 1
1
1 −1 0 ⎥
1 −1
1 0 ⎥
3 0 ⎥
1 ⎢
2 ⎢
⎢ 2
⎢ 0 −1 −3
⎥∼
⎥∼
⎥
⎢
⎣
⎣
⎦
⎦
⎣ 3 −1
1 −2 0
3 −1
1 −2 0
0 −4 −2
1 0 ⎦
4
2 −1
1 0
4
2 −1
1 0
0 −2 −5
5 0
⎡
1
0
3 ⎢
∼⎢
⎣ 0
0
⎡
1 0 −2
⎢
0 1
3
6
∼⎢
⎣ 0 0
1
0 0 10
⎤ ⎡
1
1 −1 0
1 0 −2
2
⎢ 0 1
1
3 −3 0 ⎥
3
−3
4
⎥∼⎢
−4 −2
1 0 ⎦ ⎣ 0 0 10 −11
−2 −5
5 0
0 0 −3
3
⎤ ⎡
⎤ ⎡
2 0
1 0 0
0 0
1
⎢
⎢
⎥
−3 0 ⎥
0
1
0
0
0
7 ⎢
8 ⎢ 0
⎥∼
⎥∼
−1 0 ⎦ ⎣ 0 0 1 −1 0 ⎦ ⎣ 0
−11 0
0 0 0 −1 0
0
1. P12
2. A12 (−2), A13 (−3), A14 (−4)
5. P34
6. M3 (−1/3)
⎤ ⎡
⎤
1 0 −2
2 0
0
3 −3 0 ⎥
0 ⎥
5 ⎢
⎢ 0 1
⎥∼
⎥
3 0 ⎦
0 ⎦ ⎣ 0 0 −3
0 0 10 −11 0
0
⎤ ⎡
0 0
0 0
1 0 0 0
⎢
1 0
0 0 ⎥
9 ⎢ 0 1 0 0
⎥∼
0 1 −1 0 ⎦ ⎣ 0 0 1 0
0 0
1 0
0 0 0 1
3. M2 (−1)
⎤
0
0 ⎥
⎥.
0 ⎦
0
4. A21 (−1), A23 (4), A24 (2)
7. A31 (2), A32 (−3), A34 (−10)
8. M4 (−1)
9. A43 (1)
From the last augmented matrix, it follows that the solution set to the system is given by {(0, 0, 0, 0)}.
47. The equation Ax = 0 is
2 −1
3
4
x1
x2
Reduce the augmented matrix of the system:
2 −1 0 1 1 − 12 0 2 1 − 12
∼
∼
3
4 0
3
4 0
0 11
2
1. M1 (1/2)
2. A12 (−3)
=
0
0
0
0
3
.
∼
1 − 12
0
1
3. M2 (2/11)
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0
0
4
∼
4. A21 (1/2)
1 0 0
0 1 0
.
165
From the last augmented matrix, we see that x1 = x2 = 0. Hence, the solution set is {(0, 0)}.
48. The equation Ax = 0 is
1−i
1+i
2i
−2
Reduce the augmented matrix of the system:
1 − i 2i 0 1
1
∼
1 + i −2 0
1+i
x1
x2
0
0
0
0
=
−1 + i
−2
1. M1 ( 1+i
2 )
.
2
−1 + i
0
1
0
∼
0
0
.
2. A12 (−1 − i)
It follows that x1 + (−1 + i)x2 = 0. Since x2 is a free variable, we can let x2 = t, where t is a complex
number. The solution set to the system is then given by {(t(1 − i), t) : t ∈ C}.
49. The equation Ax = 0 is
1+i
−1 + i
1 − 2i
2+i
Reduce the augmented matrix of the system:
1 + i 1 − 2i 0 1
1
∼
−1 + i 2 + i 0
−1 + i
1. M1 ( 1−i
2 )
x1
x2
=
− 1+3i
2
2+i
0
0
0
0
2
.
∼
1 − 1+3i
2
0
0
0
0
.
2. A12 (1 − i)
It follows that x1 − 1+3i
2 x2 = 0. Since x2 is a free variable, we can let x2 = r, where r is any complex
number. Thus, the solution set to the given system is {( 1+3i
2 r, r) : r ∈ C}.
50. The equation Ax = 0 is
⎡
⎤⎡
⎤ ⎡
⎤
1
2 3
x1
0
⎣ 2 −1 0 ⎦ ⎣ x2 ⎦ = ⎣ 0 ⎦ .
1
1 1
x3
0
Reduce the augmented matrix of the system:
⎤ ⎡
⎤ ⎡
⎡
1
2
3 0
1
1
2 3 0
1
2
⎣ 0
⎣ 2 −1 0 0 ⎦ ∼ ⎣ 0 −5 −6 0 ⎦ ∼
1
1 1 0
0 −1 −2 0
0
⎡
⎤ ⎡
1 0 −1 0
1 0 −1
4
5
2 0 ⎦∼⎣ 0 1
2
∼⎣ 0 1
0 0
4 0
0 0
1
1. A12 (−2), A13 (−1)
2. P23
3. M2 (−1)
⎤ ⎡
⎤
2
3 0
1
2
3 0
3
−1 −2 0 ⎦ ∼ ⎣ 0
1
2 0 ⎦
−5 −6 0
0 −5 −6 0
⎤ ⎡
⎤
0
1 0 0 0
6
0 ⎦ ∼ ⎣ 0 1 0 0 ⎦.
0 0 1 0
0
4. A21 (−2), A23 (5)
5. M3 (1/4)
6. A31 (1), A32 (−2)
From the last augmented matrix, we see that the only solution to the given system is x1 = x2 = x3 = 0:
{(0, 0, 0)}.
51. The equation Ax = 0 is
⎡
1
⎣ −1
1
⎤
⎤ ⎡
⎡
⎤ x
0
1
1 −1 ⎢ 1 ⎥ ⎢
x2 ⎥ ⎢ 0 ⎥
⎥
0 −1
2 ⎦⎢
⎣ x3 ⎦ = ⎣ 0 ⎦ .
3
2
2
0
x4
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Reduce the augmented matrix of the system:
⎡
⎤ ⎡
⎤ ⎡
⎤ ⎡
⎤
1 1
1 −1 0
1 1 1 −1 0
1 0 1 −2 0
1 0 0 −3 0
1
2
3
⎣ −1 0 −1
2 0 ⎦∼⎣ 0 1 0
1 0 ⎦∼⎣ 0 1 0
1 0 ⎦∼⎣ 0 1 0
1 0 ⎦.
1 3
2
2 0
0 2 1
3 0
0 0 1
1 0
0 0 1
1 0
1. A12 (1), A13 (−1)
2. A21 (−1), A23 (−2)
3. A31 (−1)
From the last augmented matrix, we see that x4 is a free variable. We set x4 = t, where t is a real number.
The last row of the reduced row echelon form above corresponds to the equation x3 + x4 = 0. Therefore,
x3 = −t. The second row corresponds to the equation x2 + x4 = 0, so we likewise find that x2 = −t. Finally,
from the first equation we have x1 − 3x4 = 0, so that x1 = 3t. Consequently, the solution set of the original
system is given by {(3t, −t, −t, t) : t ∈ R}.
52. The equation Ax = 0 is
⎡
2 − 3i
⎣ 3 + 2i
5−i
⎤⎡
⎤ ⎡
⎤
x1
i−1
0
−1 − i ⎦ ⎣ x2 ⎦ = ⎣ 0 ⎦ .
−2
x3
0
1+i
−1 + i
2i
Reduce the augmented matrix of this system:
⎤ ⎡
⎡
2 − 3i 1 + i
i−1 0
1
1
⎣ 3 + 2i
⎣ 3 + 2i −1 + i −1 − i 0 ⎦ ∼
5−i
2i
−2
0
5−i
1. M1 ( 2+3i
13 )
−1+5i
13
−1 + i
2i
−5−i
13
−1 − i
−2
⎤ ⎡
0
1
2
0 ⎦∼⎣ 0
0
0
−1+5i
13
0
0
−5−i
13
0
0
⎤
0
0 ⎦.
0
2. A12 (−3 − 2i), A13 (−5 + i)
−5−i
From the last augmented matrix, we see that x1 + −1+5i
13 x2 + 13 x3 = 0. Since x2 and x3 are free variables,
we can let x2 = 13r and x3 = 13s, where r and s are complex numbers. It follows that the solution set of
the system is {(r(1 − 5i) + s(5 + i), 13r, 13s) : r, s ∈ C}.
53. The equation Ax = 0 is
⎡
⎤⎡
⎤ ⎡
⎤
1
3 0
x1
0
⎣ −2 −3 0 ⎦ ⎣ x2 ⎦ = ⎣ 0 ⎦ .
1
4 0
x3
0
Reduce the augmented matrix of the system:
⎤ ⎡
⎤ ⎡
⎤ ⎡
⎤
⎡
1 3 0 0
1 3 0 0
1 0 0 0
1
3 0 0
1
2
3
⎣ 0 3 0 0 ⎦∼
⎣ 0 1 0 0 ⎦∼
⎣ 0 1 0 0 ⎦.
⎣ −2 −3 0 0 ⎦ ∼
1
4 0 0
0 1 0 0
0 3 0 0
0 0 0 0
1. A12 (2), A13 (−1)
2. P23
3. A21 (−3), A23 (−3)
From the last augmented matrix we see that the solution set of the system is {(0, 0, t) : t ∈ R}.
54. The equation Ax = 0 is
⎤
1
0
3 ⎡
⎤ ⎡
⎤
⎢ 3 −1
7 ⎥
0
⎥ x1
⎢
⎢ 2
⎣
⎦ ⎣
⎦
1
8 ⎥
⎥ x2 = 0 .
⎢
⎦
⎣ 1
1
5
x3
0
−1
1 −1
⎡
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167
Reduce the augmented matrix of the system:
⎡
⎤ ⎡
⎤ ⎡
1
0
3 0
1
1
0
3 0
⎢
⎢ 3 −1
⎥ ⎢
7 0 ⎥
⎢
⎥ 1 ⎢ 0 −1 −2 0 ⎥ 2 ⎢ 0
⎢
⎢
⎢ 2
1
8 0 ⎥
1
2 0 ⎥
⎢
⎥∼⎢ 0
⎥∼⎢ 0
⎣
⎣ 0
⎣ 1
⎦
⎦
1
5 0
0
1
2 0
−1
1 −1 0
0
1
2 0
0
1. A12 (−3), A13 (−2), A14 (−1), A15 (1)
0
1
1
1
1
2. M2 (−1)
3
2
2
2
2
⎤ ⎡
⎤
1 0 3 0
0
⎢
⎥
0 ⎥
⎥ 3 ⎢ 0 1 2 0 ⎥
⎢
⎥
0 ⎥∼⎢ 0 0 0 0 ⎥
⎥.
0 ⎦ ⎣ 0 0 0 0 ⎦
0 0 0 0
0
3. A23 (−1), A24 (−1), A25 (−1)
From the last augmented matrix, we obtain the equations x1 + 3x3 = 0 and x2 + 2x3 = 0. Since x3 is a
free variable, we let x3 = t, where t is a real number. The solution set for the given system is then given by
{(−3t, −2t, t) : t ∈ R}.
55. The equation Ax = 0 is
⎡
⎤
⎡
⎤
x1
1 −1 0 1 ⎢
0
⎥
x
⎣ 3 −2 0 5 ⎦ ⎢ 2 ⎥ = ⎣ 0 ⎦ .
⎣ x3 ⎦
−1
2 0 1
0
x4
⎡
⎤
Reduce the augmented matrix of the system:
⎤ ⎡
⎤ ⎡
⎤
⎡
1 −1 0 1 0
1 0 0 3 0
1 −1 0 1 0
1
2
⎣ 0
⎣ 3 −2 0 5 0 ⎦ ∼
1 0 2 0 ⎦ ∼ ⎣ 0 1 0 2 0 ⎦.
−1
2 0 1 0
0
1 0 2 0
0 0 0 0 0
1. A12 (−3), A13 (1)
2. A21 (1), A23 (−1)
From the last augmented matrix we obtain the equations x1 + 3x4 = 0 and x2 + 2x4 = 0. Because x3 and
x4 are free, we let x3 = t and x4 = s, where s and t are real numbers. It follows that the solution set of the
system is {(−3s, −2s, t, s) : s, t ∈ R}.
56. The equation Ax = 0 is
⎡
⎤
⎤ x
⎡
⎤
1 0 −3 0 ⎢ 1 ⎥
0
⎣ 3 0 −9 0 ⎦ ⎢ x2 ⎥ = ⎣ 0 ⎦ .
⎣ x3 ⎦
−2 0
6 0
0
x4
⎡
Reduce the augmented matrix of the system:
⎤ ⎡
⎤
⎡
1 0 −3 0 0
1 0 −3 0 0
1
⎣ 0 0
⎣ 3 0 −9 0 0 ⎦ ∼
0 0 0 ⎦.
−2 0
6 0 0
0 0
0 0 0
1. A12 (−3), A13 (2)
From the last augmented matrix we obtain x1 − 3x3 = 0. Therefore, x2 , x3 , and x4 are free variables, so
we let x2 = r, x3 = s, and x4 = t, where r, s, t are real numbers. The solution set of the given system is
therefore {(3s, r, s, t) : r, s, t ∈ R}.
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57. The equation Ax = 0 is
⎡
2+i
⎣ i
3−i
Reduce the augmented matrix of the system:
⎡
⎤ ⎡
2+i
i
3 − 2i 0
i
1
⎣ i
1 − i 4 + 3i 0 ⎦ ∼ ⎣ 2 + i
3 − i 1 + i 1 + 5i 0
3−i
⎡
1 −1 − i
3
∼ ⎣ 0 1 + 4i
0 5 + 3i
1. P12
2. M1 (−i)
⎤⎡
⎤ ⎡
⎤
3 − 2i
x1
0
4 + 3i ⎦ ⎣ x2 ⎦ = ⎣ 0 ⎦ .
1 + 5i
0
x3
i
1−i
1+i
1−i
i
1+i
4 + 3i
3 − 2i
1 + 5i
⎤ ⎡
0
1
2
0 ⎦∼⎣ 2+i
3−i
0
−1 − i
i
1+i
⎤ ⎡
⎤ ⎡
0
1 −1 − i
3 − 4i
0
3 − 4i
1 0
4
5
5+31i
⎦∼
⎣ 0 1
−7 + 3i 0 ⎦ ∼ ⎣ 0
0
1
17
−4 + 20i 0
0 5 + 3i −4 + 20i 0
0 0
⎤
⎡
⎡
⎤
0
1 0 25−32i
1 0 0 0
17
6
7
⎦∼
⎣ 0 1 0 0 ⎦.
∼ ⎣ 0 1 5+31i
0
17
0 0 1 0
0 0
1
0
3. A12 (−2 − i), A13 (−3 + i)
6. M3 (−i/10)
4. M2 ( 1−4i
17 )
3 − 3i
3 − 2i
1 + 5i
25−32i
17
5+31i
17
10i
⎤
0
0 ⎦
0
⎤
0
0 ⎦
0
5. A21 (1 + i), A23 (−5 − 3i)
7. A31 ( −25+32i
), A32 ( −5−31i
)
17
17
From the last augmented matrix above, we see that the only solution to this system is the trivial solution.
Solutions to Section 2.6
True-False Review:
(a): FALSE. An invertible matrix is also known as a nonsingular matrix.
1 1
(b): FALSE. For instance, the matrix
does not contain a row of zeros, but fails to be invertible.
2 2
(c): TRUE. If A is invertible, then the unique solution to Ax = b is x = A−1 b.
⎤
⎡
1 0
1 0 0
(d): FALSE. For instance, if A =
and B = ⎣ 0 0 ⎦, then AB = I2 , but A is not even a
0 0 1
0 1
square matrix, hence certainly not invertible.
(e): FALSE. For instance, if A = In and B = −In , then A and B are both invertible, but A + B = 0n is
not invertible.
(f ): TRUE. We have
(AB)B −1 A−1 = In
and
B −1 A−1 (AB) = In ,
and therefore, AB is invertible, with inverse B −1 A−1 .
(g): TRUE. From A2 = A, we subtract to obtain A(A−I) = 0. Left multiplying both sides of this equation
by A−1 (since A is invertible, A−1 exists), we have A − I = A−1 0 = 0. Therefore, A = I, the identity matrix.
(h): TRUE. From AB = AC, we left-multiply both sides by A−1 (since A is invertible, A−1 exists) to
obtain A−1 AB = A−1 AC. Since A−1 A = I, we obtain IB = IC, or B = C.
(c)2017 Pearson Education. Inc.
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(i): TRUE. Any 5 × 5 invertible matrix must have rank 5, not rank 4 (Theorem 2.6.6).
(j): TRUE. Any 6 × 6 matrix of rank 6 is invertible (Theorem 2.6.6).
Problems:
1. We have
−1
=
AA
2. We have
−1
AA
=
4
3
9
7
2 −1
3 −1
3. We have
a
c
b
d
7 −9
−3
4
−1
−3
1
2
=
(4)(−9) + (9)(4)
(3)(−9) + (7)(4)
(2)(−1) + (−1)(−3)
(3)(−1) + (−1)(−3)
=
1
ad − bc
(4)(7) + (9)(−3)
(3)(7) + (7)(−3)
d
−c
−b
a
(2)(1) + (−1)(2)
(3)(1) + (−1)(2)
=
1
0
=
0
1
1
0
= I2 .
0
1
= I2 .
1
a b
d −b
=
−c
a
ad − bc c d
1
ad − bc
0
=
0
ad − bc
ad − bc
1 0
=
0 1
= I2 ,
and
1
ad − bc
d
−c
−b
a
a
c
b
d
1
d −b
a b
a
c d
ad − bc −c
1
ad − bc
0
=
0
ad − bc
ad − bc
1 0
=
0 1
=
= I2 .
4. We have
⎡
⎤⎡
⎤
3 5 1
8 −29
3
19 −2 ⎦
AA−1 = ⎣ 1 2 1 ⎦ ⎣ −5
2 6 7
2 −8
1
⎡
(3)(8) + (5)(−5) + (1)(2) (3)(−29) + (5)(19) + (1)(−8)
= ⎣ (1)(8) + (2)(−5) + (1)(2) (1)(−29) + (2)(19) + (1)(−8)
(2)(8) + (6)(−5) + (7)(2) (2)(−29) + (6)(19) + (7)(−8)
⎡
⎤
1 0 0
= ⎣ 0 1 0 ⎦ = I3 .
0 0 1
5. We have
[A|I2 ] =
1 2 1
1 3 0
0
1
1
∼
1 0
1 2
0 1 −1 1
2
∼
⎤
(3)(3) + (5)(−2) + (1)(1)
(1)(3) + (2)(−2) + (1)(1) ⎦
(2)(3) + (6)(−2) + (7)(1)
3 −2
1 0
0 1 −1
1
(c)2017 Pearson Education. Inc.
= [I2 |A−1 ].
170
Therefore,
A−1 =
3 −2
−1
1
1. A12 (−1)
.
2. A21 (−2)
6. We have
[A|I2 ] =
1
1−i
1+i
1
1
0
3
0
1
∼
1
1 1+i
0 −1
∼
1 0 −1
0 1 1−i
Thus,
A
−1
=
1. A12 (−1 + i)
1
−1 + i
1+i
−1
−1
1−i
0
1
2
∼
1
1−i
1 1+i
0
1
0
−1
= [I2 |A−1 ].
1+i
−1
.
3. A21 (−1 − i)
2. M2 (−1)
7. We have
[A|I2 ] =
1
−i
i−1
2
1
0
3
∼
1
0
−i
1−i
1 0 1+i
0 1
1
−1+i
2
1+i
2
0
1
1
∼
Thus,
A−1 =
1. A12 (1 − i)
1+i
1
1
1−i
0
1
2
∼
1 −i
0
1
1
1
0
1+i
2
= [I2 |A−1 ].
−1+i
2
1+i
2
.
2. M2 (1/(1 − i))
3. A21 (i)
8. Note that AB = 02 for all 2 × 2 matrices B. Therefore, A is not invertible.
9. We have
⎡
1 −1 2 1
1 11 0
[A|I3 ] = ⎣ 2
4 −3 10 0
0
1
0
⎤ ⎡
⎤ ⎡
⎤
0
1 0 0
1 0 0
1 −1 2
1 −1 2
1
2
3 7 −2 1 0 ⎦ ∼ ⎣ 0
1 2 −4 0 1 ⎦
0 ⎦∼⎣ 0
0
1 2 −4 0 1
0
3 7 −2 1 0
1
⎡
⎤ ⎡
⎤
1
1 0 4 −3 0
1 0 0 −43 −4 13
3
4
1 ⎦ ∼ ⎣ 0 1 0 −24 −2
7 ⎦ = [I3 |A−1 ].
∼ ⎣ 0 1 2 −4 0
0 0 1 10 1 −3
0 0 1
10
1 −3
Thus,
⎡
⎤
−43 −4 13
7 ⎦.
A−1 = ⎣ −24 −2
10
1 −3
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171
1. A12 (−2), A13 (−4)
2. P23
3. A21 (1), A23 (−3)
4. A31 (−4), A32 (−2)
10. We have
⎡
3 5 1 1
[A|I3 ] = ⎣ 1 2 1 0
2 6 7 0
⎤ ⎡
0
1 2 1 0
1
0 ⎦∼⎣ 3 5 1 1
2 6 7 0
1
0
1
0
1
0
0
⎤ ⎡
0
1
2
1 0
2
0 ⎦ ∼ ⎣ 0 −1 −2 1
0
2
5 0
1
⎤
1 0
−3 0 ⎦
−2 1
⎡
⎤ ⎡
⎤ ⎡
⎤
0
1 0
2 −5 0
8 −29
3
1 2 1
1 0 −3
1 0 0
3
4
5
2 −1
3 0 ⎦∼⎣ 0 1
3 0 ⎦ ∼ ⎣ 0 1 0 −5
19 −2 ⎦ = [I3 |A−1 ].
∼ ⎣ 0 1 2 −1
0 2 5
0 0
1
0 0 1
0 −2 1
2 −8 1
2
−8
1
Thus,
⎡
⎤
8 −29
3
19 −2 ⎦ .
A−1 = ⎣ −5
2
−8
1
1. P12
2. A12 (−3), A13 (−2)
3. M2 (−1)
4. A21 (−2), A23 (−2)
5. A31 (3), A32 (−2)
11. This matrix is not invertible, because the column of zeros guarantees that the rank of the matrix is less
than three.
12. We have
⎡
4 2 −13 1
[A|I3 ] = ⎣ 2 1 −7 0
3 2
4 0
⎡
1
3
∼⎣ 0
0
1
11 0
−1 −29 0
−2 −57 1
⎤ ⎡
3 2
4 0
0
1
−7 0
0 ⎦∼⎣ 2 1
4 2 −13 1
1
0
1
0
0
1
0
⎤ ⎡
1 1
11 0
1
2
−7 0
0 ⎦∼⎣ 2 1
4 2 −13 1
0
⎤
−1 1
1 0 ⎦
0 0
⎤ ⎡
⎤ ⎡
⎤
−1
1
1
2 −1
1
1
11 0 −1
1 0 −18 0
4
5
1
29 0 −3
29 0 −3
3 −2 ⎦ ∼ ⎣ 0
2 ⎦∼⎣ 0 1
2 ⎦
0 −2 −57 1
0 0
1 1 −2
4 −4
4 −4
0
⎡
⎤
1 0 18 −34 −1
6
∼ ⎣ 0 1 0 −29 55 2 ⎦ = [I3 |A−1 ].
0 0 1
1 −2 0
Thus,
⎡
⎤
18 −34 −1
55
2 ⎦.
A−1 = ⎣ −29
1 −2
0
1. P13
2. A21 (−1)
3. A12 (−2), A13 (−4)
5. A21 (−1), A23 (2)
4. M2 (−1)
6. A31 (18), A32 (−29)
13. We have
⎡
1 2 −3 1
[A|I3 ] = ⎣ 2 6 −2 0
−1 1
4 0
0
1
0
⎤ ⎡
⎤ ⎡
1
1 2 −3
0
1 0 0
1 2 −3
1
2
4 −2 1 0 ⎦ ∼ ⎣ 0 1
0 ⎦∼⎣ 0 2
2 −1
0 3
1
1
1 0 1
0 3
1
1
(c)2017 Pearson Education. Inc.
0
1
2
0
⎤
0
0 ⎦
1
172
⎤ ⎡
⎤
3 −1 0
3 −1
0
1 0 −7
1 0 −7
4
1
1
0 ⎦∼⎣ 0 1
0 ⎦
2 −1
2 −1
∼⎣ 0 1
2
2
3
3
4
1
4 −2 1
0 0 −5
0 0
1 − 5 10 − 5
⎤
⎡
11
13
7
−5
1 0 0 −5
10
5
2 ⎦
3
1
−
∼⎣ 0 1 0
= [I3 |A−1 ].
5
10
5
3
4
1
0 0 1 −5
−5
10
⎡
3
Thus,
⎡
A−1 = ⎣
1. A12 (−2), A13 (1)
14. We have
⎡
2. M2 ( 12 )
− 13
5
3
5
4
−5
11
10
1
− 10
3
10
− 75
2
5
− 15
⎤
⎦.
4. M3 (− 15 )
3. A21 (−2), A23 (−3)
5. A31 (7), A32 (−2)
⎤ ⎡
⎤ ⎡
i
2 1 0 0
1
0 0
1
1
i
2
1 i
2
1
2
−1 2i 0 1 0 ⎦ ∼ ⎣ 0 −i −2 −1 − i 1 0 ⎦ ∼ ⎣ 0 1 −2i 1 − i
2i 5 0 0 1
0
0
1
0 0
1 −2
−2
0 1
⎡
⎤ ⎡
⎤
−i
1 0
1 0
0 −i 1 0
1 0 0
3
4
∼ ⎣ 0 1 −2i 1 − i i 0 ⎦ ∼ ⎣ 0 1 0 1 − 5i i 2i ⎦ = [I3 |A−1 ].
0 0
1 −2 0 1
0 0 1
−2
0 1
1
[A|I3 ] = ⎣ 1 + i
2
Thus,
⎡
−i
A−1 = ⎣ 1 − 5i
−2
1. A12 (−1 − i), A13 (−2)
15. We have
⎡
2
1 3 1
[A|I3 ] = ⎣ 1 −1 2 0
3
3 4 0
0
1
0
⎤
0 0
i 0 ⎦
0 1
⎤
1 0
i 2i ⎦ .
0 1
2. M2 (i)
3. A21 (−i)
4. A32 (2i)
⎤ ⎡
⎤ ⎡
0
1 −1 2 0 1 0
1 −1
2 0
1
2
1 3 1 0 0 ⎦∼⎣ 0
3 −1 1
0 ⎦∼⎣ 2
3
3 4 0 0 1
0
6 −2 0
1
⎡
⎤
0
1 0
1 −1
2
3
3 −1
1 −2 0 ⎦
∼⎣ 0
0
0
0 −2
1 1
⎤
1 0
−2 0 ⎦
−3 1
Since 2 = rank(A) < rank(A# ) = 3, we know that A−1 does not exist (we have obtained a row of zeros in
the block matrix on the left.
1. P12
16. We have
⎡
1 −1 2
3
⎢ 2
0
3
−4
[A|I4 ] = ⎢
⎣ 3 −1 7
8
1
0 3
5
2. A12 (−2), A13 (−3)
1
0
0
0
0
1
0
0
0
0
1
0
3. A23 (−2)
⎤ ⎡
⎤
0
1 0 0 0
1 −1
2
3
2 −1 −10 −2 1 0 0 ⎥
0 ⎥
1 ⎢
⎢ 0
⎥∼
⎥
⎣
⎦
0
2
1 −1 −3 0 1 0 ⎦
0
0
1
1
2 −1 0 0 1
1
(c)2017 Pearson Education. Inc.
173
⎡
⎤ ⎡
⎤
1 0 0 0
0 0 0
1
1 −1
2
3
1 0
3
5
1
1
2 −1 0 0 1 ⎥
1
2 −1 0 0
1 ⎥
2 ⎢ 0
3 ⎢
⎢ 0 1
⎥∼
⎥
∼⎢
⎣ 0
⎣
⎦
2
1 −1 −3 0 1 0
0 0 −1 −5 −1 0 1 −2 ⎦
0
2 −1 −10 −2 1 0 0
0 0 −3 −14
0 1 0 −2
⎡
⎤ ⎡
⎤
1 0
3
5
1 0 0 −10 −3 0
0 0
0
1
3 −5
1
2 −1 0
0
1 ⎥
1 −1 ⎥
4 ⎢ 0 1
5 ⎢
⎢ 0 1 0 −3 −2 0
⎥∼
⎥
∼⎢
⎣ 0 0
⎣
⎦
1
5
0 0 1
5
1 0 −1
2
1 0 −1
2 ⎦
0 0 −3 −14
0 0 0
1
0 1
0 −2
3 1 −3
4
⎡
⎤
1 0 0 0
27 10 −27
35
⎥
0
1
0
0
7
3
−8
11
6 ⎢
⎥ = [I4 |A−1 ].
∼⎢
⎣ 0 0 1 0 −14 −5
14 −18 ⎦
0 0 0 1
3
1 −3
4
Thus,
⎤
27 10 −27
35
⎢
7
3 −8
11 ⎥
⎥.
A−1 = ⎢
⎣ −14 −5
14 −18 ⎦
3
1 −3
4
⎡
1. A12 (−2), A13 (−3), A14 (−1)
4. M3 (−1)
2. P13
5. A31 (−3), A32 (−1), A34 (3)
3. A21 (1), A23 (−2), A24 (−2)
6. A41 (10), A42 (3), A43 (5)
17. We have
⎡
0 −2 −1 −3 1
⎢ 2
0
2
1 0
[A|I4 ] = ⎢
⎣ 1 −2
0
2 0
3 −1 −2
0 0
0
1
0
0
0
0
1
0
⎤ ⎡
0
1 −2
⎢ 2
0
0 ⎥
1
⎥∼⎢
0 ⎦ ⎣ 0 −2
3 −1
1
⎡
0
2 0
2
1 0
−1 −3 1
−2
0 0
0
1
0
0
1
0
0
0
⎤
0
0 ⎥
⎥
0 ⎦
1
⎤ ⎡
⎤
1 0
1 −2
0
2 0 0
1 0
1 −2
0
2 0 0
1
4
2 −3 0 1 −2 0 ⎥
− 34 0 14 − 12 0 ⎥
1
2 ⎢ 0
3 ⎢
2
⎢ 0
⎥∼
⎥
∼⎢
⎣ 0 −2 −1 −3 1 0
0 0 ⎦ ⎣ 0 −2 −1 −3 1 0
0 0 ⎦
0
5 −2 −6 0 0 −3 1
0
5 −2 −6 0 0 −3 1
⎡
⎡
⎤
⎤
1
1
1
1
1 0
1
1 0
1
0
0 0
0
0 0
2
2
2
2
1
1
1
1
− 34 0
− 12 0 ⎥
− 34 0
− 12 0 ⎥
4 ⎢ 0 1
5 ⎢
2
4
2
4
⎢ 0 1
⎥∼
⎥
∼⎢
9
1
9
9
5
⎣ 0 0
−1 0 ⎦ ⎣ 0 0 − 2 − 4 0 − 4 − 12 1 ⎦
0 −2 1
2
1
−1 0
0 0 − 92 − 94 0 − 54 − 12 1
0 0
0 − 92 1
2
⎡
⎤ ⎡
⎤
2
1
1 0 1
1 0 0
0 0 29 − 19
0 12
0
0
2
9
1 ⎥
1
⎢
− 34 0 14 − 12
0 ⎥
0 19 − 59
6 ⎢ 0 1
2
9 ⎥
⎥ 7 ⎢ 0 1 0 −1
∼⎢
1
5
2 ⎦∼⎣
1
5
2 ⎦
1
1
⎣ 0 0 1
0 18
−9
0 0 1
0 18
−9
2
9
2
9
0 0 0 − 92 1 12 −1
0
0 0 0 − 92 1 12 −1
0
⎡
⎡
⎤
⎤
2
2
2
2
1 0 0
0
1 0 0 0
0
− 19
0
− 19
9
9
9
9
1 ⎥
1 ⎥
1
5
2
1
⎢
⎢
0 1 0 −1
0
−9
0 −3
8
9 ⎢ 0 1 0 0 −9
9
9 ⎥∼
9 ⎥ = [I |A−1 ].
∼⎢
4
1
1
1
5
2 ⎦
1
2 ⎦
⎣ 0 0 1
⎣
0 18
−9
0 0 1 0
0 −9
2
9
9
3
2
2
2
1
2
1
0 0 0
1 −9 −9
0
0 0 0 1 −9 −9
0
9
9
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Thus,
⎡
0
⎢ −2
9
A−1 = ⎢
⎣ 1
9
− 29
1. P13
5. P34
2. A12 (−2), A14 (−3)
6. M3 (− 29 )
2
9
− 19
0 − 13
1
0
3
2
9
− 19
3. M2 ( 14 )
2
9
1
9
2
−9
⎤
⎥
⎥.
⎦
0
4. A21 (2), A23 (2), A24 (−5)
7. A31 (−1), A32 (− 12 )
8. M4 (− 29 )
9. A42 (1), A43 (− 12 )
18. We have
⎡
1
⎢ 3
=⎢
⎣ 0
0
⎡
1
0
2 ⎢
⎢
∼⎣
0
0
⎡
1
0
4 ⎢
⎢
∼⎣
0
0
⎤ ⎡
1
2 0 0
0
⎢ 0 −2 0 0
0 ⎥
1
⎥∼⎢
0 ⎦ ⎣ 0
0 1 65
1
0
0 7 8
⎤ ⎡
2 0
0
1 0
0 0
1
0
−2 0
0 −3 1
0 0 ⎥
0 −2
3 ⎢
⎢
⎥
∼
6
1
0
0 0
0 ⎦ ⎣ 0
0 1
5
5
0
0
0 0 − 75 1
0 0 − 25
⎤
0 0 0 −2
1
0
0
3
− 12
0
0 ⎥
1 0 0
2
⎥ = [I4 |A−1 ].
0 1 0
0
0 −4 −3 ⎦
7
0
0
− 52
0 0 1
2
2 0 0 1
4 0 0 0
0 5 6 0
0 7 8 0
Thus,
0
1
0
0
0
0
1
0
⎡
⎢
A−1 = ⎢
⎣
1. A12 (−3), M3 ( 15 )
−2
3
2
1
− 12
0
0
0
0
2. A34 (−7)
0
0
−4
7
2
1 0
−3 1
0 0
0 0
0
0
1
5
0
⎤
0
0 ⎥
⎥
0 ⎦
1
⎤
0 0
0
0 −2 1
0
0 −3 1
0 0 ⎥
⎥
1
0
0 0 −4 3 ⎦
0 0 − 75 1
0 − 25
⎤
0
0 ⎥
⎥.
3 ⎦
− 52
3. A21 (1), A13 (3)
4. M2 (− 12 ), M4 (− 52 )
−1
19. To⎡determine the third
without determining the whole ⎡inverse, we solve the
⎤
⎤ ⎡ column
⎤
⎡vector
⎤ of A
−1 −2
3
x
0
−1 −2
3 0
1
1 0 ⎦
1
1 ⎦ ⎣ y ⎦ = ⎣ 0 ⎦. The corresponding augmented matrix ⎣ −1
system ⎣ −1
−1 −2 −1 1
−1 −2 −1
z
1
⎤
⎡
0
1 2 −3
5
0 ⎦. Thus, back substitution yields z = − 14 , y = − 16 , and x = − 12
.
can be row-reduced to ⎣ 0 1 − 23
1
0 0
1 −4
⎡
⎤
−5/12
Thus, the third column vector of A−1 is ⎣ −1/6 ⎦.
−1/4
−1
20. To determine
the second
vector
⎤ ⎡ column
⎤ ⎡
⎤ of A without determining the whole⎡ inverse, we solve
⎤ the
⎡
x
2 −1 4
0
2 −1 4 0
1 2 1 ⎦ can
1 2 ⎦ ⎣ y ⎦ = ⎣ 1 ⎦. The corresponding augmented matrix ⎣ 5
linear system ⎣ 5
1 −1 3
z
0
1 −1 3 0
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⎤
0
−1
3
1 −2
0 ⎦. Thus, back-substitution yields z = −1, y = −2, and x = 1. Thus,
0
1 −1
⎡
⎤
1
the second column vector of A−1 is ⎣ −2 ⎦.
−1
7
6 20
−8
−10
2
.
21. We have A =
, b =
, and the Gauss-Jordan method yields A−1 =
2 7
2
−1
3
Therefore, we have
7
−8
−48
−10
−1
2
x=A b=
=
.
2
14
−1
3
⎡
1
be row-reduced to ⎣ 0
0
Hence, we have x1 = −48 and x2 = 14.
1 3
1
−5
3
22. We have A =
, b =
, and the Gauss-Jordan method yields A−1 =
.
2 5
3
2 −1
Therefore, we have
−5
3
1
4
−1
=
.
x=A b=
2 −1
3
−1
So we have x1 = 4 and x2 = −1.
⎡
⎤
⎡
⎤
⎡
⎤
7
5 −3
1 1 −2
−2
1 ⎦.
1 ⎦, b = ⎣ 3 ⎦, and the Gauss-Jordan method yields A−1 = ⎣ −2 −1
23. We have A = ⎣ 0 1
2
2 −1
2 4 −3
1
Therefore, we have
⎡
⎤⎡
⎤ ⎡
⎤
7
5 −3
−2
−2
1 ⎦⎣ 3 ⎦ = ⎣ 2 ⎦.
x = A−1 b = ⎣ −2 −1
2
2 −1
1
1
Hence, we have x1 = −2, x2 = 2, and x3 = 1.
1
−2i
2
4i
1
24. We have A =
,b=
, and the Gauss-Jordan method yields A−1 = 2+8i
2−i
4i
−i
−2 + i
Therefore, we have
−1
x=A
1
b=
2 + 8i
4i
−2 + i
2i
1
2
−i
1
=
2 + 8i
2 + 8i
−4 + i
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.
.
Hence, we have x1 = 1 and x2 = −4+i
2+8i .
⎡
⎤
⎡
⎡
⎤
3 4 5
−79
1
25. We have A = ⎣ 2 10 1 ⎦, b = ⎣ 1 ⎦, and the Gauss-Jordan method yields A−1 = ⎣ 12
4 1 8
38
1
Therefore, we have
⎡
⎤⎡
⎤ ⎡
⎤
−79
27
46
1
−6
x = A−1 b = ⎣ 12 −4 −7 ⎦ ⎣ 1 ⎦ = ⎣ 1 ⎦ .
38 −13 −22
1
3
Hence, we have x1 = −6, x2 = 1, and x3 = 3.
2i
1
⎤
27
46
−4 −7 ⎦.
−13 −22
176
⎡
⎤
⎤
⎡
⎤
−1
3
5
1
2
12
1 ⎣
3
3 −3 ⎦.
2 −1 ⎦, b = ⎣ 24 ⎦, and the Gauss-Jordan method yields A−1 = 12
5 −3 −1
−1
1
−36
⎡
1
26. We have A = ⎣ 1
2
Therefore, we have
−1
x=A
⎡
⎤⎡
⎤ ⎡
⎤
−1
3
5
12
−10
1 ⎣
3
3 −3 ⎦ ⎣ 24 ⎦ = ⎣ 18 ⎦ .
b=
12
5 −3 −1
−36
2
Hence, x1 = −10, x2 = 18, and x3 = 2.
27. We have
0
−1
1
0
0 −1
1
0
(0)(0) + (1)(1)
(−1)(0) + (0)(1)
(0)(−1) + (1)(0)
(−1)(−1) + (0)(0)
1
0
0
1
√
3/2 √1/2
3/2 √
−1/2
3/2
3/2
−1/2
1/2
√
√
√
√
1
(1/2)(1/2)
( 3/2)(−1/2) + √
(1/2)( √3/2)
( 3/2)(√ 3/2) + √
=
=
0
(−1/2)( 3/2) + ( 3/2)(1/2) (−1/2)(−1/2) + ( 3/2)( 3/2)
0
1
T
AA =
=
=
= I2 ,
so AT = A−1 .
28. We have
√
AAT =
= I2 ,
so AT = A−1 .
29. We have
AAT =
=
cos α
− sin α
sin α
cos α
cos α
sin α
− sin α
cos α
cos2 α + sin2 α
(− sin α)(cos α) + (cos α)(sin α)
(cos α)(− sin α) + (sin α)(cos α)
(− sin α)2 + cos2 α
=
1
0
0
1
so AT = A−1 .
30. We have
⎡
⎤
⎤
1
−2x
2x2
1
2x
2x2
1
⎣ 2x 1 − 2x2 −2x ⎦
⎣ −2x 1 − 2x2 2x ⎦
AAT =
1 + 2x2
2
2x
2x2
2x
1
−2x
1
⎤
⎡
1 + 4x2 + 4x4
0
0
1
2
4
⎦ = I3 ,
⎣
0
1
+
4x
+
4x
0
=
1 + 4x2 + 4x4
2
4
0
0
1 + 4x + 4x
1
1 + 2x2
⎡
so AT = A−1 .
31. For part 2, we have
(B −1 A−1 )(AB) = B −1 (A−1 A)B = B −1 In B = B −1 B = In ,
and for part 3, we have
(A−1 )T AT = (AA−1 )T = InT = In .
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177
32. We prove this by induction on k, with k = 1 trivial and k = 2 proven in part 2 of Theorem 2.6.10.
Assuming the statement is true for a product involving k − 1 matrices, we may proceed as follows:
−1
(A1 A2 · · · Ak )−1 = ((A1 A2 · · · Ak−1 )Ak )−1 = A−1
k (A1 A2 · · · Ak−1 )
−1
−1 −1
−1 −1
−1 −1
= A−1
k (Ak−1 · · · A2 A1 ) = Ak Ak−1 · · · A2 A1 .
In the second equality, we have applied part 2 of Theorem 2.6.10 to the two matrices A1 A2 · · · Ak−1 and Ak ,
and in the third equality, we have assumed that the desired property is true for products of k − 1 matrices.
33. Since A is skew-symmetric, we know that AT = −A. We wish to show that (A−1 )T = −A−1 . We have
(A−1 )T = (AT )−1 = (−A)−1 = −(A−1 ),
which shows that A−1 is skew-symmetric. The first equality follows from part 3 of Theorem 2.6.10, and the
second equality results from the assumption that A−1 is skew-symmetric.
34. Since A is symmetric, we know that AT = A. We wish to show that (A−1 )T = A−1 . We have
(A−1 )T = (AT )−1 = A−1 ,
which shows that A−1 is symmetric. The first equality follows from part 3 of Theorem 2.6.10, and the second
equality results from the assumption that A is symmetric.
35. We have
(In − A3 )(In + A3 + A6 + A9 ) = In (In + A3 + A6 + A9 ) − A3 (In + A3 + A6 + A9 )
= In + A3 + A6 + A9 − A3 − A6 − A9 − A12 = In − A12 = In ,
where the last equality uses the assumption that A12 = 0. This calculation shows that In − A3 and In +
A3 + A6 + A9 are inverses of one another.
36. We have
(In − A)(In + A + A2 + A3 ) = In (In + A + A2 + A3 ) − A(In + A + A2 + A3 )
= In + A + A2 + A3 − A − A2 − A3 − A4 = In − A4 = In ,
where the last equality uses the assumption that A4 = 0. This calculation shows that In − A and In + A +
A2 + A3 are inverses of one another.
37. We claim that the inverse of A15 is B 9 . To verify this, use the fact that A5 B 3 = I to observe that
A15 B 9 = A5 (A5 (A5 B 3 )B 3 )B 3 = A5 (A5 IB 3 )B 3 = A5 (A5 B 3 )B 3 = A5 IB 3 = A5 B 3 = I.
This calculation shows that the inverse of A15 is B 9 .
38. We claim that the inverse of A9 is B −3 . To verify this, use the fact that A3 B −1 = I to observe that
A9 B −3 = A3 (A3 (A3 B −1 )B −1 )B −1 = A3 (A3 IB −1 )B −1 = A3 (A3 B −1 )B −1 = A3 IB −1 = A3 B −1 = I.
This calculation shows that the inverse of A9 is B −3 .
39. We have
B = BIn = B(AC) = (BA)C = In C = C.
40. YES. Since BA = In , we know that A−1 = B (see Theorem 2.6.12). Likewise, since CA = In , A−1 = C.
Since the inverse of A is unique, it must follow that B = C.
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41. We can simply compute
1
a22 −a12
a11
a11
a21
Δ −a21
Therefore,
a12
a22
a11
a21
1
a22 a11 − a12 a21
a22 a12 − a12 a22
=
Δ −a21 a11 + a11 a21 −a21 a12 + a11 a22
1 a11 a22 − a12 a21
0
1
=
=
0
a11 a22 − a12 a21
0
Δ
a12
a22
−1
1
=
Δ
a22
−a21
−a12
a11
0
1
= I2 .
.
42. Assume that A is an invertible matrix and that Axi = bi for i = 1, 2, . . . , p (where each bi is given).
Use elementary row operations on the augmented matrix of the system to obtain the equivalence
[A|b1 b2 b3 . . . bp ] ∼ [In |c1 c2 c3 . . . cp ].
The solutions to the system can be read from the last matrix: xi = ci for each i = 1, 2, . . . , p.
43. We have
⎡
1 −1
⎣ 2 −1
1
1
⎡
1 0
2
⎣
∼ 0 1
0 0
⎤ ⎡
1 −1 2
1
1 −1
1
4
1
1
2 3 ⎦∼⎣ 0
6 −1
0
2
5 2
⎤ ⎡
0
3
1
3
1 0
3
2 −1
4 −1 ⎦ ∼ ⎣ 0 1
1
0 0
0 −2
2
⎤
1 −1
2
1
2 −1
4 −1 ⎦
5 −2
6
0
⎤
0
9 −5
0
0 −1
8 −5 ⎦ .
1
0 −2
2
Hence,
x1 = (0, −1, 0),
x2 = (9, 8, −2),
1. A12 (−2), A13 (−1)
x3 = (−5, −5, 2).
2. A21 (1), A23 (−2)
3. A31 (−3), A32 (−2)
44.
(a). Let ei denote the ith column vector of the identity matrix Im , and consider the m linear systems of
equations
Axi = ei
for i = 1, 2, . . . , m. Since rank(A) = m and each ei is a column m-vector, it follows that
rank(A# ) = m = rank(A)
and so each of the systems Axi = ei above has a solution (Note that if m < n, then there will be an infinite
number of solutions). If we let B = [x1 , x2 , . . . , xm ], then
AB = A [x1 , x2 , . . . , xm ] = [Ax1 , Ax2 , . . . , Axm ] = [e1 , e2 , . . . , em ] = In .
⎡
⎤
a d
(b). A right inverse for A in this case is a 3 × 2 matrix ⎣ b e ⎦ such that
c f
a + 3b + c
d + 3e + f
1 0
=
.
2a + 7b + 4c 2d + 7e + 4f
0 1
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Thus, we must have
a + 3b + c = 1,
d + 3e + f = 0,
2a + 7b + 4c = 0,
2d + 7e + 4f = 1.
1 3 1 1
The first and third equation comprise a linear system with augmented matrix
for a, b, and
2 7 4 0
1
1 3 1
c. The row-echelon form of this augmented matrix is
. Setting c = t, we have b = −2 − 2t
0 1 2 −2
and
a = 7 + 5t. Next, the second and fourth equation above comprise a linear system with augmented
matrix
1 3 1 0
1 3 1 0
for d, e, and f . The row-echelon form of this augmented matrix is
. Setting
2 7 4 1
0 1 2 1
f = s, we have e = 1 −
⎡
⎤ 2s and d = −3 + 5s. Thus, right inverses of A are precisely the matrices of the form
7 + 5t −3 + 5s
⎣ −2 − 2t 1 − 2s ⎦.
t
s
Solutions to Section 2.7
True-False Review:
(a): TRUE. Since every elementary matrix corresponds to a (reversible) elementary row operation, the
reverse elementary row operation will correspond to an elementary matrix that is the inverse of the original
elementary matrix.
2 0
1 0
(b): FALSE. For instance, the matrices
and
are both elementary matrices, but their
0 1
0 2
2 0
product,
, is not.
0 2
(c): FALSE. Every invertible matrix can be expressed as a product of elementary matrices. Since every
elementary matrix is invertible and products of invertible matrices are invertible, any product of elementary
matrices must be an invertible matrix.
(d): TRUE. Performing an elementary row operation on a matrix does not alter its rank, and the matrix
EA is obtained from A by performing the elementary row operation associated with the elementary matrix
E. Therefore, A and EA have the same rank.
(e): FALSE. If Pij is a permutation matrix, then Pij2 = In , since permuting the ith and jth rows of In
twice yields In . Alternatively, we can observe that Pij2 = In from the fact that Pij−1 = Pij .
1 0
1 1
(f ): FALSE. For example, consider the elementary matrices E1 =
and E2 =
. Then we
0 7
0 1
1 1
1 7
have E1 E2 =
and E2 E1 =
.
0 7
0 7
⎤
⎡
⎤
⎡
1 3 0
1 0 0
(g): FALSE. For example, consider the elementary matrices E1 = ⎣ 0 1 0 ⎦ and E2 = ⎣ 0 1 2 ⎦.
0 0 1
0 0 1
⎡
⎤
⎡
⎤
1 3 6
1 3 0
Then we have E1 E2 = ⎣ 0 1 2 ⎦ and E2 E1 = ⎣ 0 1 2 ⎦.
0 0 1
0 0 1
(h): FALSE. The only matrices we perform an LU factorization for are invertible matrices for which the
reduction to upper triangular form can be accomplished without permuting rows.
(c)2017 Pearson Education. Inc.
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(i): FALSE. The matrix U need not be a unit upper triangular matrix.
(j): FALSE. As can be seen in Example 2.7.8, a 4 × 4 matrix with LU factorization will have 6 multipliers,
not 10 multipliers.
Problems:
1.
⎡
⎤
⎡
0 1 0
0
Permutation Matrices: P12 = ⎣ 1 0 0 ⎦ , P13 = ⎣ 0
0 0 1
1
⎡
⎤
⎡
k 0 0
1
Scaling Matrices: M1 (k) = ⎣ 0 1 0 ⎦ , M2 (k) = ⎣ 0
0 0 1
0
Row Combinations:
⎡
⎤
0
0 ⎦,
1
⎤
0
0 ⎦,
1
1 0
A12 (k) = ⎣ k 1
0 0
⎡
1 k
A21 (k) = ⎣ 0 1
0 0
⎤
⎡
⎤
1
1 0 0
0 ⎦ , P23 = ⎣ 0 0 1 ⎦ .
0
0 1 0
⎤
⎡
⎤
0 0
1 0 0
k 0 ⎦ , M3 (k) = ⎣ 0 1 0 ⎦.
0 1
0 0 k
0
1
0
⎡
⎤
0
0 ⎦,
1
⎤
k
0 ⎦,
1
1 0
A13 (k) = ⎣ 0 1
k 0
⎡
1 0
A31 (k) = ⎣ 0 1
0 0
2. We have
⎡
⎤ ⎡
⎤ ⎡
⎤ ⎡
−1 −8
1 8
1
−4 −1
1
2
3
⎣ 0
3 ⎦∼⎣ 0 3 ⎦∼⎣ 0
3 ⎦∼⎣ 0
−3
7
−3 7
0
−3
7
1. A31 (−1)
2. M1 (−1)
3. A13 (3)
⎡
⎤
1 0 0
A23 (k) = ⎣ 0 1 0 ⎦ ,
0 k 1
⎡
⎤
1 0 0
A32 (k) = ⎣ 0 1 k ⎦ .
0 0 1
⎤ ⎡
8
1
4
3 ⎦∼⎣ 0
31
0
4. M2 ( 13 )
⎤ ⎡
8
1
5
1 ⎦∼⎣ 0
31
0
⎤
8
1 ⎦.
0
5. A23 (−31)
Elementary Matrices: A23 (31), M2 ( 13 ), A13 (3), M1 (−1), A31 (−1).
3. We have
3
1
5
−2
1
∼
1 −2
3
5
1. P12
2
∼
1 −2
0 11
2. A12 (−3)
3
∼
1 −2
0
1
.
1
3. M2 ( 11
)
1
Elementary Matrices: M2 ( 11
), A12 (−3), P12 .
4. We have
5
1
8
2
3 −1
1
∼
1
5
3 −1
8
2
1. P12
2
∼
1
0
2. A12 (−5)
3 −1
−7
7
3
∼
1
0
3 −1
1 −1
.
3. M2 (− 17 )
Elementary Matrices: M2 (− 17 ), A12 (−5), P12 .
5. We have
⎡
⎤ ⎡
⎤ ⎡
⎤ ⎡
⎤ ⎡
1 3
3 −1 4
1
3 2
1
3
2
1
3
2
1
2
3
4
⎣ 2
1 3 ⎦∼⎣ 2
1 3 ⎦ ∼ ⎣ 0 −5 −1 ⎦ ∼ ⎣ 0 −5 −1 ⎦ ∼ ⎣ 0 1
1
3 2
3 −1 4
0 −10 −2
0
0
0
0 0
(c)2017 Pearson Education. Inc.
2
1
5
0
⎤
⎦.
181
1. P13
2. A12 (−2), A13 (−3)
4. M2 (− 15 )
3. A23 (−2)
Elementary Matrices: M2 (− 15 ), A23 (−2), A13 (−3), A12 (−2), P13 .
6. We have
⎡
1
⎣ 2
3
2
3
4
3
4
5
⎤ ⎡
⎤ ⎡
⎤ ⎡
4
1
2
3
4
1
2
3
4
1
1
2
3
5 ⎦ ∼ ⎣ 0 −1 −2 −3 ⎦ ∼ ⎣ 0
1
2
3 ⎦∼⎣ 0
6
0 −2 −4 −6
0 −2 −4 −6
0
1. A12 (−2), A13 (−3)
2. M2 (−1)
2
1
0
⎤
4
3 ⎦.
0
3
2
0
3. A23 (2)
Elementary Matrices: A23 (2), M2 (−1), A13 (−3), A12 (−2).
7. We reduce A to the identity matrix:
1
1
2
3
1
∼
1
0
2
1
1. A12 (−1)
2
∼
1
0
0
1
.
2. A21 (−2)
1
−1
The elementary matrices corresponding to these row operations are E1 =
We have E2 E1 A = I2 , so that
A = E1−1 E2−1 =
1
1
0
1
1
0
2
1
0
1
and E2 =
1 −2
0
1
,
which is the desired expression since E1−1 and E2−1 are elementary matrices.
8. We reduce A to the identity matrix:
−2 −3 1 −2 −3 2
1
1 3 1
1 4 1
∼
∼
∼
∼
5
7
1
1
−2 −3
0 −1
0
1. A12 (2)
2. P12
3. A12 (2)
4. A21 (1)
0
1
0
1
1
0
1
−2
0
1
5
∼
1
0
0
1
1
1
,
1 −1
0
1
E5 =
1
0
1
0
0
−1
9. We reduce A to the identity matrix:
3 −4 1 −1
2 2 1 −2 3 1 −2 4 1 −2 5 1
∼
∼
∼
∼
∼
−1
2
3 −4
3 −4
0
2
0
1
0
0
1
A = E1−1 E2−1 E3−1 E4−1 E5−1 =
1
−2
.
5. M2 (−1)
The elementary matrices corresponding to these row operations are
1 0
0 1
1 0
1
E1 =
, E2 =
, E3 =
, E4 =
2 1
1 0
2 1
0
We have E5 E4 E3 E2 E1 A = I2 , so
0
−1
0
−1
,
which is the desired expression since each Ei−1 is an elementary matrix.
(c)2017 Pearson Education. Inc.
.
.
.
182
1. P12
2. M1 (−1)
4. M2 ( 12 )
3. A12 (−3)
5. A21 (2)
The elementary matrices corresponding to these row operations are
1
0 1
−1 0
1 0
, E2 =
, E3 =
, E4 =
E1 =
1 0
0 1
−3 1
0
We have E5 E4 E3 E2 E1 A = I2 , so
A = E1−1 E2−1 E3−1 E4−1 E5−1 =
0
1
1
0
−1
0
0
1
1
3
0
1
0
1
2
1
0
,
0
2
1
0
E5 =
1 −2
0
1
,
which is the desired expression since each Ei−1 is an elementary matrix.
10. We reduce A to the identity matrix:
4 2 1
4 −5 1 1
∼
∼
4 −5
0
1
4
1. P12
4
−21
3
∼
1
0
1
3. M2 (− 21
)
2. A12 (−4)
4
1
A = E1−1 E2−1 E3−1 E4−1 =
0
1
1
0
1
4
0
1
∼
1
0
0
1
.
4. A21 (−4)
The elementary matrices corresponding to these row operations are
1
0
0 1
1 0
,
, E2 =
, E3 =
E1 =
1
1 0
−4 1
0 − 21
We have E4 E3 E2 E1 A = I2 , so
4
1
0
0
−21
E4 =
1
0
1 −4
0
1
4
1
.
,
which is the desired expression since each Ei−1 is an elementary matrix.
11. We reduce A to the identity matrix:
⎤ ⎡
⎤ ⎡
⎤
⎡
⎤ ⎡
1 −1 0
1 −1 0
1 −1 0
1 −1 0
1
2
3
⎣ 2
4 2 ⎦∼⎣ 0
4 2 ⎦∼⎣ 0
4 2 ⎦
2 2 ⎦∼⎣ 0
3
1 3
0
4 3
0
0 1
3
1 3
⎤ ⎡
⎡
⎤ ⎡
⎤
1 −1 0
1 −1 0
1 0 0
4
5
6
1 0 ⎦ ∼ ⎣ 0 1 0 ⎦.
1 12 ⎦ ∼ ⎣ 0
∼⎣ 0
0
0 1
0 0 1
0
0 1
1. A12 (−2)
2. A13 (−3)
3. A23 (−1)
4. M2 ( 41 )
5. A32 (− 12 )
6. A21 (1)
The elementary matrices corresponding to these row operations are
⎡
⎤
⎡
⎤
⎡
⎤
1 0 0
1 0 0
1
0 0
1 0 ⎦,
E1 = ⎣ −2 1 0 ⎦ , E2 = ⎣ 0 1 0 ⎦ , E3 = ⎣ 0
0 0 1
−3 0 1
0 −1 1
⎤
⎡
⎡
⎡
⎤
⎤
1 0 0
1 0
0
1 1 0
E4 = ⎣ 0 14 0 ⎦ , E5 = ⎣ 0 1 − 12 ⎦ , E6 = ⎣ 0 1 0 ⎦ .
0 0 1
0 0 1
0 0
1
(c)2017 Pearson Education. Inc.
2
1
.
183
We have E6 E5 E4 E3 E2 E1 A = I3 , so
A = E1−1 E2−1 E3−1 E4−1 E5−1 E6−1
⎡
⎤⎡
⎤⎡
1 0 0
1 0 0
1
= ⎣ 2 1 0 ⎦⎣ 0 1 0 ⎦⎣ 0
0 0 1
3 0 1
0
0
1
1
⎤⎡
0
1
0 ⎦⎣ 0
1
0
0
4
0
⎤⎡
1 0
0
0 ⎦⎣ 0 1
1
0 0
⎤
1 −1 0
1 ⎦⎣
0
1 0 ⎦,
2
0
0 1
1
0
⎤⎡
which is the desired expression since each Ei−1 is an elementary matrix.
12. We reduce A to the identity matrix:
⎡
⎤ ⎡
⎤ ⎡
⎤ ⎡
⎤
0 −4 −2
1 −1
3
1 −1
3
1 −1
3
1
2
3
⎣ 1 −1
3 ⎦ ∼ ⎣ 0 −4 −2 ⎦ ∼ ⎣ 0 −4 −2 ⎦ ∼ ⎣ 0 −4 −2 ⎦
−2
2
2
−2
2
2
0
0
8
0
0
1
⎤ ⎡
⎤
⎡
⎤ ⎡
⎤ ⎡
1 −1 3
1 −1 0
1 −1 0
1 0 0
4
5
6
7
1 0 ⎦ ∼ ⎣ 0 1 0 ⎦.
∼ ⎣ 0 −4 0 ⎦ ∼ ⎣ 0 −4 0 ⎦ ∼ ⎣ 0
0
0 1
0 0 1
0
0 1
0
0 1
1. P12
2. A13 (2)
3. M3 ( 18 )
4. A32 (2)
5. A31 (−3)
6. M2 (− 14 )
7. A21 (1)
The elementary matrices corresponding to these row operations are
⎤
⎡
⎤
⎡
⎤
⎡
⎡
1 0 0
0 1 0
1 0 0
1 0
E1 = ⎣ 1 0 0 ⎦ , E2 = ⎣ 0 1 0 ⎦ , E3 = ⎣ 0 1 0 ⎦ , E4 = ⎣ 0 1
0 0 1
2 0 1
0 0
0 0 18
⎤
⎤
⎡
⎡
⎡
⎤
1
0 0
1 0 −3
1 1 0
0 ⎦ , E6 = ⎣ 0 − 14 0 ⎦ , E7 = ⎣ 0 1 0 ⎦ .
E5 = ⎣ 0 1
0 0 1
0 0
1
0
0 1
⎤
0
2 ⎦,
1
We have E7 E6 E5 E4 E3 E2 E1 A = I3 , so
A = E1−1 E2−1 E3−1 E4−1 E5−1 E6−1 E7−1
⎡
⎤⎡
⎤⎡
0 1 0
1 0 0
1
= ⎣ 1 0 0 ⎦⎣ 0 1 0 ⎦⎣ 0
0 0 1
−2 0 1
0
0
1
0
⎤⎡
0
1
0 ⎦⎣ 0
8
0
⎤⎡
0
0
1
1 −2 ⎦ ⎣ 0
0
1
0
0
1
0
⎤⎡
⎤⎡
⎤
3
1
0 0
1 −1 0
0 ⎦ ⎣ 0 −4 0 ⎦ ⎣ 0
1 0 ⎦,
1
0
0 1
0
0 1
which is the desired expression since each Ei−1 is an elementary matrix.
13. We reduce A to the identity matrix:
⎡
⎤ ⎡
⎤ ⎡
⎤ ⎡
⎤
1 2 3
1 2 3
1
2
3
1
0
3
1
2
3
⎣ 0 8 0 ⎦∼
⎣ 0 1 0 ⎦∼
⎣ 0
1
0 ⎦∼⎣ 0
1
0 ⎦
3 4 5
3 4 5
0 −2 −4
0 −2 −4
⎡
⎤ ⎡
⎤ ⎡
⎤
1 0
3
1 0 3
1 0 0
4
5
6
0 ⎦ ∼ ⎣ 0 1 0 ⎦ ∼ ⎣ 0 1 0 ⎦.
∼⎣ 0 1
0 0 −4
0 0 1
0 0 1
1. M2 ( 18 )
2. A13 (−3)
3. A21 (−2)
4. A23 (2)
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5. M3 (− 14 )
6. A31 (−3)
184
The elementary matrices corresponding to these row operations are
⎤
⎡
⎡
⎤
⎡
1 0 0
1 0 0
1
E1 = ⎣ 0 18 0 ⎦ , E2 = ⎣ 0 1 0 ⎦ , E3 = ⎣ 0
−3 0 1
0
0 0 1
⎤
⎤
⎡
⎡
⎡
1 0
0
1 0 0
1
0 ⎦ , E6 = ⎣ 0
E4 = ⎣ 0 1 0 ⎦ , E5 = ⎣ 0 1
0 2 1
0
0 0 − 14
⎤
−2 0
1 0 ⎦,
0 1
⎤
0 −3
1
0 ⎦.
0
1
We have E6 E5 E4 E3 E2 E1 A = I3 , so
A = E1−1 E2−1 E3−1 E4−1 E5−1 E6−1
⎡
⎤⎡
⎤⎡
1 0 0
1 0 0
1
= ⎣ 0 8 0 ⎦⎣ 0 1 0 ⎦⎣ 0
0 0 1
3 0 1
0
2
1
0
⎤⎡
⎤⎡
0
1
0 0
1
0 ⎦⎣ 0
1 0 ⎦⎣ 0
1
0 −2 1
0
⎤⎡
0
0
1
1
0 ⎦⎣ 0
0 −4
0
0
1
0
⎤
3
0 ⎦,
1
which is the desired expression since each Ei−1 is an elementary matrix.
14. We reduce A to the identity matrix:
2 −1 1 1
3 2 1
∼
∼
1
3
2 −1
0
1. P12
3
−7
3
1
0
∼
3. M2 (− 17 )
2. A12 (−2)
3
1
4
∼
1
0
0
1
.
4. A21 (−3)
The elementary matrices corresponding to these row operations are
1
0
0 1
1 0
,
, E2 =
, E3 =
E1 =
1 0
−2 1
0 − 17
E4 =
1 −3
0
1
.
Direct multiplication verifies that E4 E3 E2 E1 A = I2 .
15. We have
3 −2
−1
5
1
∼
3 −2
0 13
3
= U.
1. A12 ( 13 )
Hence, E1 = A12 ( 13 ). Then Equation (2.7.3) reads L = E1−1 = A12 (− 13 ) =
(2.7.2):
LU =
16. We have
2
5
3
1
1
1
− 13
∼
2
0
Then
LU =
0
1
3
− 13
2
1
5
2
0
1
3 −2
0 13
3
3 −2
−1
5
=
= U =⇒ m21 =
2
3
0 − 13
2
=
(c)2017 Pearson Education. Inc.
3
1
1
− 13
0
1
. Verifying Equation
= A.
5
=⇒ L =
2
2
5
= A.
1
5
2
0
1
.
185
1. A12 (− 52 )
17. We have
3
5
1
2
1
∼
3
0
Then
LU =
1
1
3
1
5
3
0
1
5
= U =⇒ m21 = =⇒ L =
3
3
0
1
=
1
3
3
5
1
2
1
5
3
0
1
.
= A.
1. A12 (− 53 )
18. We have
⎡
⎤ ⎡
⎤ ⎡
⎤
3 −1 2
3 −1
2
3 −1
2
1
2
⎣ 6 −1 1 ⎦ ∼ ⎣ 0
1 −3 ⎦ ∼ ⎣ 0
1 −3 ⎦ = U =⇒ m21 = 2, m31 = −1, m32 = 4.
−3
5 2
0
4
4
0
0 16
Hence,
⎡
1
L=⎣ 2
−1
0
1
4
⎤
0
0 ⎦
1
⎡
and
1
LU = ⎣ 2
−1
0
1
4
⎤⎡
⎤ ⎡
⎤
0
3 −1
2
3 −1 2
0 ⎦⎣ 0
1 −3 ⎦ = ⎣ 6 −1 1 ⎦ = A.
1
0
0 16
−3
5 2
1. A12 (−2), A13 (1)
19. We have
⎡
⎤ ⎡
5
2
1
5
1
⎣ −10 −2
3 ⎦∼⎣ 0
15
2 −3
0
⎤ ⎡
2
1
5
2
2
5 ⎦∼⎣ 0
−4 −6
0
2
2
0
2. A23 (−4)
⎤
1
5 ⎦ = U =⇒ m21 = −2, m31 = 3, m32 = −2.
4
Hence,
⎡
1
L = ⎣ −2
3
⎤
0 0
1 0 ⎦
−2 1
⎡
and
1
LU = ⎣ −2
3
⎤⎡
0 0
5
1 0 ⎦⎣ 0
−2 1
0
1. A12 (2), A13 (−3)
2
2
0
⎤ ⎡
⎤
1
5
2
1
5 ⎦ = ⎣ −10 −2
3 ⎦ = A.
4
15
2 −3
2. A23 (2)
20. We have
⎤ ⎡
⎤ ⎡
⎤
⎤ ⎡
⎡
1 −1
2
3
1 −1
2
3
1 −1
2
3
1 −1 2
3
⎢ 2
2 −1 −10 ⎥
2 −1 −10 ⎥
2 −1 −10 ⎥
0 3 −4 ⎥
1 ⎢
2 ⎢
3 ⎢
⎥∼
⎥∼
⎥ = U.
⎥∼
⎢ 0
⎢ 0
⎢ 0
⎢
⎣ 3 −1 7
2
1 −1 ⎦ ⎣ 0
0
2
9 ⎦ ⎣ 0
0
2
9 ⎦
8 ⎦ ⎣ 0
0
4
2
2
0
0
4
22
0
0
0
4
1
3 4
5
1. A12 (−2), A13 (−3), A14 (−1)
2. A23 (−1), A24 (−2)
(c)2017 Pearson Education. Inc.
3. A34 (−2)
186
Hence,
m21 = 2,
m31 = 3, m41 = 1, m32 = 1, m42 = 2, m43 = 2.
Hence,
⎡
1
⎢ 2
L=⎢
⎣ 3
1
0
1
1
2
21. We have
⎡
2 −3
⎢ 4 −1
⎢
⎣ −8
2
6
1
0
0
1
2
⎤
0
0 ⎥
⎥
0 ⎦
1
⎡
1
⎢ 2
LU = ⎢
⎣ 3
1
and
0
1
1
2
⎤
⎤ ⎡
⎤⎡
1 −1 2
3
1 −1
2
3
0
⎢
⎢
0 3 −4 ⎥
2 −1 −10 ⎥
0 ⎥
⎥ = A.
⎥=⎢ 2
⎥⎢ 0
8 ⎦
0
2
9 ⎦ ⎣ 3 −1 7
0 ⎦⎣ 0
1
3 4
5
0
0
0
4
1
0
0
1
2
⎤ ⎡
⎤ ⎡
⎤ ⎡
⎤
1
2
2 −3
1
2
2 −3
1
2
2 −3
1
2
1
1 ⎥
5 −1 −3 ⎥
5 −1 −3 ⎥
5 −1 −3 ⎥
1 ⎢
2 ⎢
3 ⎢
⎥∼
⎥∼
⎥∼
⎥ = U.
⎢ 0
⎢ 0
⎢ 0
⎦
⎦
⎦
⎣
⎣
⎣
2 −5
0 −10
6
3
0
0
4 −3
0
0
4 −3 ⎦
5
2
0
10
2 −4
0
0
4
2
0
0
0
5
1. A12 (−2), A13 (4), A14 (−3)
2. A23 (2), A24 (−2)
3. A34 (−1)
Hence,
m31 = −4,
m21 = 2,
m41 = 3,
m32 = −2,
m42 = 2,
m43 = 1.
Hence,
⎤
⎤ ⎡
⎤⎡
⎤
⎡
2 −3 1
2
2 −3
1
2
1
0 0 0
1
0 0 0
⎢
⎢
⎢
⎢ 2
1 ⎥
5 −1 −3 ⎥
1 0 0 ⎥
1 0 0 ⎥
⎥ = A.
⎥ = ⎢ 4 −1 1
⎥⎢ 0
⎥ and LU = ⎢ 2
L=⎢
⎣ −4 −2 1 0 ⎦ ⎣ 0
⎣ −4 −2 1 0 ⎦
2 2 −5 ⎦
0
4 −3 ⎦ ⎣ −8
6
1 5
2
0
0
0
5
3
2 1 1
3
2 1 1
⎡
22. We have
1
2
2
3
1
∼
1
0
2
−1
= U =⇒ m21 = 2 =⇒ L =
1
2
0
1
.
1. A12 (−2)
We now solve the triangular systems Ly = b and U x = y. From Ly = b, we obtain y =
−11
U x = y yields x =
.
7
3
−7
. Then
23. We have
⎡
⎤ ⎡
⎤ ⎡
⎤
1 −3 5
1 −3
5
1 −3
5
1
2
⎣ 3
2 2 ⎦ ∼ ⎣ 0 11 −13 ⎦ ∼ ⎣ 0 11 −13 ⎦ = U =⇒ m21 = 3, m31 = 2, m32 = 1.
2
5 2
0 11 −8
0
0
5
1. A12 (−3), A13 (−2)
2. A23 (−1)
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187
⎡
⎤
1 0 0
Hence, L = ⎣ 3 1 0 ⎦. We now solve the triangular systems Ly = b and U x = y. From Ly = b, we
2 1 1
⎡
⎤
⎡
⎤
1
3
obtain y = ⎣ 2 ⎦. Then U x = y yields x = ⎣ −1 ⎦.
−5
−1
24. We have
⎡
⎤ ⎡
2
2 2
1
1
⎣ 0
⎣ 6 3 −1 ⎦ ∼
0
−4 2
2
⎤ ⎡
⎤
2
1
2
2
1
2
−3 −4 ⎦ ∼ ⎣ 0 −3 −4 ⎦ = U =⇒ m21 = 3, m31 = −2, m32 = −2.
0 −4
0
0 −4
1. A12 (−3), A13 (2)
2. A23 (2)
⎡
⎤
1
0 0
1 0 ⎦. We now solve the triangular systems Ly = b and U x = y. From Ly = b, we
Hence, L = ⎣ 3
−2 −2 1
⎡
⎤
⎡
⎤
1
−1/12
1/3 ⎦.
obtain y = ⎣ −3 ⎦. Then U x = y yields x = ⎣
−2
1/2
25. We have
⎤ ⎡
⎤ ⎡
⎤ ⎡
⎡
4
3
0 0
4
3
0 0
4
3 0
4 3
0 0
⎥ 2 ⎢ 0 −5
⎥ 3 ⎢ 0 −5 2
⎥ 1 ⎢ 0 −5
⎢ 8 1
2
0
2
0
2
0
⎥∼⎢
⎥∼⎢
⎥∼⎢
⎢
⎣ 0 5
5
3 6 ⎦ ⎣ 0
0
5 6 ⎦ ⎣ 0
0 5
3 6 ⎦ ⎣ 0
0
0 −5 7
0
0 −5 7
0
0 0
0 0 −5 7
1. A12 (−2)
2. A23 (1)
⎤
0
0 ⎥
⎥ = U.
6 ⎦
13
3. A34 (1)
⎤
1
0
0 0
⎢ 2
1
0 0 ⎥
⎥. We
The only nonzero multipliers are m21 = 2, m32 = −1, and m43 = −1. Hence, L = ⎢
⎣ 0 −1
1 0 ⎦
0⎤ −1 1
⎡0
2
⎢ −1 ⎥
⎥
now solve the triangular systems Ly = b and U x = y. From Ly = b, we obtain y = ⎢
⎣ −1 ⎦. Then U x = y
4
⎤
⎡
677/1300
⎢ −9/325 ⎥
⎥
yields x = ⎢
⎣ −37/65 ⎦.
4/13
⎡
26. We have
2 −1
−8
3
1
∼
2 −1
0 −1
= U =⇒ m21 = −4 =⇒ L =
1. A12 (4)
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1
−4
0
1
.
188
We now solve the triangular systems
Lyi = bi ,
U xi = yi
for i = 1, 2, 3. We have 3
−4
. Then U x1 = y1 =⇒ x1 =
;
Ly1 = b1 =⇒ y1 =
11 −11 2
−6.5
. Then U x2 = y2 =⇒ x2 =
;
Ly2 = b2 =⇒ y2 =
15
−15 5
−3
Ly3 = b3 =⇒ y3 =
. Then U x3 = y3 =⇒ x3 =
.
11
−11
27. We have
⎡
−1
⎣ 3
5
⎤ ⎡
4 2
−1 4
1
1 4 ⎦ ∼ ⎣ 0 13
−7 1
0 13
⎤ ⎡
2
−1
2
10 ⎦ ∼ ⎣ 0
11
0
1. A12 (3), A13 (5)
4
13
0
⎤
2
10 ⎦ = U.
1
2. A23 (−1)
Thus, m21 = −3, m31 = −5, and m32 = 1. We now solve the triangular systems
Lyi = bi ,
U xi = yi
for i = 1, 2, 3. We have ⎡
⎤
⎡
⎤
1
−29/13
Ly1 = e1 =⇒ y1 = ⎣ 3 ⎦. Then U x1 = y1 =⇒ x1 = ⎣ −17/13 ⎦;
2
2
⎡
⎤
⎡
⎤
0
18/13
Ly2 = e2 =⇒ y2 = ⎣ 1 ⎦. Then U x2 = y2 =⇒ x2 = ⎣ 11/13 ⎦;
−1
−1
⎡
⎤
⎡
⎤
0
−14/13
Ly3 = e3 =⇒ y3 = ⎣ 0 ⎦. Then U x3 = y3 =⇒ x3 = ⎣ −10/13 ⎦.
1
1
28. Observe that if Pi is an elementary permutation matrix, then Pi−1 = Pi = PiT . Therefore, we have
−1
T
. . . P2−1 P1−1 = PkT Pk−1
. . . P2T . . . P1T = (P1 P2 . . . Pk )T = P T .
P −1 = (P1 P2 . . . Pk )−1 = Pk−1 Pk−1
29.
(a). Let A be an invertible upper triangular matrix with inverse B. Therefore, we have AB = In . Write
A = [aij ] and B = [bij ]. We will show that bij = 0 for all i > j, which shows that B is upper triangular. We
have
n
,
aik bkj = δij .
k=1
Since A is upper triangular, aik = 0 whenever i > k. Therefore, we can reduce the above summation to
n
,
aik bij = δij .
k=i
Let i = n. Then the above summation reduces to ann bnj = δnj . If j = n, we have ann bnn = 1, so
ann = 0. For j < n, we have ann bnj = 0, and therefore bnj = 0 for all j < n.
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Next let i = n − 1. Then we have
an−1,n−1 bn−1,j + an−1,n bnj = δn−1,j .
Setting j = n−1 and using the fact that bn,n−1 = 0 by the above calculation, we obtain an−1,n−1 bn−1,n−1 = 1,
so an−1,n−1 = 0. For j < n − 1, we have an−1,n−1 bn−1,j = 0 so that bn−1,j = 0.
Next let i = n − 2. Then we have an−2,n−2 bn−2,j + an−2,n−1 bn−1,j + an−2,n bnj = δn−2,j . Setting j = n − 2
and using the fact that bn−1,n−2 = 0 and bn,n−2 = 0, we have an−2,n−2 bn−2,n−2 = 1, so that an−2,n−2 = 0.
For j < n − 2, we have an−2,n−2 bn−2,j = 0 so that bn−2,j = 0.
Proceeding in this way, we eventually show that bij = 0 for all i > j.
For an invertible lower triangular matrix A with inverse B, we can either modify the preceding argument,
or we can proceed more briefly as follows: Note that AT is an invertible upper triangular matrix with inverse
B T . By the preceding argument, B T is upper triangular. Therefore, B is lower triangular, as required.
(b). Let A be an invertible unit upper triangular matrix with inverse B. Use the notations from (a). By
(a), we know that B is upper triangular. We simply must show that bjj = 0 for all j. From ann bnn = 1
(see proof of (a)), we see that if ann = 1, then bnn = 1. Moreover, from an−1,n−1 bn−1,n−1 = 1, the fact
that an−1,n−1 = 1 proves that bn−1,n−1 = 1. Likewise, the fact that an−2,n−2 bn−2,n−2 = 1 implies that if
an−2,n−2 = 1, then bn−2,n−2 = 1. Continuing in this fashion, we prove that bjj = 1 for all j.
For the last part, if A is an invertible unit lower triangular matrix with inverse B, then AT is an invertible
unit upper triangular matrix with inverse B T , and by the preceding argument, B T is a unit upper triangular
matrix. This implies that B is a unit lower triangular matrix, as desired.
30.
(a). Since A is invertible, Corollary 2.6.13 implies that both L2 and U1 are invertible. Since L1 U1 = L2 U2 ,
−1
−1
we can left-multiply by L−1
to obtain L−1
2 and right-multiply by U1
2 L1 = U 2 U 1 .
is a unit lower triangular matrix and U1−1 is an upper triangular
(b). By Problem 29, we know that L−1
2
−1
matrix. Therefore, L2 L1 is a unit lower triangular matrix and U2 U1−1 is an upper triangular matrix. Since
−1
these two matrices are equal, we must have L−1
= In . Therefore, L1 = L2 and U1 = U2 .
2 L1 = In and U2 U1
31. The system Ax = b can be written as QRx = b. If we can solve Qy = b for y and then solve Rx = y
for x, then QRx = b as desired. Multiplying Qy = b by QT and using the fact that QT Q = In , we obtain
y = QT b. Therefore, Rx = y can be replaced by Rx = QT b. Therefore, to solve Ax = b, we first determine
y = QT b and then solve the upper triangular system Rx = QT b by back-substitution.
Solutions to Section 2.8
True-False Review:
(a): FALSE. According to the given information, part (c) of the Invertible Matrix Theorem fails, while
part (e) holds. This is impossible.
(b): TRUE. This holds by the equivalence of parts (d) and (f) of the Invertible Matrix Theorem.
(c): FALSE. Part (d) of the Invertible Matrix Theorem fails according to the given information, and
therefore part (b) also fails. Hence, the equation Ax = b does not have a unique solution. But it is not
valid⎡to conclude⎤that the equation
⎡
⎤ has infinitely many solutions; it could have no solutions. For instance, if
1 0 0
0
A = ⎣ 0 1 0 ⎦ and b = ⎣ 0 ⎦, there are no solutions to Ax = b, although rank(A) = 2.
0 0 0
1
(d): FALSE. An easy counterexample is the matrix 0n , which fails to be invertible even though it is upper
triangular. Since it fails to be invertible, it cannot e row-equivalent to In , by the Invertible Matrix Theorem.
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Problems:
1. Since A is an invertible matrix, the only solution to Ax = 0 is x = 0. However, if we assume that
AB = AC, then A(B − C) = 0. If xi denotes the ith column of B − C, then xi = 0 for each i. That is,
B − C = 0, or B = C, as required.
2. If rank(A) = n, then the augmented matrix A# for the system Ax = 0 can be reduced to REF such
that each column contains a pivot except for the right-most column of all-zeros. Solving the system by
back-substitution, we find that x = 0, as claimed.
3. Since Ax = 0 has only the trivial solution, REF(A) contains a pivot in every column. Therefore, the
linear system Ax = b can be solved by back-substitution for every b in Rn . Therefore, Ax = b does have a
solution.
Now suppose there are two solutions y and z to the system Ax = b. That is, Ay = b and Az = b.
Subtracting, we find
A(y − z) = 0,
and so by assumption, y − z = 0. That is, y = z. Therefore, there is only one solution to the linear system
Ax = b.
4. If A and B are each invertible matrices, then A and B can each be expressed as a product of elementary
matrices, say
A = E1 E2 . . . Ek
and
B = E1 E2 . . . El .
Then
AB = E1 E2 . . . Ek E1 E2 . . . El ,
so AB can be expressed as a product of elementary matrices. Thus, by the equivalence of (a) and (e) in the
Invertible Matrix Theorem, AB is invertible.
5. We are assuming that the equations Ax = 0 and Bx = 0 each have only the trivial solution x = 0. Now
consider the linear system
(AB)x = 0.
Viewing this equation as
A(Bx) = 0,
we conclude that Bx = 0. Thus, x = 0. Hence, the linear equation (AB)x = 0 has only the trivial solution.
Solutions to Section 2.9
Problems:
⎤
⎤ ⎡
⎤ ⎡
13 −1
−15
0
−2 −1
⎥
⎢
⎢ 4 −1 ⎥ ⎢ 10
10 ⎥
⎥ = ⎢ −6 −11 ⎥.
⎥−⎢
1. AT − 5B = ⎢
⎦
⎣
⎦
⎣
⎣ 2
−3
20 ⎦
5 −15
5
6 −5
0
5
6
0
⎡
⎤
−3
0
⎢
2
2 ⎥
⎥
2. C T B = −5 −6 3 1 ⎢
⎣ 1 −3 ⎦ = 6 −20 .
0
1
⎡
3. Since A is not a square matrix, it is not possible to compute A2 .
8 −16 −8 −24
−3 2
1 0
11 −18 −9 −24
4. −4A − B T =
−
=
.
4
4 −20
0
0 2 −3 1
4
2 −17
−1
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191
5. We have
−2
4 2 6
−1 −1 5 0
AB =
⎤
−3
0
⎢ 2
2 ⎥
⎥ = 16
⎢
⎣ 1 −3 ⎦
6
0
1
⎡
8
−17
.
Moreover,
tr(AB) = −1.
6. We have
(AC)(AC)T =
−2
26
−2
26
4 −52
−52 676
=
.
⎤
⎡
⎤
−24
48
24
72
12
0 ⎢ 24 −24 −56 −48 ⎥
⎢ −8 −8 ⎥ −2
4 2 6
⎥.
⎢
⎥
7. (−4B)A = ⎢
⎣ −4 12 ⎦ −1 −1 5 0 = ⎣ −4 −28
52 −24 ⎦
4
4 −20
0
0 −4
⎡
8. Using Problem 5, we find that
(AB)
−1
=
−1
16
8
6 −17
1
=−
320
−17 −8
−6 16
.
⎤
−5
⎢ −6 ⎥
⎥
⎢
⎣ 3 ⎦ = [71],
1
⎡
9. We have
−5 −6
CT C =
3
1
and
tr(C T C) = 71.
10.
(a). We have
AB =
1
2
2
5
3
7
⎡
⎤
3 b
3a − 5
⎣ −4 a ⎦ =
7a − 14
a b
2a + 4b
5a + 9b
.
In order for this product to equal I2 , we require
3a − 5 = 1,
7a − 14 = 0,
2a + 4b = 0,
5a + 9b = 1.
We quickly solve this for the unique solution: a = 2 and b = −1.
(b). We have
⎡
⎤
3 −1 1
2 ⎦
BA = ⎣ −4
2
2 −1
2
5
3
7
⎡
⎤
1
1
2
2
2 ⎦.
=⎣ 0
0 −1 −1
11. We compute the (i, j)-entry of each side of the equation. We will denote the entries of AT by aTij , which
equals aji . On the left side, note that the (i, j)-entry of (AB T )T is the same as the (j, i)-entry of AB T , and
(j, i)-entry of AB T =
n
,
k=0
ajk bTki =
n
,
ajk bik =
k=0
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n
,
k=0
bik aTkj ,
192
and the latter expression is the (i, j)-entry of BAT . Therefore, the (i, j)-entries of (AB T )T and BAT are
the same, as required.
12.
(a). The (i, j)-entry of A2 is
n
,
aik akj .
k=1
(b). Assume that A is symmetric. That means that AT = A. We claim that A2 is symmetric. To see this,
note that
(A2 )T = (AA)T = AT AT = AA = A2 .
Thus, (A2 )T = A2 , and so A2 is symmetric.
13. We are assuming that A is skew-symmetric, so AT = −A. To show that B T AB is skew-symmetric, we
observe that
(B T AB)T = B T AT (B T )T = B T AT B = B T (−A)B = −(B T AB),
as required.
14. We have
A2 =
3
9
−1 −3
2
=
0
0
0
0
,
so A is nilpotent.
15. We have
⎡
0
A2 = ⎣ 0
0
and
⎡
0
A3 = A2 A = ⎣ 0
0
0
0
0
⎤
1
0 ⎦
0
0
0
0
⎤⎡
1
0
0 ⎦⎣ 0
0
0
1
0
0
⎤ ⎡
1
0
1 ⎦=⎣ 0
0
0
0
0
0
⎤
0
0 ⎦,
0
so A is nilpotent.
16. We have
17. We have
⎡
−3e−3t
⎣
6t2
A (t) =
6/t
⎡
−7t
⎢ 6t − t2 /2
B(t) dt = ⎢
⎣ t + t2 /2
0
et
1
⎤
−2 sec2 t tan t
⎦.
− sin t
−5
⎡
⎤
t3 /3
−7
1
⎢ 11/2
3t4 /4 + 2t3 ⎥
⎥ = ⎢
2
⎣ 3/2
⎦
π sin(πt/2)
0
4
e−1
t − t /4
⎤
1/3
11/4 ⎥
⎥.
2/π ⎦
3/4
18. Since A(t) is 3 × 2 and B(t) is 4 × 2, it is impossible to perform the indicated subtraction.
19. Since A(t) is 3 × 2 and B(t) is 4 × 2, it is impossible to perform the indicated subtraction.
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20. From the last equation, we see that x3 = 0. Substituting this into the middle equation, we find that
x2 = 0.5. Finally, putting the values of x2 and x3 into the first equation, we find x1 = −6 − 2.5 = −8.5.
Thus, there is a unique solution to the linear system, and the solution set is
{(−8.5, 0.5, 0)}.
21. To solve this system, we need to reduce the corresponding augmented matrix for the linear system to
row-echelon form. This gives us
⎡
⎤ ⎡
⎤ ⎡
⎤ ⎡
5 −1
2
7
7
1 11 20
1 11 20 7
1 11
1
2
3
⎣ −2
6
9
9
0 ⎦ ∼ ⎣ −2 6
0 ⎦ ∼ ⎣ 0 28 49 14 ⎦ ∼ ⎣ 0 1
−7
5 −3 −7
−7 5 −3 −7
0 82 137 42
0 82
⎡
⎤ ⎡
⎤
7
1 11
20 7
1 11 20
4
5
7/4 1/2 ⎦ ∼ ⎣ 0 1 7/4
1/2 ⎦ .
∼⎣ 0 1
0 0 −13/2 1
0 0
1 −2/13
⎤
7
20
7/4 1/2 ⎦
137 42
From the
we conclude that x3 = −2/13, and using the middle row, we can solve for x2 : we have
row,
2last
10
10
2
x2 + 74 · − 13
= 12 , so x2 = 20
26 = 13 . Finally, from the first row we can get x1 : we have x1 +11· 13 +20· − 13 =
7, and so x1 = 21
13 . So there is a unique solution:
21 10
2
.
, ,−
13 13 13
1. A21 (2)
2. A12 (2), A13 (7)
3. M2 (1/28)
4. A23 (−82)
5. M3 (−2/13)
22. To solve this system, we need to reduce the corresponding augmented matrix for the linear system to
row-echelon form. This gives us
⎡
⎤ ⎡
⎤ ⎡
1 2 −1 1
1
2 −1 1
1
1
2
⎣ 1 0
1 5 ⎦ ∼ ⎣ 0 −2
2 4 ⎦∼⎣ 0
4 4
0 12
0 −4
4 8
0
⎤ ⎡
⎤
1
1
2 −1
1 2 −1
3
1 −1 −2 ⎦ ∼ ⎣ 0 1 −1 −2 ⎦ .
−4
4
0 0
0
8
0
From this row-echelon form, we see that z is a free variable. Set z = t. Then from the middle row of the
matrix, y = t − 2, and from the top row, x + 2(t − 2) − t = 1 or x = −t + 5. So the solution set is
{(−t + 5, t − 2, t) : t ∈ R} = {(5, −2, 0) + t(−1, 1, 1) : t ∈ R}.
1. A12 (−1), A13 (−4)
2. M2 (−1/2)
3. A23 (4)
23. To solve this system, we need to reduce the corresponding augmented matrix for the linear system to
row-echelon form. This gives us
⎡
⎤ ⎡
1 −2 −1
3 0
1 −2 −1
1
⎣ −2
4
5 −5 3 ⎦ ∼ ⎣ 0
0
3
3 −6 −6
8 2
0
0 −3
⎤ ⎡
3 0
1
2
1 3 ⎦∼⎣ 0
−1 2
0
⎤ ⎡
−2 −1 3 0
1
3
0
3 1 3 ⎦∼⎣ 0
0
0 0 5
0
The bottom row of this matrix shows that this system has no solutions.
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−2 −1
0
1
0
0
⎤
3 0
1/3 1 ⎦ .
0 1
194
1. A12 (2), A13 (−3)
2. A23 (1)
3. M2 (1/3), M3 (1/3)
24. To solve this system, we need to reduce the corresponding augmented matrix for the linear system to
row-echelon form. This gives us
⎤ ⎡
1
3
3
0 −1
2 −1
1
⎢ 1
⎥ 1 ⎢ 3
3
1
−3
2
0
−1
⎢
⎥∼⎢
⎣ 4 −2 −3
6 −1
5 ⎦ ⎣ 4 −2
0
0
0
1
4 −2
0
0
⎡
1
3
1 −3
2
0
−27
−12
33
−21
3 ⎢
∼⎢
⎣ 0
28
14 −36
18
0
0
0
1
4
⎡
⎤ ⎡
1
3
1 −3
2 −1
1
⎥ 6 ⎢ 0
0
1
2
−3
−3
−6
5 ⎢
⎥ ⎢
∼⎢
⎣ 0 −27 −12 33 −21 12 ⎦ ∼ ⎣ 0
0
0
0
1
4 −2
0
⎡
⎤ ⎡
⎤
1 −3
2 −1
1
3
1 −3
2 −1
−1
2 −1
−9 −4 11 −7
1 ⎥
4 ⎥
2 ⎢
⎢ 0
⎥∼
⎥
⎣
⎦
−3
6 −1
0 −14 −7 18 −9
5
9 ⎦
0
1
4 −2
0
0
0
1
4 −2
⎤ ⎡
⎤
1
3
1 −3
2 −1
−1
⎢ 0 −27 −12 33 −21 12 ⎥
12 ⎥
4
⎥∼⎢
⎥
1
2 −3 −3 −6 ⎦
−18 ⎦ ⎣ 0
0
0
0
1
4 −2
−2
⎤ ⎡
⎤
−1
1 3 1 −3
2
3 1
−3
2
−1
−3
−6 ⎥
1 2
−3
−3
−6 ⎥
7 ⎢
⎢ 0 1 2 −3
⎥∼
⎥
25 ⎦ .
0 42 −48 −102 −150 ⎦ ⎣ 0 0 1 − 87 − 17
−
7
7
0 0
1
4
−2
0 0 0
1
4
−2
We see that x5 = t is the only free variable. Back substitution yields the remaining values:
x5 = t,
So the solution set is
1. P12
x4 = −4t − 2,
x3 = −
41 15
− t,
7
7
2 33
x2 = − − t,
7
7
2 16
x1 = − + t.
7
7
2 16
2 33
41 15
− + t, − − t, − − t, −4t − 2, t : t ∈ R
7
7
7
7
7
7
2 2 41
16 33 15
, − , − , −4, 1 + − , − , − , −2, 0 : t ∈ R .
= t
7
7
7
7 7
7
2. A12 (−3), A13 (−4)
3. M2 (3), M3 (−2)
4. A23 (1)
5. P23
6. A23 (27)
7. M3 (1/42)
25. To solve this system, we need to reduce the corresponding augmented matrix for the linear system to
row-echelon form. This gives us
⎤ ⎡
⎤ ⎡
⎤
⎡
1 1
1
1 −3
1 1
1 1 −3 6
6
1 1 1 1 −3 6
⎢ 1 1 1 2 −5 8 ⎥ 1 ⎢ 0 0
0
1 −2
0 1 −2 2 ⎥
2 ⎥
2 ⎢
⎢ 0 0
⎥∼⎢
⎥∼
⎥
⎢
⎣ 2 3 1 4 −9 17 ⎦ ⎣ 0 1 −1
2 −3
5 ⎦ ⎣ 0 1 −1 2 −3 5 ⎦
2 2 2 3 −8 14
0 0
0 −1
2 −2
0 0
0 0
0 0
⎤
⎡
1 1
1 1 −3 6
⎢
0
1
−1
2 −3 5 ⎥
3
⎥.
∼⎢
⎣ 0 0
0 1 −2 2 ⎦
0 0
0 0
0 0
From this row-echelon form, we see that x5 = t and x3 = s are free variables. Furthermore, solving this
system by back-substitution, we see that
x5 = t,
x4 = 2t + 2,
x3 = s,
x2 = s − t + 1,
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x1 = 2t − 2s + 3.
195
So the solution set is
{(2t − 2s + 3, s − t + 1, s, 2t + 2, t) : s, t ∈ R} = {t(2, −1, 0, 2, 1) + s(−2, 1, 1, 0, 0) + (3, 1, 0, 2, 0) : s, t ∈ R}.
1. A12 (−1), A13 (−2), A14 (−2)
2. A24 (1)
3. P23
26. To solve this system, we need to reduce the corresponding augmented matrix for the linear system to
row-echelon form. This gives us
1 −3
2i
1
1 −3 2i
1 1 1
1
−3
2i
2
.
∼
∼
−2i
6 2 −2
0 6 − 6i −2 −2 + 2i
0
1 − 16 (1 + i) − 13
1. A12 (2i)
1
2. M2 ( 6−6i
)
From the last augmented matrix above, we see that x3 is a free variable. Let us set x3 = t, where t is
a complex number. Then we can solve for x2 using the equation corresponding to the second row of the
row-echelon form: x2 = − 13 + 16 (1+i)t. Finally, using the first row of the row-echelon form, we can determine
that x1 = 12 t(1 − 3i). Therefore, the solution set for this linear system of equations is
1
1 1
{( t(1 − 3i), − + (1 + i)t, t) : t ∈ C}.
2
3 6
27. We reduce the corresponding linear system as follows:
1 −k 6
1
−k
1
∼
2
3 k
0 3 + 2k
6
k − 12
.
If k = − 32 , then each column of the row-reduced coefficient matrix will contain a pivot, and hence, the linear
system will have a unique solution. If, on the other hand, k = − 32 , then the system is inconsistent, because
the last row of the row-echelon form will have a pivot in the right-most column. Under no circumstances
will the linear system have infinitely many solutions.
28. First observe that if k = 0, then the second equation requires that x3 = 2, and then the first equation
requires x2 = 2. However, x1 is a free variable in this case, so there are infinitely many solutions.
Now suppose that k = 0. Then multiplying each row of the corresponding augmented matrix for the
linear system by 1/k yields a row-echelon form with pivots in the first two columns only. Therefore, the
third variable, x3 , is free in this case. So once again, there are infinitely many solutions to the system.
We conclude that the system has infinitely many solutions for all values of k.
29. Since this linear system is homogeneous, it already has at least one solution: (0, 0, 0). Therefore, it only
remains to determine the values of k for which this will be the only solution. We reduce the corresponding
matrix as follows:
⎡
⎤ ⎡
⎤ ⎡
⎤
10 k −1 0
−k 0
10k k 2
1
1/2 −1/2 0
1
2
⎣ k 1 −1 0 ⎦ ∼ ⎣ 10k 10
−10 0 ⎦ ∼ ⎣ 10k 10
−10 0 ⎦
2
2 1 −1 0
1
1/2 −1/2 0
10k k
−k 0
⎡
⎤ ⎡
⎤ ⎡
⎤
1
1/2
−1/2 0
1
1/2
−1/2 0
1 1/2 −1/2 0
3
4
5
1
−1
1
−1
0 ⎦∼⎣ 0
0 ⎦.
∼ ⎣ 0 10 − 5k 5k − 10 0 ⎦ ∼ ⎣ 0
2
2
2
0 k − 5k
0 k − 5k
0
0 k −k 0
4k
0
4k
0
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196
1. M1 (k), M2 (10), M3 (1/2)
2. P13
3. A12 (−10k), A13 (−10k)
1
4. M2 ( 10−5k
)
5. A23 (5k − k 2 )
Note that the steps above are not valid if k = 0 or k = 2 (because Step 1 is not valid with k = 0 and Step
4 is not valid if k = 2). We will discuss those special cases individually in a moment. However if k = 0, 2,
then the steps are valid, and we see from the last row of the last matrix that if k = 1, we have infinitely
many solutions. Otherwise, if k = 0, 1, 2, then the matrix has full rank, and so there is a unique solution to
the linear system.
If k = 2, then the last two rows of the original matrix are the same, and so the matrix of coefficients of
the linear system is not invertible. Therefore, the linear system must have infinitely many solutions.
If k = 0, we reduce the original linear system as follows:
⎡
⎤ ⎡
10 0 −1 0
1 0 −1/10
1
⎣ 0 1 −1 0 ⎦ ∼
⎣ 0 1
−1
2 1 −1 0
2 1
−1
⎤ ⎡
⎤ ⎡
0
1 0 −1/10 0
1 0 −1/10
2
3
−1
−1
0 ⎦∼⎣ 0 1
0 ⎦∼⎣ 0 1
0 1 −4/5 0
0 0
1/5
0
⎤
0
0 ⎦.
0
The last matrix has full rank, so there will be a unique solution in this case.
1. M1 (1/10)
2. A13 (−2)
3. A23 (−1)
To summarize: The linear system has infinitely many solutions if and only if k = 1 or k = 2. Otherwise,
the system has a unique solution.
30. To solve this system, we need to reduce the corresponding augmented matrix for the linear system to
row-echelon form. This gives us
⎡
⎤ ⎡
⎤ ⎡
⎤ ⎡
⎤
0
0
0
1 −k k 2 0
1 −k
k2
1 −k
k2
1 −k
k2
1
2
3
⎣ 1
0
k 0 ⎦∼⎣ 0
k k − k2 0 ⎦ ∼ ⎣ 0
1
−1
1
−1
1 ⎦∼⎣ 0
1 ⎦.
2
2
0
1 −1 1
0
1
−1
0
k k−k 0
0
0 2k − k −k
1
1. A12 (−1)
2. P23
3. A23 (−k)
Now provided that 2k − k 2 = 0, the system can be solved without free variables via back-substitution, and
therefore, there is a unique solution. Consider now what happens if 2k − k 2 = 0. Then either k = 0 or k = 2.
If k = 0, then only the first two columns of the last augmented matrix above are pivoted, and we have a free
variable corresponding to x3 . Therefore, there are infinitely many solutions in this case. On the other hand,
if k = 2, then the last row of the last matrix above reflects an inconsistency in the linear system, and there
are no solutions.
To summarize, the system has no solutions if k = 2, a unique solution if k = 0 and k = 2, and infinitely
many solutions if k = 0.
31. No, there are no common points of intersection. A common point of intersection would be indicated by
a solution to the linear system consisting of the equations of the three planes. However, the corresponding
augmented matrix can be row-reduced as follows:
⎡
⎤ ⎡
⎤ ⎡
⎤
1 2
1 4
4
4
1 2
1
1 2
1
1
2
⎣ 0 1 −1 1 ⎦ ∼
⎣ 0 1 −1
1 ⎦ ∼ ⎣ 0 1 −1
1 ⎦.
1 3
0 0
0 1 −1 −4
0 0
0 −5
The last row of this matrix shows that the linear system is inconsistent, and so there are no points common
to all three planes.
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197
1. A13 (−1)
2. A23 (−1)
32.
(a). We have
4 7
−2 5
1
∼
1 7/4
−2
5
1. M1 (1/4)
2
∼
1 7/4
0 17/2
2. A12 (2)
3
1
0
∼
7/4
1
.
3. M2 (2/17)
(b). We have: rank(A) = 2, since the row-echelon form of A in (a) consists two nonzero rows.
(c). We have
4 7 1
−2 5 0
0
1
1
∼
1 7/4 1/4 0
−2
5
0 1
4
∼
1. M1 (1/4)
2
∼
7/4 1/4
17/2 1/2
1
0
1 0 5/34 −7/34
0 1 1/17
2/17
2. A12 (2)
Thus,
A
−1
=
7
− 34
3
∼
1 7/4 1/4
0
1 1/17
0
2/17
3. M2 (2/17)
5
34
1
17
0
1
.
4. A21 (−7/4)
.
2
17
33.
(a). We have
2 −7
−4 14
1
∼
2 −7
0
0
1. A12 (2)
2
∼
1 −7/2
0
0
.
2. M1 (1/2)
(b). We have: rank(A) = 1, since the row-echelon form of A in (a) has one nonzero row.
(c). Since rank(A) < 2, A is not invertible.
34.
(a). We have
⎡
⎤ ⎡
⎤ ⎡
⎤ ⎡
3 −1 6
1 −1/3 2
1 −1/3
2
1 −1/3
1
2
3
⎣ 0
2 3 ⎦∼⎣ 0
2
3 ⎦∼⎣ 0
2
3 ⎦∼⎣ 0
2
3 −5 0
1 −5/3 0
0 −4/3 −2
0
0
1. M1 (1/3), M3 (1/3)
2. A13 (−1)
3. A23 (2/3)
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⎤ ⎡
2
1 −1/3
4
3 ⎦∼⎣ 0
1
0
0
0
4. M2 (1/2)
⎤
2
3/2 ⎦ .
0
198
(b). We have: rank(A) = 2, since the row-echelon form of A in (a) consists of two nonzero rows.
(c). Since rank(A) < 3, A is not invertible.
35.
(a). We have
⎡
2
⎢ 1
⎢
⎣ 0
0
1
2
0
0
0
0
3
4
⎤ ⎡
1
0
⎢ 2
0 ⎥
1
⎥∼⎢
4 ⎦ ⎣ 0
0
3
⎡
2
1
0
0
0
0
3
4
1 2
0 1
∼⎣
0 0
0 0
4 ⎢
⎢
1. P12
⎤ ⎡
⎤ ⎡
⎤
0
1
2 0
0
1
2 0
0
0 ⎥
0 ⎥
0 ⎥
2 ⎢
3 ⎢
⎥∼
⎥∼
⎥
⎢ 0 −3 0
⎢ 0 −3 0
4 ⎦ ⎣ 0
0 3
4 ⎦ ⎣ 0
0 1 −1 ⎦
3
0
0 1 −1
0
0 3
4
⎤ ⎡
⎤
0
0
1 2 0
0
⎢ 0 1 0
0
0 ⎥
0 ⎥
5
⎥∼⎢
⎥.
1 −1 ⎦ ⎣ 0 0 1 −1 ⎦
0
7
0 0 0
1
2. A12 (−2), A34 (−1)
3. P34
4. M2 (−1/3), A34 (−3)
5. M4 (1/7)
(b). We have: rank(A) = 4, since the row-echelon form of A in (a) consists of four nonzero rows.
(c). We have
⎡
2 1 0 0
⎢ 1 2 0 0
⎢
⎣ 0 0 3 4
0 0 4 3
⎡
1
0
0
0
0
1
0
0
1
2
0 −3
∼⎣
0
0
0
0
3 ⎢
⎢
⎡
1 0 0
0
0 1 0
0
∼⎣
0 0 1 −1
0 0 0
1
5 ⎢
⎢
1. P12
⎤ ⎡
⎤ ⎡
0
1 2 0 0 0 1 0 0
1
2
⎢ 2 1 0 0 1 0 0 0 ⎥ 2 ⎢ 0 −3
0 ⎥
1
⎥∼⎢
⎥∼⎢
0
0 ⎦ ⎣ 0 0 3 4 0 0 1 0 ⎦ ⎣ 0
0 0 4 3 0 0 0 1
0
0
1
⎤ ⎡
0
0 0
1 2 0
0
1
0 0
0
0
0 1 −2
0 −1/3
0 0 ⎥
4 ⎢
⎢ 0 1 0
⎥∼
1 −1 0
0 −1 1 ⎦ ⎣ 0 0 1 −1
0
3
4 0
0 0 0
7
0
1 0
0
⎤ ⎡
2/3 −1/3 0
0
1 0 0 0 2/3
⎢ 0 1 0 0 −1/3
−1/3 2/3
0
0 ⎥
6
⎥∼⎢
0
0
−1
1 ⎦ ⎣ 0 0 1 0
0
0 0 0 1
0
0
4/7 −3/7
0
0
0
1
0
2. A12 (−2), A34 (−1)
Thus,
3. P34
⎡
4. A34 (−3), M2 (−1/3)
2/3
⎢ −1/3
−1
A =⎢
⎣ 0
0
−1/3
0
2/3
0
0
−3/7
0
4/7
⎤
1
0 0
−2
0 0 ⎥
⎥
0
1 0 ⎦
0 −1 1
⎤
1
0
0
2/3
0
0 ⎥
⎥
0 −1
1 ⎦
0
4 −3
⎤
−1/3
0
0
2/3
0
0 ⎥
⎥.
0
−3/7 4/7 ⎦
0
4/7 −3/7
0
0
0
0
3
4
1 −1
0
1
0
0
5. M4 (1/7), A21 (−2)
6. A43 (1)
⎤
0
0 ⎥
⎥.
4/7 ⎦
−3/7
36.
(a). We have
⎡
⎤ ⎡
3
0
0
1
1
⎣ 0
2 −1 ⎦ ∼ ⎣ 0
1 −1
2
1
⎤ ⎡
⎤ ⎡
⎤ ⎡
⎤ ⎡
0
0
1
0
0
1
0
0
1
0 0
1
2
3
4
5
2 −1 ⎦ ∼ ⎣ 0
2 −1 ⎦ ∼ ⎣ 0 −1
2 ⎦ ∼ ⎣ 0 −1 2 ⎦ ∼ ⎣ 0
−1
2
0 −1
2
0
2 −1
0
0 3
0
(c)2017 Pearson Education. Inc.
⎤
0
0
1 −2 ⎦ .
0
1
199
1. M1 (1/3)
2. A13 (−1)
3. P23
4. A23 (2)
5. M2 (−1), M3 (1/3)
(b). We have: rank(A) = 3, since the row-echelon form of A in (a) has 3 nonzero rows.
(c). We have
⎡
3
0
0 1
⎣ 0
2 −1 0
1 −1
2 0
0
1
0
⎤ ⎡
⎤ ⎡
⎤
0
1
0
0 1/3 0 0
1
0
0 1/3 0 0
1
2
2 −1 0
2 −1
0 ⎦∼⎣ 0
1 0 ⎦∼⎣ 0
0
1 0 ⎦
1 −1
2 0
0 −1
2 −1/3 0 1
1
0 1
⎡
⎤ ⎡
1
0
0 1/3 0 0
1
4
2 −1/3 0 1 ⎦ ∼ ⎣ 0
∼ ⎣ 0 −1
0
2 −1
0
0
1 0
⎡
⎤ ⎡
0
0
1 0
0 1/3
1
5
6
0
−1 ⎦ ∼ ⎣ 0
∼ ⎣ 0 1 −2 1/3
0 0
1 −2/9 1/3 2/3
0
3
1. M1 (1/3)
2. A13 (−1)
3. P23
Hence,
4. A23 (2)
⎡
1/3
0
A−1 = ⎣ −1/9 2/3
−2/9 1/3
⎤
0 0
0 1 ⎦
1 2
0 0 1/3
−1 2 −1/3
0 3 −2/3
0 0 1/3
1 0 −1/9
0 1 −2/9
0
2/3
1/3
⎤
0
1/3 ⎦ .
2/3
5. M2 (−1), M3 (1/3)
6. A32 (2)
⎤
0
1/3 ⎦ .
2/3
37.
(a). We have
⎤ ⎡
⎡
⎤ ⎡
1
4 2
1
−2 −3 1
1
2
⎣ 1
4 2 ⎦ ∼ ⎣ −2 −3 1 ⎦ ∼ ⎣ 0
0
5 3
0
0
5 3
1. P12
2. A12 (2)
4
5
5
⎤ ⎡
2
1
3
5 ⎦∼⎣ 0
3
0
3. A23 (−1)
⎤ ⎡
4
2
1
4
5
5 ⎦∼⎣ 0
0 −2
0
4
1
0
⎤
2
1 ⎦.
1
4. M2 (1/5), M3 (−1/2)
(b). We have: rank(A) = 3, since the row-echelon form of A in (a) consists of 3 nonzero rows.
(c). We have
⎡
⎤ ⎡
⎤ ⎡
⎤
−2 −3 1 1 0 0
1
4 2 0 1 0
1 4 2 0 1 0
1
2
⎣ 1
4 2 0 1 0 ⎦ ∼ ⎣ −2 −3 1 1 0 0 ⎦ ∼ ⎣ 0 5 5 1 2 0 ⎦
0
5 3 0 0 1
0
5 3 0 0 1
0 5 3 0 0 1
⎤ ⎡
⎤
⎡
0
1 0
1
0
1 4
2
1 4 2 0
3
4
5
1
2 0 ⎦ ∼ ⎣ 0 1 1 1/5 2/5
0 ⎦
∼⎣ 0 5
0 0 −2 −1 −2 1
0 0 1 1/2
1 −1/2
⎡
⎤ ⎡
⎤
0
1/5
7/5
−1
1 0 −2 −4/5 −3/5
1 0 0
5
6
1 1/5
2/5
0 ⎦ ∼ ⎣ 0 1 0 −3/10 −3/5 1/2 ⎦ .
∼⎣ 0 1
0 0
1 1/2
0 0 1
1
−1/2
1/2
1
−1/2
(c)2017 Pearson Education. Inc.
200
1. P12
2. A12 (2)
3. A23 (−1)
Thus,
4. M2 (1/5), M3 (−1/2)
5. A21 (−4)
6. A31 (2), A32 (−1)
⎡
⎤
1/5
7/5
−1
A−1 = ⎣ −3/10 −3/5 1/2 ⎦ .
1/2
1
−1/2
38. We use the Gauss-Jordan method to find A−1 :
⎡
⎤ ⎡
⎤ ⎡
⎤
1 0 0
1
0 0
1 −1 3 1 0 0
1 −1 3
1 −1
3
1
2
⎣ 4 −3 13 0 1 0 ⎦ ∼
⎣ 0
1 1 −4 1 0 ⎦ ∼ ⎣ 0
1
1 −4
1 0 ⎦
1
1 4 0 0 1
0
2 1 −1 0 1
0
0 −1
7 −2 1
⎡
⎤ ⎡
⎤ ⎡
⎤
1 0
0
0
4
1 −1 3
1 0 4 −3 1
1 0 0 25 −7
3
4
5
1 1 −4 1
0 ⎦ ∼ ⎣ 0 1 1 −4 1
0 ⎦∼⎣ 0 1 0
3 −1
1 ⎦.
∼⎣ 0
0
0 1 −7 2 −1
0 0 1 −7 2 −1
0 0 1 −7
2 −1
1. A12 (−4), A13 (−1)
2. A23 (−2)
Thus,
3. M3 (−1)
4. A21 (1)
5. A31 (−4), A32 (−1)
⎡
⎤
25 −7
4
1 ⎦.
A−1 = ⎣ 3 −1
−7
2 −1
Now xi = A−1 ei for each i. So
⎡
⎤
25
x1 = A−1 e1 = ⎣ 3 ⎦ ,
−7
⎡
⎤
−7
x2 = A−1 e2 = ⎣ −1 ⎦ ,
2
39. We have xi = A−1 bi , where
A−1 = −
1
39
−2 −5
−7
2
⎡
⎤
4
x3 = A−1 e3 = ⎣ 1 ⎦ .
−1
.
Therefore,
1
1
1
−2 −5
1
−12
12
4
=−
=
=
,
−7
2
2
−3
3
39
39
13 1
1
1
1
−2 −5
4
−23
23
−1
=−
=
,
x2 = A b2 = −
2
3
39 −7
39 −22
39 22
x1 = A−1 b1 = −
and
x3 = A−1 b3 = −
1
39
1
39
−2 −5
−7
2
−2
5
=−
1
39
−21
24
=
1
39
21
−24
40.
(a). We have
and
(A−1 B)(B −1 A) = A−1 (BB −1 )A = A−1 In A = A−1 A = In
(B −1 A)(A−1 B) = B −1 (AA−1 )B = B −1 In B = B −1 B = In .
(c)2017 Pearson Education. Inc.
=
1
13
7
−8
.
201
Therefore,
(B −1 A)−1 = A−1 B.
(b). We have
(A−1 B)−1 = B −1 (A−1 )−1 = B −1 A,
as required.
41(a). We have B 4 = (S −1 AS)(S −1 AS)(S −1 AS)(S −1 AS) = S −1 A(SS −1 )A(SS −1 )A(SS −1 )AS = S −1 AIAIAIAS =
S −1 A4 S, as required.
41(b). We can prove this by induction on k. For k = 1, the result is B = S −1 AS, which was already
given. Now assume that B k = S −1 Ak S. Then B k+1 = BB k = S −1 AS(S −1 Ak S) = S −1 A(SS −1 )Ak S =
S −1 AIAk S = S −1 Ak+1 S, which completes the induction step.
42.
(a). We reduce A to the identity matrix:
4 7
−2 5
1
∼
7
4
1
−2
7
4
17
2
1
0
∼
5
1. M1 ( 14 )
2
3
∼
1
0
2
3. M2 ( 17
)
2. A12 (2)
7
4
4
∼
1
1
0
0
1
.
4. A21 (− 74 )
The elementary matrices corresponding to these row operations are
1
1 0
1 0
0
4
, E2 =
E1 =
,
, E3 =
2
2 1
0 1
0 17
E4 =
1 − 74
0
1
.
We have E4 E3 E2 E1 A = I2 , so that
A = E1−1 E2−1 E3−1 E4−1 =
4
0
0
1
1
−2
0
1
1
0
0
17
2
1
0
7
4
1
,
which is the desired expression since Ei−1 is an elementary matrix for each i.
(b). We can reduce A to upper triangular form by the following elementary row operation:
4
−2
7
5
1
∼
4
0
7
.
17
2
1. A12 ( 12 )
Therefore we have the multiplier m12 = − 12 . Hence, setting
L=
1
− 12
0
1
and
U=
4
0
7
17
2
,
we have the LU factorization A = LU , which can be easily verified by direct multiplication.
43.
(c)2017 Pearson Education. Inc.
202
(a). We reduce A to the identity matrix:
⎤ ⎡
⎤ ⎡
⎤ ⎡
⎡
1 2 0 0
1
2 0 0
1
2 1 0 0
⎢ 1 2 0 0 ⎥ 1 ⎢ 2 1 0 0 ⎥ 2 ⎢ 0 −3 0 0 ⎥ 3 ⎢ 0
⎥ ⎢
⎥ ⎢
⎥∼⎢
⎢
⎣ 0 0 3 4 ⎦∼⎣ 0 0 3 4 ⎦∼⎣ 0
0 3 4 ⎦ ⎣ 0
0 0 4 3
0
0 0 4 3
0
0 4 3
⎡
1 0 0
0 1 0
5 ⎢
∼⎢
⎣ 0 0 1
0 0 4
⎤ ⎡
⎤ ⎡
1 0 0
0
0
1 0 0
⎢ 0 1 0
⎥ 7 ⎢ 0 1 0
0
0 ⎥
6
⎥ ⎢
⎥ ⎢
4 ⎦∼⎣
4 ⎦∼⎣
0 0 1
0 0 1
3
3
3
0 0 0
0 0 0 − 73
1. P12
2. A12 (−2)
6. A34 (−4)
3. M2 (− 13 )
2
1
0
0
⎤ ⎡
0
1
⎢
0 ⎥
4 ⎢ 0
⎥∼
4 ⎦ ⎣ 0
3
0
⎤ ⎡
0
1
⎢ 0
0 ⎥
8
⎥ ⎢
4 ⎦∼⎣
0
3
0
1
4. A21 (−2)
7. M4 (− 37 )
0
0
3
4
8. A43 (− 43 )
1 0
⎢ 0 1
E5 = ⎢
⎣ 0 0
0 0
0
0
1
3
0
⎤
0
0 ⎥
⎥,
0 ⎦
1
⎡
1
⎢ 0
E6 = ⎢
⎣ 0
0
⎤
0
0 0
1
0 0 ⎥
⎥,
0
1 0 ⎦
0 −4 1
⎡
1
⎢ 0
E7 = ⎢
⎣ 0
0
0
1
0
0
0
0
3
4
⎤
0
0 ⎥
⎥.
0 ⎦
1
0
0
1
0
⎤
0
0
0
0 ⎥
⎥,
1
0 ⎦
0 − 37
⎤
1 −2 0 0
⎢ 0
1 0 0 ⎥
⎥
E4 = ⎢
⎣ 0
0 1 0 ⎦
0
0 0 1
⎡
⎡
⎤
1 0 0
0
⎢ 0 1 0
0 ⎥
⎥
E8 = ⎢
⎣ 0 0 1 −4 ⎦ .
3
0 0 0
1
We have
E8 E7 E 6 E 5 E4 E3 E 2 E1 A = I 4
so that
A = E1−1 E2−1 E3−1 E4−1 E5−1 E6−1 E7−1 E8−1
⎤
⎤⎡
⎤⎡
⎤⎡
⎡
1 2 0 0
1
0 0 0
1 0 0 0
0 1 0 0
⎢ 1 0 0 0 ⎥ ⎢ 2 1 0 0 ⎥ ⎢ 0 −3 0 0 ⎥ ⎢ 0 1 0 0 ⎥
⎥···
⎥⎢
⎥⎢
⎥⎢
=⎢
⎣ 0 0 1 0 ⎦⎣ 0 0 1 0 ⎦⎣ 0
0 1 0 ⎦⎣ 0 0 1 0 ⎦
0 0 0 1
0
0 0 1
0 0 0 1
0 0 0 1
⎤⎡
⎤⎡
⎡
⎤⎡
1 0 0
0
1 0
1 0 0 0
1 0 0 0
⎢ 0 1 0 0 ⎥⎢ 0 1 0 0 ⎥⎢ 0 1 0
⎥⎢ 0 1
0
⎥⎢
⎥⎢
⎥⎢
···⎢
⎣ 0 0 3 0 ⎦⎣ 0 0 1 0 ⎦⎣ 0 0 1
0 ⎦⎣ 0 0
0 0 4 1
0 0 0 1
0 0 0 − 73
0 0
0
0
1
0
⎤
0
0 ⎥
⎥
4 ⎦,
3
1
which is the desired expression since Ei−1 is an elementary matrix for each i.
(b). We can reduce A to upper triangular form by the following elementary row operations:
⎤ ⎡
⎤ ⎡
⎡
⎤
2 1 0
0
2 1 0 0
2 1 0 0
⎢ 1 2 0 0 ⎥ 1 ⎢ 0 3 0 0 ⎥ 2 ⎢ 0 3 0
0 ⎥
2
2
⎥ ⎢
⎥ ⎢
⎢
⎥.
⎣ 0 0 3 4 ⎦∼⎣ 0 0 3 4 ⎦∼⎣ 0 0 3
4 ⎦
0 0 4 3
0 0 4 3
0 0 0 − 73
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⎤
0
0 ⎥
⎥
4 ⎦
3
5. M3 ( 13 )
The elementary matrices corresponding to these row operations are
⎤
⎤
⎤
⎡
⎡
⎡
1
0 0 0
0 1 0 0
1 0 0 0
1
⎥
⎥
⎢ 1 0 0 0 ⎥
⎢
⎢
⎥ , E2 = ⎢ −2 1 0 0 ⎥ , E3 = ⎢ 0 − 3 0 0 ⎥ ,
E1 = ⎢
⎣ 0 0 1 0 ⎦
⎣ 0 0 1 0 ⎦
⎣ 0
0 1 0 ⎦
0 0 0 1
0 0 0 1
0
0 0 1
⎡
0
1
0
0
0
1
0
0
203
1. A12 (− 12 )
2. A34 (− 43 )
Therefore, the nonzero multipliers are m12 = 12 and m34 = 43 . Hence, setting
⎡
⎤
⎤
⎡
1 0 0 0
2 1 0
0
⎢ 1 1 0 0 ⎥
⎢ 0 3 0
0 ⎥
2
2
⎥,
⎥
L=⎢
and U = ⎢
⎣ 0 0 1 0 ⎦
⎣ 0 0 3
4 ⎦
0 0 43 1
0 0 0 − 73
we have the LU factorization A = LU , which can be easily verified by direct multiplication.
44.
(a). We reduce A to the identity matrix:
⎤
⎤ ⎡
⎤ ⎡
⎤ ⎡
⎡
⎤ ⎡
1 −1
2
1 −1
2
1 −1
2
1 −1
2
3
0
0
4
1
2
3
⎣ 0
2 −1 ⎦ ∼ ⎣ 0
2 −1 ⎦ ∼ ⎣ 0
2 −1 ⎦ ∼ ⎣ 0
1 − 12 ⎦
1 − 12 ⎦ ∼ ⎣ 0
3
0
0
0
3 −6
1 −1
2
0
3 −6
0
0 − 92
⎤ ⎡
⎡
⎤
⎤ ⎡
⎤ ⎡
3
1 −1
2
1 0
0
1 0
1 0 0
2
5
6
7
8
1 − 12 ⎦ ∼ ⎣ 0 1 − 12 ⎦ ∼ ⎣ 0 1 − 12 ⎦ ∼ ⎣ 0 1 0 ⎦ .
∼⎣ 0
0 0 1
0
0
1
0 0
1
0 0
1
1. P13
2. A13 (−3)
6. A21 (1)
3. M2 ( 12 )
7. A31 (− 32 )
4. A23 (−3)
8. A32 ( 12 )
5. M3 (− 29 )
The elementary matrices corresponding to these row operations are
⎤
⎤
⎤
⎡
⎡
⎤
⎡
⎡
1 0 0
0 0 1
1 0 0
1
0 0
1 0 ⎦
E1 = ⎣ 0 1 0 ⎦ , E2 = ⎣ 0 1 0 ⎦ , E3 = ⎣ 0 12 0 ⎦ , E4 = ⎣ 0
1 0 0
0 −3 1
−3 0 1
0 0 1
⎤
⎤
⎡
⎡
⎤
⎡
⎡
⎤
3
1 0
0
1 0 0
1 1 0
1 0 −2
0 ⎦ , E6 = ⎣ 0 1 0 ⎦ , E7 = ⎣ 0 1
0 ⎦ , E8 = ⎣ 0 1 12 ⎦ .
E5 = ⎣ 0 1
2
0 0 1
0 0
1
0 0 1
0 0 −9
We have
E 8 E7 E6 E 5 E4 E3 E2 E 1 A = I 3
so that
A = E1−1 E2−1 E3−1 E4−1 E5−1 E6−1 E7−1 E8−1
⎡
⎤⎡
⎤⎡
⎤⎡
⎤
0 0 1
1 0 0
1 0 0
1 0 0
= ⎣ 0 1 0 ⎦⎣ 0 1 0 ⎦⎣ 0 2 0 ⎦⎣ 0 1 0 ⎦···
1 0 0
3 0 1
0 0 1
0 3 1
⎤⎡
⎤⎡
⎤⎡
⎤
⎡
1 0
0
1 0
0
1 −1 0
1 0 32
0 ⎦⎣ 0
1 0 ⎦ ⎣ 0 1 0 ⎦ ⎣ 0 1 − 12 ⎦ ,
···⎣ 0 1
9
0
0 1
0 0 1
0 0
1
0 0 −2
which is the desired expression since Ei−1 is an elementary matrix for each i.
(b). We can reduce A to upper triangular form by the following elementary row operations:
⎤ ⎡
⎡
⎤
⎤ ⎡
3 0
0
3
0
0
3
0
0
1
2
⎣ 0
2 −1 ⎦ ∼ ⎣ 0
2 −1 ⎦ ∼ ⎣ 0 2 −1 ⎦ .
3
1 −1
2
0 −1
2
0 0
2
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204
1. A13 (− 13 )
2. A23 ( 12 )
Therefore, the nonzero multipliers are m13 = 13 and m23 = − 12 . Hence, setting
⎡
1
L=⎣ 0
1
3
0
1
− 12
⎤
0
0 ⎦
1
⎤
0
0
2 −1 ⎦ ,
3
0
2
⎡
and
3
U =⎣ 0
0
we have the LU factorization A = LU , which can be verified by direct multiplication.
45.
(a). We reduce A to the identity matrix:
⎡
⎤ ⎡
⎤ ⎡
−2 −3 1
1
4 2
1
1
2
⎣ 1
4 2 ⎦ ∼ ⎣ −2 −3 1 ⎦ ∼ ⎣ 0
0
5 3
0
5 3
0
⎡
1
∼⎣ 0
0
5
⎤ ⎡
4
2
1
6
1 −8 ⎦ ∼ ⎣ 0
0 45
0
1. P12
⎤ ⎡
4
2
1
7
1 −8 ⎦ ∼ ⎣ 0
0
1
0
2. A12 (2)
1
6. M3 ( 45
)
⎤ ⎡
4
2
1
3
5
5 ⎦∼⎣ 0
5 −3
0
⎤ ⎡
4
2
1
4
5
5 ⎦∼⎣ 0
1 −8
0
⎤ ⎡
0 34
1
8
1 −8 ⎦ ∼ ⎣ 0
0
1
0
0
1
0
3. A23 (−1)
7. A21 (−4)
4. P23
8. A32 (8)
⎤
4
2
1 −8 ⎦
5
5
⎤ ⎡
34
1
9
0 ⎦∼⎣ 0
1
0
0
1
0
⎤
0
0 ⎦.
1
5. A23 (−5)
9. A31 (−34)
The elementary matrices corresponding to these row operations are
⎡
⎤
⎡
⎤
⎡
0 1 0
1 0 0
1
E1 = ⎣ 1 0 0 ⎦ , E2 = ⎣ 2 1 0 ⎦ , E3 = ⎣ 0
0
0 0 1
0 0 1
⎤
0 0
1 0 ⎦,
−1 1
⎤
⎤
⎡
⎤
⎡
1 0 0
1 0 0
1
0 0
1 0 ⎦ , E6 = ⎣ 0 1 0 ⎦ ,
E4 = ⎣ 0 0 1 ⎦ , E5 = ⎣ 0
1
0 1 0
0 −5 1
0 0 45
⎡
⎤
⎡
⎤
⎡
⎤
1 −4 0
1 0 0
1 0 −34
1 0 ⎦ , E8 = ⎣ 0 1 8 ⎦ , E9 = ⎣ 0 1
0 ⎦.
E7 = ⎣ 0
0
0 1
0 0 1
0 0
1
⎡
We have
E9 E8 E7 E 6 E5 E4 E3 E 2 E1 A = I 3
so that
A = E1−1 E2−1 E3−1 E4−1 E5−1 E6−1 E7−1 E8−1 E9−1
⎡
⎤⎡
⎤⎡
⎤⎡
⎤
0 1 0
1 0 0
1 0 0
1 0 0
= ⎣ 1 0 0 ⎦ ⎣ −2 1 0 ⎦ ⎣ 0 1 0 ⎦ ⎣ 0 0 1 ⎦ · · ·
0 0 1
0 0 1
0 1 1
0 1 0
⎤⎡
⎤⎡
⎤⎡
⎤⎡
⎡
1 0
0
1
1 4 0
1 0 0
1 0 0
· · · ⎣ 0 1 0 ⎦ ⎣ 0 1 0 ⎦ ⎣ 0 1 0 ⎦ ⎣ 0 1 −8 ⎦ ⎣ 0
0
0 0 1
0 0
1
0 0 45
0 5 1
(c)2017 Pearson Education. Inc.
0
1
0
⎤
34
0 ⎦,
1
205
which is the desired expression since Ei−1 is an elementary matrix for each i.
(b). We can reduce A to upper triangular form by the following elementary row operations:
⎤ ⎡
⎤
⎡
⎤ ⎡
−2 −3 1
−2 −3
1
−2 −3 1
1
5 ⎦ 2 ⎣
5 ⎦
5
5
⎣ 1
4 2 ⎦∼⎣ 0
0
∼
.
2
2
2
2
0
5 3
0
5 3
0
0 −2
Therefore, the nonzero multipliers are m12 = − 12 and m23 = 2. Hence, setting
⎤
⎤
⎡
⎡
−2 −3
1
1 0 0
5
5 ⎦
and U = ⎣ 0
L = ⎣ − 12 1 0 ⎦
,
2
2
0 2 1
0
0 −2
we have the LU factorization A = LU , which can be verified by direct multiplication.
46(a). Using the distributive laws of matrix multiplication, first note that
(A+2B)2 = (A+2B)(A+2B) = A(A+2B)+2B(A+2B) = A2 +A(2B)+(2B)A+(2B)2 = A2 +2AB+2BA+4B 2 .
Thus, we have
(A + 2B)3 = (A + 2B)(A + 2B)2
= A(A + 2B)2 + 2B(A + 2B)2
= A(A2 + 2AB + 2BA + 4B 2 ) + 2B(A2 + 2AB + 2BA + 4B 2 )
= A3 + 2A2 B + 2ABA + 4AB 2 + 2BA2 + 4BAB + 4B 2 A + 8B 3 ,
as needed.
46(b). Each occurrence of B in the answer to part (a) must now be accompanied by a minus sign. Therefore,
all terms containing an odd number of Bs will experience a sign change. The answer is
(A − 2B)3 = A3 − 2A2 B − 2ABA − 2BA2 + 4AB 2 + 4BAB + 4B 2 A − 8B 3 .
47.The answer is 2k , because each term in the expansion of (A + B)k consists of a string of k matrices,
each of which is either A or B (2 possibilities for each matrix in the string). Multiplying the possibilities
for each position in the string of length k, we get 2k different strings, and hence 2k different terms in
the expansion of (A + B)k . So, for instance, if k = 4, we expect 16 terms, corresponding to the 16 strings
AAAA, AAAB, AABA, ABAA, BAAA, AABB, ABAB, ABBA, BAAB, BABA, BBAA, ABBB, BABB,
BBAB, BBBA, and BBBB. Indeed, one can verify that the expansion of (A + B)4 is precisely the sum of
the 16 terms we just wrote down.
48. We claim that
To see this, simply note that
and
A
0
0 B −1
A−1
0
0
B
A
0
0 B −1
A−1
0
−1
=
0
B
A
0
0 B −1
A−1
0
=
=
0
B
In
0
0
Im
In
0
0
Im
(c)2017 Pearson Education. Inc.
.
= In+m
= In+m .
206
49. For a 2 × 4 matrix, the leading ones can occur in 6 different positions:
1 ∗ ∗
0 1 ∗
∗
∗
1
,
0
∗
0
∗ ∗
1 ∗
1 ∗ ∗ ∗
0
,
,
0 0 0 1
0
1 ∗ ∗
0 1 ∗
0
,
0
For a 3 × 4 matrix, the leading ones can occur in 4 different positions:
⎡
⎤ ⎡
⎤ ⎡
⎤ ⎡
1 ∗ ∗ ∗
1 ∗ ∗ ∗
1 ∗ ∗ ∗
0
⎣ 0 1 ∗ ∗ ⎦,⎣ 0 1 ∗ ∗ ⎦,⎣ 0 0 1 ∗ ⎦,⎣ 0
0 0 1 ∗
0 0 0 1
0 0 0 1
0
1 ∗ ∗
0 0 1
0
,
0
0
0
1 ∗
0 1
⎤
1 ∗ ∗
0 1 ∗ ⎦
0 0 1
For a 4 × 6 matrix, the leading ones can occur in 15 different positions:
⎡
1
⎢ 0
⎢
⎣ 0
0
⎡
1
⎢ 0
⎢
⎣ 0
0
⎡
1
⎢ 0
⎢
⎣ 0
0
∗ ∗
1 ∗
0 1
0 0
∗
∗
∗
1
∗ ∗ ∗
1 ∗ ∗
0 0 1
0 0 0
∗ ∗ ∗
0 1 ∗
0 0 0
0 0 0
⎡
⎤ ⎡
⎤ ⎡
1 ∗ ∗ ∗
1 ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
⎢ 0 1 ∗ ∗ ∗ ∗ ⎥ ⎢ 0 1 ∗ ∗
1 ∗ ∗ ∗ ∗ ⎥
⎥,⎢
⎥,⎢
0 1 ∗ ∗ ∗ ⎦ ⎣ 0 0 1 ∗ ∗ ∗ ⎦ ⎣ 0 0 0 1
0 0 0 0
0 0 0 0 0 1
0 0 0 1 ∗
⎤ ⎡
⎤ ⎡
1 ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
1 ∗ ∗ ∗
⎢ 0 0 1 ∗ ∗ ∗ ⎥ ⎢ 0 0 1 ∗
1 ∗ ∗ ∗ ∗ ⎥
⎥,⎢
⎥,⎢
0 0 0 1 ∗ ⎦ ⎣ 0 0 0 1 ∗ ∗ ⎦ ⎣ 0 0 0 1
0 0 0 0 1 ∗
0 0 0 0 1
0 0 0 0
⎤ ⎡
⎤ ⎡
0 1 ∗ ∗
0 1 ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
⎢ 0 0 1 ∗ ∗ ∗ ⎥ ⎢ 0 0 1 ∗
0 0 1 ∗ ∗ ⎥
⎥,⎢
⎥,⎢
0 0 0 1 ∗ ⎦ ⎣ 0 0 0 1 ∗ ∗ ⎦ ⎣ 0 0 0 1
0 0 0 0
0 0 0 0 1 ∗
0 0 0 0 1
⎤ ⎡
⎤
⎤ ⎡
∗ ∗ ∗ ∗
0 0 1 ∗ ∗ ∗
0 1 ∗ ∗ ∗ ∗
⎥
⎥ ⎢
⎢
1 ∗ ∗ ∗ ⎥
⎥,⎢ 0 0 0 1 ∗ ∗ ⎥,⎢ 0 0 0 1 ∗ ∗ ⎥
0 0 1 ∗ ⎦ ⎣ 0 0 0 0 1 ∗ ⎦ ⎣ 0 0 0 0 1 ∗ ⎦
0 0 0 1
0 0 0 0 0 1
0 0 0 0 0 1
⎤ ⎡
1
∗ ∗
⎢ 0
∗ ∗ ⎥
⎥,⎢
∗ ∗ ⎦ ⎣ 0
0
∗ ∗
⎤ ⎡
1
∗ ∗
⎢ 0
∗ ∗ ⎥
⎥,⎢
∗ ∗ ⎦ ⎣ 0
0
0 1
⎤ ⎡
1
∗ ∗
⎢ 0
∗ ∗ ⎥
⎥,⎢
1 ∗ ⎦ ⎣ 0
0
0 1
0
⎢ 0
⎢
⎣ 0
0
1
0
0
0
∗
∗
∗
1
∗
∗
∗
0
∗
∗
∗
0
⎤
∗
∗ ⎥
⎥,
∗ ⎦
∗
⎤
∗
∗ ⎥
⎥,
∗ ⎦
1
⎤
∗
∗ ⎥
⎥,
∗ ⎦
1
For an m × n matrix with m ≤ n, the answer is the binomial coefficient
n!
n
C(n, m) =
=
.
m
m!(n − m)!
This represents n “choose” m, which is the number of ways to choose m columns from the n columns of the
matrix in which to put the leading ones. This choice then determines the structure of the matrix.
50. We claim that the inverse of A10 is B 5 . To prove this, use the fact that A2 B = I to observe that
A10 B 5 = A2 A2 A2 A2 (A2 B)BBBB = A2 A2 A2 A2 IBBBB = A2 A2 A2 (A2 B)BBB
= A2 A2 A2 IBBB = A2 A2 (A2 B)BB = A2 A2 IBB = A2 (A2 B)B = A2 IB = A2 B = I,
as required.
51. We claim that the inverse of A9 is B 6 . To prove this, use the fact that A3 B 2 = I to observe that
A9 B 6 = A3 A3 (A3 B 2 )B 2 B 2 = A3 A3 IB 2 B 2 = A3 (A3 B 2 )B 2 = A3 IB 2 = A3 B 2 = I,
as required.
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207
Chapter 3 Solutions
Solutions to Section 3.1
True-False Review:
(a): TRUE. Let A =
a
b
0
c
. Then
det(A) = ac − b0 = ac,
which is the product of the elements on the main diagonal of A.
⎡
⎤
a b c
(b): TRUE. Let A = ⎣ 0 d e ⎦. Using the schematic of Figure 3.1.1, we have
0 0 f
det(A) = adf + be0 + c00 − 0dc − 0ea − f 0b = adf,
which is the product of the elements on the main diagonal of A.
(c): FALSE. The volume of this parallelepiped is determined by the absolute value of det(A), since det(A)
could very well be negative.
(d): TRUE. There are 12 of each. The ones of even parity are (1, 2, 3, 4), (1, 3, 4, 2), (1, 4, 2, 3), (2, 1, 4, 3),
(2, 4, 3, 1), (2, 3, 1, 4), (3, 1, 2, 4), (3, 2, 4, 1), (3, 4, 1, 2), (4, 1, 3, 2), (4, 2, 1, 3), (4, 3, 2, 1), and the others are all
of odd parity.
1 0
0 0
(e): FALSE. Many examples are possible here. If we take A =
and B =
, then det(A) =
0 0
0 1
det(B) = 0, but A + B = I2 , and det(I2 ) = 1 = 0.
1 2
(f ): FALSE. Many examples are possible here. If we take A =
, for example, then det(A) =
3 4
1 · 4 − 2 · 3 = −2 < 0, even though all elements of A are positive.
(g): TRUE. In the summation that defines the determinant, each term in the sum is a product consisting
of one element from each row and each column of the matrix. But that means one of the factors in each
term will be zero, since it must come from the row containing all zeros. Hence, each term is zero, and the
summation is zero.
(h): TRUE. If the determinant of the 3 × 3 matrix [v1 , v2 , v3 ] is zero, then the volume of the parallelepiped
determined by the three vectors is zero, and this means precisely that the three vectors all lie in the same
plane.
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(i): TRUE. Applying cofactor expansion to the first row of the given 4 × 4 matrix, we see that
a1
a3
0
0
a2
a4
0
0
0
0
b1
b3
0
a4
0
= a1 0
b2
0
b4
0
b1
b3
b1
b3
= a1 a4
0
a3
b2 − a 2 0
b4
0
0
b1
b3
b2
b
− a2 a3 1
b4
b3
b2
b4
= (a1 a4 − a2 a3 )
=
a1
a3
a2
b
· 1
a4
b3
b1
b3
0
b2
b4
b2
b4
b2
.
b4
(j): TRUE. Applying cofactor expansion to the first column of the given 4 × 4 matrix, we see that
0
0
b1
b3
0
0
b2
b4
a1
a3
0
0
a2
0
a4
= b1 0
0
b4
0
= b1 b4
a1
a3
0
a1
a3
a2
0
a 4 − b3 0
0
b2
a1
a3
0
a2
a
− b3 b2 1
a4
a3
a2
a4
= (b1 b4 − b3 b2 )
a1
a3
a2
a4
0
a2
= (b1 b4 − b3 b2 )(a1 a4 − a2 a3 )
a4
=
b1
b3
b2
a
· 1
b4
a3
a2
a4
=
a1
a3
a2
b
· 1
a4
b3
b2
.
b4
Problems:
1. σ(3, 1, 4, 2) = (−1)N (3,1,4,2) = (−1)3 = −1, odd.
2. σ(2, 4, 3, 1) = (−1)N (2,4,3,1) = (−1)4 = 1, even.
3. σ(5, 4, 3, 2, 1) = (−1)N (5,4,3,2,1) = (−1)10 = 1, even.
4. σ(2, 4, 1, 5, 3) = (−1)N (2,4,1,5,3) = (−1)4 = 1, even.
5. σ(6, 1, 4, 2, 5, 3) = (−1)N (6,1,4,2,5,3) = (−1)8 = 1, even.
6. σ(6, 5, 4, 3, 2, 1) = (−1)N (6,5,4,3,2,1) = (−1)15 = −1, odd.
a
a12 7. det(A) = 11
= σ(1, 2)a11 a22 + σ(2, 1)a12 a21 = a11 a22 − a12 a21 .
a21 a22 8. a11 a23 a34 a43 a52 . This is not a possible term of an order 5 determinant, since the column indices are not
distinct.
9. a11 a25 a33 a42 a54 . All row and column indices are distinct, so this is a term of an order 5 determinant.
Further, N (1, 5, 3, 2, 4) = 4, so that σ(1, 5, 3, 2, 4) = (−1)4 = +1.
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10. a11 a32 a24 a43 a55 . This is a possible term of an order 5 determinant.
N (1, 4, 2, 3, 5) = 2 =⇒ σ(1, 4, 2, 3, 5) = (−1)2 = +1.
11. a13 a25 a31 a44 a42 . This is not a possible term of an order 5 determinant, since the row indices are not
distinct.
12. a21 a3q ap2 a43 = ap2 a21 a3q a43 . We must choose p = 1 and q = 4.
N (2, 1, 4, 3) = 2 and σ(2, 1, 4, 3) = (−1)2 = +1.
13. a13 ap4 a32 a2q = a13 a2q a32 ap4 . We must choose p = 4 and q = 1 in order for the row and column indices
to be distinct. N (3, 1, 2, 4) = 2 so that σ(3, 1, 2, 4) = (−1)2 = +1.
14. apq a34 a13 a42 . We must choose p = 2 and q = 1.
N (3, 1, 4, 2) = 3 and σ(3, 1, 4, 2) = (−1)3 = −1.
15. a3q ap4 a13 a42 . We must choose p = 2 and q = 1.
N (3, 4, 1, 2) = 4 and σ(3, 4, 1, 2) = (−1)4 = +1.
0 −2
16. det
= (0)(1) − (−2)(5) = 10.
5
1
6 −3
17. det
= (6)(−1) − (−3)(−5) = −21.
−5 −1
−4 7
18. det
= (−4)(7) − (7)(1) = −35.
1 7
2 −3
19. det
= (2)(5) − (−3)(1) = 13.
1
5
9 −8
20. det
= (9)(−3) − (−8)(−7) = −83.
−7 −3
2 −4
21. det
= (2)(0) − (−4)(−1) = −4.
−1
0
1 4
22. det
= (1)(3) − (4)(−4) = 19.
−4 3
−3
3e10
e
= (e−3 )(6e8 ) − (3e10 )(2e−5 ) = 0.
23. det
2e−5 6e8
π π2 √
√
= 2π 2 − 2π 2 = (2 − 2)π 2 .
√
24. 2 2π 25: Using the schematic in Figure 3.1.1, we have
⎡
⎤
6 −1 2
7 1 ⎦ = (6)(7)(1) + (−1)(1)(0) + (2)(−4)(3) − (0)(7)(2) − (3)(1)(6) − (1)(−4)(−1)
det ⎣ −4
0
3 1
= 42 + 0 + (−24) − 0 − 18 − 4
= −4.
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26: Using the schematic in Figure 3.1.1, we have
⎡
⎤
5 −3
0
4 −1 ⎦ = (5)(4)(−2) + (−3)(−1)(−8) + (0)(1)(2) − (−8)(4)(0) − (2)(−1)(5) − (−2)(1)(−3)
det ⎣ 1
−8
2 −2
= −40 − 24 + 0 − 0 + 10 − 6
= −60.
27: Using the schematic in Figure 3.1.1, we have
⎡
⎤
−2 −4 1
1 1 ⎦ = (−2)(1)(3) + (−4)(1)(−2) + (1)(6)(−1) − (−2)(1)(1) − (−1)(1)(−2) − (3)(6)(−4)
det ⎣ 6
−2 −1 3
= −6 + 8 − 6 + 2 − 2 + 72
= 68.
28: Using the schematic in Figure 3.1.1, we have
⎡
⎤
0 0 −3
3 ⎦ = (0)(4)(5) + (0)(3)(−2) + (−3)(0)(1) − (−2)(4)(−3) − (1)(3)(0) − (5)(0)(0)
det ⎣ 0 4
−2 1
5
= 0 + 0 + 0 − 24 − 0 − 0
= −24.
29: Using the schematic in Figure 3.1.1, we have
⎡
⎤
9 1 −7
1 ⎦ = (9)(2)(−2) + (1)(1)(−4) + (−7)(6)(0) − (−4)(2)(−7) − (0)(1)(9) − (−2)(6)(1)
det ⎣ 6 2
−4 0 −2
= −36 − 4 + 0 − 56 − 0 + 12
= −84.
30: Using the schematic in Figure 3.1.1, we have
⎡
⎤
2 −10
3
1
1 ⎦ = (2)(1)(−3) + (−10)(1)(0) + (3)(1)(8) − (0)(1)(3) − (8)(1)(2) − (−3)(1)(−10)
det ⎣ 1
0
8 −3
= −6 + 0 + 24 − 0 − 16 − 30
= −28.
31: Using the schematic in Figure 3.1.1, we have
⎡
⎤
5
4 3
9 12 ⎦ = (5)(9)(0) + (4)(12)(1) + (3)(−2)(−1) − (1)(9)(3) − (−1)(12)(5) − (0)(−2)(4)
det ⎣ −2
1 −1 0
= 0 + 48 + 6 − 27 + 60 − 0
= 87.
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32: Using the schematic in Figure 3.1.1, we have
⎡
5
det ⎣ 0
2
0
3
0
⎤
4
0 ⎦ = (5)(3)(1) + (0)(0)(2) + (4)(0)(0) − (2)(3)(4) − (0)(0)(5) − (1)(0)(0)
1
= 15 + 0 + 0 − 24 − 0 − 0
= −9.
√
e2
e−1 √π
√
√
√
33. 67 1/30 2001 = ( π)(1/30)(π 3 )+(e2 )(2001)(π)+(e−1 )( 67)(π 2 )−(π)(1/30)(e−1 )−(π 2 )(2001)( π)−
π
π2
π3 √
3
2
(π )( 67)(e ) ≈ 9601.882.
2 3 −1 1 = (2)(4)(6) + (3)(1)(3) + (−1)(1)(1) − (3)(4)(−1) − (1)(1)(2) − (6)(1)(3) = 48.
34. 1 4
3 1
6 35. The definition of the determinant for this 4 × 4 matrix includes 24 terms, but based on the positions occupied by zeros in this matrix, only one of these 24 terms will be nonzero; namely, det(A) =
σ(4, 3, 2, 1)a14 a23 a32 a41 = (−1)6 (−3)(−7)(2)(1) = 42.
36. The definition of the determinant for this 4 × 4 matrix includes 24 terms, but based on the positions occupied by zeros in this matrix, only one of these 24 terms will be nonzero; namely, det(A) =
σ(1, 2, 3, 4)a11 a22 a33 a44 = (−1)0 (4)(−2)(−6)(−3) = −144.
37. The definition of the determinant for this 4 × 4 matrix includes 24 terms, but based on the positions
occupied by zeros in this matrix, many of these terms will be zero. The nonzero terms will correspond to
the permutations that we have enumerated in the table below, along with the number of inversions, the sign,
and the value of the term before adopting the sign:
Permutation
(1, 2, 4, 3)
(1, 3, 2, 4)
(1, 3, 4, 2)
(1, 4, 2, 3)
(3, 1, 2, 4)
(3, 1, 4, 2)
(3, 2, 1, 4)
(3, 4, 1, 2)
(4, 1, 2, 3)
(4, 2, 1, 3)
(4, 3, 1, 2)
Number of Inversions
1
1
2
2
2
3
3
4
3
4
5
Sign (+/−)
−
−
+
+
+
−
−
+
−
+
−
Value of Term
72
−4
30
−32
2
−15
−12
−40
24
−144
−60
Thus, det(A) = −72 − (−4) + 30 + (−32) + 2 − (−15) − (−12) + (−40) − 24 + (−144) − (−60) = −189.
38. The definition of the determinant for this 4 × 4 matrix includes 24 terms, but based on the positions
occupied by zeros in this matrix, many of these terms will be zero. The nonzero terms will correspond to
the permutations that we have enumerated in the table below, along with the number of inversions, the sign,
and the value of the term before adopting the sign:
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Permutation
(1, 2, 3, 4)
(1, 2, 4, 3)
(1, 4, 2, 3)
(1, 4, 3, 2)
Number of Inversions
0
1
2
3
Sign (+/−)
+
−
+
−
Value of Term
−6
−20
120
−96
Thus, det(A) = −6 − (−20) + 120 − (−96) = 230.
39. The definition of the determinant for this 4 × 4 matrix includes 24 terms, but based on the positions
occupied by zeros in this matrix, many of these terms will be zero. The nonzero terms will correspond to
the permutations that we have enumerated in the table below, along with the number of inversions, the sign,
and the value of the term before adopting the sign:
Permutation
(1, 2, 3, 4)
(1, 2, 4, 3)
(2, 1, 3, 4)
(2, 1, 4, 3)
Number of Inversions
0
1
1
2
Sign (+/−)
+
−
−
+
Value of Term
−16
−24
−8
−12
Thus, det(A) = (−16) − (−24) − (−8) + (−12) = 4.
40. The definition of the determinant for this 5×5 matrix includes 5! = 120 terms, but based on the positions
occupied by zeros in this matrix, many of these terms will be zero. The nonzero terms will correspond to
the permutations that we have enumerated in the table below, along with the number of inversions, the sign,
and the value of the term before adopting the sign:
Permutation
(1, 2, 3, 4, 5)
(1, 2, 3, 5, 4)
(1, 3, 2, 4, 5)
(1, 3, 2, 5, 4)
(2, 1, 3, 4, 5)
(2, 1, 3, 5, 4)
(2, 3, 1, 4, 5)
(2, 3, 1, 5, 4)
(3, 1, 2, 4, 5)
(3, 1, 2, 5, 4)
(3, 2, 1, 4, 5)
(3, 2, 1, 5, 4)
Number of Inversions
0
1
1
2
1
2
2
3
2
3
3
4
Sign (+/−)
+
−
−
+
−
+
+
−
+
−
−
+
Value of Term
8
3
−32
−12
−32
−12
−384
−144
−48
−18
144
54
Thus, det(A) = 8−3−(−32)+(−12)−(−32)+(−12)+(−384)−(−144)+(−48)−(−18)−144+54 = −315.
41. The definition of the determinant for this 5×5 matrix includes 5! = 120 terms, but based on the positions
occupied by zeros in this matrix, many of these terms will be zero. The nonzero terms will correspond to
the permutations that we have enumerated in the table below, along with the number of inversions, the sign,
and the value of the term before adopting the sign:
Permutation
(1, 2, 3, 4, 5)
(1, 2, 3, 5, 4)
(2, 1, 3, 4, 5)
(2, 1, 3, 5, 4)
Number of Inversions
0
1
1
2
Sign (+/−)
+
−
−
+
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Value of Term
1080
1120
1620
1680
213
Thus, det(A) = 1080 − 1120 − 1620 + 1680 = 20.
42. The definition of the determinant for this 5×5 matrix includes 5! = 120 terms, but based on the positions
occupied by zeros in this matrix, many of these terms will be zero. The nonzero terms will correspond to
the permutations that we have enumerated in the table below, along with the number of inversions, the sign,
and the value of the term before adopting the sign:
Permutation
(4, 5, 3, 1, 2)
(4, 5, 3, 2, 1)
(5, 4, 3, 1, 2)
(5, 4, 3, 2, 1)
Number of Inversions
8
9
9
10
Sign (+/−)
+
−
−
+
Value of Term
−64
−192
32
96
Thus, det(A) = (−64) − (−192) − 32 + 96 = 192.
6t
e4t
e
43. We have det
= (e6t )(4e4t ) − (e4t )(6e6t ) = −2e10t .
6e6t 4e4t
44. Using the schematic in Figure 3.1.1, we have
⎡
⎤
sin t
cos t 1
det ⎣ cos t − sin t 0 ⎦ = (sin t)(− sin t)(0) + (cos t)(0)(sin t) + (1)(cos t)(− cos t)
sin t − cos t 0
− (sin t)(− sin t)(1) − (− cos t)(0)(sin t) − (0)(cos t)(cos t)
= − cos2 t + sin2 t.
2t
3t
e
e−4t 2t e 3t
3e
−4e−4t = (e2t )(3e3t )(16e−4t )+(e3t )(−4e−4t )(4e2t )+(e−4t )(2e2t )(9e3t )−(4e2t )(3e3t )(e−4t )−
45. 2e
4e2t 9e3t 16e−4t (9e3t )(−4e−4t )(e2t ) − (16e−4t )(2e2t )(e3t ) = 42et .
46. Using the schematic in Figure 3.1.1, we have
⎤
⎡ −t
e−5t
e2t
e
det ⎣ −e−t −5e−5t 2e2t ⎦ = (e−t )(−5e−5t )(4e2t ) + (e−5t )(2e2t )(e−t ) + (e2t )(−e−t )(25e−5t )
e−t
25e−5t 4e2t
− (e−t )(−5e−5t )(e2t ) − (25e−5t )(2e2t )(e−t ) − (4e2t )(−e−t )(e−5t )
= −20e−4t + 2e−4t − 25e−4t + 5e−4t − 50e−4t + 4e−4t
= −84e−4t .
47.
y1 − y1 + 4y1 − 4y1 = 8 sin 2x + 4 cos 2x − 8 sin 2x − 4 cos 2x = 0,
y2 − y2 + 4y2 − 4y2 = −8 cos 2x + 4 sin 2x + 8 cos 2x − 4 sin 2x = 0,
y3 − y3 + 4y3 − 4y3 = ex − ex + 4ex − 4ex = 0.
y1 y 2 y 3 cos 2x
sin 2x ex y1 y2 y3 = −2 sin 2x
2 cos 2x ex y1 y2 y3 −4 cos 2x −4 sin 2x ex = 2ex cos2 2x − 4ex sin 2x cos 2x + 8ex sin2 2x + 8ex cos2 2x + 4ex sin 2x cos 2x + 2ex sin2 2x = 10ex .
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48.
(a).
y1 − y1 − y1 + y1 = ex − ex − ex + ex = 0,
y2 − y2 − y2 + y2 = sinh x − cosh x − sinh x + cosh x = 0,
y3 − y3 − y3 + y3 = cosh x − sinh x − cosh x + sinh x = 0.
y1
y1
y1
y2
y2
y2
y3 ex
y3 = ex
y3 ex
cosh x
sinh x
cosh x
sinh x cosh x sinh x = ex sinh2 x + ex cosh2 x + ex sinh x cosh x − ex sinh2 x − ex cosh2 x − ex sinh x cosh x = 0.
(b). The formulas we need are
cosh x =
ex + e−x
2
and
sinh x =
ex − e−x
.
2
Adding the two equations, we find that cosh x + sinh x = ex , so that −ex + cosh x + sinh x = 0. Therefore,
we may take d1 = −1, d2 = 1, and d3 = 1.
49.
(a). S4 = {1, 2, 3, 4}
(1, 2, 3, 4)
(1, 2, 4, 3)
(1, 3, 2, 4)
(1, 3, 4, 2)
(1, 4, 2, 3)
(1, 4, 3, 2)
(2, 1, 3, 4)
(2, 1, 4, 3)
(2, 3, 1, 4)
(2, 3, 4, 1)
(2, 4, 1, 3)
(2, 4, 3, 1)
(3, 1, 2, 4)
(3, 1, 4, 2)
(3, 2, 1, 4)
(3, 2, 4, 1)
(3, 4, 1, 2)
(3, 4, 2, 1)
(4, 1, 2, 3)
(4, 1, 3, 2)
(4, 2, 1, 3)
(4, 2, 3, 1)
(4, 3, 1, 2)
(4, 3, 2, 1)
(b).
N (1, 2, 3, 4) = 0, σ(1, 2, 3, 4) = 1, even;
N (1, 3, 2, 4) = 1, σ(1, 3, 2, 4) = −1, odd;
N (1, 4, 2, 3) = 2, σ(1, 4, 2, 3) = 1, even;
N (2, 1, 3, 4) = 1, σ(2, 1, 3, 4) = −1, odd;
N (2, 3, 1, 4) = 2, σ(2, 3, 1, 4) = 1, even;
N (2, 4, 1, 3) = 3, σ(2, 4, 1, 3) = −1, odd;
N (3, 1, 2, 4) = 2, σ(3, 1, 2, 4) = 1, even;
N (3, 2, 1, 4) = 3, σ(3, 2, 1, 4) = −1, odd;
N (3, 4, 1, 2) = 4, σ(3, 4, 1, 2) = 1, even;
N (4, 1, 2, 3) = 3, σ(4, 1, 2, 3) = −1, odd;
N (4, 2, 1, 3) = 4, σ(4, 2, 1, 3) = 1, even;
N (4, 3, 1, 2) = 5, σ(4, 3, 1, 2) = −1, odd;
(c).
N (1, 2, 4, 3) = 1, σ(1, 2, 4, 3) = −1, odd;
N (1, 3, 4, 2) = 2, σ(1, 3, 4, 2) = 1, even;
N (1, 4, 3, 2) = 3, σ(1, 4, 3, 2) = −1, odd;
N (2, 1, 4, 3) = 2, σ(2, 1, 4, 3) = 1, even;
N (2, 3, 4, 1) = 3, σ(2, 3, 4, 1) = −1, odd;
N (2, 4, 3, 1) = 4, σ(2, 4, 3, 1) = 1, even;
N (3, 1, 4, 2) = 3, σ(3, 1, 4, 2) = −1, odd;
N (3, 2, 4, 1) = 4, σ(3, 2, 4, 1) = 1, even;
N (3, 4, 2, 1) = 5, σ(3, 4, 2, 1) = −1, odd;
N (4, 1, 3, 2) = 4, σ(4, 1, 3, 2) = 1, even;
N (4, 2, 3, 1) = 5, σ(4, 2, 3, 1) = −1, odd;
N (4, 3, 2, 1) = 6, σ(4, 3, 2, 1) = 1, even.
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det(A)
=
a11 a22 a33 a44 − a11 a22 a34 a43 − a11 a23 a32 a44 + a11 a23 a34 a42
+ a11 a24 a32 a43 − a11 a24 a33 a42 − a12 a21 a33 a44 + a12 a21 a34 a43
+ a12 a23 a31 a44 − a12 a23 a34 a41 − a12 a24 a31 a43 + a12 a24 a33 a41
+ a13 a21 a32 a44 − a13 a21 a34 a42 − a13 a22 a31 a44 + a13 a22 a34 a41
+ a13 a24 a31 a42 − a13 a24 a32 a41 − a14 a21 a32 a43 + a14 a21 a33 a42
+ a14 a22 a31 a43 − a14 a22 a33 a41 − a14 a23 a31 a42 + a14 a23 a32 a41
50. The definition of the determinant for this 4 × 4 matrix includes 24 terms, but based on the positions
occupied by zeros in this matrix, many of these terms will be zero. The nonzero terms will correspond to
the permutations that we have enumerated in the table below, along with the number of inversions, the sign,
and the value of the term before adopting the sign:
Permutation
(1, 2, 3, 4)
(1, 3, 2, 4)
(1, 3, 4, 2)
(1, 4, 3, 2)
(3, 2, 1, 4)
(3, 4, 1, 2)
(4, 3, 1, 2)
Number of Inversions
0
1
2
3
3
4
5
Sign (+/−)
+
−
+
−
−
+
−
Value of Term
96
8
12
−16
24
−4
10
Thus, det(A) = 96 − 8 + 12 − (−16) − 24 + (−4) − 10 = 78.
51. The definition of the determinant for this 4 × 4 matrix includes 24 terms, but based on the positions
occupied by zeros in this matrix, many of these terms will be zero. The nonzero terms will correspond to
the permutations that we have enumerated in the table below, along with the number of inversions, the sign,
and the value of the term before adopting the sign:
Permutation
(1, 3, 2, 4)
(1, 3, 4, 2)
(1, 4, 2, 3)
(1, 4, 3, 2)
(2, 1, 3, 4)
(2, 1, 4, 3)
(2, 3, 1, 4)
(2, 4, 1, 3)
(3, 1, 2, 4)
(3, 1, 4, 2)
(3, 4, 1, 2)
Number of Inversions
1
2
2
3
1
2
2
3
2
3
4
Sign (+/−)
−
+
+
−
−
+
+
−
+
−
+
Value of Term
4
−6
−2
−6
144
−72
−32
16
−84
126
−28
Thus, det(A) = −4 + (−6) + (−2) − (−6) − 144 + (−72) + (−32) − 16 + (−84) − 126 + (−28) = −508.
52. The definition of the determinant for this 5 × 5 matrix includes 5! = 120 terms, but based on the
positions occupied by zeros in this matrix, many of these terms will be zero. In fact, the only nonzero terms
will consist of permutations whose first entry is 1. Thus, a11 = 2 will be a factor in each of the nonzero
terms. The 24 permutations of {2, 3, 4, 5} will yield the same terms that we already found in Problem 50,
since the 4 × 4 submatrix consisting of the last four rows and columns is the same as the matrix in Problem
50. Therefore, the answer here is simply 2 · 78 = 156.
53. The definition of the determinant for this 5 × 5 matrix includes 5! = 120 terms, but based on the
positions occupied by zeros in this matrix, many of these terms will be zero. In fact, the only nonzero terms
(c)2017 Pearson Education. Inc.
216
will consist of permutations whose last entry is 5. Thus, a55 = −3 will be a factor in each of the nonzero
terms. The 24 permutations of {1, 2, 3, 4} will yield the same terms that we already found in Problem 51,
since the 4 × 4 submatrix consisting of the first four rows and columns is the same as the matrix in Problem
51. Therefore, the answer here is simply −3 · (−508) = 1524.
54.
(a).
det(cA)
=
ca11
ca21
=
c2 a11 a22 − c2 a12 a21 = c2 (a11 a22 − a12 a21 ) = c2 det(A).
ca12 = (ca11 )(ca22 ) − (ca12 )(ca21 )
ca22 (b).
det(cA)
,
=
σ(p1 , p2 , p3 , . . . , pn )ca1p1 ca2p2 ca3p3 · · · canpn
,
σ(p1 , p2 , p3 , . . . , pn )a1p1 a2p2 a3p3 · · · anpn
cn
=
cn det(A),
=
where each summation above runs over all permutations σ of {1, 2, 3, . . . , n}.
55.
(a):
123 = 1,
(b): Consider
132 = −1,
3 ,
3
3 ,
,
213 = −1,
231 = 1,
312 = 1,
321 = −1.
ijk a1i a2j a3k . The only nonzero terms arise when i,
j, and k are distinct. Conse-
i=1 j=1 k=1
quently,
3 ,
3
3 ,
,
ijk a1i a2j a3k
=
123 a11 a22 a33 + 132 a11 a23 a32 + 213 a12 a21 a33
=
=
+ 231 a12 a23 a31 + 312 a13 a21 a32 + 321 a13 a22 a31
a11 a22 a33 + a12 a23 a31 + a13 a21 a32 − a11 a23 a32 − a12 a21 a33 − a13 a22 a31
det(A).
i=1 j=1 k=1
56. From the given term, we have
N (n, n − 1, n − 2, . . . , 1) = 1 + 2 + 3 + · · · + (n − 1) =
n(n − 1)
,
2
because the series of (n−1) terms is just an arithmetic series which has a first term of one, common difference
of one, and last term (n − 1). Thus, σ(n, n − 1, n − 2, . . . , 1) = (−1)n(n−1)/2 .
Solutions to Section 3.2
True-False Review:
n
(a): FALSE. The determinant of the
matrix will increase by a factor of 2 . For instance, if A = I2 , then
2 0
det(A) = 1. However, det(2A) = det
= 4, so the determinant in this case increases by a factor of
0 2
four.
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(b): TRUE. In both cases, the determinant of the matrix is multiplied by a factor of c.
(c): TRUE. This follows by repeated application of Property (P8):
det(A5 ) = det(AAAAA) = (detA)(detA)(detA)(detA)(detA) = (detA)5 .
(d): TRUE. Since det(A2 ) = (detA)2 and since det(A) is a real number, det(A2 ) must be nonnegative.
(e): FALSE. The matrix is not invertible if and only if its determinant, x2 y − xy 2 = xy(x − y), is zero. For
example, if x = y = 1, the matrix is not invertible since x = y; however, neither x nor y is zero in this case.
We conclude that the statement is false.
(f ): TRUE. We have
det(AB) = (detA)(detB) = (detB)(detA) = det(BA).
Problems:
From this point on, P2 will denote an application of the property of determinants that states that if every
element in any row (column) of a matrix A is multiplied by a number c, then the determinant of the resulting
matrix is c det(A).
1.
−2
5 1 −2 5 2
1 6 3
1
=
=−
=−
5 −4
1 6
−2 5
0
6
= −17.
17
1. A12 (2)
2.
2. P12
1
1
1
2
3 2
3 2 3 1
1
2
3
2
2 −2 = 2 0
1 −1 = 2 0
6 4 = 0
0 −11 −7 0
3 −5 2 0 −11 −7 1. A12 (−2), A13 (−3)
3.
2. P2
2
3 1 −1 = 2(−18) = −36.
0 −18 3. A23 (11)
2 −1 4 2 −1 4 2 2 1 −1/2 2 3 1 −1/2
1 3
2 1 = 2 3
2 1 = 2 0
7/2 −5 = 2 · 28 = 56.
2 1 = 3
−2
0
0 8
0
0 8
0
0
8 1 4
1. A13 (1)
4.
3. A12 (2)
2
−1
4
1
2
1
−1
3 1
6 = − 2
4
12 2. P2
3. A12 (−3)
−1 2 6 −1 2 6 6 2
3
3 = − 0 5 15 = −5 · 9 0 1 3 0 9 36 0 1 4 12 −1 2 6 4
= −45 0 1 3 = (−45)(−1) = 45.
0 0 1 1. P12
2
1
1
2. A12 (2), A13 (4)
3. P2
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4. A23 (−1)
218
5.
0
1 −2 1
1
−1
0
3 = 0
2 −3
0 2
0 −3 1
0 −3 1
2
3
1 −2 = 0
1 −2 = 0
−3
0 0 −3
6 0
1. P12 , P2
6.
3
5
2
7
1 1
1
9 −6 = 5
2
1
3
1. A31 (−1)
7.
2. A13 (−2)
0 −3 1 −2 = 0.
0
0 3. A23 (3)
6 −2 1
6 −2 1
6 −2 2
3
9 −6 = 0 −21
4 = 0
1 −10 1
3
0 −11
7
0 −11
7 1 6
−2 4 = −103.
0
1
−10
=
0 0 −103 2. A12 (−5), A13 (−2)
3. A32 (−2)
4. A24 (11)
1 −1 2 4 1 −1
1 −1
1 −1
2
4 2
4 2
4 3
0
0
1 2 4 1 0
4 −4 −8 2
1 −1 −2 3
1 −1 −2 = 0
= 3 · 4 0
= −12 0
−1
1
3
2
0
5
6
0
5
6
1
0 −2 2
0
0
1 4 2 0
3
0 −6 1
0 −2 0
5
6 1 −1
1 −1
2
4 2
4 0
0
1 −1 −2 1 −1 −2 5
4
= −12 = −12 · 6 = −72.
= −12 0
1
0 0
1
0 0
0
0
0
0
0
6 0
5
6 1. A12 (−3), A13 (1), A14 (−2)
2. P2
3. P34
4. A23 (−1)
5. A34 (−5)
8. Note that in the first step below, we extract a factor of 13 from the second row of the matrix, and we
also extract a factor of 8 from the second column of the matrix.
2 32 1
2 4
1 5 −1
1
4 4 5 26 104 26 −13 1
2 1
2 −1 2
2 −1 = 13 · 8 2 1
= −104 2
2
7 2
7 56 2
7 2 7
2 7
1
1 5 −1
5
2 4
1
4 40 1
5
1
1
5
1
5 5
1
5 0 −9
0
0 −11 4
1
0
1 3
= −104 = −104(−3) 0 −11 0 −3 0 −3
0 −9
0 −6 −1 −6 0 −6 −1 −6 1 5
1 5
1
5 1
5 0 1
0 1
0
1 6
0
1 5
= 312 = 312 0 −2 0 0 0
0 0 −1
0 0 −1
0
0 0
0 −2 = 312(−1)(−2) = 624.
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219
1. P2
2. P14
3. A12 (−2), A13 (−2), A14 (−2)
0
1 −1
−1
0
1
1 −1
0
−1 −1 −1
1
0
3
= − 0
0
9.
1. P13
2
3
4
5
10.
1
0
1
2
1 −1
1 0
1 1 −1
0
1
= −
1 0
1
−1
−1 −1 −1
0 1
−1
0
1 −1
1
2 4 0
=
−
0
0
0
3 0
0 −3 −3
2. A12 (1), A14 (1)
4. P2, P23
5. A23 (9), A24 (6)
6. P34
1 −1
1 0 1 1 2 0 −1
1 2 = −
1 1 −1 1 0
0
0 −2 −1 1 −1
0
1 −1
1
2 = 9.
0 −3 −3 0
0
3 3. A23 (1), A24 (−2)
4. P34
1 2 3 5 1 2
5 3
5 2 1 0 3 1 2 2 0 3
1
2 = −
= − 0 2
3 1
4
4
3
1
−2
2 5 5 3 0 1 −1 −7 3 1 2
3
4 1 2
3
4 0 1 −1 −7 4 0 1 −1 −7 3
=
= − 3 12 1 −2 0 0
0 2
0 3
4 23 1
2 0 0
1 2
1 2
3
4 3
4 0 1 −1 −7 6 0 1 −1 −7 5
= 3
= 3 0 0
1
4 1
4 0 0
0 0
0 0
4 23 0
7 3
1
4
5
= 3 · 1 · 1 · 1 · 7 = 21.
1. CP12
11.
2. A13 (−1), A14 (−2)
3. P24
4. A23 (−2), A24 (−3)
5. P2
6. A34 (−4)
2 −1 3
−1
1 −2 −3 −4 4 2 3
4 7
1
1
2
3 1
7 2
3 2 1
7
2
3 =
−6
=
6
−2
4 −2 8
4 −2
4
8
6 6 8
6 6 −6 18 −24 −1
1 3 −4
−1
1
3 −4 1 −2 −3 −4 1 −2 −3 −4 1 −2 −3 −4 0
0 −1
0 −1
9
5
7 4
0 −8 5
0 −8 3
= 6 =
−6
=
−6
0
0
6 20 22 6 20 22 0 20 −26 0
0 −1
0 −8
0
0
5 −65 0
9
5
7
1 −2 −3 −4 1 −2 −3 −4 0 −1
0 −8 7 0 −1
0 −8 6
= 6 · 1(−1) · 5 · 234 = −7020.
= 6 =
6
0
0
5 −65 0
5 −65 0
0
0
0 20 −26 0
0 234 (c)2017 Pearson Education. Inc.
220
1. CP12 , P2
2. M1 (−1)
5. A23 (6), A24 (9)
6. P34
3. A12 (−1), A13 (−4), A14 (1)
4. P24
7. A34 (−4)
12.
7 −1 3 4 1 −1 3 2 1 −1
3
2 14
2
0
2 4 6 1
2 4 3 2
4 −2 −1 = 7 · 2 3
21
= 14 0
1
3
4
1
3
2
4 −6 −4 −7
−1
0
4 5 8 4 5 4 3
8
6 1 −1
1 −1
1 −1
3
2 3
2 3
2 0
0
0
4 −2 −1 4
1 −10 −7 5
1 −10 −7 3
= 14 = 14 = 14 0 −4 −3 0 −4 −3 0 −4 −3 0
0
0
0
3
8
6
0
3
8
6
0
0
38 27 1 −1
3
0
1
−10
6
= −56 0
1
0
0
0
38
1. P2
1 −1
2 3
2 0
7 7
1 −10
−7 = −56 · (−3/2) = 84.
= −56 3/4 0
1
3/4 0
27
0
0
0 −3/2
2. A12 (−2), A13 (−3), A14 (1)
5. A24 (−3)
6. P2
3. A23 (−1)
4. A42 (−1)
7. A34 (−38)
13.
3 7
1
1 1 −1
4 8 −1
3 7
0
8 16 −1
1
0
3
= − 0
0
0
1. P12
1
4
0
0
0
−1
4
−1
−1
−1
1
2 3 3
0 1 1
6 6 = − 4
3
9 4 8
8 12
0
2
4
7
4
1
1 0
0 4
2 = − 0
0
1 0
4
1
1 −1 0 1 0
7
1 2 3 2
8 −1 6 6 = − 0
0
7
0 9 4 0
16 −1 8 12
1 −1 0 1 4
4 2 0 4
3 6 2 4
3 9 1 8
7 8 4 1 −1 0
1 4
4 2
0 0 −1 4
2 = −(1)(4)(−1)(3)(2) = 24.
0
0 3 −1 0
0 0
2 2. A12 (−3), A13 (−4), A14 (−3), A15 (−8)
3. A23 (−1), A24 (−1), A25 (−2)
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221
14.
1
5
2
6
0
2
4
3
5
2
3
3
4
4
4
4
2
5
3
6
5
1
2
3
4
5
1
0 −6 −12 −18 −24
1
6 = 0 −1 −2 −3 −4
2
0 −7 −14 −21 −28
8
0
2
4
6
8
1
2
3
4
5
0
1
2
3
4
2
= −6 0 −1 −2 −3 −4
0 −7 −14 −21 −28
0
2
4
6
8
1
0
3
= −6 0
0
0
2
1
0
0
0
3
2
0
0
0
4
3
0
0
0
5
4
0
0
0
= −6 · 0 = 0.
1. A12 (−5), A13 (−2), A14 (−6)
2. M2 (− 16 )
3. A23 (1), A24 (7), A25 (−2)
−1
1 15. = 0; not invertible.
1 −1 2 1 = 1; invertible.
16. 3 2 −1
2 3 17. 5 −2 1 = −8; invertible.
8 −2 5 2 6 −1 1 = 14; invertible.
18. 3 5
2 0
1 19.
1 1
1
1 1
−1 1 −1
1 0
1 1 −1 −1 = 0
−1 1
1 −1 0
1
1
1 2
0
2 2
0
2 = 0 −2 −2 = 16; invertible.
0 −2 −2 2
2
0 2
2
0 1
0 2 −1 1
0
2 −1 3 −2 1
0 −2 −5
−2 −5 7 4
7
=
= 1
2 4 = 133; invertible.
2
1 6
2 0
1
2
4 −3
2 1 1 −3 4
0 0 −3
2
1 1
1
2 −3
5 2 −3
5 −1
3 −6 2
2 −3
6 1 1 −2
1. M2 (−1) 2. P7
= −
= 0; not invertible.
21. 3 −1
4 3 −1
4 2
2
1 −2
1 −2
3 −6 3 −6 20.
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222
1
2 k 1
2
k
= (3k −1)(k +4). Consequently, the system has an infinite
1
22. det(A) = 2 −k 1 = 2 −k
3
6 1
0
0 1 − 3k number of solutions if and only if k = −4 or k = 1/3 (Corollary 3.2.6).
b1
1 k
x1
, and b =
. According to Corollary
23. The system is Ax = b, where A =
, x=
x2
b2
k 4
3.2.6, the system has unique solution if and only if det(A) = 0. But, det(A) = 4 − k 2 , so that the system
has a unique solution if and only if k = ±2.
1 k 0 24. det(A) = k 1 1 = 1 + k − 1 − k 2 = k(1 − k). Consequently, the system has a unique solution if
1 1 1 and only if k = 0, 1.
25. The given system is
(1 − k)x1 +
2x1 +
x1 +
2x2 +
(1 − k)x2 +
x2 +
x3
x3
(2 − k)x3
= 0,
= 0,
= 0.
The determinant of the coefficients as a function of k is given by
1−k
2
1 1−k
1 = −(1 + k)(k − 1)(k − 4).
det(A) = 2
1
1
2−k Consequently, the system has an infinite number of solutions if and only if k = −1, k = 1, or ,k = 4.
26. From the schematic provided in Figure 3.1.1, we find that det(A) = (1)(1)(3) + (−1)(4)(0) +
(2)(3)(1) − (0)(1)(2) − (1)(4)(1) − (3)(3)(−1) = 3 + 0 + 6 − 0 − 4 + 9 = 14. Thus, by Property P10, we have
1
det(A−1 ) =
, and by Property P4, we have det(−3A) = (−3)3 · 14 = −378.
14
1 −1
1 2
27. A =
and B =
. det(A) det(B) = [3 · 1 − (−1)2][1 · 4 − 2(−2)] = 5 · 8 = 40.
2
3
−2 4
3 −2 = 3 · 16 − (−2)(−4) = 40. Hence, det(AB) = det(A) det(B).
det(AB) = −4 16 28. Using identities for hyperbolic trignometric functions, we have det(A) = (cosh x)(cosh x)−(sinh x)(sinh x) =
cosh2 x − sinh2 x = 1 and det(B) = (cosh y)(cosh y) − (sinh y)(sinh y) = cosh2 y − sinh2 y = 1. Furthermore,
we have
cosh x cosh y + sinh x sinh y cosh x sinh y + sinh x cosh y
cosh(x + y) sinh(x + y)
AB =
=
,
sinh x cosh y + cosh x sinh y sinh x sinh y + cosh x cosh y
sinh(x + y) cosh(x + y)
so that
det(AB) = cosh2 (x + y) − sinh2 (x + y) = 1.
Thus, det(AB) = det(A)det(B), and Property P9 is verified.
29. B is obtained from AT by the following elementary row operations: (1) A13 (3), (2) M1 (−2). Since
det(AT ) = det(A) = 1, and (1) leaves the determinant unchanged, we have det(B) = −2.
30. B is obtained from A by the following elementary row operations: (1) M1 (4), (2) M2 (3), (3) P12 . Thus,
det(B) = det(A) · 4 · 3 · (−1) = −12.
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31. B is obtained from A via the following sequence of four elementary row and column operations: (1) P12 ,
(2) CP12 , (3) M1 (−6), (4) M2 (3). Thus, det(B) = det(A) · (−1) · (−1) · (−6) · (3) = −18.
32. B is obtained from A by the following operations: (1) Interchange the two columns, (2) M1 (−1), (3)
A12 (−4). Now (3) leaves the determinant unchanged, and (1) and (2) each change the sign of the determinant.
Therefore, det(B) = det(A) = 1.
33. B is obtained from A via the following sequence of elementary row operations: (1) M1 (−1), (2) M2 (−2),
(3) P13 . Thus, det(B) = det(A) · (−1) · (−2) · (−1) = 12.
34. B is obtained from A by the following row operations: (1) A13 (5), (2) M2 (−4), (3) P12 , (4) P23 . Thus,
det(B) = det(A) · (−4) · (−1) · (−1) = (−6) · (−4) = 24.
35. B is obtained from A by the following operations: (1) M1 (−3), (2) A23 (−4), (3) P12 . Thus, det(B) =
det(A) · (−3) · (−1) = (−6) · (3) = −18.
36. B is obtained from AT by the following row operations: (1) M1 (2), (2) A32 (−1), (3) A13 (−1). Thus,
det(B) = det(AT ) · 2 = (−6) · (2) = −12.
37. We have det(AB T ) = det(A)det(B T ) = 5 · 3 = 15.
38. We have det(A2 B 5 ) = (detA)2 (detB)5 = 52 · 35 = 6075.
−1
39. We have det((A
2 3
B ) ) = (det(A
−1
2
3
−1
B )) = (detA
)(detB)
40. We have
det((2B)
−1
T
(AB) ) = (det((2B)
−1
T
))(det(AB) ) =
2 3
=
1 2
·3
5
det(A)det(B)
det(2B)
3
3
729
9
=
=
= 5.832.
5
125
=
5·3
3 · 24
=
5
.
16
41. We have det((5A)(2B)) = (5 · 54 )(3 · 24 ) = 150, 000.
1
.
42. We have det(B −1 A−1 ) = det(B −1 )det(A−1 ) = 13 · 15 = 15
43. We have det(2A) = 24 · 5 = 80, so that det(B −1 (2A)B) = det(B −1 )det(2A)det(B) = det(2A) = 80.
44. We have det(4B) = 44 · 3 = 768, so that det((4B)3 ) = 7683 = 452, 984, 832.
1
1 2
0
,
and A =
45. It is not true that A and B must be equal. For instance, if we take S = 2
3 4
0 1
1 4
then it is easy to verify that B = S −1 AS = 3
= A. However, using the properties of determinants,
4
2
we see that
det(B) = det(S −1 AS) = (det S −1 )(det A)(det S) =
(det A)(det S)
= det(A).
det S
46.
(a). The volume of the parallelepiped is given by |det(A)|. In this case, we have
|det(A)| = |2 + 12k + 36 − 4k − 18 − 12| = |8 + 8k|.
(b). NO. The answer does not change because the determinants of A and AT are the same, so the
volume determined by the columns of A, which is the same as the volume determined by the rows of AT , is
|det(AT )| = |det(A)|.
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(c). The matrix A is invertible if and only if det(A) = 0. That is, A is invertible if and only if 8 + 8k = 0,
if and only if k = −1.
47.
1 −1 x 1 −1
1 −1
x
x
2 2
2
0
2
= 0
3
x
3
x
1
x
−
2x
−
2x
=
4 −1 x3 0
3 x3 − 4x 0
0 x3 − x2 − 2x 1 −1
x
.
3
x(x − 2)
= 0
0
0 x(x − 2)(x + 1) We see directly that the determinant will be zero if and only if x ∈ {0, −1, 2}.
48.
αx − βy
βx + αy
βx − αy αx βx − αy −βy βx − αy =
+
αx + βy βx αx + βy αy αx + βy αx βx αx −αy −βy βx −βy −αy +
=
+
+
βy αy αx αy
βx αx βx
βy α −α β −β 2 α β 2 α β =x + xy − xy +y β α β α β
β α
α α −α β
β 2
2 α β + xy + xy = (x + y ) β α β
β α −α α −α α −α 2
2 α β − xy + xy = (x + y ) β
β β
β β α 2
2 α β = (x + y ) .
β α 49.
a1 + βb1
a2 + βb2
a3 + βb3
b1 + γc1
b2 + γc2
b3 + γc3
c1 + αa1 a1
c2 + αa2 = a2
c3 + αa3 a3
a1
= a2
a3
b1 + γc1 c1 + αa1 βb1 b1 + γc1
b2 + γc2 c2 + αa2 + βb2 b2 + γc2
b3 + γc3 c3 + αa3 βb3 b3 + γc3
b1 c 1 a 1 b1 c1 b2 c2 + αβγ b2 c2 a2 b3 c 3 a 3 b3 c 3 b1 c 1 a 1 = (1 + αβγ) b2 c2 a2 .
b3 c 3 a 3 c1 + αa1 c2 + αa2 c3 + αa3 Now if the last expression is to be zero for all ai , bi , and ci , then it must be the case that 1 + αβγ = 0; hence,
αβγ = −1.
50. Suppose A is a matrix with a row of zeros. We will use (P3) and (P7) to justify the fact that det(A) = 0.
If A has more than one row of zeros, then by (P7), since two rows of A are the same, det(A) = 0. Assume
instead that only one row of A consists entirely of zeros. Adding a nonzero row of A to the row of zeros
yields a new matrix B with two equal rows. Thus, by (P7), det(B) = 0. However, B was obtained from
A by adding a multiple of one row to another row, and by (P3), det(B) = det(A). Hence, det(A) = 0, as
required.
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51. A is orthogonal, so AT = A−1 . Using the properties of determinants, it follows that
1 = det(In ) = det(AA−1 ) = det(A) det(A−1 ) = det(A) det(AT ) = det(A) det(A).
Therefore, det(A) = ±1.
52.
(a). From the definition of determinant we have:
det(A) =
,
σ(p1 , p2 , p3 . . . , pn )a1p1 a2p2 a3p3 · · · anpn .
(52.1)
n!
If A is lower triangular, then aij = 0 whenever i < j, and therefore the only nonzero terms in (47.1) are
those with pi ≤ i for all i. Since all the pi must be distinct, the only possibility is pi = i for all i with
1 ≤ i ≤ n, and so (52.1) reduces to the single term:
det(A) = σ(1, 2, 3, . . . , n)a11 a22 a33 · · · ann .
(b).
2 −1 3
1
2 2
det(A) = 3
0 1
1
2 0
2 16
0 13
3
= − −1 −8
1
2
5 −3 −11
0
1 1 0
=
4 −1 −8
2
1 1
2
0 0 0 0 4 0
= −
1 0 −1
1
0 1
1. A41 (−5), A42 (−1), A43 (−4)
3 0 0 13 0 0 2 0 2 2 16 0 0 =
1 0 −1 −8 1 0 0 1 1
2 0 1 0 0 0 13 0 0 = −26.
−8 1 0 2 0 1
2. A31 (−3), A32 (−2)
3. P12
4. A21 (−16/13)
53. To prove this, take the determinant of both sides of AB = AC to get det(AB) = det(AC). Thus, by
Property P8, we have det(A)det(B) = det(A)det(C). Since A is invertible, det(A) = 0. Thus, we can cancel
det(A) from the last equality to obtain det(B) = det(C).
54. Note that (S −1 AS)2 = S −1 ASS −1 AS = S −1 AIAS = S −1 A2 S. Therefore,
det((S −1 AS)2 ) = det(S −1 A2 S) = det(S −1 det(A2 )det(S) = det(A2 ) = (det(A))2 ,
as required.
55. No. If A were invertible, then det(A) = 0, so that det(A3 ) = det(A) det(A) det(A) = 0.
56. Let E be an elementary matrix. There are three different possibilities for E.
(a). E permutes two rows: Then E is obtained from In by interchanging two rows of In . Since det(In ) = 1
and using Property P1, we obtain det(E) = −1.
(b). E adds a multiple of one row to another: Then E is obtained from In by adding a multiple of one row
of In to another. Since det(In ) = 1 and using Property P3, we obtain det(E) = +1.
(c). E scales a row by k: Then E is obtained from In by multiplying a row of In by k. Since det(In ) = 1
and using Property P2, we obtain det(E) = k.
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57. We have
x
0 = x1
x2
y
y1
y2
1 1 = xy1 + yx2 + x1 y2 − x2 y1 − xy2 − yx1 ,
1 which can be rewritten as
x(y1 − y2 ) + y(x2 − x1 ) = x2 y1 − x1 y2 .
Setting a = y1 − y2 , b = x2 − x1 , and c = x2 y1 − x1 y2 , we can express this equation as ax + by = c, the
equation of a line. Moreover, if we substitute x1 for x and y1 for y, we obtain a valid identity. Likewise, if we
substitute x2 for x and y2 for y, we obtain a valid identity. Therefore, we have a straight line that includes
the points (x1 , y1 ) and (x2 , y2 ).
58.
1
1
1
x
y
z
1 x
x2
x2 2
(y − x)(y + x) = (y − x)(z − x) 0 1 y + x 0 1 z+x (z + x)(z − x) 1 x
x2 3
= (y − x)(z − x) 0 1 y + x = (y − x)(z − x)(z − y) = (y − z)(z − x)(x − y).
0 0 z−y x
x2 1
1
y 2 = 0 y − x
z2 0 z − x
1. A12 (−1)A13 (−1)
2. P2
3. A23 (−1)
59. Since A is an n × n skew-symmetric matrix, AT = −A; thus,
det(AT ) = det(−A) = (−1)n det(A) = − det(A)
since n is given as odd. But by P4, det(AT ) = det(A), so det(A) = − det(A) or det(A) = 0.
60. Solving b = c1 a1 + c2 a2 + · · · + cn an for c1 a1 yields c1 a1 = b − c2 a2 − · · · − cn an .
Consequently, det(Bk ) can be written as
det(Bk ) = det([a1 , a2 , . . . , ak−1 , b, ak+1 , . . . , an ])
= det([a1 , a2 , . . . , ak−1 , (c1 a1 + c2 a2 + · · · + cn an ), ak+1 , . . . , an ])
= c1 det([a1 , . . . , ak−1 , a1 , ak+1 , . . . , an ]) + c2 det([a1 , . . . , ak−1 , a2 , ak+1 , . . . , an ]) + · · ·
+ ck det([a1 , . . . , ak−1 , ak , ak+1 , . . . , an ]) + · · · + cn det([a1 , . . . , ak−1 , an , ak+1 , . . . , an ]).
Now by P7, all except the k th determinant are zero since they have two equal columns, so that we are left
with det(Bk ) = ck det(A).
63. Using technology we find that:
1 2 3 4 a 2 1 2 3 4 3 2 1 2 3 = −192 + 88a − 8a2 = −8(a − 3)(a − 8).
4 3 2 1 2 a 4 3 2 1 Consequently, the matrix is invertible provided a = 3, 8.
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64. Using technology we find that:
1−k
3
3
4
2−k
4
1
= −(k − 6)(k + 2)2 .
1
−1 − k Consequently, the system has an infinite number of solutions
⎡ if and only if ⎤k = 6, −2.
−5
4
1
1 ⎦. This system has solution set
k = 6: In this case, the system is Bx = 0, where B = ⎣ 3 −4
3
4 −7
{t(1, 1, 1) : t ∈ R}.
⎡
⎤
3 4 1
k = −2: In this case, the system is Cx = 0, where C = ⎣ 3 4 1 ⎦. This system has solution set
3 4 1
{(r, s, −3r − 4s) : r, s ∈ R}.
65. Using technology we find that:
⎤
⎤
⎡
−2/5 1/2
0 1/10
19/10
⎢ 1/2 −1 1/2
⎢
0 ⎥
−5 ⎥
⎥ ; x = A−1 b = ⎢
⎥.
det(A) = −20; A−1 = ⎢
⎦
⎣
⎣
0 1/2 −1
1/2
1/2 ⎦
1/10
0 1/2 −2/5
12/5
⎡
Solutions to Section 3.3
True-False Review:
(a): FALSE. Because 2 + 3 = 5 is odd, the (2, 3)-cofactor is the negative of the (2, 3)-minor of the matrix.
(b): TRUE. Examining the submatrix obtained by deleting row i and column j from the matrix A, note
by examining the remaining entries from column i or from row j of A, they now comprise a column (or row)
of zeros, so this submatrix has a determinant of zero. That is, Mij = 0.
(c): TRUE. The submatrix obtained by removing the ith row and ith column of A is the matrix diag(a1 , . . . , ai−1 , ai+1 , . . . , an
whose determinant is a1 · · · ai−1 ai+1 · · · an .
⎡
⎤
1 2 3
(d): FALSE. Here is a counterexample. If we let A = ⎣ 4 5 6 ⎦, then M11 = (5)(9) − (6)(8) = −3.
7 8 9
However, in the matrix 2A, we have M11 = (10)(18) − (12)(16) = −12, which is not doubled the value of the
(1, 1)-minor of A.
(e): TRUE. This just requires a slight modification of the proof of Theorem 3.3.17. We compute
(A · adj(A))ij =
n
,
k=1
aik adj(A)kj =
n
,
aik Cjk = δij · det(A).
k=1
Therefore, A · adj(A) = det(A) · In .
(f ): TRUE. The Cofactor Expansion Theorem allows for expansion along any row or any column of the
matrix, and in all cases, the result is the determinant of the matrix.
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⎡
⎤
1 2 3
(g): FALSE. For example, let A = ⎣ 4 5 6 ⎦, and let c = 2. It is easy to see that the (1, 1)-entry of
7 8 9
adj(A) is −3. But the (1, 1)-entry of adj(2A) is −12, not −6. Therefore, the equality posed in this review
item does not generally hold. Many other counterexamples could be given as well.
⎡
⎤
⎡
⎤
⎡
⎤
1 2 3
9 8 7
10 10 10
(h): FALSE. For example, let A = ⎣ 4 5 6 ⎦ and B = ⎣ 6 5 4 ⎦. Then A + B = ⎣ 10 10 10 ⎦.
7 8 9
3 2 1
10 10 10
The (1, 1)-entry of adj(A + B) is therefore 0. However, the (1, 1)-entry of adj(A) is −3, and the (1, 1)-entry
of adj(B) is also −3. But (−3) + (−3) = 0. Many other examples abound, of course.
ae + bg af + bh
a b
e f
. Then AB =
. We compute
(i): FALSE. Let A =
and let B =
ce + dg cf + dh
c d
g h
cf + dh −(af + bh)
d −b
h −f
adj(AB) =
=
= adj(A)adj(B).
−(ce + dg)
ae + bg
−c
a
−g
e
(j): TRUE. This can be immediately deduced by substituting In for A in Theorem 3.3.17.
Problems:
From this point on, CET(col#n) will mean that the Cofactor Expansion Theorem has been applied to column
n of the determinant, and CET(row#n) will mean that the Cofactor Expansion Theorem has been applied
to row n of the determinant.
1. We have
M11 = 5,
M12 = 0,
M21 = 2,
M22 = −9
C11 = 5,
C12 = 0,
C21 = −2,
C22 = −9.
and
2. Minors: M11 = 4, M21 = −3, M12 = 2, M22 = 1;
Cofactors: C11 = 4, C21 = 3, C12 = −2, C22 = 1.
3. Minors: M11 = −9, M21 = −7, M31 = −2, M12 = 7, M22 = 1, M32 = −2, M13 = 5, M23 = 3, M33 = 2;
Cofactors: C11 = −9, C21 = 7, C31 = −2, C12 = −7, C22 = 1, C32 = 2, C13 = 5, C23 = −3, C33 = 2.
4. Minors: M11 = −5, M21 = 47, M31 = 3, M12 = 0, M22 = −2, M32 = 0, M13 = 4, M23 = −38, M33 = −2;
Cofactors: C11 = −5, C21 = −47, C31 = 3, C12 = 0, C22 = −2, C32 = 0, C13 = 4, C23 = 38, C33 = −2.
5.
3
M12 = 7
5
1
M23 = 7
5
1
4
1
3
1
0
2 6 = −4,
2 2 6 = 40,
2 3 −1 2 1 2 = 16,
M31 = 4
0
1 2 1 −1 2 1 2 = 12,
M42 = 3
7
4 6 C12 = 4, C31 = 16, C23 = −40, C42 = 12.
6. We have
9
M41 = −6
−1
0 −1
8
8 = 478,
−3
4
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M22 =
−2
0 −1
0 −3
4 = 9,
7
3
1
M23 =
−2
9 −1
0 −1
4 = 191,
7 −7
1
M43 =
−2
9 −1
4 −6
8 = −108.
0 −1
4
Therefore,
C41 = −478,
C22 = 9,
C23 = −191,
C43 = 108.
7. We have
det(A) = 6 · C12 + 9 · C22 = 6 · 4 + 9 · (−8) = −48.
1 −2 = 1 · |3| + 2 · |1| = 5.
8. 1
3 −1 2
3 −1 2 −1 2 1 4 = 3(−11) + 2(−7) + 4(−6) = −71.
+4·
+2·
9. 1 4 −2 = 3 · 1 4 3 1 3 1 3 1
4
2 1 −4 2 1 1 −4 2 −4 = −7 · 18 + 0 − 3 · 9 = −153.
3 = −7 · 10. 7 1
−3·
+
1 5 5 −2 1 −2 1 5 −2 3 1
4 1 4 1
4 2 1
2 = 3 · +
2
·
−
11. 7 1
1 2 = 3(−11) − 7(−17) + 2(−2) = 82.
3 −5 3 −5 2 3 −5 0
2 −3 0
5 = 3 · 10 + 5(−6) + 0 · 4 = 0.
12. −2
3 −5
0 1 −2
⎡
⎤
3
0 1 −2
1 −2
3
3 1
4
0
7
−2
= −2 · 0
0
7 ⎦ = −2 · (−26) − 4 · (−5) = 72.
1
3 − 4 · ⎣ 4
13. 1
3
4 0
1
1
5 −2
5 −2
1
5 −2
0 1. CET(col#4)
14. We have
det(A) = −3 · C11 + 1 · C31 = −3 · 16 + 1 · (−4) = −52.
15. We can choose to apply the Cofactor Expansion Theorem to any row or column of our choice. Choosing
row 1, we find that
det(A) = −4 · C11 + 2 · C12 + (−1) · C13 = −4 · (−18) + 2 · (−26) + (−1) · (24) = −4.
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1 0 −2 3 1 1 1 −1 = 7 − 2(−1) = 9.
1. CET(row#1)
−2·
16. 3 1 −1 = 7 2 2
5 7 2
5
−1
2 3 2 3 1 4 1
= −7 + 10 = 3.
1. CET(col#1)
1 4 = −1 · +2·
17. 0
1 4 −1 3 2 −1 3 2 −1 3 1
2 1 − 5 · −1 3 + 3 · −1 3 = 2 · 17 − 5 · 2 + 3(−7) = 3.
2 1 = 2 · 18. 5
−3 7
−3 7
2 1 3 −3 7 1. CET(col#1)
0 −2
1 −2
−2 1 0 −3 1
0 −3 = 0 · 19. 2
−2· 3 0 −1· 0
3
0
−1
3
0
1 = 0 + 6 − 6 = 0.
−3 1. CET(col#1)
20.
1 0 −1
0 0 −1
1
0 0 −1 1
0 1
0
−1
= 1 · 0 −1
0 −1 = −2 − 2 = −4.
0 − 1 · 1
−1 0 −1
0 1
1
0
1 0
1
0 1
0
1 1. CET(col#1)
21.
2 −1
5
3 1 3 1 2
5 1 2 2
1 1
1
0
−2
3
4
−2
3
= − 1
=
0 3 0
0 −1 0 2 −1 0 0
1 −1 4 1
3 −2 4 1 −1 −2 4 2
5 1 5 = 21 − 21 = 0.
+3·
=
1 −1 −1 4 1. CA32 (2)
2. CET(row#3)
22. There are two zeros in the last row of this matrix, so we will apply the Cofactor Expansion Theorem to
the last row:
det(A) = −5 · C43 + 2 · C44 = −5 ·
−4 1 −2
−4 1 −3
1 2 −6 + 2 ·
1 2 −1 = −5 · 33 + 2 · (−3) = −171.
3 0
3
3 0
2
23. By applying elementary row operations A12 (4) and A13 (2) to the matrix (which does not change the
determinant), we have
−1
3
3
−1 3 3
4 −6 −3 =
0 6 9 ,
2 −1
4
0 5 10
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and we evaluate the latter determinant by cofactor expansion along the first column to obtain
−1
3
3
4 −6 −3 = −1 · (6 · 10 − 9 · 5) = −1 · 15 = −15.
2 −1
4
24.
3
2 −3
11 1
2 −3
11 5
2
6 1
2
3
5 −5 2 0 −1 11 −27 3
5 −5 1 2
=
7
= 7
5
−3
−16 0 −9 18 −93 5
−3
−16
9 −6 27 −12 9 −6 27 −12 0 −24 54 −111 −1 11 −27 −1
11 −27 −81 150 3
4
5
= 11997.
−81 150 = − = −9 18 −93 = 0
−210 537 −24 54 −111 0 −210 537 1. A21 (−1)
2. A12 (−2), A13 (−7), A14 (−9)
4. A12 (−9), A13 (−24)
25.
3. CET(col#1)
5. CET(col#1)
2 −7
2 −1 4 3 0 −13
4 3 2 −7
−13
2 −1 5
3
1
2 3 3
1 2 2 1
5 −3 7 1 1
0 −9 = − −16
= 0 −16
= 6
6
0
−9
2
6
3
2
6
3
−10 −8 −3 4
2 −4 5 0 −10 −8 −3 2 −4 5 4
13 −2
−2 1 1 13
5 6 101 −18 4 0 −9 = 101 −18 0 = = −892.
= −16
29
14 −10 −8 −3 29 −14 0 1. A42 (−1)
2. A21 (−2), A23 (−6), A24 (−4)
5. A12 (9), A13 (3)
3. CET(col#1)
4. P2
6. CET(col#3)
26. By applying elementary row operations A21 (2), A23 (4), and A24 (1) to the matrix (which does not change
the determinant), we have
−2 0 1 1
0 4
5 1
1 2 2 0
1 2
2 0
=
,
−4 4 6 1
0 12 14 1
−1 1 0 5
0 3
2 5
and we evaluate the latter determinant by cofactor expansion along the first column to obtain
−2 0 1 1
1 2 2 0
= C21 = 31.
−4 4 6 1
−1 1 0 5
27. We proceed as follows:
−3 2 0 −1 −4
0
1 3 1
3
5
1
1
0 2 1 −3 −1 = 0
2 0 4 −2 −3
0
−1 2 6
0 −1
0
11 3
8
11
1
3 1
3
5
0
2
2 1 −3 −1 = − 0
−6 2 −8 −13
0
5 7
3
4
0
3
11
2
−6
5
1
3
5
1
3
8
11
0
3
1 −3 −1 = 0
2 −8 −13
0
7
3
4
0
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2
11
−6
5
1
3
5
1 −3
−1
3
8
11
2 −8 −13
7
3
4
232
1
0
4
=2· 0
0
0
3
1
11
−6
5
1
1
2
3
2
7
3
5
− 32 − 12
5
8
11 = 2 ·
−8 −13
3
4
1. A21 (3), A24 (−2), A25 (1)
2. P12
1
0
0
0
0
3
1
1
1
2
0 − 52
0
5
9
0
2
3. P23
3
− 32
5
− 12
1
0
6 1
49
33
= · 0
2
2
2
0
−17 −16
21
13
0
2
2
4. P 2
3
1
3
5
1
1 −3 −1
0 −5
49
33
0
5 −17 −16
0
9
21
13
5. A23 (−11), A24 (6), A25 (−5)
6. P 2
The 5 × 5 determinant remaining in the last step can be computed via cofactor expansion along the first
column, followed by cofactor expansion along the second column, which then leads to a 3 × 3 determinant
that can be computed by using the schematic in Figure 3.1.1:
−3 2 0 −1 −4
1 3 1
3
5
1
0 2 1 −3 −1 = ·
2
2 0 4 −2 −3
−1 2 6
0 −1
28.
1
0
0
0
0
3
1
3
5
1
1 −3 −1
−5
49
33
1
2302
0 −5
49
33 = ·
5 −17 −16 = −
= −1151.
2
2
0
5 −17 −16
9
21
13
0
9
21
13
5 0 0 −1
3 0 0 0 −3
0 −3 5 0 0
1 2 3
0
1 2 0 3
3
0 1 2 1 2
3
0 4 = −
1
3
0 4 = 0 1
3 0 4 1
1 −1 0 0
1 −1 0 1 0
0 −1 3 5 3 5 0
2
0 5 0 0 −1
0 −3 5
0 −3 5
0 −3
5
0 0 −9 1 −10 4 3
5 1 −10 = − = − −9 1 −10 = −9
1
3
0
4
−1 3
5
1 −3 −5 0 −1 3
5 42 0
50 42
50 6
7
= −9 1 −10 = = −170.
−26 −35 −26 0 −35 2
0
0
1
3
1. A41 (−2),A45 (−3)
5. P2
29.
2. CET(col#1)
6. A21 (−5), A23 (3)
3. A12 (−3)
4. CET(col#1)
7. CET(col#2)
0
xz 0 x + y
z x
y
z xz x + y
z 2 −x
1 −x
0
1
−1
0
1
−1
=
=
−x
1 −1 −y −1
−1
1 0
1 −y − z 0
−y − z
−1
1 −z
−z 1
−1
0 1 −1
0 xz − yz + z 2 x + y + z 0 4 xz − yz + z 2 x + y + z 3 = (x + y + z)2 .
0 0 =
= −x + y − z
−x − y − z
0
−y − z
−1 1
1. A41 (−x), A43 (1)
2. CET(col#2)
3. A32 (1), A31 (−z)
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233
30.
(a).
1 1 1 1
0
0
1 V (r1 , r2 , r3 ) = r1 r2 r3 = r1 r2 − r1 r3 − r1 r12 r22 r32 r12 r22 − r12 r32 − r12 r3 − r1 2 r2 − r1
= 2
= (r2 − r1 )(r32 − r12 ) − (r3 − r1 )(r22 − r12 )
r2 − r12 r32 − r12 = (r3 − r1 )(r2 − r1 )[(r3 + r1 ) − (r2 + r1 )] = (r2 − r1 )(r3 − r1 )(r3 − r2 ).
1. CA12 (−1), CA13 (−1)
2. CET(row#1)
(b). We use mathematical induction. The result is vacuously true when n = 1 and quickly verified when
n = 2. Suppose that the result is true when n = k − 1, for k ≥ 3, and consider
1
1
1
···
1 r1
r2
r3
···
rk 2
2
r12
r2
r3
···
rk2 .
V (r1 , r2 , . . . , rk ) = ..
..
..
.. .
.
.
. k−1
k−1
k−1
k−1 r
r
r
··· r
1
2
3
k
The determinant vanishes when rk = r1 , r2 , . . . , rk−1 , so we can write
V (r1 , r2 , . . . , rk ) = a(r1 , r2 , . . . , rk )
n−1
-
(rk − ri ),
i=1
where a(r1 , r2 , . . . , rk ) is the coefficient of rkk−1 in the expansion of V (r1 , r2 , . . . , rk ). However, using the
Cofactor Expansion Theorem along column k, we see that this coefficient is just V (r1 , r2 , . . . , rk−1 ), so by
hypothesis,
a(r1 , r2 , . . . , rk ) = V (r1 , r2 , . . . , rk−1 ) =
(rm − ri ).
1≤i<m≤n−1
Thus,
-
V (r1 , r2 , . . . , rk ) =
(rm − ri )
1≤i<m≤n−1
n−1
-
(rk − ri ) =
i=1
-
(rm − ri ).
1≤i<m≤n
Hence the result is true for n = k, so, by induction, is true for all non-negative integers n.
31. We have
det(A − λI) = det
2−λ
2
−1
4−λ
= (2 − λ)(4 − λ) + 2 = λ2 − 6λ + 10.
The quadratic formula supplies the two roots of the equation λ2 − 6λ + 10 = 0; namely, λ = 3 ± i.
32. We have
det(A − λI) = det
2−λ
3
4
13 − λ
= (2 − λ)(13 − λ) − 12 = λ2 − 15λ + 14 = (λ − 1)(λ − 14).
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The two roots are λ = 1 and λ = 14.
33. We have
det(A − λI) = det
−1 − λ
−4
2
7−λ
= (−1 − λ)(7 − λ) + 8 = λ2 − 6λ + 1.
√
The quadratic formula supplies the two roots of the equation λ2 − 6λ + 1 = 0; namely, λ = 3 ± 2 2.
34. We have
⎡
2−λ
det(A − λI) = det ⎣ −1
3
0
−6 − λ
3
⎤
0
0 ⎦ = (2 − λ)(−6 − λ)(7 − λ),
7−λ
since the matrix A − λI is lower triangular. Thus, the three roots of the equation (2 − λ)(−6 − λ)(7 − λ) = 0
are λ = 2, λ = −6, and λ = 7.
35. We have
⎡
2−λ
det(A − λI) = det ⎣ 7
7
⎤
0
7 ⎦,
7−λ
0
7−λ
7
and using cofactor expansion along the first row, we obtain
det(A − λI) = (2 − λ)
7−λ
7
7
= (2 − λ)(λ2 − 14λ) = λ(λ − 14)(2 − λ).
7−λ
The roots of λ(λ − 14)(2 − λ) = 0 are λ = 0, λ = 14, and λ = 2.
36. We have
⎡
6−λ
det(A − λI) = det ⎣ 0
−5
0
7−λ
0
⎤
−2
⎦,
0
−3 − λ
and using cofactor expansion along the second row, we obtain
det(A − λI) = (7 − λ)
6−λ
−5
−2
= (7 − λ)(λ2 − 3λ − 28) = (7 − λ)(λ − 7)(λ + 4).
−3 − λ
The roots of (7 − λ)(λ − 7)(λ + 4) = 0 are λ = 7 (repeated twice) and λ = −4.
37. We have
⎡
−5 − λ
det(A − λI) = det ⎣ −8
−5
−5
1−λ
3
⎤
0
0 ⎦,
7−λ
and using cofactor expansion along the third column, we obtain
det(A − λI) = (7 − λ)
−5 − λ
−8
−5
= (7 − λ)(λ2 + 4λ − 45) = (7 − λ)(λ + 9)(λ − 5).
1−λ
The roots of (7 − λ)(λ + 9)(λ − 5) = 0 are λ = 7, λ = −9, and λ = 5.
38. We have
⎡
−5 − λ
det(A − λI) = det ⎣ −2
−5
−1
−1 − λ
2
(c)2017 Pearson Education. Inc.
⎤
1
0 ⎦,
3−λ
235
and using cofactor expansion along the third column, we obtain
det(A−λ) = (−9−5−λ)+(3−λ)[(−5−λ)(−1−λ)−2] = −λ3 −3λ2 +10λ = −λ(λ2 +3λ−10) = −λ(λ+5)(λ−2).
The roots of −λ(λ + 5)(λ − 2) = 0 are λ = 0, λ = −5, and λ = 2.
39.
(a). det(A) = 11;
5 −4
(b). MC =
;
−1
3
5 −1
(c). adj(A) =
;
−4
3
1
5 −1
−1
(d). A =
.
3
11 −4
40.
(a). det(A) = 7;
1 −4
;
2 −1
1
2
(c). adj(A) =
;
−4 −1
1
1
2
.
(d). A−1 =
7 −4 −1
(b). MC =
41.
(a). det(A) = 0;
−6 15
;
−2 5
−6 −2
(c). adj(A) =
;
15
5
(b). MC =
(d). A−1 does not exist because det(A) = 0.
2 −3 0 2 −3 0 4 5 = 2 · 13 = 26;
1 5 = 0
(a). det(A) = 2
0 −1 2 0 −1 2 ⎡
⎤
7 −4 −2
6
4
2 ⎦;
(b). MC = ⎣
−15 −10
8
⎡
⎤
7 6 −15
(c). adj(A) = ⎣ −4 4 −10 ⎦;
−2 2
8
42.
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236
⎡
⎤
7 6 −15
1
1
⎣ −4 4 −10 ⎦.
(d). A−1 =
adj(A) =
det(A)
26
−2 2
8
−2 3 −1 −2 3 −1 4 = −2 · 4 = −8;
5 = 0 4
(a). det(A) = 2 1
0 2
3 3 0 2
⎡
⎤
−7 −6
4
4 ⎦;
(b). MC = ⎣ −11 −6
16
8 −8
⎡
⎤
−7 −11 16
8 ⎦;
(c). adj(A) = ⎣ −6 −6
4
4 −8
⎡
⎤
−7 −11 16
1
1
8 ⎦.
adj(A) = − ⎣ −6 −6
(d). A−1 =
det(A)
8
4
4 −8
43.
1 −1 2 6 0 9 (a). det(A) = 3 −1 4 = 8 0 11 = 6;
5
1 7 5 1 7 ⎡
⎤
−11 −1
8
9 −3 −6 ⎦;
(b). MC = ⎣
−2
2
2
⎡
⎤
−11
9 −2
2 ⎦;
(c). adj(A) = ⎣ −1 −3
8 −6
2
⎡
⎤ ⎡
⎤
−11
9 −2
−11/6
3/2 −1/3
1
1
2 ⎦ = ⎣ −1/6 −1/2
1/3 ⎦.
adj(A) = ⎣ −1 −3
(d). A−1 =
det(A)
6
8 −6
2
4/3
−1
1/3
44.
0
1 2 0
(a). det(A) = −1 −1 3 = 0
1 −2 1 1
⎡
⎤
5
4 3
(b). MC = ⎣ −5 −2 1 ⎦;
5 −2 1
⎡
⎤
5 −5
5
(c). adj(A) = ⎣ 4 −2 −2 ⎦;
3
1
1
⎡
5
1
1
⎣ 4
adj(A) =
(d). A−1 =
det(A)
10
3
45.
1 2 −3 4 = 10;
−2 1 ⎤
−5
5
−2 −2 ⎦.
1
1
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237
2 −3
5 0 −7
3 2
1 = 1
2
1 = 14;
(a). det(A) = 1
0
7 −1 0
7 −1 ⎡
⎤
−9
1
7
(b). MC = ⎣ 32 −2 −14 ⎦;
−13
3
7
⎡
⎤
−9
32 −13
3 ⎦;
(c). adj(A) = ⎣ 1 −2
7 −14
7
⎡
⎤ ⎡
⎤
−9
32 −13
−9/14 16/7 −13/14
1
1
⎣ 1 −2
3 ⎦ = ⎣ 1/14 −1/7
3/14 ⎦.
adj(A) =
(d). A−1 =
det(A)
14
7 −14
7
1/2
−1
1/2
46.
1 1
1
1 1
−1 1 −1
1 0
(a). det(A) = =
1 1 −1 −1 0
−1 1
1 −1 0
⎤
⎡
4 4
4
4
⎢ −4 4 −4
4 ⎥
⎥
(b). MC = ⎢
⎣ 4 4 −4 −4 ⎦;
−4 4
4 −4
⎡
⎤
4 −4
4 −4
⎢ 4
4
4
4 ⎥
⎥;
(c). adj(A) = ⎢
⎣ 4 −4 −4
4 ⎦
4
4 −4 −4
⎡
4 −4
⎢
1
1
4
⎢ 4
(d). A−1 =
adj(A) =
⎣
4
−4
det(A)
16
4
4
47.
1
1
1 2
0
2 2
0
2 0 = 16;
= 2 −2
2 −2
0 2
0 −2 2
0 −2 ⎤
4 −4
4
4 ⎥
⎥.
−4
4 ⎦
−4 −4
1 0 3
5 11 0
−2 1 1
3 4 1
=
(a). det(A) = 2 7 9
3 9 0
2 0 3 −1 0 0
⎤
⎡
84 −46 −29
81
⎢ −162
60
99 −27 ⎥
⎥;
(b). MC = ⎢
⎣
18
38 −11
3 ⎦
−30
26 130 −72
⎤
⎡
84 −162
18 −30
⎢ −46
60
38
26 ⎥
⎥;
(c). adj(A) = ⎢
⎣ −29
99 −11 130 ⎦
81 −27
3 −72
48.
18
5 11
10
3 4
=
−
6
2 7
0 −1
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0
1
9
11 0
18 18 4 1
10 = 402;
10 = − −29 0 −84 6 238
(d).
⎤
84 −162
18 −30
1
1 ⎢
60
38
26 ⎥
⎥
⎢ −46
adj(A) =
A−1 =
⎦
⎣
−29
99
−11
130
det(A)
402
81 −27
3 −72
⎤
⎡
14/67 −27/67
3/67 −5/67
⎢ −23/201
10/67
19/201 13/201 ⎥
⎥
=⎢
⎣ −29/402 33/134 −11/402 65/201 ⎦ .
27/134 −9/134
1/134 −12/67
⎡
49.
(a). Using the Cofactor Expansion Theorem along row 1 yields
det(A) = (1 + 2x2 ) + 2x(2x + 4x3 ) + 2x2 (2x2 + 4x4 )
= 8x6 + 12x4 + 6x2 + 1 = (1 + 2x2 )3 .
(b).
⎤
−(2x + 4x3 )
2x2 + 4x4
1 + 2x2
1 − 4x4
−(2x + 4x3 ) ⎦
MC = ⎣ 2x + 4x3
2
4
3
2x + 4x
2x + 4x
1 + 2x2
⎤
⎡
−2x(1 + 2x2 )
2x2 (1 + 2x2 )
1 + 2x2
= ⎣ 2x(1 + 2x2 ) (1 + 2x2 )(1 − 2x2 ) −2x(1 + 2x2 ) ⎦ ,
2x2 (1 + 2x2 )
2x(1 + 2x2 )
1 + 2x2
⎡
so that
⎡
1
1
1
⎣ −2x
adj(A) =
A−1 =
det(A)
(1 + 2x2 )2
2x2
50. det(A) =
2x
1 − 2x2
−2x
−1 −2
4
−1 −2
4
2 −1
0
2 −1 = −
0
2 −1 =
= −18 and
−8 13
3 −2
1
0 −8 13
C11 =
2 −1
= 0.
−2
1
Thus,
1
51. det(A) = 1
1
and
1
2
2
1 1
2 = 0
3 0
1
1
1
(A−1 )11 =
(adj(A))11
0
=
= 0.
det(A)
−18
1 1
1 = 1
2 1 = 2 − 1 = 1,
2 1
C23 = − 1
1 = −(2 − 1) = −1.
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⎤
2x2
2x ⎦ .
1
239
Thus,
(adj(A))32
C23
=
= −1.
det(A)
det(A)
1 0 4
1 1 1 = −
= −(−4 − 3) = 7,
3 −1 −1 0 (A−1 )32 =
2
52. det(A) = 2
3
and
0 −1 4
1
1 = 2
−1
0 3
2
1 C13 = = −2 − 3 = −5.
3 −1 Thus,
(A−1 )31 =
C13
5
(adj(A))31
=
=− .
det(A)
det(A)
7
1
0
0 0 0
1 0 1
2 −1 −1 3 −1 −1 3 2 −1
1
3
1 −1 2 =
=
det(A) = 1 −1 2 1 −1 2 0
0
1
3 0 −1
1
3 0 1
2 0 −1
−1
2 3 2 3 = 4 + 12 = 16,
= 1 −4 2 = −4 2 1
0 0 53.
and
1 1 0 1
C32 = − 2 1 3 = −(−3) −1
−1 2 0 1 = 3(2 + 1) = 9.
2 Thus,
(adj(A))23
C32
9
=
=
.
det(A)
det(A)
16
−t
−et cos(2t)
e sin(2t) et cos(2t)
, adj(A) =
, and
−et cos(2t) et sin(2t)
et sin(2t)
(A−1 )23 =
54. MC =
e−t sin(2t)
−et cos(2t)
det(A) = sin2 (2t) + cos2 (2t) = 1,
so
−t
1
e sin(2t) et cos(2t)
.
adj(A) =
A =
−et cos(2t) et sin(2t)
det(A)
−2et
2e2t −e2t
,
adj(A)
=
, and det(A) = 4e3t , so
3et
−2et
3et
1
1
2e2t −e2t
.
A−1 =
adj(A) = 3t
−2et
3et
det(A)
4e
−1
55. MC =
2e2t
−e2t
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56. We compute
⎡
−et
⎣
−et
MC =
t
e (9t + 1)
and thus,
⎡
−et
T
adj(A) = MC = ⎣ −tet
0
−tet
−tet
t
−e (1 − t)
⎤
0
−e6t (9t2 + 1) ⎦ ,
e6t (9t2 + 1)
−et
−tet
6t
−e (9t2 + 1)
⎤
et (9t + 1)
−et (1 − t) ⎦ .
e6t (9t2 + 1)
We can find det(A) via the schematic in Figure 3.1.1:
det(A) = 0 − 9t2 e4t + te4t − te4t − e4t − 0 = −e4t (9t2 + 1).
Therefore,
⎡
−et
−4t
1
e
−1
⎣ −tet
A =
· adj(A) = − 2
det(A)
9t + 1
0
⎤
et (9t + 1)
−et (1 − t) ⎦ .
e6t (9t2 + 1)
−et
−tet
6t
−e (9t2 + 1)
⎤
⎤
⎡
−e−t −te2t
3te−t −te−t −te−t
e−t
0 ⎦ , adj(A) = ⎣ −e−t
e−t
0 ⎦ , and
2t
2t
−te
0
te
0
te2t
t
e 2tet
e−2t et tet e−2t t
e−2t = 0 tet
0 = et (te−t ) = t,
det(A) = e 2tet
t
−2t −2t et
0 0 e
te 2e
⎡
3te−t
57. MC = ⎣ −te−t
−te−t
so
⎡
3te−t
1
1⎣
−1
−e−t
A =
adj(A) =
det(A)
t
−te2t
⎤
−te−t −te−t
e−t
0 ⎦.
0
te2t
⎡
⎤
⎡
⎤
−1
2 −1
−1
3 −2
3 ⎦ , adj(A) = ⎣ 2 −6
4 ⎦ . Hence,
58. MC = ⎣ 3 −6
−2
4 −2
−1
3 −2
⎡
1
A · adj(A) = ⎣ 3
4
2
4
5
⎤⎡
⎤ ⎡
3
−1
3 −2
0
5 ⎦ ⎣ 2 −6
4 ⎦=⎣ 0
6
−1
3 −2
0
0
0
0
⎤
0
0 ⎦ = 03 .
0
From Equation (3.3.4) of the text we have that, in general, A · adj(A) = det(A) · In . Since, for the given
matrix, A · adj(A) = 03 , we must have det(A) = 0.
2 −3 2 2 2 −3 = 6. Thus,
= 16, and det(B2 ) = = 7, det(B1 ) = 59. det(A) = 4
2 1 4 1
2 x1 =
det(B1 )
16
=
and
det(A)
7
x2 =
det(B2 )
6
= .
det(A)
7
Solution: (16/7, 6/7).
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60. det(A) =
1 5
1 5
= 26,
= 21, det(B1 ) =
−3 6
−4 6
x1 =
det(B1 )
26
=
det(A)
21
and
1
−3
and
det(B2 ) =
x2 =
det(B2 )
1
=− .
det(A)
21
1
= −1. Thus,
−4
Solution: (26/21, −1/21).
61.
det(A) =
−2
4
1
3 −2 −1 = −27,
4 −3
2
det(B1 ) =
−5
4
1
2 −2 −1 = 11,
1 −3
2
det(B2 ) =
−2 −5
1
3
2 −1 = 35,
4
1
2
det(B3 ) =
−2
4 −5
3 −2
2 = 17.
4 −3
1
Thus,
x1 =
det(B1 )
det(B2 )
det(B3 )
11
35
17
= − , x2 =
=
, x3 =
=− .
det(A)
27
det(A)
27
det(A)
27
Solution: (−11/27, 35/27, −17/27).
62.
3
det(A) = 1
1
4
det(B1 ) = 2
1
3
det(B2 ) = 1
1
3
det(B3 ) = 1
1
−2
1 1 −1 = 6,
0
1 −2
1 1 −1 = 9,
0
1 4
1 2 −1 = 0,
1
1 −2 4 1 2 = −3.
0 1 det(B2 )
det(B3 )
9
3
det(B1 )
= , x2 =
= 0, and x3 =
=− .
det(A)
6
det(A)
det(A)
6
Solution: (3/2, 0, −1/2).
1 −3
1 1 1 −3
7 −2 7 −2 = 4 −1 = 0
= −35 + 14 = −21 = 0.
63. det(A) = 1
7 −5 2
7 −5 1 −3 0
Thus, the system has only the trivial solution.
Thus, x1 =
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64.
1 −2
3 −1 3 −1 −5 −2
2
0
1
0 0
1
0 0
det(A) = = 1
1
0
−1
1
1
0
−1
0
1 −2
1 1 −2
1 4
−1 −1 0 −5 −2 −1 2 0 = −3,
1 −1 = − 5
= − 1
4
4
1 1 1
1 1 −2
3 −1 1 −3
3 −1 3 3 −5 −3
1 −3
2
2
0
1
0
0
1
0
= 2
=
0
1 = −11,
0
1 = 0
det(B1 ) = 0
0 −1 1
0 −1 0
0
7
2 −2 3
2 −2
3
2 −2
1 1 −2
1 3
1 1
3 0 3 −1 1 1
3 0 3 0 1
1
2 2
2 2
1
2
1
0
= 2
=
5 0 = 17,
1 2 = −4
det(B2 ) = 0 0 0 −1 1 0
1 0
3 −2 1 3 −2 1
0 3 −2
1 0 3 −2 1 1 −2 1 −1 0 −2 1 −1 2
0 2
0 0 2
0 0
det(B3 ) = =
1 0 −1 1
1 0 −1 1
0
1 3
1 1 3
1 −3
−3 −1 0 0 −2 −1 2 0 = −2(−8) = 16,
1 −1 = −2 −2
= −2 1
−3
−3
1 1 1
1 1 −2
3 1 3 −2
3 1 3
3 1 2
2
0
1
2
0
1
2
=
= − 2
1 2 det(B4 ) = 1
0 0 0
1
0 0 −1 −2 3 1
0
1 −2 3 −1
1 −2 3 −3 3 −5 −3 −5 = 6.
0 = −
= − 0 1
3
7 3 2
7
Therefore,
x1 =
17
16
11
, x2 = − , x3 = − ,
3
3
3
and x4 = −2.
Solution: (11/3, −17/3, −16/3, −2).
65.
t
e
1 e−2t t −2t 1
−t
e
=
e
det(A) = t
1 −2 = −3e ,
e −2e−2t 3 sin t
e−2t 1 −2t 3 sin t
−2t
=
e
det(B1 ) = 4 cos t −2 = −2e (3 sin t + 2 cos t),
4 cos t −2e−2t t
e 3 sin t = et 1 3 sin t = et (4 cos t − 3 sin t).
det(B2 ) = t
1 4 cos t e 4 cos t
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det(B1 )
det(B2 )
2(3 sin t + 2 cos t)
e2t (3 sin t − 4 cos t)
Thus, x1 =
and x2 =
=
=
.
t
3e
det(A)
3
det(A)
1
2 −t
e (3 sin t + 2 cos t), e2t (3 sin t − 4 cos t) .
Solution:
3
3
66.
1 4 −2
1 1 −1 −2 2 2 9 −3 −2 2 −1 −3 0 −1 −2 2 1 −16 0 det(A) = =
= −1 −3 0 = −1 −3 0 = −19,
0 −1 1
0
0 0 1 5
−1
7 1 −1
7 1 3 14
7 −2 3 −1
7 1 1 2 −2
1 1 −1 −2 2 2 5 −3 −2 2 −1 −3 0 −1 −2 2 5 −16 0 = −1 −3 0 = −1 −3 0 =
det(B2 ) = 0
0 0 0 −1 1
1 3
7 1 −3
7 1 −3
3 6
7 1 7 −2 3 −3
5 −16 = −31. Therefore x2 = det(B2 ) = 31 .
=
−1 −3 det(A)
19
67. The coefficient matrix for this linear system is
⎤
−3
1 −3 −9
⎢ 1 −2
0
4 ⎥
⎥.
A=⎢
⎣ 0
0
2
1 ⎦
1
1
0
1
⎡
By applying the cofactor expansion theorem to the third row of A, we obtain
det(A) = 2 · C33 + 1 · C34 = 2 · (−6) + 1 · 9 = −3.
To find x4 , we form
⎤
−3
1 −3 −3
⎢ 1 −2
0
1 ⎥
⎥,
B4 = ⎢
⎣ 0
0
2 −1 ⎦
1
1
0
0
⎡
and using the cofactor expansion theorem on the last row, we have
det(B4 ) = 1 · C41 + 1 · C42 = 1 · (−16) + 1 · (−3) = −19.
Thus, by Cramer’s Rule we conclude that
x4 =
det(B4 )
−19
19
=
=
.
det(A)
−3
3
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68.
b+c
a
a −a + b + c
a
a a+c
b = −a + b − c a + c
b det(A) = b
c
c
a+b 0
c
a+b −a + b + c a −a + b + c
a = 4abc,
= −c + (a + b) −a + b − c b −a + b − c a + c a
0
a a
a a
b b = b a − b + c
det(B1 ) = b a + c
c
0
a+b c
a+b c
a
a = a(a − b + c)(a + b − c),
= (a − b + c) c a+b b+c a
a b + c a − b − c
a b
b = b
0
b det(B2 ) = b
c
c a+b c
0
a+b b
b = −b(a − b − c)(a + b − c),
= −(a − b − c) c a+b b+c
a
a a
a −a + b + c
0
a + c b a + c b = det(B3 ) = b
c
0
c
c c
c
a+c b = −c(a − b − c)(a − b + c).
= (−a + b + c) c
c Case 1: If a = 0, b = 0, and c = 0, then det(A) = 0, so it follows by Theorem 3.2.5 that the given system of
equations has a unique solution. The solution in this case is given by:
det(B1 )
(a − b + c)(a + b − c)
=
,
det(A)
4bc
det(B2 )
−(a − b − c)(a + b − c)
x2 =
=
,
det(A)
4ac
det(B3 )
−(a − b − c)(a − b + c)
x3 =
=
.
det(A)
4ab
x1 =
Case 2: If a = b = 0 and c = 0, then it is easy to see from the system that the system is inconsistent. By
the symmetry of the equations, the same will be true if a = c = 0 with b = 0, or b = c = 0 with a = 0.
Case 3: Suppose a = 0 where b = 0 and c = 0, and consider the reduced row echelon form of the augmented
matrix:
⎡
⎤ ⎡
⎤ ⎡
⎤
0
b+c 0 0 0
1
0
0
1 0 0 0
⎣ c
b b b ⎦ ∼ ⎣ 0 b − c b − c b − c ⎦ ∼ ⎣ 0 1 1 1 ⎦.
b
c c c
0 c−b c−b c−b
0 0 0 0
From the last matrix we see that the system has an infinite number of solutions of the form
{(0, 1 − r, r) : r ∈ R}. By the symmetry of the three equations, it follows that the other two cases: b = 0
with a = 0 and c = 0, and c = 0 with a = 0 and b = 0 have similar forms of solutions.
69. Let B be the matrix obtained from A by adding column i to column j (i = j) in the matrix A. By
the property for columns corresponding to Property P3, we have det(B) = det(A). Cofactor expansion of B
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along column j gives
det(A) = det(B) =
n
,
(akj + aki )Ckj =
k=1
That is,
det(A) = det(A) +
n
,
akj Ckj +
k=1
n
,
n
,
aki Ckj .
k=1
aki Ckj ,
k=1
since by the Cofactor Expansion Theorem the first summation on the right-hand side is simply det(A). It
follows immediately that
n
,
aki Ckj = 0,
i = j.
k=1
73.
⎡
1.21
A = ⎣ 5.41
21.63
⎡
1.21
B2 = ⎣ 5.41
21.63
⎤
⎡
⎤
3.42 2.15
3.25 3.42 2.15
2.32 7.15 ⎦ , B1 = ⎣ 4.61 2.32 7.15 ⎦ ,
3.51 9.22
9.93 3.51 9.22
⎤
⎡
⎤
3.25 2.15
1.21 3.42 3.25
4.61 7.15 ⎦ , B3 = ⎣ 5.41 2.32 4.61 ⎦ .
9.93 9.22
21.63 3.51 9.93
From Cramer’s Rule,
x1 =
det(B1 )
det(B2 )
det(B3 )
≈ 0.25, x2 =
≈ 0.72, x3 =
≈ 0.22.
det(A)
det(A)
det(A)
74. det(A) = 32, det(B1 ) = −3218, det(B2 ) = 3207, det(B3 ) = 2896, det(B4 ) = −9682, det(B5 ) = 2414.
So,
x1 = −
1609
3207
181
4841
1207
, x2 =
, x3 =
, x4 = −
, x5 =
.
16
32
2
32
16
75. We have
(BA)ji =
n
,
bjk aki =
k=1
n
,
k=1
,
1
1
Ckj aki = δij ,
· adj(A)jk · aki =
det(A)
det(A)
where we have used Equation (3.3.4) in the last step.
Solutions to Section 3.4
Problems:
1. | − 3| = −3.
5 −1 = 5 · 7 − 3(−1) = 38.
2. 3
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k=1
246
3 5
3. −1 2
6 3
5 1
4. 6 1
14 2
7 4 = −43.
−2 4 3 = −3.
7 2.3 1.5 7.9 4.2 3.3 5.1 = 2.3 3.3
3.6
6.8 3.6 5.7 5.
4.2 5.1 4.2 3.3 5.1 + 7.9 − 1.5 6.8 5.7 5.7 6.8 3.6 = 1.035 + 16.11 − 57.828 = −40.683.
a
b
c
6.
b
c
a
c c
a = a a
b b
a −
b
c
b
b
a +
c
c
b
c a = a(bc − a2 ) − b(b2 − ac) + c(ab − c2 ) = 3abc − a3 − b3 − c3 .
8.
8 −7 −1 5 −1 2 0
8 −7 −1 3
1
0 1
5 2 0
1
0 = −1 3
=
5 −1
4 2
5 7 0
5 −1
4 2
1 −1
2 1 1 −1
29 −7 −1 29 −1 = −124.
1
0 = −1 = −1 0
8
4 8 −1
4
3
2
3
1
7.
7
1 2
2 −2 4
3 −1 5
18
9 27
9. det(A) = 11. MC =
1
10. det(A) = 2
3
2
3
1
7 1 2 3 7
3 1 2 3 16
16 0 8 12 2 −2 4 6 6 = 9
= 9 10 0 7 7 = −9 10
3
−1
5
4
4 −5
−5 0 1 3 2
1 3 6 54 14 0 −3 4 2 3 = −36 10 7 7 = −36 45 0 −14 −5 1
−5 1 3 3 14 −3 = −2196.
= 36 45 −14 7 −5
=⇒ adj(A) =
, so that
−2
3
1
7/11 −5/11
A−1 =
adj(A) =
.
−2/11
3/11
det(A)
7 −2
−5
3
3 1
1 = 0
2 0
2
3 −1 −5 = −18.
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8
7
1
12 7 3 247
⎡
⎤
⎡
⎤
5 −1 −7
5 −1 −7
5 ⎦ =⇒ adj(A) = ⎣ −1 −7
5 ⎦, so that
MC = ⎣ −1 −7
−7
5 −1
−7
5 −1
⎡
⎤
−5/18
1/18
7/18
1
7/18 −5/18 ⎦ .
adj(A) = ⎣ 1/18
A−1 =
det(A)
7/18 −5/18
1/18
3 4
7 7 −11 −38
−11 −38 = 30.
0
0
1 = − 1 = 11. det(A) = 2 6
5
20 3 14 −1 5
20 −1 ⎡
⎤
⎡
⎤
−20
5
10
−20 102 −38
5 −24
11 ⎦, so that
MC = ⎣ 102 −24 −30 ⎦ =⇒ adj(A) = ⎣
−38
11
10
10 −30
10
⎡
⎤
−2/3 17/5 −19/15
1
11/30 ⎦ .
adj(A) = ⎣ 1/6 −4/5
A−1 =
det(A)
1/3
−1
1/3
2
5 7 12. det(A) = 4 −3 2 = 116.
6
9 11 ⎡
⎤
⎡
⎤
−51 −32
54
−51
8
31
8 −20
12 ⎦ =⇒ adj(A) = ⎣ −32 −20
24 ⎦, so that
MC = ⎣
31
24 −26
54
12 −26
⎡
⎤
−51/116
2/29 31/116
1
−8/29 −5/29
6/29 ⎦ .
adj(A) = ⎣
A−1 =
det(A)
27/58
3/29 −13/58
5 −1
2 5 3 −1
4 5 = −152.
13. det(A) = 2 1 1 −1
5
9 −3 2 ⎤
⎤
⎡
⎡
−38
0
38
0
−38
32
34
2
⎢ 32
⎢
28 −124 −24 ⎥
0
28
44 −60 ⎥
⎥, so that
⎥
⎢
MC = ⎢
⎣ 38 −124 −222 130 ⎦ =⇒ adj(A) = ⎣ 34
44 −222 −16 ⎦
2 −60
130
8
0 −24 −16
8
⎤
⎡
1/4
0
−1/4
0
⎢ −4/19 −7/38
1
31/38
3/19 ⎥
⎥.
A−1 =
adj(A) = ⎢
⎣ −17/76 −11/38 111/76
2/19 ⎦
det(A)
−1/76
15/38 −65/76 −1/19
14. Applying cofactor expansion to the first row of A, we find that
det(A) = (−1) · C11 + 4 · C14 = (−1) · 2 + 4 · 14 = 54.
To apply the adjoint method, we must first compute the matrix of cofactors:
⎤
⎡
2 26 −16 14
⎢ −4 56 −22 −1 ⎥
⎥.
MC = ⎢
⎣ 12 −6
12
3 ⎦
4 −2
22
1
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Thus,
⎤
2
−4 12
4
⎢ 26
56 −6 −2 ⎥
⎥
adj(A) = MCT = ⎢
⎣ −16 −22 12 22 ⎦ .
14 −1
3
1
⎡
Therefore, we have
15. det(A) =
⎤
2
−4 12
4
1 ⎢
56 −6 −2 ⎥
⎥.
⎢ 26
A−1 =
54 ⎣ −16 −22 12 22 ⎦
14
−1
3
1
⎡
2 8
0 8
2
0
= 24, det(B1 ) =
= 24, det(B2 ) =
= −6. Thus,
−2 4
−3 4
−2 −3
x1 =
det(B1 )
24
=
=1
det(A)
24
Solution: (1, − 14 ).
4
3 5
16. A =
, B1 =
9
6 2
17. A =
cos t
sin t
x2 =
det(B2 )
6
1
=−
=− .
det(A)
24
4
3 4
, B2 =
, so that
6 9
det(B2 )
det(B1 )
37
1
=
, x2 =
=− .
x1 =
det(A)
24
det(A)
8
5
2
and
−t
sin t
e
cos t
e−t
sin t
,
B
, so that
=
, B1 =
2
3e−t − cos t
sin t 3e−t
− cos t
det(B1 )
det(B2 )
x1 =
= e−t [cos t + 3 sin t], x2 =
= e−t [sin t − 3 cos t].
det(A)
det(A)
⎤
⎡
⎤
⎡
⎡
⎤
⎤
4
1 3
5
1 3
4 5 3
4
1 5
18. A = ⎣ 2 −1 5 ⎦ , B1 = ⎣ 7 −1 5 ⎦ , B2 = ⎣ 2 7 5 ⎦ , B3 = ⎣ 2 −1 7 ⎦, so that
2
3 1
2
3 1
2
3 2
2 2 1
det(B1 )
det(B2 )
det(B3 )
1
1
21
= , x2 =
=
, x3 =
=
.
x1 =
det(A)
4
det(A)
16
det(A)
16
⎡
⎡
5
19. A = ⎣ 2
2
⎡
⎤
⎡
⎤
⎡
⎤
⎤
3 3
6
5
3
6
5 3
3
3
6
4 −7 ⎦ , B1 = ⎣ −1 4 −7 ⎦ , B2 = ⎣ 2 −1 −7 ⎦ , B3 = ⎣ 2 4 −1 ⎦, so that
4 5
9
2
4
9
2 5
4
5
9
det(B1 )
det(B2 )
det(B3 )
30
59
81
=
, x2 =
=
, x3 =
=
.
x1 =
det(A)
271
det(A)
271
det(A)
271
⎡
⎤
3.6 3.5
3.1 3.5 7.1
20. A = ⎣ 2.2 5.2 6.3 ⎦ , B1 = ⎣ 2.5 5.2
9.3 8.1
1.4 8.1 0.9
so that
⎡
x1 =
⎤
⎡
7.1
3.1
6.3 ⎦ , B2 = ⎣ 2.2
0.9
1.4
3.6
2.5
9.3
⎤
⎡
⎤
7.1
3.1 3.5 3.6
6.3 ⎦ , B3 = ⎣ 2.2 5.2 2.5 ⎦,
0.9
1.4 8.1 9.3
det(B1 )
det(B2 )
det(B3 )
= 3.77, x2 =
= 0.66, x3 =
= −1.46.
det(A)
det(A)
det(A)
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21. Since A is invertible, A−1 exists. Then,
AA−1 = In =⇒ det(AA−1 ) = det(In ) = 1 =⇒ det(A) det(A−1 ) = 1, so that det(A−1 ) =
⎡
a
22. Write A = ⎣ d
g
b
e
h
1
.
det(A)
⎤
c
f ⎦. By cofactor expansion along the first row, we have
i
det(A) = a(ei − f h) − b(di − f g) + c(dh − eg) = aei + bf g + cdh − af h − bdi − ceg.
⎡
⎤
a d g
Next, we have AT = ⎣ b e h ⎦. By cofactor expansion along the first row, we have
c f i
det(AT ) = a(ei − f h) − d(bi − ch) + g(bf − ce) = aei + bf g + cdh − af h − bdi − ceg.
Therefore, we see that det(A) = det(AT ).
23. det(2A) = 23 det(A) = 24.
24. det(A−1 ) =
1
1
= .
det(A)
3
25. det(AT B) = det(AT ) det(B) = det(A) det(B) = −12.
26. det(B 5 ) = [det(B)]5 = −1024.
27. det(B −1 AB) = det(B −1 ) det(A) det(B) =
32 = 9.
1
det(A) det(B) = det(A) = 3, so that det(B −1 AB)2 =
det(B)
28. Interchanging two columns multiplies the determinant by −1 and multiplying a column by 4 multiplies
the determinant by 4. Therefore, det(C) = −1 · 4 · det(B) = 16.
29. Multiplying a row by 3 multiplies the determinant by 3 and adding the first row to the last row does
not change the determinant. Thus, det(C) = 3 · det(A) = 9.
Solutions to Section 3.5
Additional Problems:
1.
(a). We have det(A) = (6)(1) − (−2)(6) = 18.
(b). We have
1
A∼
1
−2
1
1
2
∼
1. M1 (1/6)
1
0
1
3
= B.
2. A12 (2)
Now, det(A) = 6 det(B) = 6 · 3 = 18.
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(c). Let us use the Cofactor Expansion Theorem along the first row:
det(A) = a11 C11 + a12 C12 = (6)(1) + (6)(2) = 18.
2.
(a). We have det(A) = (−7)(−5) − (1)(−2) = 37.
(b). We have
1
1 −5
−7 −2
A∼
2
∼
1. P12
1 −5
0 −37
= B.
2. A12 (7)
Now, det(A) = −det(B) = −(−37) = 37.
(c). Let us use the Cofactor Expansion Theorem along the first row:
det(A) = a11 C11 + a12 C12 = (−7)(−5) + (−2)(−1) = 37.
3.
(a). Using Equation (3.1.2), we have
det(A) = (2)(0)(3)+(3)(2)(6)+(−5)(−4)(−3)−(−6)(0)(−5)−(−3)(2)(2)−(3)(−4)(3) = 0+36−60+−0+12+36 = 24.
(b). We have
⎡
⎤ ⎡
2
3 −5
2
2
6 −8 ⎦ ∼ ⎣ 0
A∼⎣ 0
0 −12 18
0
1
1. A12 (2), A13 (−3)
⎤
3 −5
6 −8 ⎦ = B.
0
2
2. A23 (2)
Now, det(A) = det(B) = (2)(6)(2) = 24.
(c). Let us use the Cofactor Expansion Theorem along the second row:
det(A) = (−4)(6) + (0)(36) + (2)(24) = −24 + 0 + 48 = 24.
4.
(a). Using Equation (3.1.2), we have
det(A) = (−1)(2)(−3)+(4)(2)(2)+(1)(0)(2)−(−1)(2)(2)−(4)(0)(−3)−(1)(2)(2) = 6+16+0−(−4)−0−4 = 22.
(b). We have
⎡
−1
A∼⎣ 0
0
1
⎤ ⎡
4
1
−1
2
2
2 ⎦∼⎣ 0
10 −1
0
1. A13 (2)
⎤
4
1
2
2 ⎦ = B.
0 −11
2. A23 (−5)
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Now, det(A) = det(B) = (−1)(2)(−11) = 22.
(c). Let us use the Cofactor Expansion Theorem along the first column:
det(A) = a11 C11 + a21 C21 + a31 C31 = (−1)(−10) + (0)(14) + (2)(6) = 22.
5.
(a). Of the 24 terms in the determinant expression (3.1.3), the only nonzero term is a41 a32 a23 a14 , which
has a positive sign: σ(4, 3, 2, 1) = +1. Therefore, det(A) = (−3)(1)(−5)(−2) = −30.
(b). By permuting the first and last rows, and by permuting the middle two rows, we bring A to upper
triangular form. The two permutations introduce two sign changes, so the resulting upper triangular matrix
has the same determinant as A, and it is (−3)(1)(−5)(−2) = −30.
0
0 −2 1 . This latter
(c). Cofactor expansion along the first column yields: det(A) = −(−3) · 0 −5
1 −4
1 determinant can be found by cofactor expansion along the first column: (1)[(0)(1) − (−5)(−2)] = −10.
Hence, det(A) = 3 · (−10) = −30.
6.
(a). Of the 24 terms appearing in the determinant expression (3.1.3), only terms containing the factors a11
and a44 will be nonzero (all other entries in the first column and fourth row of A are zero). Looking at entries
in the second and third rows and columns of A, we see that only the product a23 a32 is nonzero. Therefore,
the only nonzero term in the summation (3.1.3) is a11 a23 a32 a44 = (3)(1)(2)(−4) = −24. The permutation
associated with this term is (1, 3, 2, 4) which contains one inversion. Therefore, σ(1, 3, 2, 4) = −1, and so the
determinant is (−24)(−1) = 24.
(b). We have
⎡
⎤
3 −1 −2
1
2
1 −1 ⎥
1 ⎢ 0
⎥ = B.
A∼⎢
⎣ 0
0
1
4 ⎦
0
0
0 −4
1. P23
Now, det(A) = −det(B) = −(3)(2)(1)(−4) = 24.
0 1
4 (c). Cofactor expansion along the first column yields: det(A) = 3· 2 1 −1 . This latter determinant can
0 0 −4 be found by cofactor expansion along the last column: (−4)[(0)(1) − (2)(1)] = 8. Thus, det(A) = 3 · 8 = 24.
7. To obtain the given matrix from A, we add 5 times the middle row to the top row, we multiply the last
row by 3, we multiply the middle row by −1, we add a multiple of the last row to the middle row, and we
perform a row permutation. The combined effect of these operations is to multiply the determinant of A by
(1) · (3) · (−1) · (−1) = 3. Hence, the given matrix has determinant det(A) · (3) = 4 · (3) = 12.
8. To obtain the given matrix from A, we perform two row permutations, multiply a row through by −4,
and multiply a row through by 2. The combined effect of these operations is to multiply the determinant of
A by (−1)2 · (−4) · (2) = −8. Hence, the given matrix has determinant det(A) · (−8) = 4 · (−8) = −32.
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9. To obtain the given matrix from A, we permute the bottom two rows, multiply a row by 2, multiply a
row by −1, add −1 times one row to another, and then multiply each row of the matrix by 3. The combined
effect of these operations is to multiply the determinant of A by (−1)(2)(−1)(1)(3)3 = 54. Hence, the given
matrix has determinant det(A) · 54 = (4)(54) = 216.
10. To obtain the given matrix from AT , we do two row permutations, multiply a row by −1, multiply a
row by 3, and add 2 times one row to another. The combined effect of these operations is to multiply the
determinant of A by (−1)(−1)(−1)(3)(1) = −3. Hence, the given matrix has determinant det(A) · (−3) =
(4) · (−3) = −12.
2
3
11. We have det(B 2 A−1 ) = (det(B))2 det(A−1 ) = −2
= −4.5.
12. We have det(AB) = det(A)det(B) = (−2) · 3 = −6.
13. We have det((−A)3 (2B 2 )) = det(−A)3 det(2B 2 ) = (−1)3 (detA)3 · 24 · (detB)2 = (−1)3 (−2)3 · 24 · 32 =
1152.
14. We have det((A−1 B)T (2B −1 )) = det(A−1 B) · det(2B −1 ) = − 12 · 3 · 24 · 13 = −8.
15. det(A) cannot be computed.
16. det(B) cannot be computed.
17. det(C) = −18.
18. det(C T ) = −18.
19. We have AB =
⎡
8 −10
25
28
4
20. We have BA = ⎣ 1
18
, and so det(AB) = 474.
⎤
5
2
8 −13 ⎦, we have det(BA) = 0.
15
24
21. We have det(B T AT ) = det((AB)T ) = det(AB) = 474.
22. We have det(BAC) = det(BA)det(C) = 0 · (−18) = 0.
38
54
23. We have det(ACB) = det
= 4104.
133 297
⎡
⎤
1 2
1 2−1
6 0
⎣ 2 1 ⎦ =
24. Note that AAT =
, so that det(AAT ) = 6 · 21 = 126.
2
1
4
0 21
−1 4
Therefore, we conclude that det((AAT )2 ) = [det(AAT )]2 = 1262 = 15876.
⎡
⎤
⎡
⎤
2
1 4
5
2
1 2 −1
8 −13 ⎦, and using the schematic in Figure
25. Note that BA = ⎣ 5 −2 ⎦
=⎣ 1
2 1
4
4
7
18 15
24
1
· 0 = 0.
3.1.1., for example, we find that det(BA) = 0. Therefore, det(C −1 BA) = det(C −1 )det(BA) = − 18
26. Since B is a 3 × 2 matrix and C is a 3 × 3 matrix, we cannot compute BCC T .
1 −1
1 −4
, and so adj(B) =
. Since det(B) = 1,
27. The matrix of cofactors of B is MC =
−4
5
−1
5
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we have B −1 =
1 −4
−1
5
. Therefore,
(A−1 B T )−1 = (B T )−1 A = (B −1 )T A =
28.
⎤
16 −1 −5
5 −3 ⎦;
MC = ⎣ 4
−4
2 10
⎡
⎤
16
4 −4
5
2 ⎦;
adj(A) = ⎣ −1
−5 −3 10
det(A) ⎡
= 28;
⎤
4/7
1/7
−1/7
1/14 ⎦.
A−1 = ⎣ −1/28 5/28
−5/28 −3/28 5/14
29.
⎤
5
12 −11 −1
⎢ 65 −24 −23 −13 ⎥
⎥;
MC = ⎢
⎣ −25
0 −5
5 ⎦
−35
0
5 −5
⎤
⎡
5
65 −25 −35
⎢ 12 −24
0
0 ⎥
⎥;
adj(A) = ⎢
⎣ −11 −23 −5
5 ⎦
−1 −13
5 −5
det(A) = −60;
⎤
⎡
5
65 −25 −35
1 ⎢ 12 −24
0
0 ⎥
⎥.
A−1 = − ⎢
⎦
⎣
−11
−23
−5
5
60
−1 −13
5 −5
30.
⎤
88 −24 −40 −48
⎢ 32
12 −20
0 ⎥
⎥;
MC = ⎢
⎣ 16
12 −4
0 ⎦
−4
6 −2
0
⎤
⎡
88
32 16 −4
⎢ −24
12 12
6 ⎥
⎥
adj(A) = ⎢
⎣ −40 −20 −4 −2 ⎦;
−48
0
0
0
det(A) = −48;
⎤
⎡
88
32 16 −4
1 ⎢ −24
12 12
6 ⎥
⎥.
A−1 = − ⎢
48 ⎣ −40 −20 −4 −2 ⎦
−48
0
0
0
1 −1
−4
5
⎡
⎡
⎡
31.
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3
2
4
=
−2 −2
11 12
.
254
⎡
⎤
12 −12
9 −12 ⎦;
24 −27
⎤
21
24
48
9
24 ⎦;
adj(A) = ⎣ 12
−12 −12 −27
det(A) ⎡
= 9;
⎤
7/3 8/3 16/3
1
8/3 ⎦.
A−1 = ⎣ 4/3
−4/3 −4/3 −3
21
MC = ⎣ 24
48
⎡
32.
⎡
⎤
7 −2
0
2 −2 ⎦;
MC = ⎣ 0
−6
0
2
⎡
⎤
7
0 −6
2
0 ⎦;
adj(A) = ⎣ −2
0 −2
2
det(A) ⎡
= 2;
⎤
3.5
0 −3
1
0 ⎦.
A−1 = ⎣ −1
0 −1
1
⎤
4 −1 0
1 4 ⎦.
33. There are many ways to do this. One choice is to let B = ⎣ 5
0
0 10
9
⎡
⎤
1 2 3
34. FALSE. For instance, if one entry of the matrix A = ⎣ 1 2 3 ⎦ is changed, two of the rows of A
1 2 3
will still be identical, and therefore, the determinant of the resulting matrix must be zero. It is not possible
to force the determinant to equal r.
⎡
35. Note that
det(A) = 2 + 12k + 36 − 4k − 18 − 12 = 8k + 8 = 8(k + 1).
(a). Based on the calculation above, we see that A fails to be invertible if and only if k = −1.
(b). The volume of the parallelepiped determined by the row vectors of A is precisely |det(A)| = |8k + 8|.
The volume of the parallelepiped determined by the column vectors of A is the same as the volume of the
parallelepiped determined by the row vectors of AT , which is |det(AT )| = |det(A)| = |8k + 8|. Thus, the
volume is the same.
36. Note that
det(A) = 3(k + 1) + 2k + 0 − 3 − k(k + 1) − 0 = −k 2 + 4k = k(4 − k).
(a). Based on the calculation above, we see that A fails to be invertible if and only if k = 0 or k = 4.
(b). The volume of the parallelepiped determined by the row vectors of A is precisely |det(A)| = | − k 2 + 4k|.
The volume of the parallelepiped determined by the column vectors of A is the same as the volume of the
parallelepiped determined by the row vectors of AT , which is |det(AT )| = |det(A)| = | − k 2 + 4k|. Thus, the
volume is the same.
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37. Note that
det(A) = 0 + 4(k − 3) + 2k 3 − k 2 − 8k − 0 = 2k 3 − k 2 − 4k − 12.
(a). Based on the calculation above, we see that A fails to be invertible if and only if 2k 3 − k 2 − 4k − 12 = 0,
and the only real solution to this equation for k is k ≈ 2.39 (from calculator).
(b). The volume of the parallelepiped determined by the row vectors of A is precisely |det(A)| = |2k 3 − k 2 −
4k−12|. The volume of the parallelepiped determined by the column vectors of A is the same as the volume of
the parallelepiped determined by the row vectors of AT , which is |det(AT )| = |det(A)| = |2k 3 − k 2 − 4k − 12|.
Thus, the volume is the same.
38. From the assumption that AB = −BA, we can take the determinant of each side: det(AB) = det(−BA).
Hence, det(A)det(B) = (−1)n det(B)det(A) = −det(A)det(B). From this, it follows that det(A)det(B) = 0,
and therefore, either det(A) = 0 or det(B) = 0. Thus, either A or B (or both) fails to be invertible.
39. Since AAT = In , we can take the determinant of both sides to get det(AAT ) = det(In ) = 1. Hence,
det(A)det(AT ) = 1. Therefore, we have (det(A))2 = 1. We conclude that det(A) = ±1.
−3 1
40. The coefficient matrix of this linear system is A =
. We have
1 2
−3 1 = −7, det(B1 ) = 3 1 = 5, and det(B2 ) = −3 3 = −6.
det(A) = 1 2
1 2
1 1 Thus,
x1 =
Solution: (−5/7, 6/7).
det(B1 )
det(B2 )
5
6
=−
and x2 =
= .
det(A)
7
det(A)
7
⎤
2 −1 1
5 3 ⎦. We have
41. The coefficient matrix of this linear system is A = ⎣ 4
4 −3 3
2 −1 1 2 −1 1 5 3 = 32,
5 3 = 16, det(B1 ) = 0
det(A) = 4
2 −3 3 4 −3 3 2 2 1 2 −1 2 5 0 = −36.
det(B2 ) = 4 0 3 = −4, det(B3 ) = 4
4 2 3 4 −3 2 ⎡
Thus,
x1 =
det(B1 )
det(B2 )
1
= 2, x2 =
=− ,
det(A)
det(A)
4
Solution: (2, −1/4, −9/4).
and x3 =
⎡
det(B3 )
9
=− .
det(A)
4
⎤
3
1 2
42. The coefficient matrix of this linear system is A = ⎣ 2 −1 1 ⎦. We have
0
5 5
−1
3
1 2 1 2 det(A) = 2 −1 1 = −20, det(B1 ) = −1 −1 1 = −10,
−5
0
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3 −1 2 3
1 −1 det(B2 ) = 2 −1 1 = −10, det(B3 ) = 2 −1 −1 = 30.
0 −5 5 0
5 −5 Thus,
x1 =
det(B1 )
det(B2 )
1
1
= , x2 =
= ,
det(A)
2
det(A)
2
and x3 =
det(B3 )
3
=− .
det(A)
2
Solution: (1/2, 1/2, −3/2).
Chapter 4 Solutions
Solutions to Section 4.1
True-False Review:
(a): FALSE. The vectors (x, y) and (x, y, 0) do not belong to the same set, so they are not even comparable,
let alone equal to one another.
(b): TRUE. The unique additive inverse of (x, y, z) is (−x, −y, −z).
(c): TRUE. The solution set refers to collections of the unknowns that solve the linear system. Since this
system has 6 unknowns, the solution set will consist of vectors belonging to R6 .
(d): TRUE. The vector (−1) · (x1 , x2 , . . . , xn ) is precisely (−x1 , −x2 , . . . , −xn ), and this is the additive
inverse of (x1 , x2 , . . . , xn ).
(e): FALSE. There is no such name for a vector whose components are all positive.
(f ): FALSE. The correct result is
(s + t)(x + y) = (s + t)x + (s + t)y = sx + tx + sy + ty,
and in this item, only the first and last of these four terms appear.
(g): TRUE. When the vector x is scalar multiplied by zero, each component becomes zero: 0x = 0. This
is the zero vector in Rn .
(h): TRUE. This is seen geometrically from addition and subtraction of geometric vectors.
(i): FALSE. If k < 0, then kx is a vector in the third quadrant. For instance, (1, 1) lies in the first quadrant,
but (−2)(1, 1) = (−2, −2) lies in the third quadrant.
(j): TRUE. Recalling that i = (1, 0, 0), j = (0, 1, 0), and k = (0, 0, 1), we have
√
√
√
5i − 6j + 2k = 5(1, 0, 0) − 6(0, 1, 0) + 2(0, 0, 1) = (5, −6, 2),
as stated.
(k): FALSE. If the three vectors lie on the same line or the same plane, the resulting object may determine
a one-dimensional segment or two-dimensional area. For instance, if x = y = z = (1, 0, 0), then the vectors
x,y, and z rest on the segment from (0, 0, 0) to (1, 0, 0), and do not determine a three-dimensional solid
region.
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(l): FALSE. The components of kx only remain even integers if k is an integer. But, for example, if k = π,
then the components of kx are not even integers at all, let alone even integers.
Problems:
1. v1 = (−3, −12), v2 = (20, −4), v3 = (−3, −12) + (20, −4) = (17, −16).
y
x
v2
v1
(20, -4)
v3
(-3, -12)
(17, -16)
Figure 0.0.60: Figure for Problem 1
2. v1 = (6, 2), v2 = (−3, 6), v3 = (6, 2) + (−3, 6) = (3, 8).
y
(-3, 6)
(3, 8)
v3
v2
(6, 2)
v1
x
Figure 0.0.61: Figure for Problem 2
3. We have
v = −2x + 10y
= −2(5, −2, 9) + 10(−1, 6, 4)
= (−20, 64, 22),
and therefore, the additive inverse of v is
−v = (20, −64, −22).
4. v = 5(3, −1, 2, 5) − 7(−1, 2, 9, −2) = (15, −5, 10, 25) − (−7, 14, 63, −14) = (22, −19, −53, 39). Additive
inverse: −v = (−1)v = (−22, 19, 53, −39).
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5. We have
2
1
2
1
y = x + z = (1, 2, 3, 4, 5) + (−1, 0, −4, 1, 2) =
3
3
3
3
1 4 2
, , , 3, 4 .
3 3 3
6. Solving for z, we have
1
y − 2x
2
1
= (3, 0, −5) − 2(−3, 9, 9)
2
15
41
= ( , −18, − ).
2
2
z=
7. Solving for z, we have
1
1+i
x− y
2i
i
1−i
(−2 + i, 3i) + i(5, 2 − 2i)
=
2
1
= (−1 + 3i, 3 + 3i) + (5i, 2 + 2i)
2
1 13 7 7
= (− + i, + i).
2
2 2 2
z=
8. Solving for z, we have
2
3i
x+
y
1+i
1+i
3 3
= (1 − i)x + ( + i)y
2 2
z=
3 3
= (1 − i)(5 + i, 0, −1 − 2i, 1 + 8i) + ( + i)(−3, i, i, 3)
2 2
9 9
3 3
3 3 9 9
= (6 − 4i, 0, −3 − i, 9 + 7i) + (− − i, − + i, − + i, + i)
2
2
2
2
2 2 2 2
3 3
9 1 27 23
3 17
− i, − + i, − + i,
+ i .
=
2
2
2 2
2 2 2
2
9. Let x = (x1 , x2 , x3 , x4 ), y = (y1 , y2 , y3 , y4 ) be arbitrary vectors in R4 . Then
x+y
=
(x1 , x2 , x3 , x4 ) + (y1 , y2 , y3 , y4 ) = (x1 + y1 , x2 + y2 , x3 + y3 , x4 + y4 )
=
=
(y1 + x1 , y2 + x2 , y3 + x3 , y4 + x4 ) = (y1 , y2 , y3 , y4 ) + (x1 , x2 , x3 , x4 )
y + x.
10. Let x = (x1 , x2 , x3 , x4 ), y = (y1 , y2 , y3 , y4 ), z = (z1 , z2 , z3 , z4 ) be arbitrary vectors in R4 . Then
x + (y + z)
=
=
=
(x1 , x2 , x3 , x4 ) + (y1 + z1 , y2 + z2 , y3 + z3 , y4 + z4 )
(x1 + (y1 + z1 ), x2 + (y2 + z2 ), x3 + (y3 + z3 ), x4 + (y4 + z4 ))
((x1 + y1 ) + z1 , (x2 + y2 ) + z2 , (x3 + y3 ) + z3 , (x4 + y4 ) + z4 )
=
=
(x1 + y1 , x2 + y2 , x3 + y3 , x4 + y4 ) + (z1 , z2 , z3 , z4 )
(x + y) + z.
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11. Let x = (x1 , x2 , x3 ) and y = (y1 , y2 , y3 ) be arbitrary vectors in R3 , and let r, s, t be arbitrary real
numbers. Then:
1x = 1(x1 , x2 , x3 ) = (1x1 , 1x2 , 1x3 ) = (x1 , x2 , x3 ) = x.
(st)x = (st)(x1 , x2 , x3 ) = ((st)x1 , (st)x2 , (st)x3 ) = (s(tx1 ), s(tx2 ), s(tx3 )) = s(tx1 , tx2 , tx3 ) = s(tx).
r(x + y) = r(x1 + y1 , x2 + y2 , x3 + y3 ) = (r(x1 + y1 ), r(x2 + y2 ), r(x3 + y3 ))
= (rx1 + ry1 , rx2 + ry2 , rx3 + ry3 ) = (rx1 , rx2 , rx3 ) + (ry1 , ry2 , ry3 )
= rx + ry.
(s + t)x = (s + t)(x1 , x2 , x3 ) = ((s + t)x1 , (s + t)x2 , (s + t)x3 )
= (sx1 + tx1 , sx2 + tx2 , sx3 + tx3 ) = (sx1 , sx2 , sx3 ) + (tx1 , tx2 , tx3 )
= sx + tx.
12. Let x = (x1 , x2 , x3 , x4 , x5 ) and y = (y1 , y2 , y3 , y4 , y5 ) be arbitrary vectors in R5 , and let r, s, t be
arbitrary real numbers. Then:
1x = 1(x1 , x2 , x3 , x4 , x5 ) = (1x1 , 1x2 , 1x3 , 1x4 , 1x5 ) = (x1 , x2 , x3 , x4 , x5 ) = x.
(st)x = (st)(x1 , x2 , x3 , x4 , x5 )
= ((st)x1 , (st)x2 , (st)x3 , (st)x4 , (st)x5 )
= (s(tx1 ), s(tx2 ), s(tx3 ), s(tx4 ), s(tx5 ))
= s(tx1 , tx2 , tx3 , tx4 , tx5 ) = s(tx).
r(x + y) = r(x1 + y1 , x2 + y2 , x3 + y3 , x4 + y4 , x5 + y5 )
= (r(x1 + y1 ), r(x2 + y2 ), r(x3 + y3 ), r(x4 + y4 ), r(x5 + y5 ))
= (rx1 + ry1 , rx2 + ry2 , rx3 + ry3 , rx4 + ry4 , rx5 + ry5 )
= (rx1 , rx2 , rx3 , rx4 , rx5 ) + (ry1 , ry2 , ry3 , ry4 , ry5 )
= rx + ry.
(s + t)x = (s + t)(x1 , x2 , x3 , x4 , x5 )
= ((s + t)x1 , (s + t)x2 , (s + t)x3 , (s + t)x4 , (s + t)x5 )
= (sx1 + tx1 , sx2 + tx2 , sx3 + tx3 , sx4 + tx4 , sx5 + tx5 )
= (sx1 , sx2 , sx3 , sx4 , sx5 ) + (tx1 , tx2 , tx3 , tx4 , tx5 )
= sx + tx.
13. For example, if x = (2, 2) and y = (−1, −1), then x + y = (1, 1) lies in the first quadrant. If x = (1, 2)
and y = (−5, 1), then x + y = (−4, 3) lies in the second quadrant. If x = (1, 1) and y = (−2, −2), then
x + y = (−1, −1) lies in the third quadrant. If x = (2, 1) and y = (−1, −5), then x + y = (1, −4) lies in the
fourth quadrant.
Solutions to Section 4.2
True-False Review:
(a): TRUE. This is part 1 of Theorem 4.2.7.
(b): FALSE. The statement would be true if it was required that v be nonzero. However, if v = 0, then
rv = sv = 0 for all values of r and s, and r and s need not be equal. We conclude that the statement is not
true.
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(c): TRUE. From rv = sv, we subtract to obtain 0 = rv − sv = (r − s)v. By part (6) of Theorem 4.2.7,
we must have either r − s = 0 or v = 0. However, v = 0 is given. Thus, we must conclude that r − s = 0,
or r = s.
(d): FALSE. This set is not closed under scalar multiplication. In particular, if k is an irrational number
such as k = π and v is an integer, then kv is not an integer.
(e): TRUE. We have
(x + y) + ((−x) + (−y)) = (x + (−x)) + (y + (−y)) = 0 + 0 = 0,
where we have used the vector space axioms in these steps. Therefore, the additive inverse of x + y is
(−x) + (−y).
(f ): TRUE. This is part 1 of Theorem 4.2.7.
(g): TRUE. This is called the trivial vector space. Since 0 + 0 = 0 and k · 0 = 0, it is closed under addition
and scalar multiplication. Both sides of the remaining axioms yield 0, and 0 is the zero vector, and it is its
own additive inverse.
(h): FALSE. This set is not closed under addition, since 1 + 1 ∈ {0, 1}. Therefore, (A1) fails, and hence,
this set does not form a vector space. (It is worth noting that the set is also not closed under scalar
multiplication.)
(i): FALSE. This set is not closed under scalar multiplication. If k < 0 and x is a positive real number,
the result kx is a negative real number, and therefore no longer belongs to the set of positive real numbers.
(j): FALSE. Any positive element of the given set fails to have an additive inverse in the set, so Axiom
(A6) fails.
Problems:
1. If x = p/q and y = r/s, where p, q, r, s are integers (q = 0, s = 0), then x + y = (ps + qr)/(qs), which
is a rational number. Consequently, the set of all rational numbers is closed under addition. The set is not
closed under scalar multiplication since, if we multiply a rational number by an irrational number, the result
is an irrational number.
2. Let A = [aij ] and B = [bij ] be upper triangular matrices, and let k be an arbitrary real number. Then
whenever i > j, it follows that aij = 0 and bij = 0. Consequently, aij + bij = 0 whenever i > j, and kaij = 0
whenever i > j. Therefore, A + B and kA are upper triangular matrices, so the set of all upper triangular
matrices with real elements is closed under both addition and scalar multiplication.
3. In this case, S fails to be closed under addition, but it is closed under scalar multiplication.
(A1) fails: S is not closed under addition. For instance,
using the case n = 2, we can
give this specific
0 1
0 0
example (many such examples are acceptable): Let A =
and B =
. Since A is upper
0 0
1 0 0 1
triangular and B is lower triangular, both A and B belong to S. However, A + B =
is neither
1 0
upper nor lower triangular. Thus, A + B ∈ S.
(A2) holds: S is closed under scalar multiplication. To see this, suppose A ∈ S. This means A is either
upper triangular, or A is lower triangular. If A is upper triangular, then for any real constant r, rA is still
upper triangular. Thus, rA ∈ S. A similar argument holds if A is lower triangular.
4. In this case, S is closed under both addition and scalar multiplication.
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(A1) holds: S is closed under addition. To see this, suppose p(x) = a0 + a1 x + a2 x2 and q(x) =
b0 + b1 x + b2 x2 belong to S.This means that
a0 + a1 + a2 = 0
and
b0 + b1 + b2 = 0.
Then
p(x) + q(x) = (a0 + a1 x + a2 x2 ) + (b0 + b1 x + b2 x2 ) = (a0 + b0 ) + (a1 + b1 )x + (a2 + b2 )x2 ,
and since (a0 + b0 ) + (a1 + b1 ) + (a2 + b2 ) = (a0 + a1 + a2 ) + (b0 + b1 + b2 ) = 0 + 0 = 0, we conclude that
p(x) + q(x) ∈ S. Hence, S is closed under addition.
(A2) holds: S is closed under scalar multiplication. To see this, suppose p(x) is as above, and let r be a
constant. Then
r · p(x) = (ra0 ) + (ra1 )x + (ra2 )x2 ,
and since ra0 + ra1 + ra2 = r(a0 + a1 + a2 ) = r · 0 = 0, we conclude that r · p(x) ∈ S. Hence, S is closed
under scalar multiplication.
5. In this case, S is not closed under addition and not closed under scalar multiplication.
(A1) fails: S is not closed under addition. We can demonstrate with an example (many such examples can
be given). Let p(x) = 1 and let q(x) = x. Note that p(x) ∈ S and q(x) ∈ S. However, p(x)+q(x) = 1+x ∈ S.
Thus, S fails to be closed under scalar multiplication.
(A2) fails: S is not closed under scalar multiplication. We can demonstrate with an example (many such
examples can be given). With p(x) = 1 and r = 2, we have p(x) ∈ S, but r · p(x) = 2 ∈ S. Thus, S fails to
be closed under scalar multiplication.
6. In this case, S is not closed under addition and not closed under scalar multiplication.
(A1) fails: S is not closed under addition. To see this, suppose y1 and y2 are both solutions to the given
differential equation. That is,
y1 + 3y1 = 6x3 + 5
and
y2 + 3y2 = 6x3 + 5.
Then
(y1 + y2 ) + 3(y1 + y2 ) = y1 + y2 + 3y1 + 3y2 = (y1 + 3y1 ) + (y2 + 3y2 ) = (6x3 + 5) + (6x3 + 5) = 6x3 + 5,
so that y1 + y2 ∈ S. Thus, S is not closed under addition.
(A2) fails: S is not closed under scalar multiplication. To see this, let y1 be as above and let r be a
scalar. Then
(ry1 ) + 3(ry1 ) = r(y1 + 3y1 ) = r(6x3 + 5) = 6x3 + 5
for r = 1. Thus, unless r = 1, we have that ry1 ∈ S. Thus, S is not closed under scalar multiplication.
7. In this case, S is closed under both addition and scalar multiplication.
(A1) holds: S is closed under addition. To see this, suppose y1 and y2 are both solutions to the given
differential equation. That is,
y1 + 3y1 = 0 and y2 + 3y2 = 0.
Then
(y1 + y2 ) + 3(y1 + y2 ) = y1 + y2 + 3y1 + 3y2 = (y1 + 3y1 ) + (y2 + 3y2 ) = 0 + 0 = 0,
so that y1 + y2 ∈ S. Thus, S is closed under addition.
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(A2) holds: S is closed under scalar multiplication. To see this, let y1 be as above and let r be a scalar.
Then
(ry1 ) + 3(ry1 ) = r(y1 + 3y1 ) = r(0) = 0.
Thus, we have that ry1 ∈ S. Thus, S is closed under scalar multiplication.
8. In this case, S is closed under both addition and scalar multiplication.
(A1) holds: S is closed under addition. To see this, suppose x1 and x2 belong to S. Thus, Ax1 = 0 and
Ax2 = 0. Then, since
A(x1 + x2 ) = Ax1 + Ax2 = 0 + 0 = 0,
we conclude that x1 + x2 ∈ S. Thus, S is closed under addition.
(A2) holds: S is closed under scalar multiplication. To see this, let x1 be as above and let r be a scalar.
Then
A(rx1 ) = r(Ax1 ) = r(0) = 0,
so that rx1 ∈ S. Thus, S is closed under scalar multiplication.
9. In this case, S is not closed under addition, but S is closed under scalar multiplication.
1
0
0
0
(A1) fails: S is not closed under addition. As a specific example to show this, note that A =
∈S
0 0
and B =
∈ S. However, A + B = I2 , and since det(I2 ) = 1 = 0, we have det(A + B) = 0, so
0 1
that A + B ∈ S. Thus, S is not closed under addition. (Many other examples could have been used to
demonstrate this.)
(A2) holds: S is closed under scalar multiplication. Suppose that A ∈ S and r is a scalar. This means
that det(A) = 0. Using Property (P4) of determinants, we then have det(rA) = r2 det(A) = r2 · 0 = 0, so
that rA ∈ S. Thus, S is closed under scalar multiplication.
10. In this case, S is closed under neither addition nor scalar multiplication.
(A1) fails: S is not closed under addition. As a specific example to show this, note that v = (−1, 1) ∈ S
and w = (1, 1) ∈ S. However, v + w = (−1, 1) + (1, 1) = (0, 2) ∈ S. Thus, S is not closed under addition.
(Many other examples could have been used to demonstrate this.)
(A2) fails: S is not closed under scalar multiplication. As a specific example to show this, note that
w = (1, 1) ∈ S. However, 2 · w = (2, 2) ∈ S. Thus, S is not closed under scalar multiplication. (Many other
examples could have been used to demonstrate this.)
11. In this case, S is closed under neither addition nor scalar multiplication.
(A1) fails: S is not closed under addition. As a specific example to show this, note that v = (0, 1) ∈ S
and w = (1, 2) ∈ S. However, v + w = (0, 1) + (1, 2) = (1, 3) ∈ S. Thus, S is not closed under addition.
(Many other examples could have been used to demonstrate this.)
(A2) fails: S is not closed under scalar multiplication. As a specific example to show this, note that
v = (0, 1) ∈ S, while 2 · v = (0, 2) ∈ S. Thus, S is not closed under scalar multiplication. (Many other
examples could have been used to demonstrate this.)
12. In this case, S is closed under addition, but S is not closed under scalar multiplication.
(A1) holds: N is closed under addition. To see this, note that if m, n ∈ N, then m + n ∈ N since the sum
of two positive integers is another positive integer. Thus, S is closed under addition.
(A2) fails: N is not closed under scalar multiplication. To see this, note that if m ∈ N, then −1 · m =
−m ∈ N, since −m is a negative integer. Thus, S is not closed under scalar multiplication. (Many other
examples could have been used to demonstrate this.)
13. In this case, S is closed under neither addition nor scalar multiplication.
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(A1) fails: S is not closed under addition. For a specific example of this, note that p(x) = x2 ∈ S and
q(x) = x − x2 ∈ S. However, p(x) + q(x) = x2 + (x − x2 ) = x ∈ S, which shows that S not closed under
addition. (Many other examples could have been used to demonstrate this.)
(A2) fails: S is not closed under scalar multiplication. For a specific example of this, note that p(x) =
x2 ∈ S, but 0 · p(x) = 0 ∈ S. Thus, S is not closed under scalar multiplication. (Many other examples could
have been used to demonstrate this.)
14. In this case, S is closed under both addition and scalar multiplication.
(A1) holds: S is closed under addition. To see this, suppose p(x) = a1 + b1 x3 + c1 x4 and q(x) =
a2 + b2 x3 + c2 x4 belong to S. Then
p(x) + q(x) = (a1 + b1 x3 + c1 x4 ) + (a2 + b2 x3 + c2 x4 ) = (a1 + a2 ) + (b1 + b2 )x3 + (c1 + c2 )x4 ∈ S,
since the form of the final expression fits the form for elements of S. Thus, S is closed under addition.
(A2) holds: S is closed under scalar multiplication. To see this, suppose p(x) is as above, and let r be a
scalar. Then
r · p(x) = r(a1 + b1 x3 + c1 x4 ) = (ra1 ) + (rb1 )x3 + (rc1 )x4 ∈ S,
since the form of the final expression fits the form for elements of S. Thus, S is closed under scalar
multiplication.
15. Let V be R2 , i.e. {(x, y) : x, y ∈ R}. With addition and scalar multiplication as defined in the text, this
set is clearly closed under both operations.
A3: u + v = (u1 , u2 ) + (v1 , v2 ) = (u1 + v1 , u2 + v2 ) = (v1 + u1 , v2 + u2 ) = v + u.
A4: [u+v]+w = [(u1 +v1 , u2 +v2 )]+(w1 , w2 ) = ([u1 +v1 ]+w1 , [u2 +v2 ]+w2 ) = (u1 +[v1 +w1 ], u2 +[v2 +w2 ]) =
(u1 , u2 ) + [(v1 + w1 , v2 + w2 )] = u + [v + w].
A5: 0 = (0, 0) since (x, y) + (0, 0) = (x + 0, y + 0) = (x, y).
A6: If u = (a, b), then −u = (−a, −b), a, b ∈ R, since (a, b) + (−a, −b) = (a − a, b − b) = (0, 0).
Now, let u = (u1 , u2 ), v = (v1 , v2 ), where u1 , u2 , v1 , v2 ∈ R, and let r, s, t ∈ R.
A7: 1 · v = 1(v1 , v2 ) = (1 · v1 , 1 · v2 ) = (v1 , v2 ) = v.
A8: (rs)v = (rs)(v1 , v2 ) = ((rs)v1 , (rs)v2 ) = (rsv1 , rsv2 ) = (r(sv1 ), r(sv2 )) = r(sv1 , sv2 ) = r(sv).
A9: r(u + v) = r((u1 , u2 ) + (v1 , v2 )) = r(u1 + v1 , u2 + v2 ) = (r(u1 + v1 ), r(u2 + v2 )) = (ru1 + rv1 , ru2 + rv2 ) =
(ru1 , ru2 ) + (rv1 , rv2 ) = r(u1 , u2 ) + r(v1 , v2 ) = ru + rv.
A10: (r + s)u = (r + s)(u1 , u2 ) = ((r + s)u1 , (r + s)u2 ) = (ru1 + su1 , ru2 + su2 ) = (ru1 , ru2 ) + (su1 , su2 ) =
r(u1 , u2 ) + s(u1 , u2 ) = ru + su.
Thus, R2 is a vector space.
⎤
⎡
⎡
⎤
0 0
a b
⎢ 0 0 ⎥
⎥
⎣ c d ⎦, so that the additive inverse is
16. The zero vector is ⎢
⎣ 0 0 ⎦, a general element in V is A =
e f
0 0
⎡
⎤
−a −b
−A = ⎣ −c −d ⎦.
−e −f
17. The zero vector of Mm×n (R) is 0m×n , the m × n zero matrix. The additive inverse of the m × n matrix
A with (i, j)-element aij is the m × n matrix −A with (i, j)-element −aij .
18. The zero vector is p(x) = 0, a general element in V is p(x) = a0 + a1 x + a2 x2 + a3 x3 , so that the additive
inverse is −p(x) = −a0 − a1 x − a2 x2 − a3 x3 .
19. The zero vector is p(x) = 0, a general element in V is p(x) = a0 + a1 x + a2 x2 + · · · + an xn , so that the
additive inverse is −p(x) = −a0 − a1 x − a2 x2 − · · · − an xn .
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20. YES. We verify the ten axioms of a vector space.
A1: The product of two positive real numbers is a positive real number, so the set is closed under addition.
A2: Any power of a positive real number is a positive real number, so the set is closed under scalar multiplication.
A3: We have
x + y = xy = yx = y + x
for all x, y ∈ R+ , so commutativity under addition holds.
A4: We have
(x + y) + z = (xy) + z = (xy)z = x(yz) = x(y + z) = x + (y + z)
for all x, y, z ∈ R+ , so that associativity under addition holds.
A5: We claim that the zero vector in this set is the real number 1. To see this, note that 1 + x = 1x = x =
x1 = x + 1 for all x ∈ R+ .
A6: We claim that the additive inverse of the vector x ∈ R+ is x1 ∈ R+ . To see this, note that x + x1 = x x1 =
1 = x1 x = x1 + x.
A7: Note that 1 · x = x1 = x for all x ∈ R+ , so that the unit property holds.
A8: For all r, s ∈ R and x ∈ R+ , we have
(rs) · x = xrs = xsr = (xs )r = r · xs = r · (s · x),
as required for associativity of scalar multiplication.
A9: For all r ∈ R and x, y ∈ R+ , we have
r · (x + y) = r · (xy) = (xy)r = xr y r = xr + y r = r · x + r · y,
as required for distributivity of scalar multiplication over vector addition.
A10: For all r, s ∈ R and x ∈ R+ , we have
(r + s) · x = xr+s = xr xs = xr + xs = r · x + s · x,
as required for distributivity of scalar multiplication over scalar addition.
The above verification of axioms A1-A10 shows that we have a vector space structure here.
21: Since the output vectors on the right hand side of the addition operation and the scalar multiplication
operation are both in R2 , we conclude that axioms (A1) and (A2) are satisfied. Furthermore, since the
addition operation is simply the ordinary addition of vectors in R2 , all of the axioms pertaining only to
addition will hold as usual. This includes Axioms (A3), (A4), (A5), and (A6). For (A7), note that for all
(x1 , y1 ) ∈ R2 , we have
1 (x1 , y1 ) = (x1 , y1 ),
so (A7) holds. For (A8), observe that for (x1 , y1 ) ∈ R2 and scalars r, s ∈ R, we have
(rs)
(x1 , y1 ) = ((rs)x1 , y1 ) = (r(sx1 ), y1 ) = r
(sx1 , y1 ) = r
(s
(x1 , y1 )),
so that (A8) holds. For (A9), observe that for (x1 , y1 ), (x2 , y2 ) ∈ R2 and r ∈ R, we have
r
((x1 , y1 ) + (x2 , y2 )) = r
(x1 + x2 , y1 + y2 ) = (r(x1 + x2 ), y1 + y2 ),
while
r
(x1 , y1 ) + r
(x2 , y2 ) = (rx1 , y1 ) + (rx2 , y2 ) = (rx1 + rx2 , y1 + y2 ) = (r(x1 + x2 ), y1 + y2 ),
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so that (A9) holds. Finally, for (A10), observe that for scalars r, s ∈ R and (x1 , y1 ) ∈ R2 , we have
(r + s)
(x1 , y1 ) = ((r + s)x1 , y1 ) = (rx1 + sx1 , y1 ),
while
r
(x1 , y1 ) + s
(x1 , y1 ) = (rx1 , y1 ) + (sx1 , y1 ) = (rx1 + sx1 , y1 + y1 ) = (rx1 + sx1 , y1 )
for y1 = 0. Therefore, (A10) fails. Thus, this is NOT a vector space structure.
22. Axioms A1 and A2 clearly hold under the given operations.
A3: u + v = (u1 , u2 ) + (v1 , v2 ) = (u1 − v1 , u2 − v2 ) = (−(v1 − u1 ), −(v2 − u2 )) = (v1 − u1 , v2 − u2 ) = v + u.
Consequently, A3 does not hold.
A4: (u+v)+w = (u1 −v1 , u2 −v2 )+(w1 , w2 ) = ((u1 −v1 )−w1 , (u2 −v2 )−w2 ) = (u1 −(v1 +w1 ), u2 −(v2 +w2 )) =
(u1 , u2 ) + (v1 + w1 , v2 + w2 ) = (u1 , u2 ) + (v1 − w1 , v2 − w2 ) = u + (v + w). Consequently, A4 does not hold.
A5: 0 = (0, 0) since u + 0 = (u1 , u2 ) + (0, 0) = (u1 − 0, u2 − 0) = (u1 , u2 ) = u.
A6: If u = (u1 , u2 ), then −u = (u1 , u2 ) since u + (−u) = (u1 , u2 ) + (u1 , u2 ) = (u1 − u1 , u2 − u2 ) = (0, 0) = 0.
Each of the remaining axioms do not hold.
23. Axiom A5: The zero vector on R2 with the defined operation of addition is given by (1,1), for if (u1 , u2 )
is any element in R2 , then (u1 , u2 ) + (1, 1) = (u1 · 1, u2 · 1) = (u1 , u2 ). Thus, Axiom A5 holds.
Axiom A6: Suppose that (u1 , v1 ) is any element in R2 with additive inverse (a, b). From the first part of
the problem, we know that (1, 1) is the zero element, so it must be the case that (u1 , v1 ) + (a, b) = (1, 1) so
that (u1 a, v1 b) = (1, 1); hence, it follows that u1 a = 1 and v1 b = 1, but this system is not satisfied for all
(u1 , v1 ) ∈ R, namely, (0, 0). Thus, Axiom A6 is not satisfied.
24. Let A, B, C ∈ M2 (R) and r, s, t ∈ R.
A3: The addition operation is not commutative since
A + B = AB = BA = B + A.
A4: Addition is associative since
(A + B) + C = AB + C = (AB)C = A(BC) = A(B + C) = A + (B + C).
1 0
A5: I2 =
is the zero vector in M2 (R) because A + I2 = AI2 = A for all A ∈ M2 (R).
0 1
A6: We wish to determine whether for each matrix A ∈ M2 (R) we can find a matrix B ∈ M2 (R) such that
A+B = I2 (remember that we have shown in A5 that the zero vector is I2 ), equivalently, such that AB = I2 .
However, this equation can be satisfied only if A is nonsingular, therefore the axiom fails.
A7: 1 · A = A is true for all A ∈ M2 (R).
A8: (st)A = s(tA) is true for all A ∈ M2 (R) and s, t ∈ R.
A9: sA + tA = (sA)(tA) = st(AA) = (s + t)A for all s, t ∈ R and A ∈ M2 (R). Consequently, the axiom fails.
A10: rA + rB = (rA)(rB) = r2 AB = r2 (A + B). Thus, rA + rB = rA + rB for all r ∈ R, so the axiom fails.
25. M2 (R) = {A : A is a 2 × 2 real matrix}. Let A, B ∈ M2 (R) and k ∈ R.
A1: A ⊕ B = −(A + B).
A2: k · A = −kA.
A3 and A4: A ⊕ B = −(A + B) = −(B + A) = B ⊕ A. Hence, the operation ⊕ is commutative.
(A ⊕ B) ⊕ C = −((A ⊕ B) + C) = −(−(A + B) + C) = A + B − C, but
A ⊕ (B ⊕ C) = −(A + (B ⊕ C)) = −(A + (−(B + C)) = −A + B + C. Thus the operation ⊕ is not associative.
A5: An element B is needed such that A ⊕ B = A for all A ∈ M2 (R), but −(A + B) = A =⇒ B = −2A.
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Since this depends on A, there is no zero vector.
A6: Since there is no zero vector, we cannot define the additive inverse.
Let r, s, t ∈ R.
A7: 1 · A = −1A = −A = A.
A8: (st) · A = −[(st)A], but s · (t · A) = s · (−tA) = −[s(−tA)] = s(tA) = (st)A, so it follows that
(st) · A = s · (t · A).
A9: r ·(A⊕B) = −r(A⊕B) = −r(−(A+B)) = r(A+B) = rA+rB = −[(−rA)+(−rB)] = −rA⊕(−rB) =
r · A + r · B, whereas rA ⊕ rB = −(rA + rB) = −rA − rB, so this axiom fails to hold.
A10: (s+t)·A = −(s+t)A = −sA+(−tA), but s·A⊕t·A = −(sA)⊕(−tA) = −[−(sA)+(−tA)] = sA+tA,
hence (s + t) · A = s · A ⊕ t · A. We conclude that only the axioms (A1)-(A3) hold.
26. Let C2 = {(z1 , z2 ) : zi ∈ C} under the usual operations of addition and scalar multiplication.
A3 and A4: Follow from the properties of addition in C2 .
A5: (0, 0) is the zero vector in C2 since (z1 , z2 ) + (0, 0) = (z1 + 0, z2 + 0) = (z1 , z2 ) for all (z1 , z2 ) ∈ C2 .
A6: The additive inverse of the vector (z1 , z2 ) ∈ C2 is the vector (−z1 , −z2 ) for all (z1 , z2 ) ∈ C2 .
A7-A10: Follows from properties in C2 .
Thus, C2 together with its defined operations, is a complex vector space.
27. Let M2 (C) = {A : A is a 2 × 2 matrix with complex entries} under the usual operations of matrix
addition and multiplication.
A3 and A4: Follows from properties of matrix addition.
A5: The zero vector, 0, for M2 (C) is the 2 × 2 zero matrix, 02 .
A6: For each vector A = [aij ] ∈ M2 (C), the vector −A = [−aij ] ∈ M2 (C) satisfies A + (−A) = 02 .
A7-A10: Follow from properties of matrix algebra.
Hence, M2 (C) together with its defined operations, is a complex vector space.
28. Let u = (u1 , u2 , u3 ) and v = (v1 , v2 , v3 ) be vectors in C3 , and let k ∈ R.
A1: u + v = (u1 , u2 , u3 ) + (v1 , v2 , v3 ) = (u1 + v1 , u2 + v2 , u3 + v3 ) ∈ C3 .
A2: ku = k(u1 , u2 , u3 ) = (ku1 , ku2 , ku3 ) ∈ C3 .
A3 and A4: Satisfied by the properties of addition in C3 .
A5: (0, 0, 0) is the zero vector in C3 since (0, 0, 0) + (z1 , z2 , z3 ) = (0 + z1 , 0 + z2 , 0 + z3 ) = (z1 , z2 , z3 ) for all
(z1 , z2 , z3 ) ∈ C3 .
A6: (−z1 , −z2 , −z3 ) is the additive inverse of (z1 , z2 , z3 ) because (z1 , z2 , z3 ) + (−z1 , −z2 , −z3 ) = (0, 0, 0) for
all (z1 , z2 , z3 ) ∈ C3 .
Let r, s, t ∈ R.
A7: 1 · u = 1 · (u1 , u2 , u3 ) = (1u1 , 1u2 , 1u3 ) = (u1 , u2 , u3 ) = u.
A8: (st)u = (st)(u1 , u2 , u3 ) = ((st)u1 , (st)u2 , (st)u3 ) = (s(tu1 ), s(tu2 ), s(tu3 )) = s(tu1 , tu2 , tu3 ) = s(t(u1 , u2 , u3 )) =
s(tu).
A9: r(u + v) = r(u1 + v1 , u2 + v2 , u3 + v3 ) = (r(u1 + v1 ), r(u2 + v2 ), r(u3 + v3 )) = (ru1 + rv1 , ru2 + rv2 , ru3 +
rv3 ) = (ru1 , ru2 , ru3 ) + (rv1 , rv2 , rv3 ) = r(u1 , u2 , u3 ) + r(v1 , v2 , v3 ) = ru + rv.
A10: (s + t)u = (s + t)(u1 , u2 , u3 ) = ((s + t)u1 , (s + t)u2 , (s + t)u3 ) = (su1 + tu1 , su2 + tu2 , su3 + tu3 ) =
(su1 , su2 , su3 ) + (tu1 , tu2 , tu3 ) = s(u1 , u2 , u3 ) + t(u1 , u2 , u3 ) = su + tu.
Thus, C3 is a real vector space.
29. NO. If we scalar multiply a vector (x, y, z) ∈ R3 by a non-real (complex) scalar r, we will obtain the
vector (rx, ry, rz) ∈ R3 , since rx, ry, rz ∈ R.
30. Let k be an arbitrary scalar, and let v be an arbitrary vector in V . Then, using property 2 of Theorem
4.2.7, we have
k0 = k(0v) = (k0)v = 0v = 0.
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31. Assume that k is a scalar and v ∈ V such that kv = 0. If k = 0, the desired conclusion is already
reached. If, on the other hand, k = 0, then we have k1 ∈ R and
1
1
· (kv) = · 0 = 0,
k
k
or
1
· k v = 0.
k
Hence, 1·v = 0, and the unit property A7 now shows that v = 0, and again, we reach the desired conclusion.
32. We verify the axioms A1-A10 for a vector space.
A1: If a0 + a1 x + · · · + an xn and b0 + b1 x + · · · + bn xn belong to Pn (R), then
(a0 + a1 x + · · · + an xn ) + (b0 + b1 x + · · · + bn xn ) = (a0 + b0 ) + (a1 + b1 )x + · · · + (an + bn )xn ,
which again belongs to Pn (R). Therefore, Pn (R) is closed under addition.
A2: If a0 + a1 x + · · · + an xn and r is a scalar, then
r · (a0 + a1 x + · · · + an xn ) = (ra0 ) + (ra1 )x + · · · + (ran )xn ,
which again belongs to Pn (R). Therefore, Pn (R) is closed under scalar multiplication.
A3: Let p(x) = a0 + a1 x + · · · + an xn and q(x) = b0 + b1 x + · · · + bn xn belong to Pn (R). Then
p(x) + q(x) = (a0 + a1 x + · · · + an xn ) + (b0 + b1 x + · · · + bn xn )
= (a0 + b0 ) + (a1 + b1 )x + · · · + (an + bn )xn
= (b0 + a0 ) + (b1 + a1 )x + · · · + (bn + an )xn
= (b0 + b1 x + · · · + bn xn ) + (a0 + a1 x + · · · + an xn )
= q(x) + p(x),
so Pn (R) satisfies commutativity under addition.
A4: Let p(x) = a0 + a1 x + · · · + an xn , q(x) = b0 + b1 x + · · · + bn xn , and r(x) = c0 + c1 x + · · · + cn xn belong
to Pn (R). Then
[p(x) + q(x)] + r(x) = [(a0 + a1 x + · · · + an xn ) + (b0 + b1 x + · · · + bn xn )] + (c0 + c1 x + · · · + cn xn )
= [(a0 + b0 ) + (a1 + b1 )x + · · · + (an + bn )xn ] + (c0 + c1 x + · · · + cn xn )
= [(a0 + b0 ) + c0 ] + [(a1 + b1 ) + c1 ]x + · · · + [(an + bn ) + cn ]xn
= [a0 + (b0 + c0 )] + [a1 + (b1 + c1 )]x + · · · + [an + (bn + cn )]xn
= (a0 + a1 x + · · · + an xn ) + [(b0 + c0 ) + (b1 + c1 )x + · · · + (bn + cn )xn ]
= (a0 + a1 x + · · · + an xn ) + [(b0 + b1 x + · · · + bn xn ) + (c0 + c1 x + · · · + cn xn )]
= p(x) + [q(x) + r(x)],
so Pn (R) satisfies associativity under addition.
A5: The zero vector is the zero polynomial z(x) = 0 + 0 · x + · · · + 0 · xn , and it is readily verified that this
polynomial satisfies z(x) + p(x) = p(x) = p(x) + z(x) for all p(x) ∈ Pn (R).
A6: The additive inverse of p(x) = a0 + a1 x + · · · + an xn is
−p(x) = (−a0 ) + (−a1 )x + · · · + (−an )xn .
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It is readily verified that p(x) + (−p(x)) = z(x), where z(x) is defined in A5.
A7: We have
1 · (a0 + a1 x + · · · + an xn ) = a0 + a1 x + · · · + an xn ,
which demonstrates the unit property in Pn (R).
A8: Let r, s ∈ R, and p(x) = a0 + a1 x + · · · + an xn ∈ Pn (R). Then
(rs) · p(x) = (rs) · (a0 + a1 x + · · · + an xn )
= [(rs)a0 ] + [(rs)a1 ]x + · · · + [(rs)an ]xn
= r[(sa0 ) + (sa1 )x + · · · + (san )xn ]
= r[s(a0 + a1 x + · · · + an xn )]
= r · (s · p(x)),
which verifies the associativity of scalar multiplication.
A9: Let r ∈ R, let p(x) = a0 + a1 x + · · · + an xn ∈ Pn (R), and let q(x) = b0 + b1 x + · · · + bn xn ∈ Pn (R). Then
r · (p(x) + q(x)) = r · ((a0 + a1 x + · · · + an xn ) + (b0 + b1 x + · · · + bn xn ))
= r · [(a0 + b0 ) + (a1 + b1 )x + · · · + (an + bn )xn ]
= [r(a0 + b0 )] + [r(a1 + b1 )]x + · · · + [r(an + bn )]xn
= [(ra0 ) + (ra1 )x + · · · + (ran )xn ] + [(rb0 ) + (rb1 )x + · · · + (rbn )xn ]
= [r(a0 + a1 x + · · · + an xn )] + [r(b0 + b1 x + · · · + bn xn )]
= r · p(x) + r · q(x),
which verifies the distributivity of scalar multiplication over vector addition.
A10: Let r, s ∈ R and let p(x) = a0 + a1 x + · · · + an xn ∈ Pn (R). Then
(r + s) · p(x) = (r + s) · (a0 + a1 x + · · · + an xn )
= [(r + s)a0 ] + [(r + s)a1 ]x + · · · + [(r + s)an ]xn
= [ra0 + ra1 x + · · · + ran xn ] + [sa0 + sa1 x + · · · + san xn ]
= r(a0 + a1 x + · · · + an xn ) + s(a0 + a1 x + · · · + an xn )
= r · p(x) + s · p(x),
which verifies the distributivity of scalar multiplication over scalar addition.
The above verification of axioms A1-A10 shows that Pn (R) is a vector space.
Solutions to Section 4.3
True-False Review:
(a): FALSE. The null space of an m × n matrix A is a subspace of Rn , not Rm .
(b): FALSE. It is not necessarily the case that 0 belongs to the solution set of the linear system. In fact,
0 belongs to the solution set of the linear system if and only if the system is homogeneous.
(c): TRUE. If b = 0, then the line is y = mx, which is a line through the origin of R2 , a one-dimensional
subspace of R2 . On the other hand, if b = 0, then the origin does not lie on the given line, and therefore
since the line does not contain the zero vector, it cannot form a subspace of R2 in this case.
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(d): FALSE. The spaces Rm and Rn , with m < n, are not comparable. Neither of them is a subset of the
other, and therefore, neither of them can form a subspace of the other.
(e): TRUE. Choosing any vector v in S, the scalar multiple 0v = 0 still belongs to S.
(f ): FALSE. In order for a subset of V to form a subspace, the same operations of addition and scalar
multiplication must be used in the subset as used in V .
(g): FALSE. This set is not closed under addition. For instance, the point (1, 1, 0) lies in the xy-plane, the
point (0, 1, 1) lies in the yz-plane, but
(1, 1, 0) + (0, 1, 1) = (1, 2, 1)
does not belong to S. Therefore, S is not a subspace of V .
(h): FALSE. For instance, if we consider V = R3 , then the xy-plane forms a subspace of V , and the
x-axis forms a subspace of V . Both of these subspaces contain in common all points along the x-axis. Other
examples abound as well.
Problems:
1. S = {x ∈ R3 : x = (r − 2s, 3r + s, s), r, s ∈ R}.
(a). S is certainly nonempty. Let x, y ∈ S. Then for some r, s, u, v ∈ R,
x = (r − 2s, 3r + s, s) and y = (u − 2v, 3u + v, v).
Hence,
x + y = (r − 2s, 3r + s, s) + (u − 2v, 3u + v, v)
= ((r + u) − 2(s + v), 3(r + u) + (s + v), s + v)
= (a − 2b, 3a + b, b),
where a = r + u, and b = s + v. Consequently, S is closed under addition. Further, if c ∈ R, then
cx = c(r − 2s, 3r + s, s) = (cr − 2cs, 3cr + cs, cs) = (k − 2l, 3k + l, l),
where k = cr and l = cs. Therefore S is also closed under scalar multiplication. It follows from Theorem
4.3.2 that S is a subspace of R3 .
(b). The coordinates of the points in S are (x, y, z) where
x = r − 2s, y = 3r + s, z = s.
Eliminating r and s between these three equations yields 3x − y + 7z = 0.
2. S = {x ∈ R2 : x = (2k, −3k), k ∈ R}.
(a). S is certainly nonempty. Let x, y ∈ S. Then for some r, s ∈ R,
x = (2r, −3r) and y = (2s, −3s).
Hence,
x + y = (2r, −3r) + (2s, −3s) = (2(r + s), −3(r + s)) = (2k, −3k),
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where k = r + s. Consequently, S is closed under addition. Further, if c ∈ R, then
cx = c(2r, −3r) = (2cr, −3cr) = (2t, −3t),
where t = cr. Therefore S is also closed under scalar multiplication. It follows from Theorem 4.3.2 that S
is a subspace of R2 .
(b). The subspace S consists of all points lying along the line in the accompanying figure.
y
y=-3x/2
x
Figure 0.0.62: Figure for Problem 2(b)
3. YES. We can write
S = {(x, y, z) ∈ R3 : z = 3x and y = 2x} = {(x, 2x, 3x) : x ∈ R}.
Note that S is nonempty since (0, 0, 0) ∈ S. To show that the nonempty set S is a subspace of V , we will
verify that S is closed under addition and S is closed under scalar multiplication:
Closure under Addition: Let (x, 2x, 3x), (y, 2y, 3y) ∈ S. Then
(x, 2x, 3x) + (y, 2y, 3y) = (x + y, 2x + 2y, 3x + 3y) = (x + y, 2(x + y), 3(x + y)) = (t, 2t, 3t) ∈ S,
where t = x + y.
Closure under Scalar Multiplication: Let (x, 2x, 3x) ∈ S and let r be a scalar. Then
r · (x, 2x, 3x) = (rx, 2rx, 3rx) = (t, 2t, 3t) ∈ S,
where t = rx.
4. YES. S = {(x, y) ∈ R2 : 3x + 2y = 0}. S = ∅ since (0, 0) ∈ S.
Closure under Addition: Let (x1 , x2 ), (y1 , y2 ) ∈ S. Then 3x1 + 2x2 = 0 and 3y1 + 2y2 = 0, so 3(x1 + y1 ) +
2(x2 + y2 ) = 0, which implies that (x1 + y1 , x2 + y2 ) ∈ S.
Closure under Scalar Multiplication: Let a ∈ R and (x1 , x2 ) ∈ S. Then 3x1 + 2x2 = 0 =⇒ a(3x1 + 2x2 ) =
a · 0 =⇒ 3(ax1 ) + 2(ax2 ) = 0, which shows that (ax1 , ax2 ) ∈ S.
Thus, S is a subspace of R2 by Theorem 4.3.2.
5. NO. S = {(x1 , 0, x3 , 2) : x1 , x3 ∈ R} is not a subspace of R4 because it is not closed under addition. To
see this, let (a, 0, b, 2), (c, 0, d, 2) ∈ S. Then (a, 0, b, 2) + (c, 0, d, 2) = (a + c, 0, b + d, 4) ∈
/ S. Note that S also
fails the zero vector check.
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/ S since 0 + 0 + 0 = 1.
6. NO. S = {(x, y, z) ∈ R3 : x + y + z = 1} is not a subspace of R3 because (0, 0, 0) ∈
/ S.
7. NO. S = {u ∈ R2 : Au = b, A is a fixed m × n matrix} is not a subspace of Rn since 0 ∈
8. NO. S = {(x, y) ∈ R2 : x2 − y 2 = 0} is not a subspace of R2 , since it is not closed under addition, as we
now observe: If (x1 , y1 ), (x2 , y2 ) ∈ S, then (x1 , y1 ) + (x2 , y2 ) = (x1 + x2 , y1 + y2 ).
(x1 + x2 )2 − (y1 + y2 )2 = x21 + 2x1 x2 + x22 − (y12 + 2y1 y2 + y22 )
= (x21 − y12 ) + (x22 − y22 ) + 2(x1 x2 − y1 y2 )
= 0 + 0 + 2(x1 x2 − y1 y2 ) = 0, in general.
/ S.
Thus, (x1 , y1 ) + (x2 , y2 ) ∈
9. NO. S = {A ∈ M2 (R) : det(A) = 1} is not a subspace of M2 (R). To see this, let k ∈ R be a scalar and
let A ∈ S. Then det(kA) = k 2 det(A) = k 2 · 1 = k 2 = 1, unless k = ±1. Note also that det(A) = 1 and
det(B) = 1 does not imply that det(A + B) = 1.
10. YES. S = {A = [aij ] ∈ Mn (R) : aij = 0 whenever i < j}. Note that S = ∅ since 0n ∈ S. Now let
A = [aij ] and B = [bij ] be lower triangular matrices. Then aij = 0 and bij = 0 whenever i < j. Then
aij + bij = 0 and caij = 0 whenever i < j. Hence A + B = [aij + bij ] and cA = [caij ] are also lower triangular
matrices. Therefore S is closed under addition and scalar multiplication. Consequently, S is a subspace of
M2 (R) by Theorem 4.3.2.
/ S.
11. NO. S = {A ∈ Mn (R) : A is invertible} is not a subspace of Mn (R) because 0n ∈
12. YES. We can write
S=
or
S=
: a+b+c+d=0
a
c
b
d
a
c
b
−(a + b + c)
∈V
: a, b, c ∈ R .
Note that S is nonempty since the zero matrix belongs to S. To show that the nonempty set S is a
subspace of V , we will verify that S is closed under addition and S is closed under scalar multiplication. In
this case, we will use the first descriptionof S, although
either of them can
be used.
a b
e f
∈ S. Therefore, a + b + c + d = 0
Closure under Addition: Let A =
∈ S and B =
c d
g h
and e + f + g + h = 0. Now,
a b
e f
a+e b+f
A+B =
+
=
,
c d
g h
c+g d+h
and since
(a + e) + (b + f ) + (c + g) + (d + h) = (a + b + c + d) + (e + f + g + h) = 0 + 0 = 0,
we conclude that A + B ∈ S, as needed.
Closure under Scalar Multiplication: Let A be as above and let r be a scalar. Then
a b
ra rb
rA = r
=
,
c d
rc rd
and since
ra + rb + rc + rd = r(a + b + c + d) = r(0) = 0,
we conclude that rA ∈ S, as needed.
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13. YES. We can write
⎧⎡
⎨ a
S= ⎣ c
⎩
e
⎤
b
d ⎦∈V
f
⎧⎡
⎨
S= ⎣
⎩
or
a
c
−(a + c)
⎫
⎬
: a + c + e = 0 and b + d + f = 0
⎭
⎫
⎤
b
⎬
⎦ : a, b, c, d ∈ R .
d
⎭
−(b + d)
Note that S is nonempty since the zero matrix belongs to S. To show that the nonempty set S is a
subspace of V , we will verify that S is closed under addition and S is closed under scalar multiplication. In
this case, we will use the first description ⎡of S, although
either of ⎡
them can⎤be used.
⎤
a b
u v
Closure under Addition: Let A = ⎣ c d ⎦ ∈ S and B = ⎣ w x ⎦ ∈ S. Therefore,
y z
e f
a + c + e = b + d + f = u + w + y = v + x + z = 0.
Now,
⎡
a
A+B =⎣ c
e
⎤ ⎡
b
u
d ⎦+⎣ w
f
y
⎤ ⎡
v
a+u
x ⎦=⎣ c+w
z
e+y
⎤
b+v
d + x ⎦,
f +z
and since
(a + u) + (c + w) + (e + y) = (a + c + e) + (u + w + y) = 0 + 0 = 0
and
(b + v) + (d + x) + (f + z) = (b + d + f ) + (v + x + z) = 0 + 0 = 0,
we conclude that A + B ∈ S, as needed.
Closure under Scalar Multiplication: Let A be as above and let r be a scalar. Then
⎡
⎤ ⎡
⎤
a b
ra rb
rA = r ⎣ c d ⎦ = ⎣ rc rd ⎦ ,
e f
re rf
and since
ra + rc + re = r(a + c + e) = r(0) = 0 and rb + rd + rf = r(b + d + f ) = r(0) = 0,
we conclude that rA ∈ S, as needed.
14. NO. We can write
S=
a b
d e
c
f
∈V
: a + b + c = 10 and d + e + f = 10 .
Note that S is not a subspace of V , for example, by the Zero Vector Check:
15. YES. S = {A ∈ M2 (R) : AT = A}. S = ∅ since 02 ∈ S.
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0
0
0
0
0
∈ S.
273
Closure under Addition: If A, B ∈ S, then (A + B)T = AT + B T = A + B, which shows that A + B ∈ S.
Closure under Scalar Multiplication: If r ∈ R and A ∈ S, then (rA)T = rAT = rA, which shows that rA ∈ S.
Consequently, S is a subspace of M2 (R) by Theorem 4.3.2.
16. YES. We can write
S = {f ∈ V
: f (a) = 5 · f (b)}.
Note that the zero function f (x) = 0 for all x belongs to S, so that S is nonempty. To show that the
nonempty set S is a subspace of V , we will verify that S is closed under addition and S is closed under scalar
multiplication.
Closure under Addition: Suppose that f and g belong to S. Therefore, f (a) = 5f (b) and g(a) = 5g(b).
Hence,
(f + g)(a) = f (a) + g(a) = 5f (b) + 5g(b) = 5(f (b) + g(b)) = 5 · (f + g)(b),
which shows that f + g ∈ S.
Closure under Scalar Multiplication: Suppose that f belongs to S and r is a scalar. Therefore
(rf )(a) = rf (a) = r(5(f (b)) = 5(rf (b)) = 5 · (rf )(b),
which shows that rf ∈ S.
17. NO. S = {f ∈ V : f (a) = 1}, where V is the vector space of all real-valued functions defined on [a, b].
We claim that S is not a subspace of V .
Not Closed under Addition: If f, g ∈ S, then (f + g)(a) = f (a) + g(a) = 1 + 1 = 2 = 1, which shows that
f +g ∈
/ S.
It can also be shown that S is not closed under scalar multiplication.
18. YES. S = {f ∈ V : f (−x) = f (x) for all x ∈ R}. Note that S = ∅ since the zero function O(x) = 0 for
all x belongs to S. Let f, g ∈ S. Then
(f + g)(−x) = f (−x) + g(−x) = f (x) + g(x) = (f + g)(x)
and if c ∈ R, then
(cf )(−x) = cf (−x) = cf (x) = (cf )(x),
so f + g and c · f belong to S. Therefore, S is closed under addition and scalar multiplication. Therefore, S
is a subspace of V by Theorem 4.3.2.
19. YES. S = {p ∈ P2 (R) : p(x) = ax2 + b, a, b ∈ R}. Note that S = ∅ since p(x) = 0 belongs to S.
Closure under Addition: Let p, q ∈ S. Then for some a1 , a2 , b1 , b2 ∈ R,
p(x) = a1 x2 + b1 and q(x) = a2 x2 + b2 .
Hence,
(p + q)(x) = p(x) + q(x) = (a1 + a2 )x2 + b1 + b2 = ax2 + b,
where a = a1 + a2 and b = b1 + b2 , so that S is closed under addition.
Closure under Scalar Multiplication: If k ∈ R, then
(kp)(x) = kp(x) = ka1 x2 + kb1 = cx2 + d,
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where c = ka1 and d = kb1 , so that S is also closed under scalar multiplication.
It follows from Theorem 4.3.2 that S is a subspace of P2 (R).
20. NO. S = {p ∈ P2 (R) : p(x) = ax2 + 1, a ∈ R}. We claim that S is not closed under addition:
Not Closed under Addition: Let p, q ∈ S. Then for some a1 , a2 ∈ R,
p(x) = a1 x2 + 1 and q(x) = a2 x2 + 1.
Hence,
(p + q)(x) = p(x) + q(x) = (a1 + a2 )x2 + 2 = ax2 + 2
where a = a1 + a2 . Consequently, p + q ∈
/ S, and therefore S is not closed under addition. It follows that S
is not a subspace of P2 (R).
21. YES. S = {y ∈ C 2 (I) : y +2y −y = 0}. Note that S is nonempty since the function y = 0 belongs to S.
Closure under Addition: Let y1 , y2 ∈ S.
(y1 + y2 ) + 2(y1 + y2 ) − (y1 + y2 ) = y1 + y2 + 2(y1 + y2 ) − y1 − y2
= y1 + 2y1 − y1 + y2 + 2y2 − y2
Thus, y1 + y2 ∈ S.
= (y1 + 2y1 − y1 ) + (y2 + 2y2 − y2 )
= 0 + 0 = 0.
Closure under Scalar Multiplication: Let k ∈ R and y1 ∈ S.
(ky1 ) + 2(ky1 ) − (ky1 ) = ky1 + 2ky1 − ky1 = k(y1 + 2y1 − y1 ) = k · 0 = 0, which shows that ky1 ∈ S.
Hence, S is a subspace of V by Theorem 4.3.2.
22. NO. S = {y ∈ C 2 (I) : y + 2y − y = 1}. S is not a subspace of V .
We show that S fails to be closed under addition (one can also verify that it is not closed under scalar
multiplication, but this is unnecessary if one shows the failure of closure under addition):
Not Closed under Addition: Let y1 , y2 ∈ S.
(y1 + y2 ) + 2(y1 + y2 ) − (y1 + y2 ) = y1 + y2 + 2(y1 + y2 ) − y1 − y2
= (y1 + 2y1 − y1 ) + (y2 + 2y2 − y2 ) Thus, y1 + y2 ∈
/ S.
= 1 + 1 = 2 = 1.
Or alternatively:
Not Closed under Scalar Multiplication: Let k ∈ R and y1 ∈ S.
((ky1 ) + 2(ky1 ) − (ky1 ) = ky1 + 2ky1 − ky1 = k(y1 + 2y1 − y1 ) = k · 1 = k = 1, unless k = 1. Therefore,
ky1 ∈
/ S unless k = 1.
23. The null space of A is the subspace of R2 consisting of points (x, y) with x + 4y = 0. The matrix A is
already in row-echelon form. Setting y = t to be a free variable, then x = −4t. Hence,
nullspace(A) = {(−4t, t) : t ∈ R}.
24. The null space of A is the subspace of R3 consisting of points (x, y, z) with x − 3y + 2z = 0. The matrix
A is already in row-echelon form. Setting y = t and z = r to be free variables, then x = 3t − 2r. Hence,
nullspace(A) = {(3t − 2r, t, r) : r, t ∈ R}.
25. The null space of A is the subspace of R2 consisting of points (x, y) with 2x − 4y = 0, x + 2y = 0, and
−3x − 5y = 0. A short series of elementary row operations provides the following row-echelon form for A:
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⎡
⎤
1 2
⎣ 0 1 ⎦. Since there are no free variables here, we conclude that the null space of A contains no nonzero
0 0
vectors. Therefore,
nullspace(A) = {(0, 0)}.
26. The null space of A is the subspace of R4 consisting of points (x, y, z, w) with x + 2y + 3z + 4w = 0 and
5x + 6y+ 7z + 8w = 0. A short series of elementary row operations provides the following row-echelon form
1 2 3 4
for A:
. Therefore, we take free variables z = t and w = r. Hence, y = −2t − 3r. Thus,
0 1 2 3
x = −2(−2t − 3r) − 3t − 4r = t + 2r. Therefore,
nullspace(A) = {(t + 2r, −2t − 3r, t, r) : r, t ∈ R}.
⎡
⎤
1
−2 ⎦. nullspace(A) = {x ∈ R3 : Ax = 0}. The reduced row echelon form of the
4
⎡
⎤
1 0 0 0
augmented matrix of the system Ax = 0 is ⎣ 0 1 0 0 ⎦. Consequently, the linear system Ax = 0 has
0 0 1 0
only the trivial solution (0, 0, 0), so nullspace(A) = {(0, 0, 0)}.
⎡
⎤
1 3 −2
1
6 ⎦. nullspace(A) = {x ∈ R3 : Ax = 0}. The REDUCED ROW ECHELON
28. A = ⎣ 3 10 −4
2 5 −6 −1
⎡
⎤
1 0 −8 −8 0
2
3 0 ⎦, so that nullspace(A) =
FORM of the augmented matrix of the system Ax = 0 is ⎣ 0 1
0 0
0
0 0
{(8r + 8s, −2r − 3s, r, s) : r, s ∈ R}.
⎡
⎤
1
i −2
4i −5 ⎦. nullspace(A) = {x ∈ C3 : Ax = 0}. The REDUCED ROW ECHELON
29. A = ⎣ 3
−1 −3i
i
⎡
⎤
1 0 0 0
FORM of the augmented matrix of the system Ax = 0 is ⎣ 0 1 0 0 ⎦. Consequently, the linear system
0 0 1 0
Ax = 0 has only the trivial solution (0, 0, 0), so nullspace(A) = {(0, 0, 0)}.
1 −2
27. A = ⎣ 4 −7
−1
3
30. Since the zero function y(x) = 0 for all x ∈ I is not a solution to the differential equation, the set of all
solutions does not contain the zero vector from C 2 (I), hence it is not a vector space at all and cannot be a
subspace.
31.
(a). As an example, we can let V = R2 . If we take S1 to be the set of points lying on the x-axis and S2
to be the set of points lying on the y-axis, then it is readily seen that S1 and S2 are both subspaces of V .
However, the union of these subspaces is not closed under addition. For instance, the points (1, 0) and (0, 1)
lie in S1 ∪ S2 , but (1, 0) + (0, 1) = (1, 1) ∈ S1 ∪ S2 . Therefore, the union S1 ∪ S2 does not form a subspace
of V .
(b). Since S1 and S2 are both subspaces of V , they both contain the zero vector. It follows that the zero
vector is an element of S1 ∩ S2 , hence this subset is nonempty. Now let u and v be vectors in S1 ∩ S2 , and let
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c be a scalar. Then u and v are both in S1 and both in S2 . Since S1 and S2 are each subspaces of V , u + v
and cu are vectors in both S1 and S2 , hence they are in S1 ∩ S2 . This implies that S1 ∩ S2 is a nonempty
subset of V , which is closed under both addition and scalar multiplication. Therefore, S1 ∩ S2 is a subspace
of V .
(c). Note that S1 is a subset of S1 + S2 (every vector v ∈ S1 can be written as v + 0 ∈ S1 + S2 ), so S1 + S2
is nonempty.
Closure under Addition: Now let u and v belong to S1 + S2 . We may write u = u1 + u2 and v = v1 + v2 ,
where u1 , v1 ∈ S1 and u2 , v2 ∈ S2 . Then
u + v = (u1 + u2 ) + (v1 + v2 ) = (u1 + v1 ) + (u2 + v2 ).
Since S1 and S2 are closed under addition, we have that u1 + v1 ∈ S1 and u2 + v2 ∈ S2 . Therefore,
u + v = (u1 + u2 ) + (v1 + v2 ) = (u1 + v1 ) + (u2 + v2 ) ∈ S1 + S2 .
Hence, S1 + S2 is closed under addition.
Closure under Scalar Multiplication: Next, let u ∈ S1 + S2 and let c be a scalar. We may write u = u1 + u2 ,
where u1 ∈ S1 and u2 ∈ S2 . Thus,
c · u = c · u1 + c · u2 ,
and since S1 and S2 are closed under scalar multiplication, c · u1 ∈ S1 and c · u2 ∈ S2 . Therefore,
c · u = c · u 1 + c · u 2 ∈ S1 + S2 .
Hence, S1 + S2 is closed under scalar multiplication.
Solutions to Section 4.4
True-False Review:
(a): TRUE. By its very definition, when a linear span of a set of vectors is formed, that span becomes
closed under addition and under scalar multiplication. Therefore, it is a subspace of V .
(b): FALSE. In order to say that S spans V , it must be true that all vectors in V can be expressed as a
linear combination of the vectors in S, not simply “some” vector.
(c): TRUE. Every vector in V can be expressed as a linear combination of the vectors in S, and therefore,
it is also true that every vector in W can be expressed as a linear combination of the vectors in S. Therefore,
S spans W , and S is a spanning set for W .
(d): FALSE. To illustrate this, consider V = R2 , and consider the spanning set {(1, 0), (0, 1), (1, 1)}. Then
the vector v = (2, 2) can be expressed as a linear combination of the vectors in S in more than one way:
v = 2(1, 1)
and
v = 2(1, 0) + 2(0, 1).
Many other illustrations, using a variety of different vector spaces, are also possible.
(e): TRUE. To say that a set S of vectors in V spans V is to say that every vector in V belongs to span(S).
So V is a subset of span(S). But of course, every vector in span(S) belongs to the vector space V , and so
span(S) is a subset of V . Therefore, span(S) = V .
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(f ): FALSE. This is not necessarily the case. For example, the linear span of the vectors (1, 1, 1) and
(2, 2, 2) is simply a line through the origin, not a plane.
(g): FALSE. There are vector spaces that do not contain finite spanning sets. For instance, if V is the
vector space consisting of all polynomials with coefficients in R, then since a finite spanning set could not
contain polynomials of arbitrarily large degree, no finite spanning set is possible for V .
(h): FALSE. To illustrate this, consider V = R2 , and consider the spanning set S = {(1, 0), (0, 1), (1, 1)}.
The proper subset S = {(1, 0), (0, 1)} is still a spanning set for V . Therefore, it is possible for a proper
subset of a spanning set for V to still be a spanning set for V .
⎡
⎤
a b c
(i): TRUE. The general matrix ⎣ 0 d e ⎦ in this vector space can be written as aE11 + bE12 + cE13 +
0 0 f
dE22 + eE23 + f E33 , and therefore the matrices in the set {E11 , E12 , E13 , E22 , E23 , E33 } span the vector
space.
(j): FALSE. For instance, it is easily verified that {x2 , x2 + x, x2 + 1} is a spanning set for P2 (R), and yet,
it contains only polynomials of degree 2.
(k): FALSE. For instance, consider m = 2 and n = 3. Then one spanning set for R2 is {(1, 0), (0, 1), (1, 1), (2, 2)},
which consists of four vectors. On the other hand, one spanning set for R3 is {(1, 0, 0), (0, 1, 0), (0, 0, 1)},
which consists of only three vectors.
(l): TRUE. This is explained in True-False Review Question (g) above.
Problems:
1. NO. This single vector only spans all of its multiples, which is a line in R2 . Any point in R2 that does
not lie along the lie through the origin with parallel vector (5, −1) does not lie in the span of this set.
2. YES. {(1, −1), (2, −2), (2, 3)}. Since v1 = (1, −1), and v2 = (2, 3) are noncolinear, the given set of
vectors does span R2 . (See the comment immediately preceding Example 4.4.3 in the text.)
3. NO. The given set of vectors does not span R2 since it does not contain two nonzero and non-collinear
vectors.
4. NO. The three vectors in the given set are all collinear. Consequently, the set of vectors does not span
R2 .
1 2
4 5. YES. Since −1 5 −2 = −23 = 0, the given vectors are not coplanar, and therefore span R3 .
1 3
1 6. YES. We compute
1
−2
1
2
4 3 −1 = −7 = 0,
1
2 which shows that the given vectors are not coplanar. Therefore, they span all of R3 .
2
3 1 7. NO. Since −1 −3 1 = 0, the vectors are coplanar, and therefore the given set does not span R3 .
4
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8. NO. We compute
1
2
3
4
5
6
7 8 = 0,
9 which shows that the given vectors are coplanar. Therefore, they only span the plane in which they lie,
which is not all of R3 .
−4 5 6 9. Since 1 1 0 = 0, the vectors are coplanar, and therefore the given set does not span R3 .The linear
3 6 2 span of the vectors is those points (x, y, z) for which the system
c1 (−4, 1, 3) + c2 (5, 1, 6) + c3 (6, 0, 2) = (x, y, z)
⎤
⎤
⎡
y
−4 5 6 x
1 1 0
⎦.
−3y + z
is consistent. Reducing the augmented matrix ⎣ 1 1 0 y ⎦ of this system yields ⎣ 0 3 2
3 6 2 z
0 0 0 x + 13y − 3z
This system is consistent if and only if x + 13y − 3z = 0. Consequently, the linear span of the given set of
vectors consists of all points lying on the plane with equation x + 13y − 3z = 0.
1 3 4 10. Since 2 4 5 = 0, the vectors are coplanar, and therefore the given set does not span R3 . The
3 5 6 linear span of the vectors is those points (x, y, z) for which the system
⎡
c1 (1, 2, 3) + c2 (3, 4, 5) + c3 (4, 5, 6) = (x, y, z)
⎤
⎤
⎡
x
1 3 4 x
1 3 4
2x − y ⎦.
is consistent. Reducing the augmented matrix ⎣ 2 4 5 y ⎦ of this system yields ⎣ 0 2 3
3 5 6 z
0 0 0 x − 2y + z
This system is consistent if and only if x − 2y + z = 0. Consequently, the linear span of the given set of
vectors consists of all points lying on the plane with the equation x − 2y + z = 0.
⎡
11. Let (x1 , x2 ) ∈ R2 and a, b ∈ R.
(x1 , x2 ) = av1 + bv2 = a(2, −1) + b(3, 2) = (2a, −a) + (3b, 2b) = (2a + 3b, −a + 2b). It follows that
2x1 − 3x2
x1 + 2x2
2a + 3b = x1 and −a + 2b = x2 , which implies that a =
and b =
so (x1 , x2 ) =
7
7
x1 + 2x2
31
9
2x1 − 3x2
v1 +
v2 . {v1 , v2 } spans R2 , and in particular, (5, −7) =
v1 − v2 .
7
7
7
7
12. Let (x1 , x2 ) ∈ R2 and a, b ∈ R.
(x1 , x2 ) = av1 + bv2 = a(1, −5) + b(6, 3) = (a, −5a) + (6b, 3b) = (a + 6b, −5a + 3b). It follows that
x1 − 2x2
5x1 + x2
a + 6b = x1 and −5a + 3b = x2 , which implies that a =
and b =
. Thus,
11
33
x1 − 2x2
5x1 + x2
v1 +
v2 .
(x1 , x2 ) =
11
33
Hence, {v1 , v2 } spans R2 .
13. Let (x, y, z) ∈ R3 and a, b, c ∈ R.
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(x, y, z) = av1 + bv2 + cv3 = a(1, −3, 2) + b(1, 0, −1) + c(1, 2, −4)
= (a, −3a, 2a) + (b, 0, −b) + (c, 2c, −4c)
= (a + b + c, −3a + 2c, 2a − b − 4c).
Therefore,
a + b + c = x,
−3a + 2c = y,
2a − b − 4c = z.
Substituting x = 9, y = 8, and z = 7, we obtain the system
−3a + 2c = 8, 2a − b − 4c = 7.
⎡
⎤
1
1
1 9
0
2 8 ⎦. Row-reducing this augmented matrix
The augmented matrix for this system is ⎣ −3
2 −1 −4 7
⎤
⎡
9
1 1 1
56
35 ⎦. Back substitution now yields c = −24, b = 155
yields ⎣ 0 3 5
3 and a = − 3 . Therefore, we have
0 0 1 −24
155
(9, 8, 7) = − 56
3 v1 + 3 v2 − 24v3 .
a + b + c = 9,
14. Let (x, y, z) ∈ R3 and a, b, c ∈ R.
(x, y, z) = av1 + bv2 + cv3 = a(−1, 3, 2) + b(1, −2, 1) + c(2, 1, 1)
= (−a, 3a, 2a) + (b, −2b, b) + (2c, c, c)
= (−a + b +
⎧ 2c, 3a − 2b + c, 2a + b + c).
⎪ a + b + 2c = x
⎨
These equalities result in the system: 3a − 2b + c = y
⎪
⎩
2a + b + c = z
Upon solving for a, b, and c we obtain
a=
−x − 5y + 7z
7x + 3y − z
−3x + y + 5z
,b=
, and c =
.
16
16
16
Consequently, {v1 , v2 , v3 } spans R3 , and
−3x + y + 5z
−x − 5y + 7z
7x + 3y − z
(x, y, z) =
v1 +
v2 +
v3 .
16
16
16
15. Let (x, y) ∈ R2 and a, b, c ∈ R.
(x, y) = av1 + bv2 + cv3 = a(1, 1) + b(−1, 2) + c(1, 4)
= (a, a) + (−b, 2b) + (c, 4c)
= (a − b + c, a + 2b + 4c).
a−b+c=x
These equalities result in the system:
a + 2b + 4c = y
Solving the system we find that
a=
2x + y − 6c
y − x − 3c
, and b =
3
3
where c is a free real variable. It follows that
y − x − 3c
2x + y − 6c
v1 +
v2 + cv3 ,
(x, y) =
3
3
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which implies that the vectors v1 , v2 , and v3 span R2 . Moreover, if c = 0, then
(x, y) =
2x + y
3
v1 +
y−x
3
v2 ,
so R2 = span{v1 , v2 } also.
16. We can write v = (c1 , c2 , c2 − 2c1 ) = c1 (1, 0, −2) + c2 (0, 1, 1), which shows that S is spanned by
{(1, 0, −2), (0, 1, 1)}.
17. v = (c1 , c2 , c2 − 2c1 , c1 − 2c2 ) = (c1 , 0, −c1 , c1 ) + (0, c2 , c2 , −2c2 ) = c1 (1, 0, −1, 1) + c2 (0, 1, 1, −2). Thus,
{(1, 0, −1, 1), (0, 1, 1, −2)} spans S.
0 α
0 −1
0 −1
18. S = A ∈ M2 (R) : A =
, α ∈ R = A ∈ M2 (R) : A = α
= span
.
−α 0
1
0
1
0
19.
(a). S = ∅ since
0
0
0
0
∈ S.
Closure under Addition: Let x, y ∈ S. Then,
x+y =
x12
x22
x11
0
+
y12
y22
y11
0
=
x11 + y11
0
x12 + y12
x22 + y22
,
which implies that x + y ∈ S.
Closure under Scalar Multiplication: Let r ∈ R and x ∈ S. Then
rx = r
x11
0
x12
x22
=
rx11
0
rx12
rx22
,
which implies thatrx ∈ S.
Consequently, S is a subspace of M2 (R) by Theorem 4.3.2.
1 0
0 1
0 0
a11 a12
= a11
+ a12
+ a22
.
(b). A =
0 1
0 a22 0 0
0 0
1 0
0 1
0 0
Therefore, S = span
,
,
.
0 0
0 0
0 1
a
b
20. A typical element of S can be written in the form v =
. Therefore, we can write
c −(a + b + c)
v=a
1
0
0 −1
+b
0
0
1
−1
1
−1
0
,
1
+c
0
1
0
−1
which shows that S is spanned by
1
0
0
−1
0
,
0
0
−1
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281
⎡
⎤
a
b
−(a + b)
c
d
−(c + d) ⎦. There21. A typical element of S can be written in the form v = ⎣
−(a + c) −(b + d) a + b + c + d
fore, we can write
⎡
⎤
⎡
⎤
⎡
⎤
⎡
⎤
1 0 −1
0
1 −1
0 0
0
0
0
0
0 ⎦ + b⎣ 0
0
0 ⎦ + c ⎣ 1 0 −1 ⎦ + d ⎣ 0
1 −1 ⎦ ,
v = a⎣ 0 0
−1 0
1
0 −1
1
−1 0
1
0 −1
1
which shows that S is spanned by
⎧⎡
⎤ ⎡
⎤ ⎡
1 0 −1
0
1 −1
0
⎨
⎣ 0 0
0 ⎦,⎣ 0
0
0 ⎦,⎣ 1
⎩
−1 0
1
0 −1
1
−1
⎤ ⎡
⎤⎫
0
0
0
0
0 ⎬
0 −1 ⎦ , ⎣ 0
1 −1 ⎦ .
⎭
0
1
0 −1
1
⎡
a
22. A typical element of S can be written in the form v = ⎣ b
c
⎡
1
v = a⎣ 0
0
0
0
0
⎤
⎡
0
0
0 ⎦+b⎣ 1
0
0
1
0
0
⎤
⎡
0
0
0 ⎦+c⎣ 0
0
1
which shows that S is spanned by
⎧⎡
⎤ ⎡
⎤ ⎡
0 1 0
0 0
⎨ 1 0 0
⎣ 0 0 0 ⎦,⎣ 1 0 0 ⎦,⎣ 0 0
⎩
0 0 0
0 0 0
1 0
0
0
0
b
d
e
⎤
c
e ⎦. Therefore, we can write
f
⎤
⎡
1
0
0 ⎦+d⎣ 0
0
0
0
1
0
⎤ ⎡
1
0
0 ⎦,⎣ 0
0
0
⎤ ⎡
0
0
0 ⎦,⎣ 0
0
0
0
1
0
⎤
⎡
0
0
0 ⎦+e⎣ 0
0
0
0
0
1
0
0
1
⎤
⎡
0
0
1 ⎦+f ⎣ 0
0
0
⎤ ⎡
0
0
1 ⎦,⎣ 0
0
0
0
0
0
0
0
0
⎤
0
0 ⎦,
1
⎤⎫
0 ⎬
0 ⎦ .
⎭
1
23. x − 2y − z = 0 =⇒ x = 2y + z, so v ∈ R3 .
=⇒ v = (2y + z, y, z) = (2y, y, 0) + (z, 0, z) = y(2, 1, 0) + z(1, 0, 1).
Therefore S = {v ∈ R3 : v = a(2, 1, 0) + b(1, 0, 1), a, b ∈ R}, hence {(2, 1, 0), (1, 0, 1)} spans S.
24. Note that S consists precisely of the constant polynomials: S = {c : c ∈ R}. Thus, S can be spanned
by {1}.
25. In Problem 23 in Section 4.3, we found that nullspace(A) = {(−4t, t) : t ∈ R}. Hence, nullspace(A) =
{t(−4, 1) : t ∈ R}. Thus, a spanning set for nullspace(A) is {(−4, 1)}.
26. In Problem 24 in Section 4.3, we found that nullspace(A) = {(3t − 2r, t, r) : r, t ∈ R}. Hence,
nullspace(A) = {t(3, 1, 0)+r(−2, 0, 1) : r, t ∈ R}. Thus, a spanning set for nullspace(A) is {(3, 1, 0), (−2, 0, 1)}.
27. In Problem 25 in Section 4.3, we found that nullspace(A) = {(0, 0)}, so it can be spanned by {(0, 0)}.
28. In Problem 26 in Section 4.3, we found that nullspace(A) = {(t + 2r, −2t − 3r, t, r) : r, t ∈ R}.
Hence, nullspace(A) = {t(1, −2, 1, 0) + r(2, −3, 0, 1) : r, t ∈ R}. Thus, a spanning set for nullspace(A) is
{(1, −2, 1, 0), (2, −3, 0, 1)}.
29. In Problem 27 in Section 4.3, we found that nullspace(A) = {(0, 0, 0)}, so it can be spanned by {(0, 0, 0)}.
30. In Problem 28 in Section 4.3, we found that nullspace(A) = {(8r + 8s, −2r − 3s, r, s) : r, s ∈ R}.
Hence, nullspace(A) = {r(8, −2, 1, 0) + s(8, −3, 0, 1) : r, s ∈ R}. Thus, a spanning set for nullspace(A) is
{(8, −2, 1, 0), (8, −3, 0, 1)}.
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31. In Problem 29 in Section 4.3, we found that nullspace(A = {(0, 0, 0)}, so it can be spanned by {(0, 0, 0)}.
⎡
⎤
1 2 3
5
2 ⎦. nullspace(A) = {x ∈ R4 : Ax = 0}. The REDUCED ROW ECHELON FORM
32. A = ⎣ 1 3 4
2 4 6 −1
⎡
⎤
1 0 1 0 0
of the augmented matrix of this system is ⎣ 0 1 1 0 0 ⎦. Consequently, nullspace(A) = {r(1, 1, −1, 0) :
0 0 0 1 0
r ∈ R} = span{(1, 1, −1, 0)}.
⎤ ⎡
⎡
⎤⎡
⎤
1 2 3
x1
0
33. nullspace(A) = {x ∈ R3 : Ax = 0}. Ax = 0 =⇒ ⎣ 1 3 4 ⎦ ⎣ x2 ⎦ = ⎣ 0 ⎦.
2 4 6
x3
0
Performing Gauss-Jordan elimination on the augmented matrix of the system we obtain:
⎤ ⎡
⎤ ⎡
⎤
1 2 3 0
1 0 1 0
1 2 3 0
⎣ 1 3 4 0 ⎦ ∼ ⎣ 0 1 1 0 ⎦ ∼ ⎣ 0 1 1 0 ⎦.
2 4 6 0
0 0 0 0
0 0 0 0
⎡
From the last matrix we find that x1 = −r and x2 = −r where x3 = r, with r ∈ R. Consequently,
x = (x1 , x2 , x3 ) = (−r, −r, r) = r(−1, −1, 1).
Thus,
nullspace(A) = {x ∈ R3 : x = r(−1, −1, 1), r ∈ R} = span{(−1, −1, 1)}.
34 Let v ∈ span{v1 , v2 } and a, b ∈ R.
v = av1 + bv2 = a(1, −1, 2) + b(2, −1, 3) = (a, −a, 2a) + (2b, −b, 3b) = (a + 2b, −a − b, 2a + 3b).
Thus, span{v1 , v2 } = {v ∈ R3 : v = (a + 2b, −a − b, 2a + 3b), a, b ∈ R}.
Geometrically, span{v1 , v2 } is the plane through the origin determined by the two given vectors. The plane
has parametric equations x = a + 2b, y = −a − b, and z = 2a + 3b. If a, b, and c are eliminated from the
equations, then the resulting Cartesian equation is given by x − y − z = 0.
35. Let v ∈ span{v1 , v2 } and a, b ∈ R.
v = av1 + bv2 = a(1, 2, −1) + b(−2, −4, 2) = (a, 2a, −a) + (−2b, −4b, 2b) = (a − 2b, 2a − 4b, −a + 2b)
= (a − 2b)(1, 2, −1) = k(1, 2, −1), where k = a − 2b.
Thus, span{v1 , v2 } = {v ∈ R3 : v = k(1, 2, −1), k ∈ R}. Geometrically, span{v1 , v2 } is the line through the
origin determined by the vector (1, 2, −1).
36. Let v ∈ span{v1 , v2 , v3 } and a, b, c ∈ R.
v = av1 + bv2 + cv3 = a(1, 1, −1) + b(2, 1, 3) + c(−2, −2, 2) = (a, a, −a) + (2b, b, 3b) + (−2c, −2c, 2c)
= (a + 2b − 2c, a + b − 2c, −a + 3b + 2c).
⎧
⎪
⎨ a + 2b − 2c = x
a + b − 2c = y
−a + 3b + 2c = z
Performing Gauss-Jordan elimination on the augmented matrix of the system, we obtain:
⎤ ⎡
⎤ ⎡
⎤
⎡
x
2y − x
1 2 −2 x
1
2 −2
1 0 −2
⎦.
⎣ 1 1 −2 y ⎦ ∼ ⎣ 0 −1
0 y−x ⎦∼⎣ 0 1
0
x−y
−1 3
2 z
0
5
0 x+z
0 0
0 5y − 4x + z
Assuming that v = (x, y, z) and using the last ordered triple, we obtain the system:
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⎩
283
It is clear from the last matrix that the subspace, S, of R3 is a plane through (0, 0, 0) with Cartesian equation
4x − 5y − z = 0. Moreover, {v1 , v2 } also spans the subspace S since
v = av1 + bv2 + cv3 = a(1, 1, −1) + b(2, 1, 3) + c(−2, −2, 2) = a(1, 1, −1) + b(2, 1, 3) − 2c(1, 1, −1)
= (a − 2c)(1, 1, −1) + b(2, 1, 3) = dv1 + bv2 where d = a − 2c ∈ R.
37. If v ∈ span{v1 , v2 } then there exist a, b ∈ R such that
v = av1 + bv2 = a(1, −1, 2) + b(2, 1, 3) = (a, −a, 2a) + (2b, b, 3b) = (a + 2b, −a + b, 2a + 3b).
⎧
⎪
⎨ a + 2b = 3
−a + b = 3
Hence, v = (3, 3, 4) is in span{v1 , v2 } provided there exists a, b ∈ R satisfying the system:
⎪
⎩
2a + 3b = 4
Solving this system we find that a = −1 and b = 2.
Consequently, v = −v1 + 2v2 so that (3, 3, 4) ∈ span{v1 , v2 }.
38. If v ∈ span{v1 , v2 } then there exist a, b ∈ R such that
v = av1 + bv2 = a(−1, 1, 2) + b(3, 1, −4) = (−a, a, 2a) + (3b, b, −4b) = (−a + 3b, a + b, 2a − 4b).
⎧
⎪
⎨ −a + 3b = 5
a+b= 3
Hence v = (5, 3, −6) is in span{v1 , v2 } provided there exists a, b ∈ R satisfying the system:
⎪
⎩
2a − 4b = −6
Solving this system we find that a = 1 and b = 2.
Consequently, v = v1 + 2v2 so that (5, 3, −6) ∈ span{v1 , v2 }.
39. If v ∈ span{v1 , v2 } then there exist a, b ∈ R such that
v = av1 + bv2 = (3a, a, 2a) + (−2b, −b, b) = (3a − 2b, a − b, 2a + b).
⎧
⎪
⎨ 3a − 2b = 1
a−b= 1
Hence v = (1, 1, −2) is in span{v1 , v2 } provided there exists a, b ∈ R satisfying the system:
⎪
⎩
2a + b = −2
⎤
⎤
⎡
⎡
1
1
3 −2
1 −1
1 −2 ⎦, it follows that the system has no solution. Hence it
1 ⎦ reduces to ⎣ 0
Since ⎣ 1 −1
2
1 −2
0
0
2
must be the case that (1, 1, −2) ∈
/ span{v1 , v2 }.
40. If p ∈ span{p1 , p2 } then there exist a, b ∈ R such that p(x) = ap1 (x) + bp2 (x), so p(x) = 2x2 − x + 2 is
in span{p1 , p2 } provided there exist a, b ∈ R such that
2x2 − x + 2 = a(x − 4) + b(x2 − x + 3)
= ax − 4a + bx2 − bx + 3b
= bx2 + (a − b)x + (3b − 4a).
Equating like coefficients and solving, we find that a = 1 and b = 2.
Thus, 2x2 − x + 2 = 1 · (x − 4) + 2 · (x2 − x + 3) = p1 (x) + 2p2 (x) so p ∈ span{p1 , p2 }.
41. Let A ∈ span{A1 , A2 , A3 } and c1 , c2, c3 ∈ R.
1 −1
0 1
3 0
+ c2
+ c3
A = c1 A1 + c2 A2 + c3 A3 = c1
2
0
−2 1
1 2
−c1 + c2
0
c2
3c3 0
c1 + 3c3
c1 −c1
+
+
=
.
=
c3 2c3
2c1
0
−2c2 c2
2c1 − 2c2 + c3 c2 + 2c3
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Therefore span{A1 , A2 , A3 } =
A ∈ M2 (R) : A =
42. Let A ∈ span{A1 , A2 } and
a, b ∈ R.
1 2
−2
+b
A = aA1 + bA2 = a
−1
3
1
c1 + 3c3
2c1 − 2c2 + c3
−c1 + c2
c2 + 2c3
.
1
a 2a
−2b
b
a − 2b 2a + b
=
+
=
.
−1
−a 3a
b −b
−a + b 3a − b
a − 2b 2a + b
. Now, to determine whether B ∈ span{A1 , A2 },
−a + b 3a − b
So span{A1 , A2 } = A ∈ M2 (R) : A =
a − 2b 2a + b
3 1
let
=
. This implies that a = 1 and b = −1, thus B ∈ span{A1 , A2 }.
−a + b 3a − b
−2 4
43. (a). The general vector in span{f, g} is of the form h(x) = c1 cosh x + c2 sinh x where c1 , c2 ∈ R.
(b). Let h ∈ S and c1 , c2 ∈ R. Then
h(x) = c1 f (x) + c2 g(x) = c1 cosh x + c2 sinh x
ex + e−x
ex − e−x
c1 ex
c1 e−x
c2 ex
c2 e−x
+ c2
=
+
+
−
2
2
2
2
2
2
c1 + c2 x c1 − c2 −x
x
−x
=
e +
e = d1 e + d2 e
2
2
c1 + c2
c1 − c2
where d1 =
and d2 =
. Therefore S =span{ex , e−x }.
2
2
44. The origin in R3 .
= c1
45. All points lying on the line through the origin with direction v1 .
46. All points lying on the plane through the origin containing v1 and v2 .
47. If v1 = v2 = 0, then the subspace is the origin in R3 . If at least one of the vectors is nonzero, then the
subspace consists of all points lying on the line through the origin in the direction of the nonzero vector.
48. Suppose that S is a subset of S . We must show that every vector in span(S) also belongs to span(S ).
Every vector v that lies in span(S) can be expressed as v = c1 v1 + c2 v2 + · · · + ck vk , where v1 , v2 , . . . , vk
belong to S. However, since S is a subset of S , v1 , v2 , . . . , vk also belong to S , and therefore, v belongs to
span(S ). Thus, we have shown that every vector in span(S) also lies in span(S ).
49.
Proof of =⇒: We begin by supposing that span{v1 , v2 , v3 } = span{v1 , v2 }. Since v3 ∈ span{v1 , v2 , v3 },
our supposition implies that v3 ∈ span{v1 , v2 }, which means that v3 can be expressed as a linear combination
of the vectors v1 and v2 .
Proof of ⇐=: Now suppose that v3 can be expressed as a linear combination of the vectors v1 and v2 . We
must show that span{v1 , v2 , v3 } = span{v1 , v2 }, and we do this by showing that each of these subsets is a
subset of the other. Since it is clear that span{v1 , v2 } is a subset of span{v1 , v2 , v3 }, we focus our attention
on proving that every vector in span{v1 , v2 , v3 } belongs to span{v1 , v2 }. To see this, suppose that v belongs
to span{v1 , v2 , v3 }, so that we may write v = c1 v1 + c2 v2 + c3 v3 . By assumption, v3 can be expressed as a
linear combination of v1 and v2 , so that we may write v3 = d1 v1 + d2 v2 . Hence, we obtain
v = c1 v1 + c2 v2 + c3 v3 = c1 v1 + c2 v2 + c3 (d1 v1 + d2 v2 ) = (c1 + c3 d1 )v1 + (c2 + c3 d2 )v2 ∈ span{v1 , v2 },
as required.
Solutions to Section 4.5
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(a): FALSE. For instance, consider the vector space V = R2 . Here are two different minimal spanning sets
for V :
{(1, 0), (0, 1)} and {(1, 0), (1, 1)}.
Many other examples of this abound.
(b): TRUE. We have seven column vectors, and each of them belongs to R5 . Therefore, the number
of vectors present exceeds the number of components in those vectors, and hence they must be linearly
dependent.
(c): FALSE. For instance, the 7 × 5 zero matrix, 07×5 , does not have linearly independent columns.
(d): TRUE. Any linear dependencies within the subset also represent linear dependencies within the original, larger set of vectors. Therefore, if the nonempty subset were linearly dependent, then this would require
that the original set is also linearly dependent. In other words, if the original set is linearly independent,
then so is the nonempty subset.
(e): TRUE. This is stated in Theorem 4.5.23.
(f ): TRUE. If we can write v = c1 v1 + c2 v2 + · · · + ck vk , then {v, v1 , v2 , . . . , vk } is a linearly dependent
set.
(g): TRUE. This is a rephrasing of the statement in True-False Review Question (d) above.
(h): FALSE. None of the vectors (1, 0), (0, 1), and (1, 1) in R2 are proportional to each other, and yet,
they form a linearly dependent set of vectors.
(i): FALSE. The illustration given in part (c) of Example 4.5.24 gives an excellent case-in-point here.
Problems:
1. This set is linearly independent by Theorem 4.5.5.
2. {(1, −1), (1, 1)}. These vectors are elements of R2 . Since there are two vectors, and the dimension
of R2
1 1 = 0. Now
is two, Corollary 4.5.17 states that the vectors will be linearly dependent if and only if −1
1
1 1 −1 1 = 2 = 0. Consequently, the given vectors are linearly independent.
3. {(2, −1), (3, 2), (0, 1)}. These vectors are elements of R2 , but since there are three vectors, the vectors
are linearly dependent by Corollary 4.5.17. Let v1 = (2, −1), v2 = (3, 2), v3 = (0, 1). We now determine a
dependency relationship. The condition
c1 v 1 + c2 v2 + c3 v 3 = 0
requires 2c1 + 3c2 = 0 and −c1 + 2c2 + c3 = 0.
1 0 − 37
2
0 1
7
Hence the system has solution c1 = 3r, c2 = −2r, c3 = 7r. Consequently, 3v1 − 2v2 + 7v3 = 0.
The REDUCED ROW ECHELON FORM of the augmented matrix of this system is
3
4.
1, −1), (1, 1, 1)}. These vectors are elements of R .
{(1, −1, 0), (0,
1
0 1 −1
1 1 = 3 = 0, so by Corollary 4.5.17, the vectors are linearly independent.
0 −1 1 5. {(1, 2, 3), (1, −1, 2), (1, −4, 1)}. These vectors are elements of R3 .
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.
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1
1
1 2 −1 −4 = 0, so by Corollary 4.5.17, the vectors are linearly dependent. Let v1 = (1, 2, 3), v2 =
3
2
1 (1, −1, 2), v3 = (1, −4, 1). We determine a dependency relation. The condition c1 v1 + c2 v2 + c3 v3 = 0
requires
c1 + c2 + c3 = 0, 2c1 − c2 − 4c3 = 0, 3c1 + 2c2 + c3 = 0.
⎡
⎤
1 0 −1 0
2 0 ⎦.
The REDUCED ROW ECHELON FORM of the augmented matrix of this system is ⎣ 0 1
0 0
0 0
Hence c1 = r, c2 = −2r, c3 = r, and so v1 − 2v2 + v3 = 0.
6. Given {(−2, 4, −6), (3, −6, 9)}. The vectors are linearly dependent because 3(−2, 4, −6) + 2(3, −6, 9) =
(0, 0, 0), which gives a linear dependency relation.
Alternatively, let a, b ∈ R and observe that
a(−2, 4, −6) + b(3, −6, 9) = (0, 0, 0) =⇒ (−2a, 4a, −6a) + (3b, −6b, 9b) = (0, 0, 0).
⎧
⎪
⎨ −2a + 3b = 0
4a − 6b = 0
The last equality results in the system:
⎪
⎩
−6a + 9b = 0
⎤
⎡
1 − 32 0
0 0 ⎦, which
The REDUCED ROW ECHELON FORM of the augmented matrix of this system is ⎣ 0
0
0 0
3
implies that a = b. Thus, the given set of vectors is linearly dependent.
2
7. {(1, −1, 2), (2, 1, 0)}. Let a, b ∈ R.
a(1, −1, 2) + b(2, 1, 0) = (0, 0, 0) =⇒ (a,
⎧ −a, 2a) + (2b, b, 0) = (0, 0, 0) =⇒ (a + 2b, −a + b, 2a) = (0, 0, 0). The
⎪ a + 2b = 0
⎨
−a + b = 0 . Since the only solution of the system is a = b = 0, it
last equality results in the system:
⎪
⎩
2a = 0
follows that the vectors are linearly independent.
8. {(−1, 1, 2), (0, 2, −1), (3, 1, 2), (−1, −1, 1)}. These vectors are elements of R3 .
Since there are four vectors, it follows from Corollary 4.5.17 that the vectors are linearly dependent. Let
v1 = (−1, 1, 2), v2 = (0, 2, −1), v3 = (3, 1, 2), v4 = (−1, −1, 1). Then
c1 v 1 + c2 v 2 + c3 v3 + c4 v 4 = 0
requires
−c1 + 3c3 − c4 = 0, c1 + 2c2 + c3 − c4 = 0, 2c1 − c2 + 2c3 + c4 = 0.
⎡
2
1 0 0
5
⎢
⎢
3
The REDUCED ROW ECHELON FORM of the augmented matrix of this system is ⎢
⎢ 0 1 0 −5
⎣
0 0 1 − 15
so that
c1 = −2r, c2 = 3r, c3 = r, c4 = 5r. Hence, −2v1 + 3v2 + v3 + 5v4 = 0.
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⎤
⎥
⎥
0 ⎥
⎥
⎦
0
287
9. {(1, −1, 2, 3), (2, −1, 1, −1), (−1, 1, 1, 1)}. Let a, b, c ∈ R.
a(1, −1, 2, 3) + b(2, −1, 1, −1) + c(−1, 1, 1, 1) = (0, 0, 0, 0)
=⇒ (a, −a, 2a, 3a) + (2b, −b, b, −b) + (−c, c, c, c) = (0, 0, 0, 0)
=⇒ (a + 2b − c, −a − b + c, 2a + b + c, 3a − b + c) = (0, 0, 0, 0).
⎧
a + 2b − c = 0
⎪
⎪
⎪
⎨ −a − b + c = 0
. The REDUCED ROW ECHELON FORM of the
The last equality results in the system:
⎪
2a + b + c = 0
⎪
⎪
⎩
3a − b + c = 0
⎡
⎤
1 0 0 0
⎢ 0 1 0 0 ⎥
⎥
augmented matrix of this system is ⎢
⎣ 0 0 1 0 ⎦. Consequently, a = b = c = 0. Thus, the given set of
0 0 0 0
vectors is linearly independent.
10. {(2, −1, 0, 1), (1, 0, −1, 2), (0, 3, 1, 2), (−1, 1, 2, 1)}. These vectors are elements in R4 .
By
CET (row 1), we obtain:
2
1 0 −1 0 3 1 −1 3 1 −1
0 3 −1
0
3
1
= 2 −1 1 2 − 0 1 2 + 0 −1 1 = 21 = 0.
0 −1 1
2 2 2 1 1 2 1 1
2 2 1
2 2
1 Thus, it follows from Corollary 4.5.17 that the vectors are linearly independent.
1 4 7 11. Since 2 5 8 = 0, the given set of vectors is linearly dependent in R3 . Further, since the vectors
3 6 9 are not collinear, it follows that span{v1 , v2 , v3 } is a plane 3-space.
2
1 −3 3 −9 = 0, the given set of vectors is linearly dependent.
12. (a). Since −1
5 −4 12 (b). By inspection, we see that v3 = −3v2 . Hence v2 and v3 are collinear and therefore span{v2 , v3 } is the
line through the origin that has the direction of v2 . Further, since v1 is not proportional to either of these
vectors, it does not lie along the same line, hence v1 is not in span{v2 , v3 }.
1 0 1 1 0
1 2 k − 1 = (3 − k)(k + 4). Now, the determinant is zero when
13. 1 2 k = 0 2 k − 1 = k 6−k k k 6 0 k 6−k k = 3 or k = −4. Consequently, by Corollary 4.5.17, the vectors are linearly dependent if and only if k = 3
or k = −4.
14. {(1, 0, 1, k), (−1, 0, k, 1), (2, 0, 1, 3)}. These vectors are elements in R4 .
Let a, b, c ∈ R.
a(1, 0, 1, k) + b(−1, 0, k, 1) + c(2, 0, 1, 3) = (0, 0, 0, 0)
=⇒ (a, 0, a, ka) + (−b, 0, kb, b) + (2c, 0, c, 3c) = (0, 0, 0, 0)
=⇒ (a − b + 2c, 0, a + kb + c, ka + b + 3c) ⎧
= (0, 0, 0, 0).
⎪
⎨ a − b + 2c = 0
a + kb + c = 0 . Evaluating the determinant of the coefficient
The last equality results in the system:
⎪
⎩
ka + b + 3c = 0
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matrix, we obtain
1
1
k
−1 2 1
k 1 = 1
1 3 k
0
k+1
k+1
0 −1 = 2(k + 1)(2 − k).
3 − 2k Consequently, the system has only the trivial solution, hence the given set of vectors are linearly independent
if and only if k = 2, −1.
15. {(1, 1, 0, −1), (1, k, 1, 1), (2, 1, k, 1), (−1, 1, 1, k)}. These vectors are elements in R4 .
Let a, b, c, d ∈ R.
a(1, 1, 0, −1) + b(1, k, 1, 1) + c(2, 1, k, 1) + d(−1, 1, 1, k) = (0, 0, 0, 0)
=⇒ (a, a, 0, −a) + (b, kb, b, b) + (2c, c, kc, c) + (−d, d, d, kd) = (0, 0, 0, 0)
=⇒ (a + b + 2c − d, a + kb + c + d, b + kc + d, −a + b + c + kd) = (0, 0, 0, 0).
⎧
a + b + 2c − d = 0
⎪
⎪
⎪
⎨ a + kb + c + d = 0
The last equality results in the system:
. By Corollary 3.2.6, this system has the
⎪
b + kc + d = 0
⎪
⎪
⎩
−a + b + c + kd = 0
trivial solution if and only if the determinant of the coefficient matrix is zero. Evaluating the determinant
of
we obtain:
the coefficient matrix,
1 1 2 −1 1
1
2
−1 2 0 −k 2 + k − 1 3 − k k − 1 −1
0 k − 1 −1
1 k 1
2
1
= 1
=
k
1 = 1
k
1 0 1 k
1
k
1 1 0
2
3 k−1
0
3 − 2k
k−3 −1 1 1
2
3 k−1 k 0
2
2
k −k+1 1 k −k+1 k−3 = (k − 3)(k − 1)(k + 2).
= (k − 3) =
3 − 2k
k−3 3 − 2k
1 For the original set of vectors to be linearly independent, we need a = b = c = 0. We see that this condition
will be true provided that k ∈ R such that k = 3, 1, −2.
1 1
2 −1
3 6
0 0
16. Let a, b, c ∈ R. a
+b
+c
=
0 1
0
1
0 4
0 0
⎧
⎪
⎨ a + 2b + 3c = 0
a + 2b + 3c a − b + 6c
0 0
a − b + 6c = 0 .
=⇒
=
. The last equation results in the system:
0
a + b + 4c
0 0
⎪
⎩
a + b + 4c = 0
⎤
⎡
1 0
5 0
The REDUCED ROW ECHELON FORM of the augmented matrix of this system is ⎣ 0 1 −1 0 ⎦,
0 0
0 0
which implies that the system has an infinite number of solutions. Consequently, the given matrices are
linearly dependent.
2 −1
−1 2
0 0
17. Let a, b ∈ R. a
+b
=
3
4
1 3
0 0
⎧
2a − b = 0
⎪
⎪
⎪
⎨ 3a + b = 0
2a − b −a + 2b
0 0
. This system
=⇒
=
. The last equation results in the system:
3a + b 4a + 3b
0 0
⎪ −a + 2b = 0
⎪
⎪
⎩
4a + 3b = 0
has only the trivial solution a = b = 0, thus the given matrices are linearly independent in M2 (R).
1 0
−1 1
2 1
0 0
18. Let a, b, c ∈ R. a
+b
+c
=
1 2
2 1
5 7
0 0
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⎧
a − b + 2c = 0
⎪
⎪
⎪
⎨
a
+ 2b + 5c = 0
0 0
a − b + 2c
b+c
.
. The last equation results in the system:
=
=⇒
0 0
a + 2b + 5c 2a + b + 7c
⎪
2a + b + 7c = 0
⎪
⎪
⎩
b+c=0
⎡
⎤
1 0 3 0
⎢ 0 1 1 0 ⎥
⎥
The REDUCED ROW ECHELON FORM of the augmented matrix of this system is ⎢
⎣ 0 0 0 0 ⎦, which
0 0 0 0
implies that the system has an infinite number of solutions. Consequently, the given matrices are linearly
dependent.
19. Let a, b ∈ R. ap1 (x) + bp2 (x) = 0 =⇒ a(1 − x) + b(1 + x) = 0 =⇒ (a + b) + (−a + b)x = 0.
a+b=0
Equating like coefficients, we obtain the system:
. Since the only solution to this system is
−a + b = 0
a = b = 0, it follows that the given vectors are linearly independent.
20. Let a, b ∈ R. ap1 (x) + bp2 (x) = 0 =⇒ a(2 + 3x) + b(4 + 6x) = 0 =⇒ (2a + 4b) + (3a + 6b)x = 0.
2a + 4b = 0
Equating like coefficients, we obtain the system:
. The REDUCED ROW ECHELON FORM
3a + 6b = 0
1 2 0
, which implies that the system has an infinite number
of the augmented matrix of this system is
0 0 0
of solutions. Thus, the given vectors are linearly dependent.
21. Let c1 , c2 , c3 ∈ R. c1 p1 (x) + c2 p2 (x) + c3 p3 (x) = 0 =⇒ c1 (1 − 3x2 ) + c2 (2x + x2 ) + c3 (5) = 0. Equating
like coefficients, we obtain
c1 + 5c3 = 0, 2c2 = 0, −3c1 + c2 = 0.
It is easy to solve this system of equations to find that c1 = c2 = c3 = 0. Therefore, the given set of vectors
is linearly independent.
22. In Section 4.6, we will see that four vectors in P2 (R) must be linearly dependent, with no calculations
necessary. However, for now we proceed as follows. Let c1 , c2 , c3 , c4 ∈ R. c1 p1 (x) + c2 p2 (x) + c3 p3 (x) +
c4 p4 (x) = 0 =⇒ c1 (3x + 5x2 ) + c2 (1 + x + x2 ) + c3 (2 − x) + c4 (1 + 2x2 ) = 0. Equating like coefficients, we
obtain
c2 + 2c3 + c4 = 0, 3c1 + c2 − c3 = 0, 5c1 + c2 + 2c4 = 0.
⎡
⎤
0 1
2 1 0
This linear system has augmented matrix A# = ⎣ 3 1 −3 0 0 ⎦. Because the matrix A# only has
5 1
0 2 0
three rows, its RREF can have at most three pivots, which implies that at least one of the columns A will
be unpivoted. The resulting presence of free variables implies that this system has an infinite number of
solutions for c1 , c2 , c3 , and c4 . Thus, the given set of vectors is linearly dependent.
23. Let c1 , c2 ∈ R. c1 p1 (x) + c2 p2 (x) = 0 =⇒ c1 (a + bx) + c2 (c + dx) = 0 =⇒ (ac1 + cc2 ) + (bc1 + dc2 )x = 0.
ac1 + cc2 = 0
Equating like coefficients, we obtain the system:
. The determinant of the matrix of coefbc1 + dc2 = 0
a c = ad − bc. Consequently, the system has just the trivial solution, and hence p1 (x) and
ficients is b d p2 (x) are linearly independent if and only if ad − bc = 0.
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24. Since cos 2x = cos2 x−sin2 x, f1 (x) = f3 (x)−f2 (x) so it follows that f1 , f2 , and f3 are linearly dependent
in C ∞ (−∞, ∞).
25. Let v1 = (1, 2, 3), v2 = (−3, 4, 5), v3 = (1, − 43 , − 53 ). By inspection, we see that v2 = −3v3 . Further,
since v1 and v2 are not proportional they are linearly independent. Consequently, {v1 , v2 } is a linearly
independent set of vectors and span{v1 , v2 } =span{v1 , v2 , v3 }.
26. Let v1 = (3, 1, 5), v2 = (0, 0, 0), v3 = (1, 2, −1), v4 = (−1, 2, 3). Since v2 = 0, it is certainly true
that span{v1 , v3 v4 } =span{v1 , v2 , v3 v4 }. Further, since det[v1 , v3 , v4 ] = 42 = 0, {v1 , v3 , v4 } is a linearly
independent set.
27. Since we have four vectors in R3 , the given set is linearly dependent. We could determine the specific linear dependency between the vectors to find a linearly independent
subset, but in this case, if we just take any
1
1 3 three of the vectors, say (1, −1, 1), (1, −3, 1), (3, 1, 2), then −1 −3 1 = 2 = 0, so that these vectors are
1
1 2 linearly independent. Consequently, span{(1, −1, 1), (1, −3, 1), (3, 1, 2)} = span{(1, 1, 1), (1, −1, 1), (1, −3, 1), (3, 1, 2)}.
28. Let
v1 = (1, 1, −1, 1),v2 = (2, −1, 3, 1), v3 = (1, 1, 2, 1), v4 = (2, −1, 2, 1).
1
2 1
2 1 −1 1 −1 = 0, the set {v1 , v2 , v3 , v4 } is linearly dependent. We now determine the linearly
Since 3 2
2 −1
1
1 1
1 dependent relationship. The REDUCED ROW ECHELON FORM of the augmented matrix corresponding
to the system
c1 (1, 1, −1, 1) + c2 (2, −1, 3, 1) + c3 (1, 1, 2, 1) + c4 (2, −1, 2, 1) = (0, 0, 0, 0)
⎤
1
1 0 0
0
3
⎢ 0 1 0
1 0 ⎥
⎥
is ⎢
⎣ 0 0 1 − 1 0 ⎦, so that c1 = −r, c2 = −3r, c3 = r, c4 = 3r, where r is a free variable. It follows
3
0 0 0
0 0
that a linearly dependent relationship between the given set of vectors is
⎡
−v1 − 3v2 + v3 + 3v4 = 0
so that
v1 = −3v2 + v3 + 3v4 .
Consequently, span{v2 , v3 , v4 } = span{v1 , v2 , v3 , v4 }, and {v2 , v3 , v4 } is a linearly independent set.
1 2
−1 2
3 2
29. Let A1 =
, A2 =
, A3 =
. Then
3 4
5 7
1 1
c1 A1 + c2 A2 + c3 A3 = 02
requires that
c1 − c2 + 3c3 = 0, 2c1 + 2c2 + 2c3 = 0, 3c1 + 5c2 + c3 = 0, 4c1 + 7c2 + c3 = 0.
⎡
⎤
1 0
2 0
⎢ 0 1 −1 0 ⎥
⎥.
The REDUCED ROW ECHELON FORM of the augmented matrix of this system is ⎢
⎣ 0 0
0 0 ⎦
0 0
0 0
Consequently, the matrices are linearly dependent. Solving the system gives c1 = −2r, c2 = c3 = r. Hence,
a linearly dependent relationship is −2A1 + A2 + A3 = 02 .
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30. We first determine whether the given set of polynomials is linearly dependent. Let
p1 (x) = 2 − 5x, p2 (x) = 3 + 7x, p3 (x) = 4 − x.
Then
c1 (2 − 5x) + c2 (3 + 7x) + c3 (4 − x) = 0
requires
2c1 + 3c2 + 4c3 = 0 and
− 5c1 + 7c2 − c3 = 0.
This system has solution (−31r, −18r, 29r), where r is a free variable. Consequently, the given set of polynomials is linearly dependent, and a linearly dependent relationship is
−31p1 (x) − 18p2 (x) + 29p3 (x) = 0,
or equivalently,
1
[31p1 (x) + 18p2 (x)].
29
Hence, the linearly independent set of vectors {2 − 5x, 3 + 7x} spans the same subspace of P1 (R) as that
spanned by {2 − 5x, 3 + 7x, 4 − x}.
p3 (x) =
31. We first determine whether the given set of polynomials is linearly dependent. Let
p1 (x) = 2 + x2 , p2 (x) = 4 − 2x + 3x2 , p3 (x) = 1 + x.
Then
c1 (2 + x2 ) + c2 (4 − 2x + 3x2 ) + c3 (1 + x) = 0
leads to the system
2c1 + 4c2 + c3 = 0, −2c2 + c3 = 0, c1 + 3c2 = 0.
This system has solution (−3r, r, 2r) where r is a free variable. Consequently, the given set of vectors is
linearly dependent, and a specific linear relationship is
−3p1 (x) + p2 (x) + 2p3 (x) = 0,
or equivalently,
p2 (x) = 3p1 (x) − 2p3 (x).
Hence, the linearly independent set of vectors {2 + x2 , 1 + x} spans the same subspace of P2 (R) as that
spanned by the given set of vectors.
1 x x2 32. W [f1 , f2 , f3 ](x) = 0 1 2x = 2. Since W [f1 , f2 , f3 ](x) = 0 on I, it follows that the functions are
0 0 2 linearly independent on I.
sin x
W [f1 , f2 , f3 ](x) = cos x
− sin x
33.
cos x
− sin x
− cos x
sin x
tan x
= cos x
sec2 x
2 tan x sec2 x 0
cos x
− sin x
0
tan x
sec2 x
tan x(2 sec2 x + 1) = − tan x(2 sec2 x + 1).
W [f1 , f2 , f3 ](x) is not always zero over I, so the vectors are linearly independent by Theorem 4.5.23.
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1 3x x2 − 1 3 2x = 6 = 0 on I. Consequently, {f1 , f2 , f3 } is a linearly
2x = 34. W [f1 , f2 , f3 ](x) = 0 3
0 2 0 0
2
independent set on I by Theorem 4.5.23.
2x
e
1 1
1 0
1 e3x
e−x 0 2x
3e3x −e−x = e4x 2 3 −1 = e4x 2 1 −3 = 12e4x . Since the
35. W [f1 , f2 , f3 ](x) = 2e
4 9
4e2x 9e3x
4 5 −3 1 e−x Wronskian is never zero, the functions are linearly independent on (−∞, ∞).
36. On [0, ∞), f2 = 7f1 , so that the functions are linearly dependent on
this interval. Therefore W [f1 , f2 ](x) =
3x3 7x2 = −21x4 = 0. Since the Wronskian is
0 for x ∈ [0, ∞). However, on (−∞, 0), W [f1 , f2 ](x) = 9x2 14x not zero for all x ∈ (−∞, ∞), the functions are linearly independent on that interval.
1 x 2x − 1 = 0. By inspection, we see that f3 = 2f2 − f1 , so that the functions
2
37. W [f1 , f2 , f3 ](x) = 0 1
0 0
0
are linearly dependent on (−∞, ∞).
38. We show that the Wronskian (the determinant can be computed by cofactor expansion along the first
row) is identically zero:
x
−x
e
x e −x cosh x e
x −e−x sinh x = −(cosh x + sinh x) − (cosh x − sinh x) + 2 cosh x = 0.
e
e
cosh x Thus, the Wronskian is identically zero on (−∞, ∞). Furthermore, {f1 , f2 , f3 } is a linearly dependent set
because
1
1
1
1
1
ex + e−x
1
= 0 for all x ∈ I.
− f1 (x) − f2 (x) + f3 (x) = − ex − e−x + cosh x = − ex − e−x +
2
2
2
2
2
2
2
39. We show that the Wronskian is identically zero for f1 (x) = ax3 and f2 (x) = bx3 , which covers the
functions in this problem as a special case:
ax3
bx3 5
5
3ax2 3bx2 = 3abx − 3abx = 0.
Next, let a, b ∈ R.
If x ≥ 0, then
af1 (x) + bf2 (x) = 0 =⇒ 2ax3 + 5bx3 = 0 =⇒ (2a + 5b)x3 = 0 =⇒ 2a + 5b = 0.
If x ≤ 0, then
af1 (x) + bf2 (x) = 0 =⇒ 2ax3 − 3bx3 = 0 =⇒ (2a − 3b)x3 = 0 =⇒ 2a − 3b = 0.
Solving the resulting system, we obtain a = b = 0; therefore, {f1 , f2 } is a linearly independent set of vectors
on (−∞, ∞).
40. (a). When x > 0, f2 (x) = 1 and when x < 0, f2 (x) = −1; thus f2 (0) does not exist, which implies that
f2 ∈
/ C 1 (−∞, ∞).
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(b). Let a, b ∈ R. On the interval (−∞, 0), ax + b(−x) = 0, which has more than the trivial solution for a
and b. Thus, {f1 , f2 } is a linearly dependent set of vectors on (−∞, 0).
On the interval [0, ∞), ax + bx = 0 =⇒ a + b = 0, which has more than the trivial solution for a and b.
Therefore {f1 , f2 } is a linearly dependent set of vectors on [0, ∞).
a−b=0
On the interval (−∞, ∞), a and b must satisfy: ax + b(−x) = 0 and ax + bx = 0, that is,
. Since
a+b=0
this system has only a = b = 0 as its solution, {f1 , f2 } is a linearly independent set of vectors on (−∞, ∞).
y
y = f2(x) = -x
f1(x) = - f2(x) on (-', 0)
y = f1(x) = x = f2(x)
f1(x) = f2(x) on (0, ')
x
y = f1(x) = x
Figure 0.0.63: Figure for Problem 40
41. Let a, b ∈ R.
ax + bx = 0 if x = 0
(a + b)x = 0
=⇒
a(0) + b(1) = 0 if x = 0
b=0
=⇒ a = 0 and b = 0 so {f1 , f2 } is a linearly independent set on I.
af1 (x) + bf2 (x) = 0 =⇒
42. Let a, b, c ∈ R and x ∈ (−∞, ∞).
af1 (x) + bf2 (x) + cf3 (x) = 0 =⇒
a(x − 1) + b(2x) + c(3) = 0 for x ≥ 1
2a(x − 1) + b(2x) + c(3) = 0 for x < 1
(a + 2b)x + (−a + 3c) = 0
a + 2b = 0 and − a + 3c = 0
=⇒
(2a + 2b)x + (−2a + 3c) = 0
2a + 2b = 0 and − 2a + 3c = 0.
Since the only solution to this system of equations is a = b = c = 0, it follows that the given functions are
linearly independent on (−∞, ∞).
The domain space may be divided into three types of intervals: (1) interval subsets of (−∞, 1), (2) interval
subset of [1, ∞), (3) intervals containing 1 where 1 is not an endpoint of the intervals.
=⇒
For intervals of type (3):
Intervals such as type (3) are treated as above [with domain space of (−∞, ∞)]: vectors are linearly independent.
For intervals of type (1):
a(2(x − 1)) + b(2x) + c(3) = 0 =⇒ (2a + 2b)x + (−2a + 3c) = 0 =⇒ 2a + 2b = 0, and −2a + 3c = 0.
Since this system has three variables with only two equations, the solution to the system is not unique, hence
intervals of type (1) result in linearly dependent vectors.
For intervals of type (2):
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a(x−1)+b(2x)+c(3) = 0 =⇒ a+2b = 0 and −a+3c = 0. As in the last case, this system has three variables
with only two equations, so it must be the case that intervals of type (2) result in linearly dependent vectors.
43. (a). Let f0 (x) = 1, f1 (x)
= x2 , f3 (x) = x3 .
= x, f2 (x)
1 x x2 x 3 0 1 2x 3x2 = 12 = 0. Hence, {f1 , f2 , f3 , f4 } is linearly independent on
Then W [f1 , f2 , f3 , f4 ](x) = 6x 0 0 2
0 0 0
6 any interval.
1 x x2 · · ·
xn
n−1
0 1 2x · · ·
nx
0 0 2 · · · n(n − 1)xn−2 (b). W [f0 , f1 , f2 , . . . , fn ](x) = .
.. ..
..
..
. .
.
.
0 0 0 ···
n!
The matrix corresponding to this determinant is upper triangular, so the value of the determinant is given
by the product of all of the diagonal entries. W [f0 , f1 , f2 , . . . , fn ](x) = 1 · 1 · 2 · 6 · 24 · · · n!, which is not zero
regardless of the actual domain of x. Consequently, the functions are linearly independent on any interval.
44. (a). Let f1 (x) = er1 x , f2 (x) = er2 x and f 3 (x) = er3 x . Then
e r1 x
1
e r2 x
er3 x r1 x r2 x r3 x r1 x
r2 x
r3 x r2 e
r3 e
= e e e r1
W [f1 , f2 , f3 ](x) = r1 e
r12 er1 x r22 er2 x r32 er3 x r12
1
r2
r22
1 r3 r32 = e(r1 +r2 +r3 )x (r3 − r1 )(r3 − r2 )(r2 − r1 ).
If ri = rj for i = j, then W [f1 , f2 , f3 ](x) is never zero, and hence the functions are linearly independent on
any interval. If, on the other hand, ri = rj with i = j, then fi − fj = 0, so that the functions are linearly
dependent. Thus, r1 , r2 , r3 must all be different in order that f1 , f2 , f3 are linearly independent.
(b).
e r1 x
r1 e r 1 x
r12 er1 x
W [f1 , f2 , f3 , . . . , fn ](x) = ..
.
n−1
r
e r1 x
e r2 x
r2 e r 2 x
r22 er2 x
..
.
···
···
···
ern x rn ern x rn2 ern x .. . rn−1 ern x r2n−1 er2 x · · ·
n
1
1
···
r1
r
···
2
2
2
r1 x r 2 x
r n x r1
r
···
2
= e e ···e
..
..
.
n−1 . n−1
r
r
···
1
1
2
n−1 r
1
rn
rn2
..
.
n
∗
= er1 x er2 x · · · ern x V (r1 , r2 , . . . , rn )
= e r1 x e r2 x · · · e rn x
(rm − ri ).
1≤i<m≤n
Now from the last equality, we see that if ri = rj for i = j, where i, j ∈ {1, 2, 3, . . . , n}, then W [f1 , f2 , . . . , fn ](x)
is never zero, and hence the functions are linearly independent on any interval. If, on the other hand, ri = rj
with i = j, then f1 − fj = 0, so that the functions are linearly dependent. Thus {f1 , f2 , . . . , fn } is a linearly
independent set if and only if all the ri are distinct for i ∈ {1, 2, 3, . . . , n}.
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(*Note: V (r1 , r2 , . . . , rn ) is the n × n Vandermonde determinant. See Problem 30 in Section 3.3).
45. Let a, b ∈ R. Assume that av + bw = 0. Then a(αv1 + v2 ) + b(v1 + αv2 ) = 0 which implies that
(αa + b)v1 + (a + bα)v2 = 0. Now since it is given that v1 and v2 are linearly independent,
αa + b = 0 and
α 1 = 0. That is, if and
a + bα = 0. This system has only the trivial solution for a and b if and only if 1 α only if α2 − 1 = 0, or α = ±1. Therefore, the vectors are linearly independent provided that α = ±1.
46. It is given that v1 and v2 are linearly independent. Let u1 = a1 v1 + b1 v2 , u2 = a2 v1 + b2 v2 , and
u3 = a3 v1 + b3 v2 where a1 , a2 , a3 , b1 , b2 , b3 ∈ R. Let c1 , c2 , c3 ∈ R. Then
c1 u 1 + c2 u 2 + c3 u 3 = 0
=⇒ c1 (a1 v1 + b1 v2 ) + c2 (a2 v1 + b2 v2 ) + c3 (a3 v1 + b3 v2 ) = 0
=⇒ (c1 a1 + c2 a2 + c3 a3 )v1 + (c1 b1 + c2 b2 + c3 b3 )v2 = 0.
c1 a1 + c2 a2 + c3 a3 = 0
Now since v1 and v2 are linearly independent:
There are an infinite number
c1 b1 + c2 b2 + c3 b3 = 0.
of solutions to this homogeneous system since there are three unknowns but only two equations. Hence,
{u1 , u2 , u3 } is a linearly dependent set of vectors.
47. Given that {v1 , v2 , . . . , vm } is a linearly independent set of vectors in a vector space V , and
uk =
m
,
aik vi , k ∈ {1, 2, . . . , n}.
(47.1)
i=1
(a). Let ck ∈ R, k ∈ {1, 2, . . . , n}. Using system (47.1) and
n
,
ck uk = 0, we obtain:
k=1
n
,
k=1
ck
m
,
aik vi
i=1
= 0 ⇐⇒
n
m
,
,
i=1
aik ck
vi = 0.
k=1
Since the vi for each i ∈ {1, 2, . . . , m} are linearly independent,
n
,
aik ck = 0, 1 ≤ i ≤ m.
(47.2)
k=1
But this is a system of m equations with n unknowns c1 , c2 , . . . , cn . Since n > m, the system has more
unknowns than equations, and so has nontrivial solutions. Thus, {u1 , u2 , . . . , un } is a linearly dependent
set.
(b). If m = n, then the system (47.2) has a trivial solution ⇐⇒ the coefficient matrix of the system is
invertible ⇐⇒ det[aij ] = 0.
(c). If n < m, then the homogeneous system (47.2) has just the trivial solution if and only if rank(A) = n.
Recall that for a homogeneous system, rank(A# ) = rank(A).
(d). Corollary 4.5.17.
48. We assume that
c1 (Av1 ) + c2 (Av2 ) + · · · + cn (Avn ) = 0.
Our aim is to show that c1 = c2 = · · · = cn = 0. We manipulate the left side of the above equation as
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follows:
c1 (Av1 ) + c2 (Av2 ) + · · · + cn (Avn ) = 0
A(c1 v1 ) + A(c2 v2 ) + · · · + A(cn vn ) = 0
A(c1 v1 + c2 v2 + · · · + cn vn ) = 0.
Since A is invertible, we can left multiply the last equation by A−1 to obtain
c1 v1 + c2 v2 + · · · + cn vn = 0.
Since {v1 , v2 , . . . , vn } is linearly independent, we can now conclude directly that c1 = c2 = · · · = cn = 0, as
required.
49. Assume that c1 v1 + c2 v2 + c3 v3 = 0. We must show that c1 = c2 = c3 = 0. Let us suppose for the
moment that c3 = 0. In that case, we can solve the above equation for v3 :
v3 = −
c1
c2
v1 − v2 .
c3
c3
However, this contradicts the assumption that v3 does not belong to span{v1 , v2 }. Therefore, we conclude
that c3 = 0. Our starting equation therefore reduces to c1 v1 + c2 v2 = 0. Now the assumption that {v1 , v2 }
is linearly independent shows that c1 = c2 = 0. Therefore, c1 = c2 = c3 = 0, as required.
50. Assume that c1 v1 + c2 v2 + · · · + ck vk + ck+1 vk+1 = 0. We must show that c1 = c2 = · · · = ck+1 = 0.
Let us suppose for the moment that ck+1 = 0. In that case, we can solve the above equation for vk+1 :
vk+1 = −
c1
c2
ck
v1 −
v2 − · · · −
vk .
ck+1
ck+1
ck+1
However, this contradicts the assumption that vk+1 does not belong to span{v1 , v2 , . . . , vk }. Therefore, we
conclude that ck+1 = 0. Our starting equation therefore reduces to c1 v1 + c2 v2 + · · · + ck vk = 0. Now
the assumption that {v1 , v2 , . . . , vk } is linearly independent shows that c1 = c2 = · · · = ck = 0. Therefore,
c1 = c2 = · · · = ck = ck+1 = 0, as required.
51. Let {v1 , v2 , . . . , vk } be a set of vectors with k ≥ 2. Suppose that vk can be expressed as a linear
combination of {v1 , v2 , . . . , vk−1 }. We claim that
span{v1 , v2 , . . . , vk } = span{v1 , v2 , . . . , vk−1 }.
Since every vector belonging to span{v1 , v2 , . . . , vk−1 } evidently belongs to span{v1 , v2 , . . . , vk }, we focus
on showing that every vector in span{v1 , v2 , . . . , vk } also belongs to span{v1 , v2 , . . . , vk−1 }:
Let v ∈ span{v1 , v2 , . . . , vk }. We therefore may write
v = c1 v1 + c2 v2 + · · · + ck vk .
By assumption, we may write vk = d1 v1 + d2 v2 + · · · + dk−1 vk−1 . Therefore, we obtain
v = c1 v1 + c2 v2 + · · · + ck vk
= c1 v1 + c2 v2 + · · · + ck−1 vk−1 + ck (d1 v1 + d2 v2 + · · · + dk−1 vk−1 )
= (c1 + ck d1 )v1 + (c2 + ck d2 )v2 + · · · + (ck−1 + ck dk−1 )vk−1
∈ span{v1 , v2 , . . . , vk−1 }.
This shows that every vector belonging to span{v1 , v2 , . . . , vk } also belongs to span{v1 , v2 , . . . , vk−1 }, as
needed.
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52. We first prove part 1 of Proposition 4.5.8. Suppose that we have a set {u, v} of two vectors in a vector
space V . If {u, v} is linearly dependent, then we have
cu + dv = 0,
where c and d are not both zero. Without loss of generality, suppose that c = 0. Then we have
d
u = − v,
c
so that u and v are proportional. Conversely, if u and v are proportional, then v = cu for some constant c.
Thus, cu − v = 0, which shows that {u, v} is linearly dependent.
For part 2 of Proposition 4.5.8, suppose the zero vector 0 belongs to a set S of vectors in a vector space
V . Then 1 · 0 is a linear dependency among the vectors in S, and therefore S is linearly dependent.
53. Suppose that {v1 , v2 , . . . , vk } spans V and let v be any vector in V . By assumption, we can write
v = c 1 v 1 + c2 v 2 + · · · + ck vk ,
for some constants c1 , c2 , . . . , ck . Therefore,
c1 v 1 + c2 v 2 + · · · + ck vk − v = 0
is a linear dependency among the vectors in {v, v1 , v2 , . . . , vk }. Thus, {v, v1 , v2 , . . . , vk } is linearly dependent.
54. Let S = {p1 , p2 , . . . , pk } and assume without loss of generality that the polynomials are listed in
decreasing order by degree:
deg(p1 ) > deg(p2 ) > · · · > deg(pk ).
To show that S is linearly independent, assume that
c1 p1 + c2 p2 + · · · + ck pk = 0.
We wish to show that c1 = c2 = · · · = ck = 0. We require that each coefficient on the left side of the above
equation is zero, since we have 0 on the right-hand side. Since p1 has the highest degree, none of the terms
c2 p2 ,c3 p3 , . . . , ck pk can cancel the leading coefficient of p1 . Therefore, we conclude that c1 = 0. Thus, we
now have
c2 p2 + c3 p3 + · · · + ck pk = 0,
and we can repeat this argument again now to show successively that c2 = c3 = · · · = ck = 0.
Solutions to Section 4.6
(a): FALSE. It is not enough that S spans V . It must also be the case that S is linearly independent.
(b): FALSE. For example, R2 is not a subspace of R3 , since R2 is not even a subset of R3 .
(c): TRUE. Any set of two non-proportional vectors in R2 will form a basis for R2 .
(d): FALSE. We have dim[Pn (R)] = n + 1 and dim[Rn ] = n.
(e): FALSE. For example, if V = R2 , then the set S = {(1, 0), (2, 0), (3, 0)}, consisting of 3 > 2 vectors,
fails to span V , a 2-dimensional vector space.
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(f ): TRUE. We have dim[P3 (R)] = 4, and so any set of more than four vectors in P3 (R) must be linearly
dependent (a maximal linearly independent set in a 4-dimensional vector space consists of four vectors).
(g): FALSE. For instance, the two vectors 1 + x and 2 + 2x in P3 (R) are linearly dependent.
(h): FALSE. If the differential equation is not homogeneous, such as, for instance y + y = 1, then the zero
function is not a solution, so the set of all solutions to the differential equation fails the zero vector check
and does not form a vector space.
(i): FALSE. Only linearly independent sets with fewer than n vectors can be extended to a basis for V .
(j): TRUE. We can build such a subset by choosing vectors from the set as follows. Choose v1 to be any
vector in the set. Now choose v2 in the set such that v2 ∈ span{v1 }. Next, choose v3 in the set such that
v3 ∈ span{v1 , v2 }. Proceed in this manner until it is no longer possible to find a vector in the set that
is not spanned by the collection of previously chosen vectors. This point will occur eventually, since V is
finite-dimensional. Moreover, the chosen vectors will form a linearly independent set, since each vi is chosen
from outside span{v1 , v2 , . . . , vi−1 }. Thus, the set we obtain in this way is a linearly independent set of
vectors that spans V , hence forms a basis for V .
(k): FALSE. The set of all 3 × 3 upper triangular matrices forms a 6-dimensional subspace of M3 (R), not
a 3-dimensional subspace. One basis is given by {E11 , E12 , E13 , E22 , E23 , E33 }.
Problems:
1. NO. Since dim[R2 ] = 2, any basis must contain exactly two vectors. Here, we are only given one vector,
so this set cannot form a basis.
2
2
2. YES. dim[R
] = 2. There are two vectors, so if they are to form a basis for R , they need to be linearly
1 −1 = 2 = 0. This implies that the vectors are linearly independent, hence they form a
independent: 1
1 basis for R2 .
3
3
3. YES. dim[R
] = 3. There are three vectors, so if they are to form a basis for R , they need to be linearly
1
3
1 1 = 13 = 0. This implies that the vectors are linearly independent, hence they
independent: 2 −1
1
2 −1 3
form a basis for R .
3
4. NO. dim[R 3 ] = 3. There are
three vectors, so if they are to form a basis for R , they need to be linearly
1
2
3
5 11 = 0. This implies that the vectors are linearly dependent, hence they do not
independent: −1
1 −2 −5 form a basis for R3 .
5. NO. dim[R4 ] = 4. We need 4 linearly independent vectors in order to span R4 . However, there are only
3 vectors in this set. Thus, the vectors cannot be a basis for R4 .
4
6. YES. dim[R
vectors,
so if they are to form a basis for R3 , they need to be linearly
] = 4. There are four
1
2 −1
2 2 −1
2 −7 0 −5 1
−1 2 −3 1
1
1 −1 0 −1
2 −3 1 = −11.
1 = 3 1
independent: = = 3 1
3
1
1 3
1
1 −5 0 −2 0
−5 0 −2 0
2 −1 −2
0 −5
2 0 −2 Since this determinant is nonzero, the given vectors are linearly independent. Consequently, they form a
basis for R4 .
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7. NO. Since dim[R3 ] = 3, any basis must contain exactly three vectors. Here, we are given four vectors,
so this set cannot form a basis.
8. dim[R4 ] = 4. There are four
vectors, so if they are to form a basis for R3 , they need to be linearly
0 1 0 k −1 1 0 0 0 k −1 0 1 0 = − 0 1 2 − −1 1 0 = −(−1 + 2k) − (−k 2 ) = k 2 − 2k + 1 =
independent: 0
1
1
2
k 0 1 k 0 1 k 0 0 1 (k − 1)2 = 0 when k = 1. Thus, the vectors will form a basis for R4 provided k = 1.
9. NO. Since dim[P1 (R)] = 2, any basis must contain exactly two vectors. Here, we are given three vectors,
so this set cannot form a basis.
10. YES. We will check whether or not S is linearly independent. Suppose c1 (1 − 3x2 ) + c2 (2x + 5x2 ) +
c3 (1 − x + 3x2 ) = 0. Equating like powers of x on each side, we have
c1 + c3 = 0,
2c2 − c3 = 0,
−3c1 + 5c2 + 3c3 = 0.
It is easy
to verify that this linear system has only the trivial solution (the determinant of the coefficient
1 0
1 matrix, 0 2 −1 = 17 = 0, for instance, is nonzero). That is, c1 = c2 = c3 = 0. Thus, S is linearly
−3 5
3 independent. By Corollary 4.6.13, we conclude that S is a basis.
11. YES. We will check whether or not S is linearly independent. Suppose c1 (5x2 )+c2 (1+6x)+c3 (−3−x2 ) =
0. Equating like powers of x on each side, we have
c2 − 3c3 = 0,
6c2 = 0,
5c1 − c3 = 0.
It is easy to verify that this linear system has only the trivial solution (the second equation implies c2 = 0,
from which the first implies that c3 = 0, and then the third implies that c1 = 0). That is, c1 = c2 = c3 = 0.
Thus, S is linearly independent. By Corollary 4.6.13, we conclude that S is a basis.
12. NO. Since dim[P2 (R)] = 3, any basis must contain exactly three vectors. Here, we are given four
vectors, so this set cannot form a basis.
13. NO. Since dim[P3 (R)] = 4, any basis must contain exactly four vectors. Here, we are given three
vectors, so this set cannot form a basis.
14. YES. We will check whether or not S is linearly independent. Suppose
c1 (1 + x + 2x3 ) + c2 (2 + x + 3x2 − x3 ) + c3 (−1 + x + x2 − 2x3 ) + c4 (2 − x + x2 + 2x3 ) = 0.
Equating like powers of x, we obtain th e system
c1 + 2c2 − c3 + 2c4 = 0, c1 + c2 + c3 − c4 = 0, 3c2 + c3 + c4 = 0, 2c1 − c2 − 2c3 + 2c4 = 0.
⎡
⎤
1
2 −1
2 0
⎢ 1
1
1 −1 0 ⎥
⎥. One can check, by using Gaussian
The augmented matrix for this linear system is ⎢
⎣ 0
3
1
1 0 ⎦
2 −1 −2
2 0
Elimination or determinants, that the coefficient matrix for this linear system is invertible. Therefore, we
conclude that c1 = c2 = c3 = c4 = 0. Thus, S is linearly independent. By Corollary 4.6.13, we conclude that
S is a basis.
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15. NO. Since dim[M2 (R)] = 4, any basis must contain exactly four vectors. Here, we are given five vectors,
so this set cannot form a basis.
16. YES. We will check whether or not S is linearly independent. Suppose
−2 −8
0 1
−5
0
3 −2
0
c1
+ c2
+ c3
+ c4
=
1
4
−1 1
5 −4
4 −1
0
0
0
.
The corresponding linear system of equations is
−2c1 − 5c3 + 3c4 = 0, −8c1 + c2 − 2c4 = 0, c1 − c2 + 5c3 + 4c4 = 0, 4c1 + c2 − 4c3 − c4 = 0.
⎤
⎡
−2
0 −5
3
⎢ −8
1
0 −2 ⎥
⎥. Using any method to compute the
The matrix of coefficients for this linear system is ⎢
⎣ 1 −1
5
4 ⎦
4
1 −4 −1
determinant of this coefficient matrix shows that it is nonzero, and hence the coefficient matrix is invertible.
Thus, c1 = c2 = c3 = c4 = 0. Thus, S is linearly independent. By Corollary 4.6.13, we conclude that S is a
basis.
17. NO. Since dim[M32 (R)] = 6, any basis must contain exactly six vectors. Here, we are given five vectors,
so this set cannot form a basis.
1 1 1
2 2 2
18. NO. By inspection, note that S contains two proportional vectors,
and
.
1 1 1
2 2 2
Therefore, S is linearly dependent, and hence, S cannot form a basis.
19. A row-echelon form of A is given by 1 −9/8 3/8 3/8 −5/8 , which has four unpivoted columns.
Therefore, the linear system Ax = 0 has four free variables. We conclude that the dimension of the null
space of A is 4.
1
3
x1
0
1
3 0
20. Ax = 0 =⇒
∼
=
. The augmented matrix for this system is
−2 −6
−2 −6 0
x2
0
1 3 0
; thus, x1 + 3x2 = 0, or x1 = −3x2 . Let x2 = r so that x1 = −3r where r ∈ R. Consequently,
0 0 0
S = {x ∈ R2 : x = r(−3, 1), r ∈ R} = span{(−3, 1)}. It follows that {(−3, 1)} is a basis for S and dim[S] = 1.
⎤
⎤
⎡
⎤⎡
⎡
0 0 0
x1
0
21. Ax = 0 =⇒ ⎣ 0 0 0 ⎦ ⎣ x2 ⎦ = ⎣ 0 ⎦. The REDUCED ROW ECHELON FORM of the augx
0 1 0
0
⎤
⎡3
0 1 0 0
mented matrix for this system is ⎣ 0 0 0 0 ⎦. We see that x2 = 0, and x1 and x3 are free variables:
0 0 0 0
x1 = r and x2 = s. Hence, (x1 , x2 , x3 ) = (r, 0, s) = r(1, 0, 0) + s(0, 0, 1), so that the solution set of the system
is S = {x ∈ R3 : x = r(1, 0, 0) + s(0, 0, 1), r, s ∈ R}. Therefore we see that {(1, 0, 0), (0, 0, 1)} is a basis for S
and dim[S] = 2.
⎡
⎤⎡
⎤
⎤
⎡
1 −1
4
x1
0
3 −2 ⎦ ⎣ x2 ⎦ = ⎣ 0 ⎦. The REDUCED ROW ECHELON FORM of the
22. Ax = 0 =⇒ ⎣ 2
1
2 −2
0
x
⎤
⎡ 3
1 0
2 0
augmented matrix for this system is ⎣ 0 1 −2 0 ⎦. If we let x3 = r then (x1 , x2 , x3 ) = (−2r, 2r, r) =
0 0
0 0
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r(−2, 2, 1), so that the solution set of the system is S = {x ∈ R3 : x = r(−2, 2, 1), r ∈ R}. Therefore we see
that {(−2, 2, 1)} is a basis for S and dim[S] = 1.
⎤
⎤⎡
⎡
⎤
⎡
x1
1 −1 2 3
0
⎥
⎢ 2 −1 3 4 ⎥ ⎢ x2 ⎥
⎢
⎥⎢
⎥ = ⎢ 0 ⎥. The REDUCED ROW ECHELON FORM of the
23. Ax = 0 =⇒ ⎢
⎣ 1
⎣ 0 ⎦
0 1 1 ⎦ ⎣ x3 ⎦
0
3 −1 4 5 ⎡ x4
⎤
1 0
1
1 0
⎢ 0 1 −1 −2 0 ⎥
⎥. If we let x3 = r, x4 = s then x2 = r + 2s and
augmented matrix for this system is ⎢
⎣ 0 0
0
0 0 ⎦
0 0
0
0 0
x1 = −r − s. Hence, the solution set of the system is S = {x ∈ R4 : x = r(−1, 1, 1, 0) + s(−1, 2, 0, 1), r, s ∈
R} = span{(−1, 1, 1, 0), (−1, 2, 0, 1)}. Further, the vectors v1 = (−1, 1, 1, 0), v2 = (−1, 2, 0, 1) are linearly
independent since c1 v1 + c2 v2 = 0 =⇒ c1 (−1, 1, 1, 0) + c2 (−1, 2, 0, 1) = (0, 0, 0, 0) =⇒ c1 = c2 = 0.
Consequently, {(−1, 1, 1, 0), (−1, 2, 0, 1)} is a basis for S and dim[S] = 2.
24. If we let y = r and z = s where r, s ∈ R, then x = 3r − s. It follows that any ordered triple in S can be
written in the form:
(x, y, z) = (3r − s, r, s) = (3r, r, 0) + (−s, 0, s) = r(3, 1, 0) + s(−1, 0, 1), where r, s ∈ R. If we let v1 = (3, 1, 0)
and v2 = (−1, 0, 1), then S = {v ∈ R3 : v = r(3, 1, 0) + s(−1, 0, 1), r, s ∈ R} = span{v1 , v2 }; moreover, v1
and v2 are linearly independent for if a, b ∈ R and av1 + bv2 = 0, then
a(3, 1, 0) + b(−1, 0, 1) = (0, 0, 0),
which implies that (3a, a, 0) + (−b, 0, b) = (0, 0, 0), or (3a − b, a, b) = (0, 0, 0). In other words, a = b = 0.
Since {v1 , v2 } spans S and is linearly independent, it is a basis for S. Also, dim[S] = 2.
25.
S = {x ∈ R3 : x = (r, r − 2s, 3s − 5r), r, s ∈ R}
= {x ∈ R3 : x = (r, r, −5r) + (0, −2s, 3s), r, s ∈ R}
= {x ∈ R3 : x = r(1, 1, −5) + s(0, −2, 3), r, s ∈ R}.
Thus, S = span{(1, 1, −5), (0, −2, 3)}. The vectors v1 = (1, 1, −5) and v2 = (0, −2, 3) are linearly independent for if a, b ∈ R and av1 + bv2 = 0, then
a(1, 1, −5)+b(0, −2, 3) = (0, 0, 0) =⇒ (a, a, −5a)+(0, −2b, 3b) = (0, 0, 0) =⇒ (a, a−2b, 3b−5a) = (0, 0, 0) =⇒ a = b = 0.
It follows that {v1 , v2 } is a basis for S and dim[S] = 2.
26. We claim that S = {1, x, x2 , x3 } is a basis for P3 (R). We will verify that S is linearly independent and
that S spans P3 (R).
S is linearly independent: Assume c0 (1) + c1 (x) + c2 (x2 ) + c3 (x3 ) = 0. It follows immediately from
this equation that c0 = c1 = c2 = c3 = 0. Therefore, S is linearly independent.
S spans P3 (R): Consider an arbitrary polynomial p(x) = a + bx + cx2 + dx3 ∈ P3 (R). We can write p(x)
as a linear combination of vectors in S in the obvious way: p(x) = a(1) + b(x) + c(x2 ) + d(x3 ). Therefore, S
spans P3 (R).
Since S is linearly independent and spans P3 (R), we conclude that S forms a basis for P3 (R). Since S
contains four vectors, we conclude that dim[P3 (R)] = 4.
27. Many acceptable bases are possible here. One example is
S = {x3 , x3 + 1, x3 + x, x3 + x2 }.
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All of the polynomials in this set have degree 3. We verify that S is a basis: Assume that
c1 (x3 ) + c2 (x3 + 1) + c3 (x3 + x) + c4 (x3 + x2 ) = 0.
Thus,
(c1 + c2 )x3 + c4 x2 + c3 x + c2 = 0,
from which we quickly see that c1 = c2 = c3 = c4 = 0. Thus, S is linearly independent. Since P3 (R) is
4-dimensional, we can now conclude from Corollary 4.6.13 that S is a basis for P3 (R).
a b
28. S = {A ∈ M2 (R) : A =
, a, b, c ∈ R}. Each vector in S can be written as
0 c
1 0
0 1
0 0
A=a
+b
+c
,
0 0
0 0
0 1
1 0
0 1
0 0
so that S = span
,
,
. Since the vectors in this set are clearly linearly inde0 0
0 0
0 1
1 0
0 1
0 0
pendent, it follows that a basis for S is
,
,
, and therefore dim[S] = 3.
0 0
0 0
0 1
a
b
29. S = {A ∈ M2 (R) : A =
, a, b, c ∈ R}. Each vector in S can be written as
c −a
1
0
0 1
0 0
A=a
+b
+c
,
0 −1
0 0
1 0
1
0
0 1
0 0
so that S = span
,
,
. Since this set of vectors is clearly linearly independent,
0 −1
0 0
1 0
1
0
0 1
0 0
it follows that a basis for S is
,
,
, and therefore dim[S] = 3.
0 −1
0 0
1 0
30. We see directly that v3 = 2v1 . Let v be an arbitrary vector in S. Then
v = c1 v1 + c2 v2 + c3 v3 = (c1 + 2c3 )v1 + c2 v2 = d1 v1 + d2 v2 ,
where d1 = (c1 + 2c3 ) and d2 = c2 . Hence S = span{v1 , v2 }. Further, v1 and v2 are linearly independent
for if a, b ∈ R and av1 + bv2 = 0, then
a(1, 0, 1) + b(0, 1, 1) = (0, 0, 0) =⇒ (a, 0, a) + (0, b, b) = (0, 0, 0) =⇒ (a, b, a + b) = (0, 0, 0) =⇒ a = b = 0.
Consequently, {v1 , v2 } is a basis for S and dim[S] = 2.
f1 (x) − f2 (x)
ex − e−x
31. f3 depends on f1 and f2 since sinh x =
. Thus, f3 (x) =
.
2
2
x
e
e−x = −2 = 0 for all x ∈ R, so {f1 , f2 } is a linearly independent set. Thus, {f1 , f2 }
W [f1 , f2 ](x) = x
e
−e−x is a basis for S and dim[S] = 2.
32. The given set of matrices
is linearly
dependent
because it contains the zero vector. Consequently, the
1 3
−1 4
5 −6
, A2 =
, A3 =
span the same subspace of M2 (R) as that
matrices A1 =
−1 2
1 1
−5
1
spanned by the original set. We now determine whether {A1 , A2 , A3 } is linearly independent. The vector
equation:
c1 A1 + c2 A2 + c3 A3 = 02
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leads to the linear system
c1 − c2 + 5c3 = 0, 3c1 + 4c2 − 6c3 = 0, −c1 + c2 − 5c3 = 0, 2c1 + c2 + c3 = 0.
This system has solution (−2r, 3r, r), where r is a free variable. Consequently, {A1 , A2 , A3 } is linearly
dependent with linear dependency relation
−2A1 + 3A2 + A3 = 0,
or equivalently,
A3 = 2A1 − 3A2 .
It follows that the set of matrices {A1 , A2 } spans the same subspace of M2 (R) as that spanned by the original
set of matrices. Further {A1 , A2 } is linearly independent by inspection, and therefore it is a basis for the
subspace.
33. (a). We must show that every vector (x, y) ∈ R2 can be expressed as a linear combination of v1 and
v2 . Mathematically, we express this as
c1 (1, 1) + c2 (−1, 1) = (x, y)
which implies that
c1 − c2 = x
and
c1 + c2 = y.
Adding the equations here, we obtain 2c1 = x + y or
c1 =
1
(x + y).
2
Now we can solve for c2 :
1
(y − x).
2
Therefore, since we were able to solve for c1 and c2 in terms of x and y, we see that the system of equations
is consistent for all x and y. Therefore, {v1 , v2 } spans R2 .
1 −1 = 2 = 0, so the vectors are linearly independent.
(b). 1
1 c2 = y − c1 =
(c). We can draw this conclusion from part (a) alone by using Theorem 4.6.12 or from part (b) alone by
using Theorem 4.6.10.
34. (a). We must show that every vector (x, y) ∈ R2 can be expressed as a linear combination of v1 and
v2 . Mathematically, we express this as
c1 (2, 1) + c2 (3, 1) = (x, y)
which implies that
2c1 + 3c2 = x
and c1 + c2 = y.
From this, we can solve for c1 and c2 in terms of x and y:
c1 = 3y − x
and
c2 = x − 2y.
Therefore, since we were able to solve for c1 and c2 in terms of x and y, we see that the system of equations
is consistent for all x and y. Therefore, {v1 , v2 } spans R2 .
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2
(b). 1
3 = −1 = 0, so the vectors are linearly independent.
1 (c). We can draw this conclusion from part (a) alone by using Theorem 4.6.12 or from part (b) alone by
using Theorem 4.6.10.
35. dim[R3 ] = 3. There
so if they are linearly independent, then they are a basis of R3 by
are 3 vectors,
0 3
6 Theorem 4.6.10. Since 6 0 −3 = 81 = 0, the given vectors are linearly independent, and so {v1 , v2 , v3 }
3 3
0 3
is a basis for R .
36. dim[P2 (R)] = 3. There are 3 vectors, so {p1 , p2 , p3 } may be a basis for P2 (R) depending on α. To be a
basis, the set of vectors must be linearly independent. Let a, b, c ∈ R. Then ap1 (x) + bp2 (x) + cp3 (x) = 0
=⇒ a(1 + αx2 ) + b(1 + x + x2 ) + c(2 + x) = 0
=⇒ (a +⎧b + 2c) + (b + c)x + (aα + b)x2 = 0. Equating like coefficients in the last equality, we obtain the
⎪
⎨ a + b + 2c = 0
b + c = 0 Reduce the augmented matrix of this system.
system:
⎪
⎩
aα + b = 0.
⎡
1
⎣ 0
α
⎤ ⎡
1 2 0
1
1 1 0 ⎦∼⎣ 0
1 0 0
0
1
1
α−1
2
1
2α
⎤ ⎡
0
1 1
2
1
0 ⎦∼⎣ 0 1
0 0 α+1
0
⎤
0
0 ⎦.
0
For this system to have only the trivial solution, the last row of the matrix must not be a zero row. This
means that α = −1. Therefore, for the given set of vectors to be linearly independent (and thus a basis), α
can be any value except −1.
37. dim[P2 (R)] = 3. There are 3 vectors, so in order to form a basis for P2 (R), they must be linearly
independent. Since we are dealing with functions, we will use the Wronskian.
1+x
W [p1 , p2 , p3 ](x) = 1
0
x(x − 1)
2x − 1
2
1 + 2x2 1 + x
= 1
4x
0
4
x(x − 1)
2x − 1
2
1 + 2x = −2.
2
0
Since the Wronskian is nonzero, {p1 , p2 , p3 } is linearly independent, and hence forms a basis for P2 (R).
1 x 3x2 −1 2
38. W [p0 , p1 , p2 ](x) = 0 1
3x = 3 = 0, so {p0 , p1 , p2 } is a linearly independent set. Since
0 0
3
dim[P2 (R)] = 3, it follows that {p0 , p1 , p2 } is a basis for P2 (R).
39.
(a). Suppose
that a, b,c, d ∈ R.
−1 1
1 3
1 0
0 −1
a
+b
+c
+d
= 02
0 1
−1 0
1 2
2
3
=⇒
−a + b + c
−b + c + 2d
⎧
−a + b + c = 0
⎪
⎪
⎪
⎨ a + 3b − d = 0
a + 3b − c
= 02 . The last matrix results in the system:
The
a + 2c + 3d
⎪
−b + c + 2d = 0
⎪
⎪
⎩
a + 2c + 3d = 0.
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⎡
⎤
1 −1 −1
0 0
⎢ 0
4
1 −1 0 ⎥
⎥,
REDUCED ROW ECHELON FORM of the augmented matrix of this system is ⎢
⎣ 0
0
5
7 0 ⎦
0
0
0
1 0
which implies that a = b = c = d = 0. Hence, the given set of vectors is linearly independent. Since
dim[M2 (R)] = 4, it follows that {A1 , A2 , A3 , A4 } is a basis for M2 (R).
5 6
(b). We wish to express the vector
in the form
7 8
5
7
6
8
=a
−1
0
1
1
+b
1
−1
3
0
+c
1
1
0
2
+d
0 −1
2
3
.
Matching entries on each side of this equation (upper left, upper right, lower left, and lower right), we obtain
a linear system with augmented matrix
⎡
−1
1 1
0
⎢ 1
3
0
−1
⎢
⎣ 0 −1 1
2
1
0 2
3
⎤
5
6 ⎥
⎥.
7 ⎦
8
55
56
Solving this system by Gaussian elimination, we find that a = − 34
3 , b = 12, c = − 3 , and d = 3 . Thus, we
have
34 −1 1
55 1 0
56 0 −1
5 6
1 3
−
=−
+ 12
+
.
7 8
0 1
1 2
2
3
−1 0
3
3
3
40.
⎡
⎤
⎤ x
⎤
⎡
1
1 −1
1 ⎢ 1 ⎥
0
x2 ⎥ ⎣
⎦
5 −6 ⎦ ⎢
(a). Ax = 0 =⇒ ⎣ 2 −3
⎣ x3 ⎦ = 0 . The augmented matrix for this linear system is
5
0
2 −3
0
x4
⎡
⎡
⎤ ⎡
1
1 −1
1 0
1
1
⎣ 2 −3
5 −6 0 ⎦ ∼ ⎣ 0
5
0
2 −3 0
0
⎤ ⎡
⎤ ⎡
⎤
1 −1
1 0
1
1 −1
1 0
1 1
−1 1 0
2
3
−5
7 −8 0 ⎦ ∼ ⎣ 0 −5
7 −8 0 ⎦ ∼ ⎣ 0 1 −1.4 1.6 0 ⎦ .
−5
7 −8 0
0
0
0
0 0
0 0
0 0 0
1. A12 (−2), A13 (−5)
2. A23 (−1)
3. M2 (− 15 )
We see that x3 = r and x4 = s are free variables, and therefore nullspace(A) is 2-dimensional. Now we
can check directly that Av1 = 0 and Av2 = 0, and since v1 and v2 are linearly independent (they are
non-proportional), they therefore form a basis for nullspace(A) by Corollary 4.6.13.
(b). An arbitrary vector (x, y, z, w) in nullspace(A) can be expressed as a linear combination of the basis
vectors:
(x, y, z, w) = c1 (−2, 7, 5, 0) + c2 (3, −8, 0, 5),
where c1 , c2 ∈ R.
41.
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⎡
a
c
(a). An arbitrary matrix in S takes the form ⎣
−a − c
⎡
1
= a⎣ 0
−1
⎤
⎡
0 −1
0
0
0 ⎦ + b⎣ 0
0
1
0
⎤
−a − b
⎦
−c − d
a+b+c+d
b
d
−b − d
⎤
⎡
1 −1
0
0
0 ⎦ + c⎣ 1
−1
1
−1
Therefore, we have the following basis for S:
⎧⎡
⎤ ⎡
⎤ ⎡
0
1 −1
0
1 0 −1
⎨
⎣ 0 0
0
0 ⎦,⎣ 1
0 ⎦,⎣ 0
⎩
0 −1
1
−1
−1 0
1
⎤
⎡
⎤
0
0
0
0
0
0 −1 ⎦ + d ⎣ 0
1 −1 ⎦ .
0
1
0 −1
1
⎤ ⎡
0
0
0
0 −1 ⎦ , ⎣ 0
0
1
0
⎤⎫
0
0 ⎬
1 −1 ⎦ .
⎭
−1
1
From this, we see that dim[S] = 4.
(b). We see that each of the matrices
⎡
⎤ ⎡
1 0 0
0 1
⎣ 0 0 0 ⎦,⎣ 0 0
0 0 0
0 0
⎤ ⎡
0
0
0 ⎦,⎣ 0
0
0
0
0
0
⎤ ⎡
1
0
0 ⎦,⎣ 1
0
0
0
0
0
⎤ ⎡
0
0
0 ⎦,⎣ 0
0
1
0
0
0
⎤
0
0 ⎦
0
has a different row or column that does not sum to zero, and thus none of these matrices belong to S, and
they are linearly independent from one another. Therefore, supplementing the basis for S in part (a) with
the five matrices here extends the basis in (a) to a basis for M3 (R).
42.
(a). A typical element in S has the form
⎡
a
d
A=⎣
−(a + d)
⎡
1
= a⎣ 0
−1
0
0
0
⎤ ⎡
⎤ ⎡
0
0
1 0
0
0 ⎦+b ⎣ 0
0 0 ⎦+c ⎣ 0
0
0 −1 0
0
Therefore, a basis for S is
⎧⎡
⎤ ⎡
⎤ ⎡
1 0 0
0
1 0
0
⎨
⎣ 0 0 0 ⎦,⎣ 0
0 0 ⎦,⎣ 0
⎩
−1 0 0
0 −1 0
0
b
e
−(b + e)
⎤
c
⎦
f
−(c + f )
⎤ ⎡
0
1
0
0
0 ⎦+d ⎣ 1
0 −1
−1
⎤ ⎡
0
1
0
0
0 ⎦,⎣ 1
0 −1
−1
0
0
0
0
0
0
⎤ ⎡
⎤
⎡
0
0
0 0
0
0 ⎦+e ⎣ 0
1 0 ⎦+f ⎣ 0
0
0 −1 0
0
⎤ ⎡
⎤ ⎡
0
0
0 0
0
0 ⎦,⎣ 0
1 0 ⎦,⎣ 0
0
0 −1 0
0
⎤
0
0
0
1 ⎦.
0 −1
⎤⎫
0
0 ⎬
0
1 ⎦ .
⎭
0 −1
Thus, dim[S] = 6.
(b). To extend the basis for S in part (a) to a basis for V , note that dim[V ] = 9, so that three additional
matrices must be added. We can add one vector for which the first column fails to sum to zero, one vector
for which the second column fails to sum to zero, and ⎡
one vector for
the third
⎤ which
⎡
⎤ column
⎡ fails to ⎤sum to
1 0 0
0 1 0
0 0 1
zero. The easiest way to do this is to add the vectors ⎣ 0 0 0 ⎦, ⎣ 0 0 0 ⎦, and ⎣ 0 0 0 ⎦. It is
0 0 0
0 0 0
0 0 0
possible to express each of the matrices Eij (with 1 ≤ i, j ≤ 3) by using these three matrices together with
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the six members of the basis for S found in part (a). Therefore, we see that the nine matrices formed span
V.
a b
∈ Sym2 (R); a, b, c ∈ R. This vector can be represented as
43. Let A ∈ Sym2 (R) so A = AT . A =
b c
1 0
0 1
0 0
a linear combination of
,
,
:
1 0
0 1
0 0
1 0
0 1
0 0
1 0
0 1
0 0
A=a
+b
+c
. Since
,
,
is a linearly independent
0 0
1 0
0 1
0 0
1 0
0 1
a basis forSym2 (R). Thus, dim[Sym
Let B ∈ Skew2 (R) so B =
set that also spansSym2 (R), it is 2 (R)] = 3. 0
b
0
1
0
1
=b
, where b ∈ R. The set
is linearly independent
−B T . Then B =
−b 0
−1 0
−1 0
0 1
. Consequently, dim[Skew2 (R)] = 1. Now,
and spans Skew2 (R) so that a basis for Skew2 (R) is
−1 0
dim[Sym2 (R)] = 4, and hence dim[Sym2 (R)] + dim[Skew2 (R)] = 3 + 1 = 4 = dim[M2 (R)].
44. We know that dim[Mn (R)] = n2 . Let S ∈ Symn (R) and let [Sij ] be the matrix with ones in the (i, j)
and (j, i) positions and zeroes elsewhere. Then the general n × n symmetric matrix can be expressed as:
S = a11 S11 + a12 S12 +a13 S13 + · · · + a1n S1n
+ a22 S22 +a23 S23 + · · · + a2n S2n
+ ···
+ an−1 n−1 Sn−1 n−1 + an−1 n Sn−1 n + ann Snn .
We see that S has been resolved into a linear combination of
n(n + 1)
linearly independent matrices, which therefore form a basis for
n + (n − 1) + (n − 2) + · · · + 1 =
2
n(n + 1)
.
Symn (R); hence dim[Sym2 (R)] =
2
Now let T ∈ Skewn (R) and let [Tij ] be the matrix with one in the (i, j)-position, negative one in the (j, i)position, and zeroes elsewhere, including the main diagonal. Then the general n × n skew-symmetric matrix
can be expressed as:
T = a12 T12 + a13 T13 +a14 T14 + · · · + a1n T1n
+ a23 T23 +a24 T24 + · · · + a2n T2n
+ ···
+ an−1 n Tn−1 n
(n − 1)n
We see that T has been resolved into a linear combination of (n−1)+(n−2)+(n−3)+· · ·+2+1 =
2
(n − 1)n
.
linearly independent vectors, which therefore form a basis for Skewn (R); hence dim[Skewn (R)] =
2
Consequently, using these results, we have
n(n + 1) (n − 1)n
+
= n2 = dim[Mn (R)].
dim[Symn (R)] + dim[Skew2 (R)] =
2
2
45. (a). S is a two-dimensional subspace of R3 . Consequently, any two linearly independent vectors
lying in this subspace determine a basis for S. By inspection we see, for example, that v1 = (4, −1, 0)
and v2 = (3, 0, 1) both lie in the plane. Further, since they are not proportional, these vectors are linearly
independent. Consequently, a basis for S is {(4, −1, 0), (3, 0, 1)}.
(b). To extend the basis obtained in Part (a) to obtain a basis for R3 , we require one more vector that
does not lie in S. For example, since v3 = (1, 0, 0) does not lie on the plane it is an appropriate vector.
Consequently, a basis for R3 is {(4, −1, 0), (3, 0, 1), (1, 0, 0)}.
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46. Each vector in S can be written as
a b
1
=a
b a
0
0
1
+b
0
1
1
0
.
1 0
0 1
Consequently, a basis for S is given by the linearly independent set
,
. To extend this
0 1
1 0
1 0
0 1
basis to M2 (R), we can choose, for example, the two vectors
and
. Then the linearly
0 0
0 0
1 0
0 1
1 0
0 1
independent set
,
,
,
is a basis for M2 (R).
0 1
1 0
0 0
0 0
47. The vectors in S can be expressed as
(2a1 + a2 )x2 + (a1 + a2 )x + (3a1 − a2 ) = a1 (2x2 + x + 3) + a2 (x2 + x − 1),
and since {2x2 + x + 3, x2 + x − 1} is linearly independent (these polynomials are non-proportional) and
certainly span S, they form a basis for S. To extend this basis to V = P2 (R), we must include one additional
vector (since P2 (R) is 3-dimensional). Any polynomial that is not in S will suffice. For example, x ∈ S,
since x cannot be expressed in the form (2a1 + a2 )x2 + (a1 + a2 )x + (3a1 − a2 ), since the equations
2a1 + a2 = 0,
a1 + a2 = 1,
3a1 − a2 = 0
are inconsistent. Thus, the extension we use as a basis for V here is {2x2 + x + 3, x2 + x − 1, x}. Many other
correct answers can also be given here.
48. Let y(x) = erx , so that y (x) = rerx and y (x) = r2 erx . Substituting into the given differential equation,
we obtain
y + 2y − 3y = r2 erx + 2rerx − 3erx = erx (r2 + 2r − 3).
From
erx (r2 + 2r − 3) = 0,
we conclude that r2 + 2r − 3 = 0. Factoring, we obtain (r + 3)(r − 1) = 0, so that r = −3 and r = 1.
Thus, y1 (x) = e−3x and y2 (x) = ex are two non-proportional solutions. Since the solution space is a
two-dimensional vector space, we conclude that a basis is given by {e−3x , ex }.
49. Let y(x) = erx , so that y (x) = rerx and y (x) = r2 erx . Substituting into the given differential equation,
we obtain
y + 6y = r2 erx + 6rerx = erx (r2 + 6r).
From
erx (r2 + 6r) = 0,
we conclude that r2 + 6r = 0. Factoring, we obtain r(r + 6) = 0, so that r = 0 and r = −6. Thus, y1 (x) = 1
and y2 (x) = e−6x are two non-proportional solutions. Since the solution space is a two-dimensional vector
space, we conclude that a basis is given by {1, e−6x }.
50. Let y(x) = erx , so that y (x) = rerx and y (x) = r2 erx . Substituting into the given differential equation,
we obtain
y − 2y = r2 erx − 2erx = erx (r2 − 2).
From
erx (r2 − 2) = 0,
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√
√
√
√
we conclude that
r2 −2 = 0. Factoring, we obtain (r+ 2)(r− 2) = 0, so that r = ± 2. Thus, y1 (x) = e 2x
√
and y2 (x) = e− 2x are two non-proportional solutions.
Since the solution space is a two-dimensional vector
√
√
2x − 2x
space, we conclude that a basis is given by {e
,e
}.
51. A general vector in S has the form y(x) = c(sin 4x + 5 cos 4x), where c is a real constant. Since the
solution space is a two-dimensional vector space, we need to add one additional solution to the differential
equation that is not proportional to y1 (x) = sin 4x + 5 cos 4x. There are many valid choices. One of the
easiest is y2 (x) = sin 4x. Therefore, our extended basis is given by {sin 4x + 5 cos 4x, sin 4x}.
52. Since S is a basis for Pn−1 (R), S contains n vectors. Therefore, S ∪{xn } is a set of n+1 vectors, which is
precisely the dimension of Pn (R). Moreover, xn does not lie in Pn−1 (R) = span(S), and therefore, S ∪ {xn }
is linearly independent by Problem 50 in Section 4.5. By Corollary 4.6.13, we conclude that S ∪ {xn } is a
basis for Pn (R).
53. Since S is a basis for Pn−1 (R), S contains n vectors. Therefore, S ∪ {p} is a set of n + 1 vectors, which
is precisely the dimension of Pn (R). Moreover, p does not lie in Pn−1 (R) = span(S), and therefore, S ∪ {p}
is linearly independent by Problem 50 in Section 4.5. By Corollary 4.6.13, we conclude that S ∪ {p} is a
basis for Pn (R).
54.
(a). Let ek denote the kth standard basis vector. Then a basis for Cn with scalars in R is given by
{e1 , e2 , . . . , en , ie1 , ie2 , . . . , ien },
and the dimension is 2n.
(b). Using the notation in part (a), a basis for Cn with scalars in C is given by
{e1 , e2 , . . . , en },
and the dimension is n.
Solutions to Section 4.7
True-False Review:
(a): TRUE. This is the content of Theorem 4.7.1. The existence of such a linear combination comes from
the fact that a basis for V must span V , and the uniqueness of such a linear combination follows from the
linear independence of the vectors comprising a basis.
(b): TRUE. This follows from the Equation [v]B = PB←C [v]C , which is just Equation (4.7.6) with the
roles of B and C reversed.
(c): TRUE. The number of columns in the change-of-basis matrix PC←B is the number of vectors in B,
while the number of rows of PC←B is the number of vectors in C. Since all bases for the vector space V
contain the same number of vectors, this implies that PC←B contains the same number of rows and columns.
(d): TRUE. If V is an n-dimensional vector space, then
PC←B PB←C = In = PB←C PC←B ,
which implies that PC←B is invertible.
(e): TRUE. This follows from the linearity properties:
[v − w]B = [v + (−1)w]B = [v]B + [(−1)w]B = [v]B + (−1)[w]B = [v]B − [w]B .
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(f ): FALSE. It depends on the order in which the vectors in the bases B and C are listed. For instance, if
2
we consider the bases B = {(1, 0), (0, 1)} and C = {(0,
1), (1, 0)} for R ,then although B and C contain the
1
0
while [v]C =
.
same vectors, if we let v = (1, 0), then [v]B =
0
1
2
(g): FALSE. For instance, if we consider the bases B = {(1,
0),(0, 1)} and C = {(0, 1), (1, 0)} for R , and
1
= [w]C .
if we let v = (1, 0) and w = (0, 1), then v = w, but [v]B =
0
(h): TRUE. If B = {v1 , v2 , . . . , vn }, then the column vector [vi ]B is the ith standard basis vector (1 in
the ith position and zeroes elsewhere). Thus, for each i, the ith column of PB←B consists of a 1 in the ith
position and zeroes elsewhere. This describes precisely the identity matrix.
Problems:
1. Write
(27, 6) = c1 (7, −1) + c2 (−9, −2).
Then
7c1 − 9c2 = 27 and
− c1 − 2c2 = 6.
Solving this system of equations gives c1 = 0 and c2 = −3. Thus,
0
[v]B =
.
−3
2. Write
(5, −10) = c1 (2, −2) + c2 (1, 4).
Then
2c1 + c2 = 5
and
− 2c1 + 4c2 = −10.
Solving this system of equations gives c1 = 3 and c2 = −1. Thus,
3
[v]B =
.
−1
3. Write
(8, −2) = c1 (−1, 3) + c2 (3, 2).
Then
−c1 + 3c2 = 8
and
3c1 + 2c2 = −2.
Solving this system of equations gives c1 = −2 and c2 = 2. Thus,
−2
[v]B =
.
2
4. Write
(−9, 1, −8) = c1 (1, 0, 1) + c2 (1, 1, −1) + c3 (2, 0, 1).
Then
c1 + c2 + 2c3 = −9
and
c2 = 1
and
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311
Solving this system of equations gives c1 = −4, c2 = 1, and c3 = −3. Thus,
⎡
⎤
−4
[v]B = ⎣ 1 ⎦ .
−3
5. Write
(1, 7, 7) = c1 (1, −6, 3) + c2 (0, 5, −1) + c3 (3, −1, −1).
Then
c1 + 3c3 = 1
and
− 6c1 + 5c2 − c3 = 7
and
3c1 − c2 − c3 = 7.
Using Gaussian elimination to solve this system of equations gives c1 = 4, c2 = 6, and c3 = −1. Thus,
⎡
⎤
4
[v]B = ⎣ 6 ⎦ .
−1
6. Write
(1, 7, 7) = c1 (3, −1, −1) + c2 (1, −6, 3) + c3 (0, 5, −1).
Then
3c1 + c2 = 1
and
− c1 − 6c2 + 5c3 = 7
and
− c1 + 3c2 − c3 = 7.
Using Gaussian elimination to solve this system of equations gives c1 = −1, c2 = 4, and c3 = 6. Thus,
⎡
⎤
−1
[v]B = ⎣ 4 ⎦ .
6
7. Write
(5, 5, 5) = c1 (−1, 0, 0) + c2 (0, 0, −3) + c3 (0, −2, 0).
Then
−c1 = 5
− 2c3 = 5
and
Therefore, c1 = −5, c2 = − 53 , and c3 = − 52 . Thus,
and
− 3c2 = 5.
⎤
−5
[v]B = ⎣ −5/3 ⎦ .
−5/2
8. Write
⎡
−4x2 + 2x + 6 = c1 (x2 + x) + c2 (2 + 2x) + c3 (1).
Equating the powers of x on each side, we have
c1 = −4
and
c1 + 2c2 = 2
and
2c2 + c3 = 6.
Solving this system of equations, we find that c1 = −4, c2 = 3, and c3 = 0. Hence,
⎤
⎡
−4
[p(x)]B = ⎣ 3 ⎦ .
0
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9. Write
15 − 18x − 30x2 = c1 (5 − 3x) + c2 (1) + c3 (1 + 2x2 ).
Equating the powers of x on each side, we have
5c1 + c2 + c3 = 15
and
− 3c1 = −18
2c3 = −30.
and
Solving this system of equations, we find that c1 = 6, c2 = 0, and c3 = −15. Hence,
⎡
⎤
6
0 ⎦.
[p(x)]B = ⎣
−15
10. Write
4 − x + x2 − 2x3 = c1 (1) + c2 (1 + x) + c3 (1 + x + x2 ) + c4 (1 + x + x2 + x3 ).
Equating the powers of x on each side, we have
c1 + c2 + c3 + c4 = 4
c2 + c3 + c4 = −1
and
and
c3 + c4 = 1
c4 = −2.
and
Solving this system of equations, we find that c1 = 5, c2 = −2, c3 = 3, and c4 = −2. Hence,
⎤
⎡
5
⎢ −2 ⎥
⎥
[p(x)]B = ⎢
⎣ 3 ⎦.
−2
11. Write
8 + x + 6x2 + 9x3 = c1 (x3 + x2 ) + c2 (x3 − 1) + c3 (x3 + 1) + c4 (x3 + x).
Equating the powers of x on each side, we have
−c2 + c3 = 8
and
c4 = 1
and
c1 = 6
and
c1 + c2 + c3 + c4 = 9.
Solving this system of equations, we find that c1 = 6, c2 = −3, c3 = 5, and c4 = 1. Hence
⎤
⎡
6
⎢ −3 ⎥
⎥
[p(x)]B = ⎢
⎣ 5 ⎦.
1
12. Write
−3 −2
−1
2
= c1
1
1
1
1
+ c2
1
1
1
0
+ c3
1
0
1
0
+ c4
1
0
0
0
.
Equating the individual entries of the matrices on each side of this equation (upper left, upper right, lower
left, and lower right, respectively) gives
c1 + c2 + c3 + c4 = −3
and
c1 + c2 + c3 = −2
and
c1 + c2 = −1
and
c1 = 2.
Solving this system of equations, we find that c1 = 2, c2 = −3, c3 = −1, and c4 = −1. Thus,
⎤
⎡
2
⎢ −3 ⎥
⎥
[A]B = ⎢
⎣ −1 ⎦ .
−1
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13. Write
−10
16
−15 −14
= c1
2 −1
3
5
+ c2
0
−1
4
1
+ c3
1
1
1
1
+ c4
3 −1
2
5
.
Equating the individual entries of the matrices on each side of this equation (upper left, upper right, lower
left, and lower right, respectively) gives
2c1 + c3 + 3c4 = −10, −c1 + 4c2 + c3 − c4 = 16, 3c1 − c2 + c3 + 2c4 = −15, 5c1 + c2 + c3 + 5c4 = −14.
Solving this system of equations, we find that c1 = −2, c2 = 4, c3 = −3, and c4 = −1. Thus,
⎤
⎡
−2
⎢ 4 ⎥
⎥
[A]B = ⎢
⎣ −3 ⎦ .
−1
14. Write
5
7
6
8
= c1
−1
0
1
1
+ c2
1 3
−1 0
+ c3
1
1
0
2
+ c4
0 −1
2
3
.
Equating the individual entries of the matrices on each side of this equation (upper left, upper right, lower
left, and lower right, respectively) gives
−c1 + c2 + c3 = 5
and
c1 + 3c2 − c4 = 6
and
− c2 + c3 + 2c4 = 7
and
c1 + 2c3 + 3c4 = 8.
Solving this system of equations, we find that c1 = −34/3, c2 = 12, c3 = −55/3, and c4 = 56/3. Thus,
⎤
⎡
−34/3
⎢ 12 ⎥
⎥
[A]B = ⎢
⎣ −55/3 ⎦ .
56/3
15. Write
(x, y, z) = c1 (0, 6, 3) + c2 (3, 0, 3) + c3 (6, −3, 0).
Then
6c1 − 3c3 = y and 3c1 + 3c2 = z.
⎤
⎡
3 3
0 z
The augmented matrix for this linear system is ⎣ 6 0 −3 y ⎦. We can reduce this to row-echelon form
0 3
6 z
⎤
⎡
z/3
1 1 0
⎦. Solving this system by back-substitution gives
x/3
as ⎣ 0 1 2
0 0 9 y + 2x − 2z
3c2 + 6c3 = x
c1 =
1
2
1
x+ y− z
9
9
9
and
and
1
2
4
c2 = − x − y + z
9
9
9
and
Hence, denote the ordered basis {v1 , v2 , v3 } by B, we have
⎤
⎡ 1
2
1
9x + 9y − 9z
[v]B = ⎣ − 19 x − 29 y + 49 z ⎦ .
2
1
2
9x + 9y − 9z
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c3 =
2
1
2
x + y − z.
9
9
9
314
16. Write
a0 + a1 x + a2 x2 = c1 (1 + x) + c2 x(x − 1) + c3 (1 + 2x2 ).
Equating powers of x on both sides of this equation, we have
c 1 + c3 = a0
c 1 − c2 = a1
c2 + 2c3 = a2 .
⎤
⎡
1
0 1 a0
The augmented matrix corresponding to this system of equations is ⎣ 1 −1 0 a1 ⎦. We can reduce this
0
1 2 a2
⎤
⎡
a0
1 0 1
⎦. Thus, solving by back-substitution, we have
a2
to row-echelon form as ⎣ 0 1 2
0 0 1 −a0 + a1 + a2
c1 = 2a0 − a1 − a2
and
and
and
c2 = 2a0 − 2a1 − a2
and
c3 = −a0 + a1 + a2 .
Hence, relative to the ordered basis B = {p1 , p2 , p3 }, we have
⎤
⎡
2a0 − a1 − a2
[p(x)]B = ⎣ 2a0 − 2a1 − a2 ⎦ .
−a0 + a1 + a2
17. Let v1 = (9, 2) and v2 = (4, −3). Setting
(9, 2) = c1 (2, 1) + c2 (−3, 1)
3
and solving, we find c1 = 3 and c2 = −1. Thus, [v1 ]C =
. Next, setting
−1
(4, −3) = c1 (2, 1) + c2 (−3, 1)
−1
and solving, we find c1 = −1 and c2 = −2. Thus, [v2 ]C =
. Therefore,
−2
3 −1
PC←B =
.
−1 −2
18. Let v1 = (−5, −3) and v2 = (4, 28). Setting
(−5, −3) = c1 (6, 2) + c2 (1, −1)
−1
and solving, we find c1 = −1 and c2 = 1. Thus, [v1 ]C =
. Next, setting
1
(4, 28) = c1 (6, 2) + c2 (1, −1)
4
and solving, we find c1 = 4 and c2 = −20. Thus, [v2 ]C =
. Therefore,
−20
−1
4
PC←B =
.
1 −20
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19. Let v1 = (2, −5, 0), v2 = (3, 0, 5), and v3 = (8, −2, −9). Setting
(2, −5, 0) = c1 (1, −1, 1) + c2 (2, 0, 1) + c3 (0, 1, 3)
⎡
⎤
4
and solving, we find c1 = 4, c2 = −1, and c3 = −1. Thus, [v1 ]C = ⎣ −1 ⎦. Next, setting
−1
(3, 0, 5) = c1 (1, −1, 1) + c2 (2, 0, 1) + c3 (0, 1, 3)
⎡
⎤
1
and solving, we find c1 = 1, c2 = 1, and c3 = 1. Thus, [v2 ]C = ⎣ 1 ⎦. Finally, setting
1
(8, −2, −9) = c1 (1, −1, 1) + c2 (2, 0, 1) + c3 (0, 1, 3)
⎡
⎤
−2
and solving, we find c1 = −2, c2 = 5, and c3 = −4. Thus, [v3 ]C = ⎣ 5 ⎦. Therefore,
−4
⎡
⎤
4 1 −2
5 ⎦.
PC←B = ⎣ −1 1
−1 1 −4
20. Let v1 = (−7, 4, 4), v2 = (4, 2, −1), and v3 = (−7, 5, 0). Setting
(−7, 4, 4) = c1 (1, 1, 0) + c2 (0, 1, 1) + c3 (3, −1, −1)
⎡
⎤
0
and solving, we find c1 = 0, c2 = 5/3 and c3 = −7/3. Thus, [v1 ]C = ⎣ 5/3 ⎦. Next, setting
−7/3
(4, 2, −1) = c1 (1, 1, 0) + c2 (0, 1, 1) + c3 (3, −1, −1)
and solving, we find c1 = 0, c2 = 19/3 and c3 = −7/3. Setting
(4, 2, −1) = c1 (1, 1, 0) + c2 (0, 1, 1) + c3 (3, −1, −1)
⎡
⎤
3
and solving, we find c1 = 3, c2 = −2/3, and c3 = 1/3. Thus, [v2 ]C = ⎣ −2/3 ⎦. Finally, setting
1/3
(−7, 5, 0) = c1 (1, 1, 0) + c2 (0, 1, 1) + c3 (3, −1, −1)
⎡
⎤
5
and solving, we find c1 = 5, c2 = −4, and c3 = −4. Hence, [v3 ]C = ⎣ −4 ⎦. Therefore,
−4
⎤
⎡
0
3
5
PC←B = ⎣ 53 − 23 −4 ⎦ .
1
−4
− 73
3
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21. Let v1 = 7 − 4x and v2 = 5x. Setting
7 − 4x = c1 (1 − 2x) + c2 (2 + x)
3
. Next, setting
and solving, we find c1 = 3 and c2 = 2. Thus, [v1 ]C =
2
5x = c1 (1 − 2x) + c2 (2 + x)
−2
and solving, we find c1 = −2 and c2 = 1. Hence, [v2 ]C =
. Therefore,
1
PC←B =
3 −2
2
1
.
22. Let v1 = −4 + x − 6x2 , v2 = 6 + 2x2 , and v3 = −6 − 2x + 4x2 . Setting
−4 + x − 6x2 = c1 (1 − x + 3x2 ) + c2 (2) + c3 (3 + x2 )
⎡
⎤
−1
and solving, we find c1 = −1, c2 = 3, and c3 = −3. Thus, [v1 ]C = ⎣ 3 ⎦. Next, setting
−3
6 + 2x2 = c1 (1 − x + 3x2 ) + c2 (2) + c3 (3 + x2 )
⎡
⎤
0
and solving, we find c1 = 0, c2 = 0, and c3 = 2. Thus, [v2 ]C = ⎣ 0 ⎦. Finally, setting
2
−6 − 2x + 4x2 = c1 (1 − x + 3x2 ) + c2 (2) + c3 (3 + x2 )
⎡
⎤
2
and solving, we find c1 = 2, c2 = −1, and c3 = −2. Thus, [v3 ]C = ⎣ −1 ⎦. Therefore,
−2
⎡
−1
PC←B = ⎣ 3
−3
⎤
0
2
0 −1 ⎦ .
2 −2
23. Let v1 = −2 + 3x + 4x2 − x3 , v2 = 3x + 5x2 + 2x3 , v3 = −5x2 − 5x3 , and v4 = 4 + 4x + 4x2 . Setting
−2 + 3x + 4x2 − x3 = c1 (1 − x3 ) + c2 (1 + x) + c3 (x + x2 ) + c4 (x2 + x3 )
⎤
⎡
0
⎢ −2 ⎥
⎥
and solving, we find c1 = 0, c2 = −2, c3 = 5, and c4 = −1. Thus, [v1 ]C = ⎢
⎣ 5 ⎦. Next, setting
−1
3x + 5x2 + 2x3 = c1 (1 − x3 ) + c2 (1 + x) + c3 (x + x2 ) + c4 (x2 + x3 )
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⎤
0
⎢ 0 ⎥
⎥
and solving, we find c1 = 0, c2 = 0, c3 = 3, and c4 = 2. Thus, [v2 ]C = ⎢
⎣ 3 ⎦. Next, solving
2
⎡
−5x2 − 5x3 = c1 (1 − x3 ) + c2 (1 + x) + c3 (x + x2 ) + c4 (x2 + x3 )
⎤
⎡
0
⎢ 0 ⎥
⎥
and solving, we find c1 = 0, c2 = 0, c3 = 0, and c4 = −5. Thus, [v3 ]C = ⎢
⎣ 0 ⎦. Finally, setting
−5
4 + 4x + 4x2 = c1 (1 − x3 ) + c2 (1 + x) + c3 (x + x2 ) + c4 (x2 + x3 )
⎤
⎡
2
⎢ 2 ⎥
⎥
and solving, we find c1 = 2, c2 = 2, c3 = 2, and c4 = 2. Thus, [v4 ]C = ⎢
⎣ 2 ⎦. Therefore,
2
⎤
⎡
0 0
0 2
⎢ −2 0
0 2 ⎥
⎥.
PC←B = ⎢
⎣ 5 3
0 2 ⎦
−1 2 −5 2
24. Let v1 = 2 + x2 , v2 = −1 − 6x + 8x2 , and v3 = −7 − 3x − 9x2 . Setting
2 + x2 = c1 (1 + x) + c2 (−x + x2 ) + c3 (1 + 2x2 )
⎡
⎤
3
and solving, we find c1 = 3, c2 = 3, and c3 = −1. Thus, [v1 ]C = ⎣ 3 ⎦. Next, solving
−1
−1 − 6x + 8x2 = c1 (1 + x) + c2 (−x + x2 ) + c3 (1 + 2x2 )
⎡
⎤
−4
and solving, we find c1 = −4, c2 = 2, and c3 = 3. Thus, [v2 ]C = ⎣ 2 ⎦. Finally, solving
3
−7 − 3x − 9x2 = c1 (1 + x) + c2 (−x + x2 ) + c3 (1 + 2x2 )
⎡
⎤
−2
and solving, we find c1 = −2, c2 = 1, and c3 = −5. Thus, [v3 ]C = ⎣ 1 ⎦. Therefore,
−5
⎡
⎤
3 −4 −2
2
1 ⎦.
PC←B = ⎣ 3
−1
3 −5
25. Let v1 =
1
0
0 −1
3 5
−2 −4
, v2 =
, v3 =
, and v4 =
. Setting
−1 −2
3
0
0 0
0
0
1
0
1 1
1 1
1 1
1 0
= c1
+ c2
+ c3
+ c4
−1 −2
1 1
1 0
0 0
0 0
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⎤
−2
⎢ 1 ⎥
⎥
and solving, we find c1 = −2, c2 = c3 = c4 = 1. Thus, [v1 ]C = ⎢
⎣ 1 ⎦. Next, setting
1
0 −1
1 1
1 1
1 1
1 0
c1
+ c2
+ c3
+ c4
3
0
1 1
1 0
0 0
0 0
⎤
⎡
0
⎢ 3 ⎥
⎥
and solving, we find c1 = 0, c2 = 3, c3 = −4, and c4 = 1. Thus, [v2 ]C = ⎢
⎣ −4 ⎦. Next, setting
1
3 5
1 1
1 1
1 1
1 0
c1
+ c2
+ c3
+ c4
0 0
1 1
1 0
0 0
0 0
⎤
⎡
0
⎢ 0 ⎥
⎥
and solving, we find c1 = 0, c2 = 0, c3 = 5, and c4 = −2. Thus, [v3 ]C = ⎢
⎣ 5 ⎦. Finally, setting
−2
−2 −4
1 1
1 1
1 1
1 0
c1
+ c2
+ c3
+ c4
0
0
1 1
1 0
0 0
0 0
⎤
⎡
0
⎢ 0 ⎥
⎥
and solving, we find c1 = 0, c2 = 0, c3 = −4, and c4 = 2. Thus, [v4 ]C = ⎢
⎣ −4 ⎦. Therefore, we have
2
⎤
⎡
−2
0
0
0
⎥
⎢ 1
3
0
0
⎥.
PC←B = ⎢
⎣ 1 −4
5 −4 ⎦
1
1 −2
2
⎡
26. Let v1 = E12 , v2 = E22 , v3 = E21 , and v4 = E11 . We see by inspection that
⎤
⎤
⎤
⎤
⎡
⎡
⎡
⎡
0
1
0
0
⎢ 0 ⎥
⎢ 0 ⎥
⎢ 0 ⎥
⎢ 1 ⎥
⎥
⎥
⎥
⎥
⎢
⎢
⎢
[v1 ]C = ⎢
⎣ 0 ⎦ , [v2 ]C = ⎣ 0 ⎦ , [v3 ]C = ⎣ 1 ⎦ , [v4 ]C = ⎣ 0 ⎦ .
1
0
0
0
Therefore,
⎡
0
⎢ 0
PC←B = ⎢
⎣ 0
1
1
0
0
0
0
0
1
0
⎤
0
1 ⎥
⎥.
0 ⎦
0
27. We could simply compute the inverse of the matrix obtained in Problem 17. For instructive purposes,
however, we proceed directly. Let w1 = (2, 1) and w2 = (−3, 1). Setting
(2, 1) = c1 (9, 2) + c2 (4, −3)
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and solving, we obtain c1 = 2/7 and c2 = −1/7. Thus, [w1 ]B =
2/7
−1/7
. Next, setting
(−3, 1) = c1 (9, 2) + c2 (4, −3)
−1/7
. Therefore,
and solving, we obtain c1 = −1/7 and c2 = −3/7. Thus, [w2 ]B =
−3/7
⎤
⎡ 2
− 17
7
⎦.
PB←C = ⎣
1
3
−7 −7
28. We could simply compute the inverse of the matrix obtained in Problem 18. For instructive purposes,
however, we proceed directly. Let w1 = (6, 2) and w2 = (1, −1). Setting
(6, 2) = c1 (−5, −3) + c2 (4, 28)
−5/4
. Next, setting
and solving, we obtain c1 = −5/4 and c2 = −1/16. Thus, [w1 ]B =
−1/16
(1, −1) = c1 (−5, −3) + c2 (4, 28)
−1/4
. Therefore,
and solving, we obtain c1 = −1/4 and c2 = −1/16. Thus, [w2 ]B =
−1/16
⎤
⎡
− 14
− 54
⎦.
PB←C = ⎣
1
1
− 16
− 16
29. We could simply compute the inverse of the matrix obtained in Problem 19. For instructive purposes,
however, we proceed directly. Let w1 = (1, −1, 1), w2 = (2, 0, 1), and w3 = (0, 1, 3). Setting
(1, −1, 1) = c1 (2, −5, 0) + c2 (3, 0, 5) + c3 (8, −2, −9)
⎡
⎤
1/5
and solving, we find c1 = 1/5, c2 = 1/5, and c3 = 0. Thus, [w1 ]B = ⎣ 1/5 ⎦. Next, setting
0
(2, 0, 1) = c1 (2, −5, 0) + c2 (3, 0, 5) + c3 (8, −2, −9)
⎡
⎤
−2/45
and solving, we find c1 = −2/45, c2 = 2/5, and c3 = 1/9. Thus, [w2 ]B = ⎣ 2/5 ⎦. Finally, setting
1/9
(0, 1, 3) = c1 (2, −5, 0) + c2 (3, 0, 5) + c3 (8, −2, −9)
⎡
⎤
−7/45
and solving, we find c1 = −7/45, c2 = 2/5, and c3 = −1/9. Thus, [w3 ]B = ⎣ 2/5 ⎦. Therefore,
−1/9
⎡ 1
⎤
2
7
− 45
− 45
5
⎢
⎥
⎢ 1
⎥
2
2 ⎥
⎢
PB←C = ⎢ 5
5
5 ⎥.
⎣
⎦
1
1
0
−9
9
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30. We could simply compute the inverse of the matrix obtained in Problem 21. For instructive purposes,
however, we proceed directly. Let w1 = 1 − 2x and w2 = 2 + x. Setting
1 − 2x = c1 (7 − 4x) + c2 (5x)
and solving, we find c1 = 1/7 and c2 = −2/7. Thus, [w1 ]B =
1/7
−2/7
. Setting
2 + x = c1 (7 − 4x) + c2 (5x)
and solving, we find c1 = 2/7 and c2 = 3/7. Thus, [w2 ]B =
⎡
PB←C = ⎣
1
7
2
7
− 27
3
7
2/7
3/7
. Therefore,
⎤
⎦.
31. Referring to Problem 23, we have
⎡
0
⎢ −2
−1
PB←C = (PC←B ) = ⎢
⎣ 5
−1
⎤−1 ⎡
0
0 2
1/2
⎢ −7/6
0
0 2 ⎥
⎥ =⎢
⎣ −11/30
3
0 2 ⎦
2 −5 2
1/2
−1/2
5/6
13/30
0
⎤
0
0
1/3
0 ⎥
⎥.
2/15 −1/5 ⎦
0
0
32. We could simply compute the inverse of the matrix obtained in Problem 26. For instructive purposes,
however, we proceed directly. Let w1 = E22 , w2 = E11 , w3 = E21 , and w4 = E12 . We see by inspection
that
⎤
⎤
⎤
⎤
⎡
⎡
⎡
⎡
0
0
0
1
⎢ 1 ⎥
⎢ 0 ⎥
⎢ 0 ⎥
⎢ 0 ⎥
⎥
⎥
⎥
⎥
⎢
⎢
⎢
[w1 ]B = ⎢
⎣ 0 ⎦ , [w2 ]B = ⎣ 0 ⎦ , [w3 ]B = ⎣ 1 ⎦ , [w4 ]B = ⎣ 0 ⎦ .
0
1
0
0
Therefore
⎡
0
⎢ 1
PB←C = ⎢
⎣ 0
0
0
0
0
1
0
0
1
0
⎤
1
0 ⎥
⎥.
0 ⎦
0
33. We first compute [v]B and [v]C directly. Setting
(−5, 3) = c1 (9, 2) + c2 (4, −3)
⎡
and solving, we obtain c1 = −3/35 and c2 = −37/35. Thus, [v]B = ⎣
(−5, 3) = c1 (2, 1) + c2 (−3, 1)
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− 35
− 37
35
⎤
⎦. Setting
321
⎡
and solving, we obtain c1 = 4/5 and c2 = 11/5. Thus, [v]C = ⎣
PC←B =
3 −1
−1 −2
4
5
11
5
⎤
⎦. Now, according to Problem 17,
, so
PC←B [v]B =
3 −1
−1 −2
⎡
⎣
3
− 35
− 37
35
⎤
⎡
⎦=⎣
4
5
11
5
⎤
⎦ = [v]C ,
which confirms Equation (4.7.6).
34. We first compute [v]B and [v]C directly. Setting
(−1, 2, 0) = c1 (−7, 4, 4) + c2 (4, 2, −1) + c3 (−7, 5, 0)
⎡ 3 ⎤
43
⎢
⎥
⎢ 12 ⎥
⎥
and solving, we obtain c1 = 3/43, c2 = 12/43, and c3 = 10/43. Thus, [v]B = ⎢
⎢ 43 ⎥. Setting
⎣
⎦
10
43
(−1, 2, 0) = c1 (1, 1, 0) + c2 (0, 1, 1) + c3 (3, −1, −1)
⎡
⎤
2
and solving, we obtain c1 = 2, c2 = −1, and c3 = −1. Thus, [v]C = ⎣ −1 ⎦. Now, according to Problem
−1
⎤
⎡
⎡
⎡
⎤
⎤⎡ 3 ⎤
0
3
5
0
3
5
2
43
⎥
⎢
⎢
⎢
⎥
⎥
⎥⎢
⎥
⎢
⎢ 5
⎢
⎥
⎥
⎥⎢
2
⎥, so PC←B [v]B = ⎢ 5 − 2 −4 ⎥ ⎢ 12 ⎥ = ⎢ −1 ⎥ = [v]C , which
−
−4
20, PC←B = ⎢
3
3
⎥
⎢
⎢ 3
⎢ 3
⎥
⎥ ⎢ 43 ⎥
⎦
⎣
⎣
⎣
⎦
⎦
⎦⎣
1
1
7
10
−1
−4
−4
− 73
−
3
3
3
43
confirms Equation (4.7.6).
35. We first compute [v]B and [v]C directly. Setting
6 − 4x = c1 (7 − 4x) + c2 (5x)
⎤
⎡
6
and solving, we obtain c1 = 6/7 and c2 = −4/35. Thus, [v]B = ⎣
7
4
− 35
⎦. Next, setting
6 − 4x = c1 (1 − 2x) + c2 (2 + x)
⎡ 14 ⎤
and solving, we obtain c1 = 14/5 and c2 = 8/5. Thus, [v]C = ⎣
PC←B =
3 −2
2
1
5
8
5
⎦. Now, according to Problem 21,
, so
⎡
PC←B [v]B = ⎣
3
2
−2
1
⎤⎡
⎦⎣
6
7
4
− 35
⎤
⎡ 14 ⎤
⎦=⎣
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8
5
⎦ = [v]C ,
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which confirms Equation (4.7.6).
36. We first compute [v]B and [v]C directly. Setting
5 − x + 3x2 = c1 (−4 + x − 6x2 ) + c2 (6 + 2x2 ) + c3 (−6 − 2x + 4x2 )
⎡
⎤
1
and solving, we obtain c1 = 1, c2 = 5/2, and c3 = 1. Thus, [v]B = ⎣ 52 ⎦. Next, setting
1
5 − x + 3x2 = c1 (1 − x + 3x2 ) + c2 (2) + c3 (3 + x2 )
⎡
⎤
1
and solving, we obtain c1 = 1, c2 = 2, and c3 = 0. Thus, [v]C = ⎣ 2 ⎦. Now, according to Problem 22,
0
⎡
⎤
−1 0
2
PC←B = ⎣ 3 0 −1 ⎦, so
−3 2 −2
⎤ ⎡
⎤⎡
⎤
1
−1 0
2
1
PC←B [v]B = ⎣ 3 0 −1 ⎦ ⎣ 52 ⎦ = ⎣ 2 ⎦ = [v]C ,
−3 2 −2
0
1
⎡
which confirms Equation (4.7.6).
37. We first compute [v]B and [v]C directly. Setting
1
0
0 −1
3
−1 −1
+ c2
+ c3
= c1
−1 −2
3
0
0
−4
5
5
0
+ c4
−2 −4
0
0
⎤
−5/2
⎢ −13/6 ⎥
⎥
and solving, we obtain c1 = −5/2, c2 = −13/6, c3 = 37/6, and c4 = 17/2. Thus, [v]B = ⎢
⎣ 37/6 ⎦. Next,
17/2
setting
−1 −1
1 1
1 1
1 1
1 0
= c1
+ c2
+ c3
+ c4
−4
5
1 1
1 0
0 0
0 0
⎤
⎡
5
⎢ −9 ⎥
⎥
and solving, we find c1 = 5, c2 = −9, c3 = 3, and c4 = 0. Thus, [v]C = ⎢
⎣ 3 ⎦. Now, according to Problem
0
⎤
⎡
−2
0
0
0
⎢ 1
3
0
0 ⎥
⎥, and so
25, PC←B = ⎢
⎣ 1 −4
5 −4 ⎦
1
1 −2
2
⎡
⎤
⎤ ⎡
⎤⎡
5
−5/2
−2
0
0
0
⎥
⎥ ⎢
⎢
⎢ 1
3
0
0 ⎥
⎥ ⎢ −13/6 ⎥ = ⎢ −9 ⎥ = [v]C ,
PC←B [v]B = ⎢
⎦
⎣
⎦
⎣
⎣ 1 −4
3 ⎦
37/6
5 −4
0
17/2
1
1 −2
2
⎡
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which confirms Equation (4.7.6).
38. Write x = a1 v1 + a2 v2 + · · · + an vn . We have
cx = c(a1 v1 + a2 v2 + · · · + an vn ) = (ca1 )v1 + (ca2 )v2 + · · · + (can )vn .
Hence,
⎡
⎤
⎤
a1
ca1
⎢ a2 ⎥
⎢ ca2 ⎥
⎢
⎢
⎥
⎥
[cx]B = ⎢ . ⎥ = c ⎢ . ⎥ = c[x]B .
⎣ .. ⎦
⎣ .. ⎦
⎡
can
an
39. We must show that {v1 , v2 , . . . , vn } is linearly independent and spans V .
Check linear independence: Assume that
c1 v1 + c2 v2 + · · · + cn vn = 0.
We wish to show that c1 = c2 = · · · = cn = 0. Now by assumption, the zero vector 0 can be uniquely written
as a linear combination of the vectors in {v1 , v2 , . . . , vn }. Since
0 = 0 · v1 + 0 · v2 + · · · + 0 · vn ,
we therefore conclude that
c1 = c2 = · · · = cn = 0,
as needed.
Check spanning property: Let v be an arbitrary vector in V . By assumption, it is possible to express v
(uniquely) as a linear combination of the vectors in {v1 , v2 , . . . , vn }; say
v = a1 v 1 + a2 v 2 + · · · + an v n .
Therefore, v lies in span{v1 , v2 , . . . , vn }. Since v is an arbitrary member of V , we conclude that {v1 , v2 , . . . , vn }
spans V .
40. Let B = {v1 , v2 , . . . , vn } and let C = {vσ(1) , vσ(2) , . . . , vσ(n) }, where σ is a permutation of the set
{1, 2, . . . , n}. We will show that PC←B contains exactly one 1 in each row and column, and zeroes elsewhere
(the argument for PB←C is essentially identical, or can be deduced from the fact that PB←C is the inverse
of PC←B ).
Let i be in {1, 2, . . . , n}. The ith column of PC←B is [vi ]C . Suppose that ki ∈ {1, 2, . . . , n} is such that
σ(ki ) = i. Then [vi ]C is a column vector with a 1 in the ki th position and zeroes elsewhere. Since the values
k1 , k2 , . . . , kn are distinct, we see that each column of PC←B contains a single 1 (with zeroes elsewhere) in a
different position from any other column. Hence, when we consider all n columns as a whole, each position
in the column vector must have a 1 occurring exactly once in one of the columns. Therefore, PC←B contains
exactly one 1 in each row and column, and zeroes elsewhere.
Solutions to Section 4.8
True-False Review:
(a): TRUE. Note that rowspace(A) is a subspace of Rn and colspace(A) is a subspace of Rm , so certainly
if rowspace(A) = colspace(A), then Rn and Rm must be the same. That is, m = n.
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(b): FALSE. A basis for the row space of A consists of the nonzero row vectors of any row-echelon form of
A.
(c): FALSE. The nonzero column vectors of the original matrix A that correspond to the nonzero column
vectors of a row-echelon form of A form a basis for colspace(A).
(d): TRUE. Both rowspace(A) and colspace(A) have dimension equal to rank(A), the number of nonzero
rows in a row-echelon form of A. Equivalently, their dimensions are both equal to the number of pivots
occurring in a row-echelon form of A.
(e): TRUE. For an invertible n × n matrix, rank(A) = n. That means there are n nonzero rows in a rowechelon form of A, and so rowspace(A) is n-dimensional. Therefore, we conclude that rowspace(A) = Rn .
(f ): TRUE. This follows immediately from the (true) statements in True-False Review Questions 4-5 above.
Problems:
1 − 16
. Consequently, a basis for rowspace(A) is {(1, − 16 )}, whereas a
1. A row-echelon form for A is
0
0
basis for colspace(A) is {(6, 12)}. Rowspace(A) consists of points on the line in the xy-plane with equation
y = − 16 x, while colspace(A) consists of points on the line in the xy-plane with equation y = 2x.
1 −2
2. A row-echelon form of A is
. Consequently, a basis for rowspace(A) is {(1, −2)}, whereas a
0
0
basis for colspace(A) is {(1, −3)}.
3. (a). n = 5; a basis for rowspace(A) is given by {(1, 2, 3, 4, 5)}.
(b). m = 1; a basis for colspace(A) is given by {1}.
4. (a). n = 1; a basis for rowspace(A) is given by {1}.
(b). m = 3; a basis for colspace(A) is given by {(−3, 1, 7)}.
1 1 −3 2
. Consequently, a basis for rowspace(A) is
5. (a). n = 4; a row-echelon form of A is
0 1 −2 1
{(1, 1, −3, 2), (0, 1, −2, 1)}.
(b). m = 2; using the row-echelon form found in part (a), we can immediately find a basis for colspace(A):
{(1, 3), (1, 4)}.
⎡
⎤
1 2 3
6. (a). n = 3; a row-echelon form of A is ⎣ 0 1 2 ⎦. Consequently, a basis for rowspace(A) is
0 0 0
{(1, 2, 3), (0, 1, 2)}.
(b). m = 3; using the row-echelon form found in part (a), we can immediately find a basis for colspace(A):
{(1, 5, 9), (2, 6, 10)}.
⎤
⎡
0 1 13
7. (a). n = 3; a row-echelon form of A is ⎣ 0 0 0 ⎦. Consequently, a basis for rowspace(A) is {(0, 1, 13 )}.
0 0 0
(b). m = 3; using the row-echelon form found in part (a), we can immediately find a basis for colspace(A):
{(3, −6, 12)}.
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⎡
1
⎢ 0
⎢
8. (a). n = 4; a row-echelon form of A is ⎣
0
0
{(1, 2, −1, 3), (0, 0, 0, 1)}.
⎤
2 −1 3
0
0 1 ⎥
⎥. Consequently, a basis for rowspace(A) is
0
0 0 ⎦
0
0 0
(b). m = 4; using the row-echelon form found in part (a), we can immediately find a basis for colspace(A):
{(1, 3, 1, 5), (3, 5, −1, 7)}.
⎡
⎤
1 −1
2
3
2 −4
3 ⎦ (Note: This is not row-echelon form, but it is not
9. (a). n = 4; we can reduce A to ⎣ 0
0
0
6 −13
necessary to bring the leading nonzero element in each row to 1.). Consequently, a basis for rowspace(A) is
{(1, −1, 2, 3), (0, 2, −4, 3), (0, 0, 6, −13)}.
(b). m = 3; using the row-echelon form found in part (a), we can immediately find a basis for colspace(A):
{(1, 1, 3), (−1, 1, 1), (2, −2, 4)}.
⎡
⎤
1 −1
2
1 ⎦. A row-echelon form of
10. (a). We determine a basis for the row space of the matrix ⎣ 5 −4
7 −5 −4
⎡
⎤
1 −1
2
1 −9 ⎦. Consequently, a basis for the subspace spanned by the given vectors is
this matrix is ⎣ 0
0
0
0
{(1, −1, 2), (0, 1, −9)}.
⎤
⎡
1
5
7
(b). We determine a basis for the column space of the matrix A = ⎣ −1 −4 −5 ⎦. A row-echelon form
2
1 −4
⎡
⎤
1 5 7
for this matrix is ⎣ 0 1 2 ⎦. Since the first two columns contain the pivots, we may take as a basis for
0 0 0
colspace(A), and hence as a basis of R3 spanned by the given set of vectors, the set {(1, −1, 2), (5, −4, 1)}.
⎤
⎡
1 3
3
⎢ 1 5 −1 ⎥
⎥. A row-echelon form of
11. (a). We determine a basis for the rowspace of the matrix ⎢
⎣ 2 7
4 ⎦
1 4
1
⎤
⎡
1 3
3
⎢ 0 1 −2 ⎥
⎥. Consequently, a basis for the subspace spanned by the given vectors is
this matrix is ⎢
⎣ 0 0
0 ⎦
0 0
0
{(1, 3, 3), (0, 1, −2)}.
⎡
⎤
1
1 2 1
5 7 4 ⎦. A row-echelon form
(b). We determine a basis for the column space of the matrix A = ⎣ 3
3 −1 4 1
⎡
⎤
1 1
2
1
for this matrix is ⎣ 0 1 1/2 1/2 ⎦. Since the first two columns contain the pivots, we may take as a basis
0 0
0
0
for colspace(A), and hence as a basis of R3 spanned by the given set of vectors, the set {(1, 3, 3), (1, 5, −1)}.
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⎡
⎤
1 1 −1
2
3 −4 ⎦. A row-echelon form of
12. (a). We determine a basis for the rowspace of the matrix ⎣ 2 1
1 2 −6 10
⎡
⎤
1 1 −1 2
this matrix is ⎣ 0 1 −5 8 ⎦. Consequently, a basis for the subspace spanned by the given vectors is
0 0
0 0
{(1, 1, −1, 2), (0, 1, −5, 8)}.
⎤
⎡
1
2
1
⎢ 1
1
2 ⎥
⎥. A row-echelon form
(b). We determine a basis for the column space of the matrix A = ⎢
⎣ −1
3 −6 ⎦
2 −4 10
⎤
⎡
1 2
1
⎢ 0 1 −1 ⎥
⎥. Since the first two columns contain the pivots, we may take as a basis for
⎢
for this matrix is ⎣
0 0
0 ⎦
0 0
0
colspace(A), and hence as a basis of R4 spanned by the given set of vectors, the set {(1, 1, −1, 2), (2, 1, 3, −4)}.
⎤
⎡
1 4 1 3
⎢ 2 8 3 5 ⎥
⎥
13. (a). We determine a basis for the rowspace of the matrix ⎢
⎣ 1 4 0 4 ⎦. A row-echelon form of
2 8 2 6
⎤
⎡
1 4 1
3
⎢ 0 0 1 −1 ⎥
⎥. Consequently, a basis for the subspace spanned by the given vectors is
this matrix is ⎢
⎣ 0 0 0
0 ⎦
0 0 0
0
{(1, 4, 1, 3), (0, 0, 1, −1)}.
⎤
⎡
1 2 1 2
⎢ 4 8 4 8 ⎥
⎥
(b). We determine a basis for the column space of the matrix A = ⎢
⎣ 1 3 0 2 ⎦. A row-echelon form for
3 5 4 6
⎡
⎤
1 2
1 2
⎢ 0 1 −1 0 ⎥
⎥. Since the first two columns contain the pivots, we may take as a basis for
this matrix is ⎢
⎣ 0 0
0 0 ⎦
0 0
0 0
colspace(A), and hence as a basis of R4 spanned by the given set of vectors, the set {(1, 4, 1, 3), (2, 8, 3, 5)}.
⎡
⎤
1 2 4
14. (a). A row-echelon form of A is ⎣ 0 1 1 ⎦. Consequently, a basis for rowspace(A) is
0 0 0
{(1, 2, 4), (0, 1, 1)}, whereas a basis for colspace(A) is {(1, 5, 3), (2, 11, 7)}.
(b). We see that both of these subspaces are two-dimensional, and therefore each corresponds geometrically
to a plane. By inspection, we see that the two basis vectors for rowspace(A) satisfy the equations 2x+y −z =
0, and therefore rowspace(A) corresponds to the plane with this equation. Similarly, we see that the two
basis vectors for colspace(A) satisfy the equations 2x − y + z = 0, and therefore colspace(A) corresponds to
the plane with this equations.
0 1
15. Many examples are possible here. One possibility is A =
. We see that rowspace(A) consists
0 0
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of all vectors of the form (0, y), where y ∈ R. However, colspace(A) consists of all vectors of the form (x, 0),
where x ∈ R. There are no nonzero vectors in common to both
and column space of A.
the row space
1
1
Here is a second example that the reader can verify: A =
. We have rowspace(A) = {(r, r) :
−1 −1
r ∈ R}, while colspace(A) = {(r, −r) : r ∈ R}. Thus, rowspace(A) and colspace(A) have no nonzero vectors
in common.
1 1
16. If A =
, colspace(A) is spanned by (1, 2), but if we permute the two rows of A, we obtain a
2 2
new matrix whose column space is spanned by (2, 1). On the other hand, if we multiply the first row by 2,
then we obtain a new matrix whose column space is spanned by (2, 2). And if we add −2 times the first row
to the second row, we obtain a new matrix whose column space is spanned by (1, 0). Therefore, in all cases,
colspace(A) is altered by the row operations performed.
17. Since colspace(A) is a subspace of Rm and nullspace(A) is a subspace of Rn , if colspace(A) =
nullspace(A), we must have m = n.
18. To the contrary, if A is invertible, then nullspace(A) = {0}. The assumption that rowspace(A) =
nullspace(A) therefore implies that rowspace(A) = {0}. In other words, A = 0n . But the zero matrix is not
invertible, so this is a contradiction. Hence, we conclude that A cannot be invertible.
Solutions to Section 4.9
True-False Review:
(a): FALSE. For example, consider the 7 × 3 zero matrix, 07×3 . We have rank(07×3 ) = 0, and therefore
by the Rank-Nullity Theorem, nullity(07×3 ) = 3. But |m − n| = |7 − 3| = 4. Many other examples can be
given. In particular, provided that m > 2n, the m × n zero matrix will show the falsity of the statement.
(b): FALSE. In this case, rowspace(A) is a subspace of R9 , hence it cannot possibly be equal to R7 . The
correct conclusion here should have been that rowspace(A) is a 7-dimensional subspace of R9 .
(c): TRUE. By the Rank-Nullity Theorem, rank(A) = 7, and therefore, rowspace(A) is a 7-dimensional
subspace of R7 . Hence, rowspace(A) = R7 .
0 1
(d): FALSE. For instance, the matrix A =
is an upper triangular matrix with two zeros appearing
0 0
on the main diagonal. However, since rank(A) = 1, we also have nullity(A) = 1.
(e): TRUE. An invertible matrix A must have nullspace(A) = {0}, but if colspace(A) is also {0}, then A
would be the zero matrix, which is certainly not invertible.
1 0
0 1
(f ): FALSE. For instance, if we take A =
and B =
, then nullity(A)+ nullity(B) =
0 0
0 0
1 1
1 + 1 = 2, but A + B =
, and nullity(A + B) = 1.
0 0
1 0
0 0
(g): FALSE. For instance, if we take A =
and B =
, then nullity(A)·nullity(B) = 1 · 1 =
0 0
0 1
1, but AB = 02 , and nullity(AB) = 2.
(h): TRUE. If x belongs to the nullspace of B, then Bx = 0. Therefore, (AB)x = A(Bx) = A0 = 0,
so that x also belongs to the nullspace of AB. Thus, nullspace(B) is a subspace of nullspace(AB). Hence,
nullity(B) ≤ nullity(AB), as claimed.
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(i): TRUE. If y belongs to nullspace(A), then Ay = 0. Hence, if Axp = b, then
A(y + xp ) = Ay + Axp = 0 + b = b,
which demonstrates that y + xp is also a solution to the linear system Ax = b.
Problems:
1. By inspection, we can see that nullspace(A) = {0}, the trivial subspace of R. Hence, nullity(A) = 0. We
also see by inspection that rank(A) = 1. Thus,
rank(A) + nullity(A) = 1 + 0 = 1 = n,
which verifies the Rank-Nullity Theorem.
2. The matrix is already in row-echelon form. A vector (x, y, z, w) in nullspace(A) must satisfy x−6z−w = 0.
We see that y,z, and w are free variables, and
⎫
⎧⎡
⎧⎡
⎤ ⎡
⎤
⎤⎫
⎤ ⎡
0
6z + w
1 ⎪
6
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎬
⎨⎢
⎨⎢
⎢ 0 ⎥ ⎢ 0 ⎥⎬
⎥
1 ⎥
y
⎥
⎢
⎥
⎢
⎢
⎥
⎥
⎢
nullspace(A) = ⎣
⎦ : y, z, w ∈ R⎪ = span ⎪⎣ 0 ⎦ , ⎣ 1 ⎦ , ⎣ 0 ⎦⎪ .
z
⎪
⎪
⎪
⎪
⎪
⎭
⎭
⎩
⎩
0
w
1
0
Therefore, nullity(A) = 3. Moreover, since this row-echelon form contains one nonzero row, rank(A) = 1.
Since the number of columns of A is 4 = 1 + 3, the Rank-Nullity Theorem is verified.
3. We bring A to row-echelon form:
REF(A) =
1 − 12
0
0
.
A vector (x, y) in nullspace(A) must satisfy x − 12 y = 0. Setting y = 2t, we get x = t. Therefore,
t
1
nullspace(A) =
: t ∈ R = span
.
2t
2
Therefore, nullity(A) = 1. Since REF(A) contains one nonzero row, rank(A) = 1. Since the number of
columns of A is 2 = 1 + 1, the Rank-Nullity Theorem is verified.
4. We bring A to row-echelon form:
⎡
1
REF(A) = ⎣ 0
0
⎤
1 −1
1
7 ⎦.
0
1
Since there are no unpivoted columns, there are no free variables in the associated homogeneous linear
system, and so
nullspace(A) = {0}.
Therefore, nullity(A) = 0. Since REF(A) contains three nonzero rows, rank(A) = 3. Since the number of
columns of A is 3 = 3 + 0, the Rank-Nullity Theorem is verified.
5. We bring A to row-echelon form:
⎤
1 4 −1 3
1 1 ⎦.
REF(A) = ⎣ 0 1
0 0
0 0
⎡
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A vector (x, y, z, w) in nullspace(A) must satisfy x+4y−z+3w = 0 and y+z+w = 0. We see that z and w are
free variables. Set z = t and w = s. Then y = −z−w = −t−s and x = z−4y−3w = t−4(−t−s)−3s = 5t+s.
Therefore,
⎧⎡
⎫
⎧⎡
⎤⎫
⎤
⎤ ⎡
1 ⎪
5t + s
5
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨⎢
⎬
⎨⎢
⎥⎬
⎥ ⎢
−t − s ⎥
⎥ : s, t ∈ R = span ⎢ −1 ⎥ , ⎢ −1 ⎥ .
nullspace(A) = ⎢
⎦
⎣
⎣ 1 ⎦ ⎣ 0 ⎦⎪
t
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎭
⎩
⎭
⎩
1
s
0
Therefore, nullity(A) = 2. Moreover, since REF(A) contains two nonzero rows, rank(A) = 2. Since the
number of columns of A is 4 = 2 + 2, the Rank-Nullity Theorem is verified.
6. Since all rows (or columns) of this matrix are proportional to the first one, rank(A) = 1. Since A has two
columns, we conclude from the Rank-Nullity Theorem that
nullity(A) = 2 − rank(A) = 2 − 1 = 1.
7. The first and last rows of A are not proportional, but the middle rows are proportional to the first row.
Therefore, rank(A) = 2. Since A has five columns, we conclude from the Rank-Nullity Theorem that
nullity(A) = 5 − rank(A) = 5 − 2 = 3.
8. Since the second and third columns are not proportional and the first column is all zeros, we have
rank(A) = 2. Since A has three columns, we conclude from the Rank-Nullity Theorem that
nullity(A) = 3 − rank(A) = 3 − 2 = 1.
9. This matrix (already in row-echelon form) has one nonzero row, so rank(A) = 1. Since it has four
columns, we conclude from the Rank-Nullity Theorem that
nullity(A) = 4 − rank(A) = 4 − 1 = 3.
⎡
1
10. The augmented matrix for this linear system is ⎣ 2
1
matrix to row-echelon form:
⎡
1 3 −1
⎣ 0 1 11
0 0
0
⎤
3 −1 4
7
9 11 ⎦. We quickly reduce this augmented
5 21 10
⎤
4
3 ⎦.
0
A solution (x, y, z) to the system will have a free variable corresponding to the third column: z = t. Then
y + 11t = 3, so y = 3 − 11t. Finally, x + 3y − z = 4, so x = 4 + t − 3(3 − 11t) = −5 + 34t. Thus, the solution
set is
⎧⎡
⎫ ⎧ ⎡
⎫
⎤
⎤ ⎡
⎤
34
−5
⎨ −5 + 34t
⎬ ⎨
⎬
⎣ 3 − 11t ⎦ : t ∈ R = t ⎣ −11 ⎦ + ⎣ 3 ⎦ : t ∈ R .
⎩
⎭ ⎩
⎭
t
1
0
⎡
⎤
⎡
⎤
−5
34
Observe that xp = ⎣ 3 ⎦ is a particular solution to Ax = b, and that ⎣ −11 ⎦ forms a basis for
0
1
nullspace(A). Therefore, the set of solution vectors obtained does indeed take the form (4.9.3).
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⎤
1 −1 2 3 6
11. The augmented matrix for this linear system is ⎣ 1 −2 5 5 13 ⎦. We quickly reduce this aug2 −1 1 4 5
mented matrix to row-echelon form:
⎤
⎡
6
1 −1
2
3
⎣ 0
1 −3 −2 −7 ⎦ .
0
0
0
0
0
⎡
A solution (x, y, z, w) to the system will have free variables corresponding to the third and fourth columns:
z = t and w = s. Then y − 3z − 2w = −7 requires that y = −7 + 3t + 2s, and x − y + 2z + 3w = 6 requires
that x = 6 + (−7 + 3t + 2s) − 2t − 3s = −1 + t − s. Thus, the solution set is
⎫
⎧⎡
⎫ ⎧ ⎡
⎤
⎤ ⎡
⎤
⎡
⎤
−1
−1
1
−1 + t − s
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎬
⎨⎢
⎬ ⎨ ⎢
⎥
⎥ ⎢
⎥
⎢
⎥
⎢ −7 + 3t + 2s ⎥ : s, t ∈ R = t ⎢ 3 ⎥ + s ⎢ 2 ⎥ + ⎢ −7 ⎥ : s, t ∈ R .
⎣ 0 ⎦ ⎣ 0 ⎦
⎦
⎣
t
⎪
⎪ ⎣ 1 ⎦
⎪
⎪
⎪
⎪
⎪
⎭
⎩
⎩
⎭ ⎪
0
1
0
s
⎤
−1
⎢ −7 ⎥
⎥
Observe that xp = ⎢
⎣ 0 ⎦ is a particular solution to Ax = b, and that
0
⎡
⎧⎡
⎤⎫
⎤ ⎡
−1 ⎪
1
⎪
⎪
⎪
⎨⎢
⎥⎬
⎥ ⎢
⎢ 3 ⎥,⎢ 2 ⎥
⎦ ⎣ 0 ⎦⎪
⎣
⎪
⎪
⎪ 1
⎭
⎩
1
0
is a basis for nullspace(A). Therefore, the set of solution vectors obtained does indeed take the form (4.9.3).
⎡
⎤
1
1 −2 −3
⎢ 3 −1 −7
2 ⎥
⎥. We quickly reduce this augmented
12. The augmented matrix for this linear system is ⎢
⎣ 1
1
1
0 ⎦
2
2 −4 −6
matrix to row-echelon form:
⎤
⎡
−3
1 1 −2
1
⎦.
⎣ 0 1
− 11
4
4
0 0
1
1
There are no free variables in the solution
⎧⎡set (since
⎤⎫ none of the first three columns is unpivoted), and we find
2 ⎬
⎨
the solution set by back-substitution: ⎣ −3 ⎦ . It is easy to see that this is indeed a particular solution:
⎩
⎭
1
⎡
⎤
2
xp = ⎣ −3 ⎦. Since the row-echelon form of A has three nonzero rows, rank(A) = 3. Thus, nullity(A) = 0.
1
Hence, nullspace(A) = {0}. Thus, the only term in the expression (4.9.3) that appears in the solution is xp ,
and this is precisely the unique solution we obtained in the calculations above.
13. By inspection, we see that a particular solution to this (homogeneous) linear system is xp = 0. We
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quickly reduce this augmented matrix to row-echelon form:
⎤
⎡
1 1 −1 5 0
⎣ 0 1 −1 7 0 ⎦ .
2
2
0 0
0 0 0
A solution (x, y, z, w) to the system will have free variables corresponding to the third and fourth columns:
z = t and w = s. Then y − 12 z + 72 w = 0 requires that y = 12 t − 72 s, and x + y − z + 5w = 0 requires that
x = t − ( 12 t − 72 s) − 5s = 12 t − 32 s. Thus, the solution set is
⎫ ⎧ ⎡ 1 ⎤
⎫
⎧⎡ 1
⎤
⎡ 3 ⎤
−2
t − 32 s
⎪
⎪
⎪
⎪
2
2
⎪
⎪
⎪
⎪
⎬ ⎨ ⎢ 1 ⎥
⎬
⎨⎢ 1
7
⎥
⎢ −7 ⎥
t
−
s
2
⎢ 2
⎥ : s, t ∈ R = t ⎢ 2 ⎥ + s ⎢ 2 ⎥ : s, t ∈ R .
⎣
⎦
⎣ 0 ⎦
t
⎪ ⎣ 1 ⎦
⎪
⎪
⎪
⎪
⎪
⎪
⎩
⎭ ⎪
⎭
⎩
s
0
1
⎡ 1 ⎤
⎡ 3 ⎤
−2
2
⎢ 1 ⎥
⎢ −7 ⎥
2 ⎥
⎢ 2 ⎥
Since the vectors ⎢
⎣ 1 ⎦ and ⎣ 0 ⎦ form a basis for nullspace(A), our solutions to take the proper form
0
1
given in (4.9.3).
14. By the Rank-Nullity Theorem, rank(A) = 7 − nullity(A) = 7 − 4 = 3, and hence, colspace(A) is
3-dimensional. But since A has three rows, colspace(A) is a subspace of R3 . Therefore, since the only
3-dimensional subspace of R3 is R3 itself, we conclude that colspace(A) = R3 . Now rowspace(A) is also
3-dimensional, but it is a subspace of R5 . Therefore, it is not accurate to say that rowspace(A) = R3 .
15. By the Rank-Nullity Theorem, rank(A) = 4 − nullity(A) = 4 − 0 = 4, so we conclude that rowspace(A) is
4-dimensional. Since rowspace(A) is a subspace of R4 (since A contains four columns), and it is 4-dimensional,
we conclude that rowspace(A) = R4 . Although colspace(A) is 4-dimensional, colspace(A) is a subspace of
R6 , and therefore it is not accurate to say that colspace(A) = R4 .
16. If rowspace(A) = nullspace(A), then we know that rank(A) = nullity(A). Therefore, rank(A)+ nullity(A)
must be even. But rank(A)+ nullity(A) is the number of columns of A. Therefore, A contains an even number
of columns.
17. We know that rank(A) + nullity(A) = 7. But since A only has five rows, rank(A) ≤ 5. Therefore, nullity(A) ≥ 2. However, since nullspace(A) is a subspace of R7 , nullity(A) ≤ 7. Therefore, 2 ≤
nullity(A) ≤ 7. There are
⎤ many examples of a 5 × 7 matrix A with nullity(A) = 2; one example is
⎡
1 0 0 0 0 0 0
⎢ 0 1 0 0 0 0 0 ⎥
⎥
⎢
⎢ 0 0 1 0 0 0 0 ⎥. The only 5 × 7 matrix with nullity(A) = 7 is 05×7 , the 5 × 7 zero matrix.
⎥
⎢
⎣ 0 0 0 1 0 0 0 ⎦
0 0 0 0 1 0 0
18. We know that rank(A) + nullity(A) = 8. But since A only has three rows, rank(A) ≤ 3. Therefore, nullity(A) ≥ 5. However, since nullspace(A) is a subspace of R8 , nullity(A) ≤ 8. Therefore, 5 ≤
nullspace(A) ≤ 8. There are
⎡
⎤ many examples of a 3 × 8 matrix A with nullity(A) = 5; one example is
1 0 0 0 0 0 0 0
⎣ 0 1 0 0 0 0 0 0 ⎦. The only 3 × 8 matrix with nullity(A) = 8 is 03×8 , the 3 × 8 zero matrix.
0 0 1 0 0 0 0 0
19. If Bx = 0, then ABx = A0 = 0. This observation shows that nullspace(B) is a subspace of
nullspace(AB). On the other hand, if ABx = 0, then Bx = (A−1 A)Bx = A−1 (AB)x = A−1 0 = 0, so
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Bx = 0. Therefore, nullspace(AB) is a subspace of nullspace(B). As a result, since nullspace(B) and
nullspace(AB) are subspaces of each other, they must be equal: nullspace(AB) = nullspace(B). Therefore,
nullity(AB) = nullity(B).
Solutions to Section 4.10
True-False Review:
(a): TRUE. This is the equivalence of statements (a) and (m) in the Invertible Matrix Theorem.
1 2 3
(b): TRUE. Here is an example to show that this is possible: A =
. The two rows of A are
1 2 4
non-proportional, so they form a linearly independent set. However, the three columns of A include two
identical columns, so of course the columns for a linearly dependent set of vectors.
(c): FALSE. If the matrix has n linearly independent rows, then by the equivalence of (a) and (j) in the
Invertible Matrix Theorem, such a matrix would be invertible. But if that were so, then by part (m) of the
Invertible Matrix Theorem, such a matrix would have to have n linearly independent columns.
(d): FALSE. An n × n matrix A with det(A) = 0 is not invertible by part (g) of the Invertible Matrix
Theorem. Therefore, by the equivalence of (a) and (l) in the Invertible Matrix Theorem, the columns of A
do not form a basis for Rn .
(e): TRUE. If rowspace(A) = Rn , then by the equivalence of (a) and (n) in the Invertible Matrix Theorem,
A is not invertible. Therefore, A is not row-equivalent to the identity matrix. Since B is row-equivalent to
A, then B is not row-equivalent to the identity matrix, and therefore, B is not invertible. Hence, by part
(k) of the Invertible Matrix Theorem, we conclude that colspace(B) = Rn .
(f ): FALSE. If nullspace(A) = {0}, then A is invertible by the equivalence of (a) and (c) in the Invertible
Matrix Theorem. Since E is an elementary matrix, it is also invertible. Therefore, EA is invertible. By part
(g) of the Invertible Matrix Theorem, det(EA) = 0, contrary to the statement given.
(g): FALSE. The matrix [A|B] has 2n columns, but only n rows, and therefore rank([A|B]) ≤ n. Hence,
by the Rank-Nullity Theorem, nullity([A|B]) ≥ n > 0.
(h): TRUE. The first and third rows are proportional, and hence statement (m) in the Invertible Matrix
Theorem is false. Therefore, by the equivalence with (a), the matrix is not invertible.
⎤
⎡
0 1 0 0
⎢ 1 0 0 0 ⎥
⎥
(i): FALSE. For instance, the matrix ⎢
⎣ 0 0 0 1 ⎦ is of the form given, but satisfies any (and all) of
0 0 1 0
the statements of the Invertible Matrix Theorem.
⎡
⎤
1 2 1
(j): FALSE. For instance, the matrix ⎣ 3 6 2 ⎦ is of the form given, but has a nonzero determinant,
1 0 0
and so by part (g) of the Invertible Matrix Theorem, it is invertible.
Solutions to Section 4.11
Problems:
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1. Write v = (a1 , a2 , a3 , a4 , a5 ) ∈ R5 . Then we have
(r + s)v = (r + s)(a1 , a2 , a3 , a4 , a5 )
= ((r + s)a1 , (r + s)a2 , (r + s)a3 , (r + s)a4 , (r + s)a5 )
= (ra1 + sa1 , ra2 + sa2 , ra3 + sa3 , ra4 + sa4 , ra5 + sa5 )
= (ra1 , ra2 , ra3 , ra4 , ra5 ) + (sa1 , sa2 , sa3 , sa4 , sa5 )
= r(a1 , a2 , a3 , a4 , a5 ) + s(a1 , a2 , a3 , a4 , a5 )
= rv + sv.
2. Write v = (a1 , a2 , a3 , a4 , a5 ) and w = (b1 , b2 , b3 , b4 , b5 ) in R5 . Then we have
r(v + w) = r((a1 , a2 , a3 , a4 , a5 ) + (b1 , b2 , b3 , b4 , b5 ))
= r(a1 + b1 , a2 + b2 , a3 + b3 , a4 + b4 , a5 + b5 )
= (r(a1 + b1 ), r(a2 + b2 ), r(a3 + b3 ), r(a4 + b4 ), r(a5 + b5 ))
= (ra1 + rb1 , ra2 + rb2 , ra3 + rb3 , ra4 + rb4 , ra5 + rb5 )
= (ra1 , ra2 , ra3 , ra4 , ra5 ) + (rb1 , rb2 , rb3 , rb4 , rb5 )
= r(a1 , a2 , a3 , a4 , a5 ) + r(b1 , b2 , b3 , b4 , b5 )
= rv + rw.
3. YES. To show that V forms a vector space, we will show that it is a subspace of P2 (R) by appealing to
Theorem 4.3.2. Note that V is non-empty since the polynomial p(x) = 0 belongs to V . Next, we show that
V is closed under addition and scalar multiplication:
Closure under Addition: Suppose p(x) and q(x) belong to V . Then we have p(3) = q(3) = 0 and
p (5) = q (5) = 0. Hence,
(p + q)(3) = p(3) + q(3) = 0 + 0 = 0
and
(p + q) (5) = p (5) + q (5) = 0 + 0 = 0.
Therefore, p(x) + q(x) belongs to V .
Closure under Scalar Multiplication: Let k be a scalar and let p(x) be as above. Then
(kp)(3) = k · p(3) = k · 0 = 0
and
(kp) (5) = k · p (5) = k · 0 = 0.
Therefore, k · p(x) belongs to V .
4. NO. This set of polynomials is not closed under scalar multiplication. For example, the polynomial
p(x) = 2x belongs to the set, but 13 p(x) = 23 x does not belong to the set (since 23 is not an even integer).
5. YES. This set of polynomials forms a subspace of the vector space P5 (R). To confirm this, we will check
that this set is closed under addition and scalar multiplication:
Closure under Addition: Let p(x) = a0 + a1 x + a4 x4 + a5 x5 and q(x) = b0 + b1 x + b4 x4 + b5 x5 be polynomials
in the set under consideration (their x2 and x3 terms are zero). Then
p(x) + q(x) = (a0 + b0 ) + (a1 + b1 )x + · · · + (a4 + b4 )x4 + (a5 + b5 )x5
is again in the set (since it still has no x2 or x3 terms). So closure under addition holds.
Closure under Scalar Multiplication: Let p(x) = a0 + a1 x + a4 x4 + a5 x5 be in the set, and let k be a scalar.
Then
kp(x) = (ka0 ) + (ka1 )x + (ka4 )x4 + (ka5 )x5 ,
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which is again in the set (since it still has no x2 or x3 terms). So closure under scalar multiplication holds.
6. NO. We can see immediately that the zero vector (0, 0, 0) is not a solution to this linear system (the first
equation is not satisfied by the zero vector), and therefore, we know at once that this set cannot be a vector
space.
7. YES. The set of solutions to this linear system forms a subspace of R3 . To confirm this, we will check
that this set is closed under addition and scalar multiplication:
Closure under Addition: Let (a1 , a2 , a3 ) and (b1 , b2 , b3 ) be solutions to the linear system. This means that
4a1 − 7a2 + 2a3 = 0,
5a1 − 2a2 + 9a3 = 0
4b1 − 7b2 + 2b3 = 0,
5b1 − 2b2 + 9b3 = 0.
and
Adding the equations on the left, we get
4(a1 + b1 ) − 7(a2 + b2 ) + 2(a3 + b3 ) = 0,
so the vector (a1 + b1 , a2 + b2 , a3 + b3 ) satisfies the first equation in the linear system. Likewise, adding the
equations on the right, we get
5(a1 + b1 ) − 2(a2 + b2 ) + 9(a3 + b3 ) = 0,
so (a1 + b1 , a2 + b2 , a3 + b3 ) also satisfies the second equation in the linear system. Therefore, (a1 + b1 , a2 +
b2 , a3 + b3 ) is in the solution set for the linear system, and closure under addition therefore holds.
Closure under Scalar Multiplication: Let (a1 , a2 , a3 ) be a solution to the linear system, and let k be a scalar.
We have
4a1 − 7a2 + 2a3 = 0,
5a1 − 2a2 + 9a3 = 0,
and so, multiplying both equations by k, we have
k(4a1 − 7a2 + 2a3 ) = 0,
k(5a1 − 2a2 + 9a3 ) = 0,
or
4(ka1 ) − 7(ka2 ) + 2(ka3 ) = 0,
5(ka1 ) − 2(ka2 ) + 9(ka3 ) = 0.
Thus, the vector (ka1 , ka2 , ka3 ) is a solution to the linear system, and closure under scalar multiplication
therefore holds.
1 1
−1 1
8. NO. This set is not closed under addition. For example, the vectors
and
both
1 1
1 1
belong to the set (their entries are all nonzero), but
1 1
−1 1
0 2
+
=
,
1 1
1 1
2 2
which does not belong to the set (some entries are zero, and some are nonzero). So closure under addition
fails, and therefore, this set does not form a vector space.
1 2
9. YES. The set of 2 × 2 real matrices that commute with C =
forms a subspace of M2 (R). To
2 2
confirm this, we will check that this set is closed under addition and scalar multiplication:
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Closure under Addition: Let A and B be 2 × 2 real matrices that commute with C. That is, AC = CA and
BC = CB. Then
(A + B)C = AC + BC = CA + CB = C(A + B),
so A + B commutes with C, and therefore, closure under addition holds.
Closure under Scalar Multiplication: Let A be a 2 × 2 real matrix that commutes with C, and let k be a
scalar. Then since AC = CA, we have
(kA)C = k(AC) = k(CA) = C(kA),
so kA is still in the set. Thus, the set is also closed under scalar multiplication.
10. YES. The set of functions f : [0, 1] → [0, 1] such that f (0) = f 41 = f 12 = f 34 = f (1) = 0 is
a subspace of the vector space of all functions [0, 1] → [0, 1]. We confirm this by checking that this set is
closed under addition and scalar multiplication:
Closure under Addition: Let g and h be functions such that
1
1
3
g(0) = g
=g
=g
= g(1) = 0
4
2
4
and
1
1
3
h(0) = h
=h
=h
= h(1) = 0.
4
2
4
Now
(g + h)(0) = g(0) + h(0) = 0 + 0 = 0,
1
1
1
=g
+h
= 0 + 0 = 0,
(g + h)
4
4
4
1
1
1
=g
+h
= 0 + 0 = 0,
(g + h)
2
2
2
3
3
3
=g
+h
= 0 + 0 = 0,
(g + h)
4
4
4
(g + h)(1) = g(1) + h(1) = 0 + 0 = 0.
Thus, g + h belongs to the set, and so the set is closed under addition.
Closure under Scalar Multiplication: Let g be a function with
1
1
3
g(0) = g
=g
=g
= g(1) = 0,
4
2
4
and let k be a scalar. Then
(kg)(0) = k · g(0) = k · 0 = 0,
1
1
(kg)
=k·g
= k · 0 = 0,
4
4
1
1
=k·g
= k · 0 = 0,
(kg)
2
2
3
3
=k·g
= k · 0 = 0,
(kg)
4
4
(kg)(1) = k · g(1) = k · 0 = 0.
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Thus, kg belongs to the set, and so the set is closed under scalar multiplication.
11. NO. This set is not closed under addition. For example, the function f defined by f (x) = x belongs to
the set, and the function g defined by
x, if x ≤ 1/2
g(x) =
0, if x > 1/2
belongs to the set. But
(f + g)(x) =
2x,
x,
if x ≤ 1/2
if x > 1/2.
Then f + g : [0, 1] → [0, 1], but for 0 < x ≤ 12 , |(f + g)(x)| = 2x > x, so f + g is not in the set. Therefore,
the set is not closed under addition.
12. NO. This set is not closed under addition. For example, if we let
1 0
0 1
A=
and
B=
,
0 1
0 0
then A2 = A is symmetric, and B 2 = 02 is symmetric, but
2 1 1
1
=
(A + B)2 =
0 1
0
2
1
is not symmetric, so A + B is not in the set. Thus, the set in question is not closed under addition.
13. We must check each of the vector space axioms (A1)-(A10).
Axiom (A1): Assume that (a1 , a2 ) and (b1 , b2 ) belong to V . Then a2 , b2 > 0. Hence,
(a1 , a2 ) + (b1 , b2 ) = (a1 + b1 , a2 b2 ) ∈ V,
since a2 b2 > 0. Thus, V is closed under addition.
Axiom (A2): Assume that (a1 , a2 ) ∈ V , and let k be a scalar. Note that since a2 > 0, the expression
ak2 > 0 for every k ∈ R. Hence, k(a1 , a2 ) = (ka1 , ak2 ) ∈ V , thereby showing that V is closed under scalar
multiplication.
Axiom (A3): Let (a1 , a2 ), (b1 , b2 ) ∈ V . We have
(a1 , a2 ) + (b1 , b2 ) = (a1 + b1 , a2 b2 )
= (b1 + a1 , b2 a2 )
= (b1 , b2 ) + (a1 , a2 ),
as required.
Axiom (A4): Let (a1 , a2 ), (b1 , b2 ), (c1 , c2 ) ∈ V . We have
((a1 , a2 ) + (b1 , b2 )) + (c1 , c2 ) = (a1 + b1 , a2 b2 ) + (c1 , c2 )
= ((a1 + b1 ) + c1 , (a2 b2 )c2 )
= (a1 + (b1 + c1 ), a2 (b2 c2 ))
= (a1 , a2 ) + (b1 + c1 , b2 c2 )
= (a1 , a2 ) + ((b1 , b2 ) + (c1 , c2 )),
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as required.
Axiom (A5): We claim that (0, 1) is the zero vector in V . To see this, let (b1 , b2 ) ∈ V . Then
(0, 1) + (b1 , b2 ) = (0 + b1 , 1 · b2 ) = (b1 , b2 ).
Since this holds for every (b1 , b2 ) ∈ V , we conclude that (0, 1) is the zero vector.
Axiom (A6): We claim that the additive inverse of a vector (a1 , a2 ) in V is the vector (−a1 , a−1
2 ) (Note that
a−1
>
0
since
a
>
0.)
To
check
this,
we
compute
as
follows:
2
2
−1
(a1 , a2 ) + (−a1 , a−1
2 ) = (a1 + (−a1 ), a2 a2 ) = (0, 1).
Axiom (A7): We have
1 · (a1 , a2 ) = (a1 , a12 ) = (a1 , a2 )
for all (a1 , a2 ) ∈ V .
Axiom (A8): Let (a1 , a2 ) ∈ V , and let r and s be scalars. Then we have
(rs)(a1 , a2 ) = ((rs)a1 , ars
2 )
= (r(sa1 ), (as2 )r )
= r(sa1 , as2 )
= r(s(a1 , a2 )),
as required.
Axiom (A9): Let (a1 , a2 ) and (b1 , b2 ) be members of V , and let r be a scalar. We have
r((a1 , a2 ) + (b1 , b2 )) = r(a1 + b1 , a2 b2 )
= (r(a1 + b1 ), (a2 b2 )r )
= (ra1 + rb1 , ar2 br2 )
= (ra1 , ar2 ) + (rb1 , br2 )
= r(a1 , a2 ) + r(b1 , b2 ),
as required.
Axiom (A10): Let (a1 , a2 ) ∈ V , and let r and s be scalars. We have
(r + s)(a1 , a2 ) = ((r + s)a1 , ar+s
2 )
= (ra1 + sa1 , ar2 as2 )
= (ra1 , ar2 ) + (sa1 , as2 )
= r(a1 , a2 ) + s(a1 , a2 ),
as required.
14. We must show that S is closed under addition and closed under scalar multiplication:
Closure under Addition: Let (a, 2a ) and (b, 2b ) be elements of S. Now consider the sum of these elements:
(a, 2a ) + (b, 2b ) = (a + b, 2a 2b ) = (a + b, 2a+b ) ∈ S,
which shows that S is closed under addition.
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Closure under Scalar Multiplication: Let (a, 2a ) be an element of S, and let k be a scalar. Then
k(a, 2a ) = (ka, (2a )k ) = (ka, 2ka ) ∈ S,
which shows that S is closed under scalar multiplication.
Thus, S is a subspace of V .
15. Note that 3(1, 2) = (3, 23 ) = (3, 8), so the second vector is a multiple of the first one under the vector
space operations of V from Problem 13. Therefore, {(1, 2), (3, 8)} is linearly dependent.
16. We show that S = {(1, 4), (2, 1)} is linearly independent and spans V .
S is linearly independent: Assume that
c1 (1, 4) + c2 (2, 1) = (0, 1).
This can be written
(c1 , 4c1 ) + (2c2 , 1c2 ) = (0, 1)
or
(c1 + 2c2 , 4c1 ) = (0, 1).
In order for 4c1 = 1, we must have c1 = 0. And then in order for c1 + 2c2 = 0, we must have c2 = 0.
Therefore, S is linearly independent.
S spans V : Consider an arbitrary vector (a1 , a2 ) ∈ V , where a2 > 0. We must find constants c1 and c2 such
that
c1 (1, 4) + c2 (2, 1) = (a1 , a2 ).
Thus,
(c1 , 4c1 ) + (2c2 , 1c2 ) = (a1 , a2 )
or
(c1 + 2c2 , 4c1 ) = (a1 , a2 ).
Hence,
c1 + 2c2 = a1
and
4c 1 = a 2 .
From the second equation, we conclude that
c1 = log4 (a2 ).
Thus, from the first equation,
1
(a1 − log4 (a2 )).
2
Hence, since we were able to find constants c1 and c2 in order that
c2 =
c1 (1, 4) + c2 (2, 1) = (a1 , a2 ),
we conclude that {(1, 4), (2, 1)} spans V .
17. This really hinges on whether or not the given vectors are linearly dependent or linearly independent.
If we assume that
c1 (2 + x2 ) + c2 (4 − 2x + 3x2 ) + c3 (1 + x) = 0,
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then
(2c1 + 4c2 + c3 ) + (−2c2 + c3 )x + (c1 + 3c2 )x2 = 0.
Thus, we have
− 2c2 + c3 = 0
2c1 + 4c2 + c3 = 0
Since the matrix of coefficients
c1 + 3c2 = 0.
⎡
⎤
2
4 1
⎣ 0 −2 1 ⎦
1
3 0
fails to be invertible, we conclude that there will be non-trivial solutions for c1 , c2 , and c3 . Thus, the
polynomials are linearly dependent. We can therefore remove a vector from the set without decreasing the
span. We remove 1 + x, leaving us with {2 + x2 , 4 − 2x + 3x2 }. Since these polynomials are not proportional,
they are now linearly independent, and hence, they are a basis for their span. Hence,
dim{2 + x2 , 4 − 2x + 3x2 , 1 + x} = 2.
18. NO. This set is not closed under scalar multiplication. For example, (1, 1) belongs to S, but 2 · (1, 1) =
(2, 2) does not belong to S.
19. NO. This set is not closed under scalar multiplication. For example, (1, 1) belongs to S, but 2 · (1, 1) =
(2, 2) does not belong to S.
20. NO. The zero vector (zero matrix) is not an orthogonal matrix. Any subspace must contain a zero
vector.
21. YES. We show that S is closed under addition and closed under scalar multiplication.
Closure under Addition: Assume that f and g belong to S. Thus, f (a) = 2f (b) and g(a) = 2g(b). We must
show that f + g belongs to S. We have
(f + g)(a) = f (a) + g(a) = 2f (b) + 2g(b) = 2[f (b) + g(b)] = 2(f + g)(b),
so f + g ∈ S. So S is closed under addition.
Closure under Scalar Multiplication: Assume that f belongs to S and k is a scalar. Thus, f (a) = 2f (b).
Moreover,
(kf )(a) = kf (a) = k(2f (b)) = 2(kf (b)) = 2(kf )(b),
so kf ∈ S. Thus, S is closed under scalar multiplication.
22. YES. We show that S is closed under addition and closed under scalar multiplication.
Closure under Addition: Assume that f and g belong to S. Thus,
b
b
f (x)dx = 0
and
g(x)dx = 0.
a
Hence,
a
b
b
(f + g)(x)dx =
a
b
f (x)dx +
a
g(x)dx = 0 + 0 = 0,
a
so f + g ∈ S.
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Closure under Scalar Multiplication: Assume that f belongs to S and k is a scalar. Thus,
b
f (x)dx = 0.
a
Therefore,
b
b
(kf )(x)dx =
a
b
f (x)dx = k · 0 = 0,
kf (x)dx = k
a
a
so kf ∈ S.
23. YES. We show that S is closed under addition and scalar multiplication.
Closure under Addition: Assume that
⎡
a1
⎣ c1
e1
⎤ ⎡
b1
a2
d1 ⎦ , ⎣ c 2
f1
e2
⎤
b2
d2 ⎦ ∈ S.
f2
Thus,
a 1 + b1 = c 1 + f 1 ,
a1 − c 1 = e 1 − f 1 − d 1 ,
a 2 + b2 = c 2 + f 2 ,
a2 − c 2 = e 2 − f 2 − d 2 .
Hence,
(a1 + a2 ) − (c1 + c2 ) = (e1 + e2 ) − (f1 + f2 ) − (d1 + d2 ).
(a1 + a2 ) + (b1 + b2 ) = (c1 + c2 ) + (f1 + f2 )
and
Thus,
⎤ ⎡
b2
a1 + a2
d2 ⎦ = ⎣ c 1 + c 2
f2
e1 + e2
⎡
a1
⎣ c1
e1
⎤ ⎡
b1
a2
d1 ⎦ + ⎣ c 2
f1
e2
⎤
b 1 + b2
d1 + d2 ⎦ ∈ S,
f1 + f2
which means S is closed under addition.
Closure under Scalar Multiplication: Assume that
⎡
a
⎣ c
e
⎤
b
d ⎦∈S
f
and k is a scalar. Then we have a + b = c + f and a − c = e − f − d. Thus, ka + kb = kc + kf and
ka − kc = ke − kf − kd. Therefore,
⎤ ⎡
⎡
⎤
a b
ka kb
k ⎣ c d ⎦ = ⎣ kc kd ⎦ ∈ S,
e f
ke kf
so S is closed under scalar multiplication.
24. (a) NO, (b) YES. Since 3 vectors are required to span R3 , S cannot span V . However, since the
vectors are not proportional, they are linearly independent.
⎡
⎤
6 −3
2
1
1 ⎦,
25. (a) YES, (b) YES. If we place the three vectors into the columns of a 3 × 3 matrix ⎣ 1
1 −8 −1
we observe that the matrix is invertible. Hence, its columns are linearly independent. Since we have 3
linearly independent vectors in the 3-dimensional space R3 , we have a basis for R3 .
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26. (a) NO, (b) YES. Since we have only 3 vectors in a 4-dimensional vector space, they cannot possibly
span R4 . To check linear independence, we place the vectors into the columns of a matrix:
⎤
⎡
6 1
1
⎢ −3 1 −8 ⎥
⎥
⎢
⎣ 2 1 −1 ⎦ .
0 0
0
The
rows for the same invertible matrix as in the previous problem, so the reduced row-echelon form
⎡ first three ⎤
1 0 0
⎢ 0 1 0 ⎥
⎥
is ⎢
⎣ 0 0 1 ⎦, so there are no free variables, and hence, the column vectors form a linearly independent
0 0 0
set.
27. (a) YES, (b) NO. Since (0, 0, 0) is a member of S, S cannot be linearly independent. However, the set
{(10, −6, 5), (3, −3, 2), (6, 4, −1)} is linearly independent (these vectors can be made to form a 3 × 3 matrix
that is invertible), and thus, S must span at least a 3-dimensional space. Since dim[R3 ] = 3, we know that
S spans R3 .
28. (a) YES, (b) YES. Consider the linear equation
c1 (2x − x3 ) + c2 (1 + x + x2 ) + 3c3 + c4 x = 0.
Then
(c2 + 3c3 ) + (2c1 + c2 + c4 )x + c2 x2 − c1 x3 = 0.
From the latter equation, we see that c1 = c2 = 0 (looking at the x2 and x3 coefficients) and thus, c3 = c4 = 0
(looking at the constant term and x coefficient). Thus, c1 = c2 = c3 = c4 = 0, and hence, S is linearly
independent. Since we have four linearly independent vectors in the 4-dimensional vector space P3 (R), we
conclude that these vectors also span P3 (R).
29. (a) NO, (b) NO. The set S only contains four vectors, although dim[P4 (R)] = 5, so it is impossible
for S to span P4 (R). Alternatively, simply note that none of the polynomials in S contain an x3 term.
To check linear independence, consider the equation
c1 (x4 + x + 2 + 1) + c2 (x2 + x + 1) + c3 (x + 1) + c4 (x4 + 2x + 3) = 0.
Rearranging this, we have
(c1 + c4 )x4 + (c1 + c2 )x2 + (c2 + c3 + 2c4 )x + (c1 + c2 + 3c4 ) = 0,
and so we look to solve the linear system with augmented matrix
⎡
⎤ ⎡
⎤ ⎡
⎤
1 1 1 3 0
1
1
1
3 0
1 1 1 3 0
⎢ 0 1 1 2 0 ⎥ ⎢ 0
⎢
⎥
1
1
2 0 ⎥
⎢
⎥∼⎢
⎥ = ⎢ 0 1 1 2 0 ⎥,
⎣ 1 1 0 0 0 ⎦ ⎣ 0
0 −1 −3 0 ⎦ ⎣ 0 0 1 3 0 ⎦
1 0 0 1 0
0 −1 −1 −2 0
0 0 0 0 0
where we have added −1 times the first row to each of the third and fourth rows in the first step, and zeroed
at the last row in the second step. We see that c4 is a free variable, which means that a nontrivial solution
to the linear system exists, and therefore, the original vectors are linearly dependent.
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30. (a) NO, (b) YES. The vector space M2×3 (R) is 6-dimensional, and since only four vectors belong
to S, S cannot possibly span M2×3 (R). On the other hand, if we form the linear system with augmented
matrix
⎡
⎤
−1 3 −1 −11 0
⎢ 0 2 −2 −6 0 ⎥
⎢
⎥
⎢ 0 1 −3 −5 0 ⎥
⎢
⎥,
⎢ 0 1
3
1 0 ⎥
⎢
⎥
⎣ 1 2
2 −2 0 ⎦
1 3
1 −5 0
row reduction shows that each column of a row-echelon form contains a pivot, and therefore, the vectors are
linearly independent.
31. (a) NO, (b) NO. Since M2 (R) is only 4-dimensional and S contains 5 vectors, S cannot possibly be
linearly independent. Moreover, each matrix in S is symmetric, and therefore, only symmetric matrices can
be found in the span(S). Thus, S fails to span M2 (R).
32. Assume that {v1 , v2 , v3 } is linearly independent, and that v4 does not lie in span{v1 , v2 , v3 }. We will
show that {v1 , v2 , v3 , v4 } is linearly independent. To do this, assume that
c1 v1 + c2 v2 + c3 v3 + c4 v4 = 0.
We must prove that c1 = c2 = c3 = c4 = 0.
If c4 = 0, then we rearrange the above equation to show that
v4 = −
c1
c2
c3
v1 − v2 − v3 ,
c4
c4
c4
which implies that v4 ∈ span{v1 , v2 , v3 }, contrary to our assumption. Therefore, we know that c4 = 0.
Hence the equation above reduces to
c1 v1 + c2 v2 + c3 v3 = 0,
and the linear independence of {v1 , v2 , v3 } now implies that c1 = c2 = c3 = 0. Therefore, c1 = c2 = c3 =
c4 = 0, as required.
33. Write A = [a1 , a2 , . . . , an ].
We first assume that AT A is invertible, and we show that {a1 , a2 , . . . , an } is linearly independent. Suppose
that
c1 a1 + c2 a2 + · · · + cn an = 0.
⎡
⎤
c1
⎢ c2 ⎥
⎢
⎥
We can write this as Ac = 0, where c = ⎢ . ⎥. Left multiplying both sides of Ac = 0 by AT , we obtain
⎣ .. ⎦
cn
AT Ac = AT 0 = 0.
From the assumption that AT A is invertible, it therefore follows at once that c = 0. That is, c1 = c2 =
· · · = cn = 0, which shows that {a1 , a2 , . . . , an } is linearly independent.
Conversely, assume that the columns of A are linearly independent. To show that AT A is invertible, we
will show that nullspace(AT A) = {0}. To do this, suppose that AT Ax = 0. We will show that x = 0. To
do this, left multiply both sides of AT Ax = 0 by xT to get
xT AT Ax = xT 0 = 0,
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or
(Ax)T (Ax) = 0.
This means that the sum⎡ of the
⎤ squares of the entries of the column vector Ax is zero. This means that
x1
⎢ x2 ⎥
⎢
⎥
Ax = 0. If we write x = ⎢ . ⎥, then
⎣ .. ⎦
xn
x1 a1 + x2 a2 + · · · + xn an = 0.
Since the columns of A are linearly independent, we conclude that x1 = x2 = · · · = xn = 0. Thus, x = 0, as
desired.
34. (a). Clearly, S is a nonempty subset of M4 (R), so we can apply Theorem 4.3.2 and verify that S is
closed under addition and scalar multiplication:
Closure under Addition: Suppose that A and B belong to S. This implies that AT = −A and
T
B = −B. Therefore,
(A + B)T = AT + B T = (−A) + (−B) = −(A + B),
which proves that A + B belongs to S.
Closure under Scalar Multiplication: Let k be a real constant, and let A be as above. Then
(k · A)T = k · AT = k · (−A) = −(k · A),
which proves that k · A belongs to S.
⎤
0
a
b c
⎢ −a
0
d e ⎥
⎥. Since we have six free variables, we know
(b). A typical element in S has the form ⎢
⎣ −b −d
0 f ⎦
−c −e −f 0
that dim[S] = 6. To obtain a basis, we will use the notation Eij to denote the matrix whose (i, j)-element
is 1, and all other entries are 0. Then we obtain the following basis for S by extracting the free variables:
⎡
{E12 − E21 , E13 − E31 , E14 − E41 , E23 − E32 , E24 − E42 , E34 − E43 }.
(c). Since dim[M4 (R)] = 16 and dim[S] = 6, we need 10(= 16 − 6) additional matrices in M4 (R) to extend
the basis from part (b). Each matrix added must be outside of S and must not be expressible in terms of
any of the other matrices in the basis. There are many acceptable ways to do this, but the most natural is
to add the following 10 matrices:
E12 , E13 , E14 , E23 , E24 , E34 , E11 , E22 , E33 , E44 .
35. (a). Clearly, S is a nonempty subset of M4 (R), so we can apply Theorem 4.3.2 and verify that S is
closed under addition and scalar multiplication:
Closure under Addition: Suppose that A and B belong to S. Write the elements A and B in general
form as
⎡
⎤
a1
b1
c1
−(a1 + b1 + c1 )
⎢
⎥
d1
e1
f1
−(d1 + e1 + f1 )
⎥
A=⎢
⎣
⎦
g1
h1
i1
−(g1 + h1 + i1 )
−(a1 + d1 + g1 ) −(b1 + e1 + h1 ) −(c1 + f1 + i1 ) a1 + b1 + c1 + d1 + e1 + f1 + g1 + h1 + i1
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and
⎡
a2
⎢
d2
B=⎢
⎣
g2
−(a2 + d2 + g2 )
b2
e2
h2
−(b2 + e2 + h2 )
c2
f2
i2
−(c2 + f2 + i2 )
⎤
−(a2 + b2 + c2 )
⎥
−(d2 + e2 + f2 )
⎥.
⎦
−(g2 + h2 + i2 )
a 2 + b2 + c 2 + d 2 + e 2 + f 2 + g 2 + h 2 + i 2
Using these forms for A and B, it is easy to see that the entries in A + B will have all rows and all columns
summing to zero. Hence, A + B belongs to S.
Closure under Scalar Multiplication: Let k be a real constant, and let A be as above. Then it is
easy to see by using the form of A given above that, upon scaling all entries through by the constant k, the
new matrix k · A still has row sums and column sums of zero. Thus, k · A belongs to S.
(b). Using the typical form of elements of S (see the matrix A in part (a)), we have nine free variables, so
that dim[S] = 9. To obtain a basis, we will use the notation Eij to denote the matrix whose (i, j)-element
is 1, and all other entries are 0. Then we obtain the following basis for S by extracting the free variables:
{E11 − E14 − E41 + E44 , E12 − E14 − E42 + E44 , E13 − E14 − E43 + E44 ,
E21 − E24 − E41 + E44 , E22 − E24 − E42 + E44 , E23 − E24 − E43 + E44 ,
E31 − E34 − E41 + E44 , E32 − E34 − E42 + E44 , E33 − E34 − E43 + E44 }.
(c). Since dim[M4 (R)] = 16 and dim[S] = 9, we need 7(= 16 − 9) additional matrices in M4 (R) to extend
the basis from part (b). Each matrix added must be outside of S and must not be expressible in terms of
any of the other matrices in the basis. There are many acceptable ways to do this, but one choice is
E11 , E12 , E21 , E13 , E31 , E14 , E41 .
36. (a). We must verify the axioms (A1)-(A10) for a vector space:
Axiom (A1): Assume that (v1 , w1 ) and (v2 , w2 ) belong to V ⊕ W . Since v1 +V v2 ∈ V and w1 +W w2 ∈ W ,
then the sum
(v1 , w1 ) + (v2 , w2 ) = (v1 +V v2 , w1 +W w2 )
lies in V ⊕ W .
Axiom (A2): Assume that (v, w) belongs to V ⊕ W , and let k be a scalar. Since k ·V v ∈ V and k ·W w ∈ W ,
the scalar multiplication
k · (v, w) = (k ·V v, k ·W w)
lies in V ⊕ W .
For the remainder of the axioms, we will omit the ·V and ·W notations. They are to be understood.
Axiom (A3): Assume that (v1 , w1 ), (v2 , w2 ) ∈ V ⊕ W . Then
(v1 , w1 ) + (v2 , w2 ) = (v1 + v2 , w1 + w2 ) = (v2 + v1 , w2 + w1 ) = (v2 , w2 ) + (v1 , w1 ),
as required.
Axiom (A4): Assume that (v1 , w1 ), (v2 , w2 ), (v3 , w3 ) ∈ V ⊕ W . Then
((v1 , w1 ) + (v2 , w2 )) + (v3 , w3 ) = (v1 + v2 , w1 + w2 ) + (v3 , w3 )
= ((v1 + v2 ) + v3 , (w1 + w2 ) + w3 )
= (v1 + (v2 + v3 ), w1 + (w2 + w3 ))
= (v1 , w1 ) + (v2 + v3 , w2 + w3 )
= (v1 , w1 ) + ((v2 , w2 ) + (v3 , w3 )),
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as required.
Axiom (A5): We claim that the zero vector in V ⊕ W is (0V , 0W ), where 0V is the zero vector in the vector
space V and 0W is the zero vector in the vector space W . To check this, let (v, w) ∈ V ⊕ W . Then
(0V , 0W ) + (v, w) = (0V + v, 0W + w) = (v, w),
which confirms that (0V , 0W ) is the zero vector for V ⊕ W .
Axiom (A6): We claim that the additive inverse of the vector (v, w) ∈ V ⊕ W is the vector (−v, −w), where
−v is the additive inverse of v in the vector space V and −w is the additive inverse of w in the vector space
W . We check this:
(v, w) + (−v, −w) = (v + (−v), w + (−w)) = (0V , 0W ),
as required.
Axiom (A7): For every vector (v, w) ∈ V ⊕ W , we have
1 · (v, w) = (1 · v, 1 · w) = (v, w),
where in the last step we have used the fact that Axiom (A7) holds in each of the vector spaces V and W .
Axiom (A8): Let (v, w) be a vector in V ⊕ W , and let r and s be scalars. Using the fact that Axiom (A8)
holds in V and W , we have
(rs)(v, w) = ((rs)v, (rs)w)
= (r(sv), r(sw))
= r(sv, sw)
= r(s(v, w)).
Axiom (A9): Let (v1 , w1 ) and (v2 , w2 ) be vectors in V ⊕ W , and let r be a scalar. Then
r((v1 , w1 ) + (v2 , w2 )) = r(v1 + v2 , w1 + w2 )
= (r(v1 + v2 ), r(w1 + w2 ))
= (rv1 + rv2 , rw1 + rw2 )
= (rv1 , rw1 ) + (rv2 , rw2 )
= r(v1 , w1 ) + r(v2 , w2 )),
as required.
Axiom (A10): Let (v, w) ∈ V ⊕ W , and let r and s be scalars. Then
(r + s)(v, w) = ((r + s)v, (r + s)w)
= (rv + sv, rw + sw)
= (rv, rw) + (sv, sw)
= r(v, w) + s(v, w),
as required.
(b). We show that {(v, 0) : v ∈ V } is a subspace of V ⊕ W , by checking closure under addition and closure
under scalar multiplication:
Closure under Addition: Suppose (v1 , 0) and (v2 , 0) belong to {(v, 0) : v ∈ V }, where v1 , v2 ∈ V . Then
(v1 , 0) + (v2 , 0) = (v1 + v2 , 0) ∈ {(v, 0) : v ∈ V },
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which shows that the set is closed under addition.
Closure under Scalar Multiplication: Suppose (v, 0) ∈ {(v, 0) : v ∈ V } and k is a scalar. Then k(v, 0) = (kv, 0)
is again in the set. Thus, {(v, 0) : v ∈ V } is closed under scalar multiplication.
Therefore, {(v, 0) : v ∈ V } is a subspace of V ⊕ W .
(c). Let {v1 , v2 , . . . , vn } be a basis for V , and let {w1 , w2 , . . . , wm } be a basis for W . We claim that
S = {(vi , 0) : 1 ≤ i ≤ n} ∪ {(0, wj ) : 1 ≤ j ≤ m}
is a basis for V ⊕ W . To show this, we will verify that S is a linearly independent set that spans V ⊕ W :
Check that S is linearly independent: Assume that
c1 (v1 , 0) + c2 (v2 , 0) + · · · + cn (vn , 0) + d1 (0, w1 ) + d2 (0, w2 ) + · · · + dm (0, wm ) = (0, 0).
We must show that
c1 = c2 = · · · = cn = d1 = d2 = · · · = dm = 0.
Adding the vectors on the left-hand side, we have
(c1 v1 + c2 v2 + · · · + cn vn , d1 w1 + d2 w2 + · · · + dm wm ) = (0, 0),
so that
c1 v 1 + c 2 v2 + · · · + cn vn = 0
and
d1 w1 + d2 w2 + · · · + dm wm = 0.
Since {v1 , v2 , . . . , vn } is linearly independent,
c1 = c2 = · · · = cn = 0,
and since {w1 , w2 , . . . , wm } is linearly independent,
d1 = d2 = · · · = dm = 0.
Thus, S is linearly independent.
Check that S spans V ⊕ W : Let (v, w) ∈ V ⊕ W . We must express (v, w) as a linear combination of the
vectors in S. Since {v1 , v2 , . . . , vn } spans V , there exist scalars c1 , c2 , . . . , cn such that
v = c1 v1 + c2 v2 + · · · + cn vn ,
and since {w1 , w2 , . . . , wm } spans W , there exist scalars d1 , d2 , . . . , dm such that
w = d1 w 1 + d2 w 2 + · · · + dm w m .
Then
(v, w) = c1 (v1 , 0) + c2 (v2 , 0) + · · · + cn (vn , 0) + d1 (0, w1 ) + d2 (0, w2 ) + · · · + dm (0, wm ).
Therefore, (v, w) is a linear combination of vectors in S, so S spans V ⊕ W .
Therefore, S is a basis for V ⊕ W . Since S contains n + m vectors, dim[V ⊕ W ] = n + m.
37. There are many examples here. One such example is S = {x3 , x3 − x2 , x3 − x, x3 − 1}, a basis for P3 (R)
whose vectors all have degree 3. To see that S is a basis, note that S is linearly independent, since if
c1 x3 + c2 (x3 − x2 ) + c3 (x3 − x) + c4 (x3 − 1) = 0,
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then
(c1 + c2 + c3 + c4 )x3 − c2 x2 − c3 x − c4 = 0,
and so c1 = c2 = c3 = c4 = 0. Since S is a linearly independent set of 4 vectors and dim[P3 (R)] = 4, S is a
basis for P3 (R).
38. Let A be an m × n matrix. By the Rank-Nullity Theorem,
dim[colspace(A)] + dim[nullspace(A)] = n.
Since, by assumption, colspace(A) = nullspace(A) = r, n = 2r must be even.
39. The ith row of the matrix A is bi (c1 c2 . . . cn ). Therefore, each row of A is a multiple of the first row,
and so rank(A) = 1. Thus, by the Rank-Nullity Theorem, nullity(A) = n − 1.
1 2
40. A row-echelon form of A is given by
. Thus, a basis for the rowspace of A is given by {(1, 2)},
0 0
−3
, and a basis for the nullspace of A is given by
a basis for the columnspace of A is given by
−6
−2
. All three subspaces are one-dimensional.
1
⎡
⎤
1 −6 −2
0
1 1/3 5/21 ⎦. Thus, a basis for the rowspace of A is
41. A row-echelon form of A is given by ⎣ 0
0
0
0
0
⎧⎡
⎤ ⎡
⎤⎫
6 ⎬
⎨ −1
5
)}, and a basis for the column space of A is given by ⎣ 3 ⎦ , ⎣ 3 ⎦ . For
{(1, −6, −2, 0), (0, 1, 13 , 21
⎩
⎭
7
21
the nullspace, we observe that the equations corresponding to the row-echelon form of A can be written as
x − 6y − 2z = 0
and
5
1
y + z + w = 0.
3
21
5
t and x = − 10
Set w = t and z = s. Then y = − 13 s − 21
7 t. Thus,
nullspace(A) =
−
10
1
5
t, − s − t, s, t
7
3
21
: s, t ∈ R
=
10
5
1
t − , − , 0, 1 + s 0, − , 1, 0
.
7
21
3
5
1
Hence, a basis for the nullspace of A is given by {(− 10
7 , − 21 , 0, 1), (0, − 3 , 1, 0)}.
⎤
⎡
1 −2.5 −5
⎢ 0
1 1.3 ⎥
⎥. Thus, a basis for the rowspace of A is given by
42. A row-echelon form of A is given by ⎢
⎣ 0
0
1 ⎦
0
0
0
{(1, −2.5, −5), (0, 1, 1.3), (0, 0, 1)}, and a basis for the columnspace of A is given by
⎧⎡
⎤⎫
⎤ ⎡
⎤ ⎡
0
3 ⎪
−4
⎪
⎪
⎪
⎨⎢
⎥ ⎢
⎥ ⎢
⎥⎬
⎢ 0 ⎥ , ⎢ 10 ⎥ , ⎢ 13 ⎥ .
⎣ 6 ⎦ ⎣ 5 ⎦ ⎣ 2 ⎦⎪
⎪
⎪
⎪
⎩
⎭
10
−2
5
Moreover, we have that the nullspace of A is 0-dimensional, and so the basis is empty.
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⎤
1 0
2
2
1
⎢ 0 1 −1 −4 −3 ⎥
⎥. Thus, a basis for the rowspace of A is
43. A row-echelon form of A is given by ⎢
⎣ 0 0
1
4
3 ⎦
0 0
0
1
0
given by
{(1, 0, 2, 2, 1), (0, 1, −1, −4, −3), (0, 0, 1, 4, 3), (0, 0, 0, 1, 0)},
⎡
a basis for the columnspace of A is given by
⎧⎡
⎤ ⎡
⎤⎫
⎤ ⎡
⎤ ⎡
5
2 ⎪
3
5
⎪
⎪
⎪
⎨⎢
⎥ ⎢
⎥⎬
⎥ ⎢
⎥ ⎢
⎢ 1 ⎥,⎢ 0 ⎥,⎢ 2 ⎥,⎢ 2 ⎥ .
⎣ 1 ⎦ ⎣ 1 ⎦ ⎣ 1 ⎦ ⎣ −2 ⎦⎪
⎪
⎪
⎪
⎭
⎩
0
−2
−2
−4
For the nullspace, if the variables corresponding the columns are (x, y, z, u, v), then the row-echelon form
tells us that v = t is a free variable, u = 0, z = −3t, y = 0, and x = 5t. Thus,
nullspace(A) = {(5t, 0, −3t, 0, t) : t ∈ R} = {t(5, 0, −3, 0, 1) : t ∈ R},
and so a basis for the nullspace of A is {(5, 0, −3, 0, 1)}.
44. Any of the conditions (a)-(p) appearing in the Invertible Matrix Theorem would be appropriate at this
point in the text.
Chapter 5 Solutions
Solutions to Section 5.1
(a): FALSE. Many counterexamples exist. For instance, the vectors v = (1, 1) and w = (1, 0) in R2 are
linearly independent, but they are not orthogonal.
(b): FALSE. We have
kv, kw = kv, kw = kkw, v = k 2 w, v = k 2 v, w,
where we have used the axioms of an inner product to carry out these steps. Therefore, the result conflicts
with the given statement, which must therefore be a false statement.
(c): TRUE. We have
c1 v1 + c2 v2 , w = c1 v1 , w + c2 v2 , w = c1 v1 , w + c2 v2 , w = c1 · 0 + c2 · 0 = 0.
(d): TRUE. We have
x + y, x − y = x, x − x, y + y, x − y, y = x, x − y, y = ||x||2 − ||y||2 .
This will be negative if and only if ||x||2 < ||y||2 , and since ||x|| and ||y|| are nonnegative real numbers,
||x||2 < ||y||2 if and only if ||x|| < ||y||.
(e): FALSE. For example, if V is the inner product space of integrable functions on (−∞, ∞), then the
formula
b
f, g =
f (t)g(t)dt
a
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is a valid any product for any choice of real numbers a < b. See also Problems 18 and 27 in this section, in
which a “non-standard” inner product on R2 is given.
(f ): TRUE. The angle between the vectors −2v and −2w is
cos θ =
(−2v) · (−2w)
(−2)2 (v · w)
v·w
=
=
,
|| − 2v|||| − 2w||
(−2)2 ||v||||w||
||v||||w||
and this is the angle between the vectors v and w.
(g): FALSE. This definition of p, q will not satisfy the requirements of an inner product. For instance, if
we take p = x, then p, p = 0, but p = 0.
Problems:
π
1. v, w = 0, so that cos θ = 0. That is, θ = rad.
2
√
√
2. v, w = 8, ||v|| = 3 3, ||w|| = 7. Hence,
cos θ =
π
π
3. f, g =
0
v, w
8
= √ =⇒ θ ≈ 0.95 radians.
||v||||w||
3 21
x sin xdx = π, ||f || =
2
sin xdx
0
3 π
π
π
2
. Hence,
=
x dx =
, ||g|| =
2
3
0
√
π
6
cos θ = 1/2 3 1/2 =
≈ 0.68 radians.
π
π
π
2
3
4.
π/2
f, g =
π/2
sin x(2 cos x+4)dx = 2
0
π/2
sin x cos xdx+4
0
0
||f || = f, f =
||g|| = g, g =
0
(4 cos2 x + 16 cos x + 16)dx =
0
cos θ = √
b
π/2
sin2 xdx =
π/2
Thus,
5.
π/2
sin xdx = (sin2 x)|0 −4 cos x|0
π
π √
4 · + 16 · 1 + 16 · = 9π + 16.
4
2
5·2
√
=⇒ θ ≈ 0.559 rad.
π · 9π + 16
b
xm · xn dx =
= (1−0)−4(0−1) = 5,
√
π
π
=
,
4
2
1
(bm+n+1 − am+n+1 ),
m+n+1
a
a
b
1
2m
x dx =
||f || = f, f =
(b2m+1 − a2m+1 ),
2m + 1
a
f, g =
π/2
xm+n dx =
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||g|| = g, g =
b
x2n dx =
a
Thus,
θ = cos
−1
1
(b2n+1 − a2n+1 ).
2n + 1
√
√
2m + 1 2n + 1(bm+n+1 − am+n+1 )
√
√
.
(m + n + 1) b2m+1 − a2m+1 b2n+1 − a2n+1
6. v, w
i)(−1 − i) + (3 − 2i)(1 + 3i) + (4 + i)(3 + i) = 19 + 11i. ||v|| =
= (2 + √
||w|| = w, w = 22.
√
v, v = 35.
7.
v, w = (6 − 3i)(−i) + (4)(−2i) + (−2 + 5i)(−3i) + (3i)(−4i) = (−3 − 6i) + (−8i) + (15 + 6i) + 12 = 24 − 8i,
√
√
√
||v|| = v, v = (6 − 3i)(6 + 3i) + (4)(4) + (−2 + 5i)(−2 − 5i) + (3i)(−3i) = 45 + 16 + 29 + 9 = 99 = 3 11,
√
√
||w|| = w, w = (i)(−i) + (2i)(−2i) + (3i)(−3i) + (4i)(−4i) = 1 + 4 + 9 + 16 = 30.
8. We can
show that this mapping fails the first property of Definition 5.1.3 with an example. Let A =
0 1
. Then the given mapping dictates that A, A = 0 even though A = 0. On the other hand, the
0 0
remaining properties (2) - (4) in Definition 5.1.3 do in fact hold. With A and B as given in the problem, k
c11 c12
, we can verify this as follows:
a constant, and C =
c21 c22
A, B = a11 b11 = b11 a11 = B, A,
kA, B = (ka11 )b11 = k(a11 b11 ) = kA, B,
A + B, C = (a11 + b11 )c11 = a11 c11 + b11 c11 = A, C + B, C.
9. We can
show that this mapping fails the first property of Definition 5.1.3 with an example. Let A =
0 1
. Then the given mapping dictates that A, A = det(A2 ) = det(02 ) = 0 even though A = 0.
0 0
Property (2) in Definition 5.1.3 holds, since
A, B = det(AB) = (det A)(det B) = (det B)(det A) = det(BA) = B, A.
In general, property (3) fails. Here is an example to illustrate this: 2I2 , I2 = det(2I2 · I2 ) = 4 but
2I2 , I2 = 2det(I2 · I2 ) = 2. Finally, for property (4), consider the example A = B = C = I2 . We have
A + B, C = 2I2 , I2 = 4, but A, C + B, C = 1 + 1 = 2. Thus, property (4) fails. Hence, we conclude
that, in general, only property (2) holds for the mapping in this problem.
1
0
, then according to the given mapping, we have
10. Property (1) fails. For instance, if we set A =
0 −1
A, A = −2, violating the requirement that v, v ≥ 0 for all vectors v. Now we consider the remaining
properties. First, we see that property (2) in Defintion 5.1.3 holds:
A, B = a11 b22 + a12 b21 + a21 b12 + a22 b11 = b11 a22 + b12 a21 + b21 a12 + b22 a11 = B, A.
Next, observe that property (3) in Definition 5.1.3 holds:
kA, B = (ka11 )b22 + (ka12 )b21 + (ka21 )b12 + (ka22 )b11 = k(a11 b22 + a12 b21 + a21 b12 + a22 b11 ) = kA, B.
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Finally, observe that property (4) in Definition 5.1.3 holds:
A + B, C = (a11 + b11 )c22 + (a12 + b12 )c21 + (a21 + b21 )c12 + (a22 + b22 )c11
= (a11 c22 + a12 c21 + a21 c12 + a22 c11 ) + (b11 c22 + b12 c21 + b21 c12 + b22 c11 )
= A, C + B, C.
In conclusion, only property (1) fails in this example.
11. Let A, B, C ∈ M2 (R).
(1): A, A = a211 + a212 + a221 + a222 ≥ 0, and A, A = 0 ⇐⇒ a11 = a12 = a21 = a22 = 0 ⇐⇒ A = 0.
(2): A, B = a11 b11 + a12 b12 + a21 b21 + a22 b22 = b11 a11 + b12 a12 + b21 a21 + b22 a22 = B, A.
(3): Let k ∈ R.
kA, B = ka11 b11 + ka12 b12 + ka21 b21 + ka22 b22 = k(a11 b11 + a12 b12 + a21 b21 + a22 b22 ) = kA, B.
(4):
(A + B), C = (a11 + b11 )c11 + (a12 + b12 )c12 + (a21 + b21 )c21 + (a22 + b22 )c22
= (a11 c11 + b11 c11 ) + (a12 c12 + b12 c12 ) + (a21 c21 + b21 c21 ) + (a22 c22 + b22 c22 )
= (a11 c11 + a12 c12 + a21 c21 + a22 c22 ) + (b11 c11 + b12 c12 + b21 c21 + b22 c22
= A, C + B, C.
√
√
12. A, B = 13, ||A|| = 33, ||B|| = 7.
13. A,
B = 2 · 3 +(−1)1 + 3(−1) + 5 · 2 = 12.
√
||A|| = A, A = 2 · 2 + (−1)(−1) + 3 · 3 + 5 · 5 = 39.
√
||B|| = B, B = 3 · 3 + 1 · 1 + (−1)(−1) + 2 · 2 = 15.
14. Let p1 , p2 , p3 ∈ P1 (R) where p1 (x) = a + bx, p2 (x) = c + dx, and p3 (x) = e + f x.
Define
p1 , p2 = ac + bd.
(14.1)
The properties 1 through 4 of Definition 5.1.3 must be verified.
(1): p1 , p1 = a2 + b2 ≥ 0 and p1 , p1 = 0 ⇐⇒ a = b = 0 ⇐⇒ p1 (x) = 0.
(2): p1 , p2 = ac + bd = ca + db = p2 , p1 .
(3): Let k ∈ R.
kp1 , p2 = kac + kbd = k(ac + bd) = kp1 , p2 .
(4):
p1 + p2 , p3 = (a + c)e + (b + d)f = ae + ce + bf + df
= (ae + bf ) + (ce + df )
= p1 , p3 + p2 , p3 .
Hence, the mapping defined by (14.1) is an inner product in P1 (R).
15. NO. Property (1) in Definition 5.1.3 fails. For instance, consider the function defined by
0,
if 0 ≤ x ≤ 1/2 ,
f (x) =
x − 12 , if 1/2 ≤ x ≤ 1 .
Clearly f = 0. However, f, f =
1/2
0
[f (x)]2 dx = 0.
16. YES. We verify the four properties in Definition 5.1.3. For property (1), note that
1
x[f (x)]2 dx ≥ 0,
f, f =
0
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since on the interval 0 ≤ x ≤ 1, we have that x[f (x)]2 ≥ 0, so a positive function is being integrated.
However, in order for f, f = 0, we must have f (x) = 0 for all x ∈ [0, 1].
For property (2), we observe that
f, g =
1
1
xf (x)g(x)dx =
0
0
xg(x)f (x) = g, f .
For property (3), we observe that
kf, g =
1
1
x(kf )(x)g(x)dx = k
0
xf (x)g(x)dx = kf, g.
0
For property (4), we observe that
f + g, h =
1
1
x(f (x) + g(x))h(x)dx =
0
1
xf (x)h(x)dx +
0
0
xg(x)h(x)dx = f, h + g, h.
Since properties (1) - (4) have all been verified, we conclude that the given mapping does form a valid inner
product on V .
17. NO. Property (1) in Definition 5.1.3 fails. To see an explicit example, observe that
x, x =
0
x3 dx =
−1
1 40
1
1
x |−1 = (0 − 1) = − < 0.
4
4
4
Many other illustrations could also suffice here.
18. Property 1:
u, u = 2u1 u1 + u1 u2 + u2 u1 + 2u2 u2 = 2u21 + 2u1 u2 + 2u22 = (u1 + u2 )2 + u21 + u22 ≥ 0.
u, u = 0 ⇐⇒ (u1 + u2 )2 + u21 + u22 = 0 ⇐⇒ u1 = 0 = u2 ⇐⇒ u = 0.
Property 2:
u, v = 2u1 v1 + u1 v2 + u2 v1 + 2u2 v2 = 2u2 v2 + u2 v1 + u1 v2 + 2u1 v1 = v, u.
Property 3:
ku, v = k(2u1 v1 + u1 v2 + u2 v1 + 2u2 v2 ) = 2ku1 v1 + ku1 v2 + ku2 v1 + 2ku2 v2
= 2(ku1 )v1 + (ku1 )v2 + (ku2 )v1 + 2(ku2 )v2 = ku, v
= 2ku1 v1 + ku1 v2 + ku2 v1 + 2ku2 v2
= 2u1 (kv1 ) + u1 (kv2 ) + u2 (kv1 ) + 2u2 (kv2 ) = u, kv.
Property 4:
(u + v), w = (u1 + v1 , u2 + v2 ), (w1 , w2 )
= 2(u1 + v1 )w1 + (u1 + v1 )w2 + (u2 + v2 )w1 + 2(u2 + v2 )w2
= 2u1 w1 + 2v1 w1 + u1 w2 + v1 w2 + u2 w1 + v2 w1 + 2u2 w2 + 2v2 w2
= 2u1 w1 + u1 w2 + u2 w1 + 2u2 w2 + 2v1 w1 + v1 w2 + v2 w1 + 2v2 w2
= u, w + v, w.
Therefore u, v = 2u1 v1 + u1 v2 + u2 v1 + 2u2 v2 defines an inner product on R2 .
19. (a). Using the defined inner product:
v, w = (1, 0), (−1, 2) = 2 · 1(−1) + 1 · 2 + 0(−1) + 2 · 0 · 2 = 0.
(b). Using the standard inner product:
v, w = (1, 0), (−1, 2) = 1(−1) + 0 · 2 = −1 = 0.
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20. (a). Using the defined inner product:
v, w = 2 · 2 · 3 + 2 · 6 + (−1)3 + 2(−1)6 = 9 = 0.
(b). Using the standard inner product:
v, w = 2 · 3 + (−1)6 = 0.
21. (a). Using the defined inner product:
v, w = 2 · 1 · 2 + 1 · 1 + (−2) · 2 + 2 · (−2) · 1 = −3 = 0.
(b). Using the standard inner product:
v, w = 1 · 2 + (−2)(1) = 0.
22. (a). Show symmetry: v, w = w, v.
v, w = (v1 , v2 ), (w1 , w2 ) = v1 w1 − v2 w2 = w1 v1 − w2 v2 = (w1 , w2 ), (v1 , v2 ) = w, v.
(b). Show kv, w = kv, w = v, kw.
Note that kv = k(v1 , v2 ) = (kv1 , kv2 ) and kw = k(w1 , w2 ) = (kw1 , kw2 ).
kv, w = (kv1 , kv2 ), (w1 , w2 ) = (kv1 )w1 − (kv2 )w2 = k(v1 w1 − v2 w2 ) = k(v1 , v2 ), (w1 , w2 ) = kv, w.
Also,
v, kw = (v1 , v2 ), (kw1 , kw2 ) = v1 (kw1 ) − v2 (kw2 ) = k(v1 w1 − v2 w2 ) = k(v1 , v2 ), (w1 , w2 ) = kv, w.
(c). Show (u + v), w = u, w + v, w.
Let w = (w1 , w2 ) and note that u + v = (u1 + v1 , u2 + v2 ).
(u + v), w = (u1 + v1 , u2 + v2 ), (w1 , w2 ) = (u1 + v1 )w1 − (u2 + v2 )w2
= u1 w 1 + v 1 w 1 − u2 w 2 − v 2 w 2 = u1 w 1 − u2 w 2 + v 1 w 1 − v 2 w 2
= (u1 , v2 ), (w1 , w2 ) + (v1 , v2 ), (w1 , w2 ) = u, w + v, w.
Property 1 fails since, for example, u, u < 0 whenever |u2 | > |u1 |.
23. v, v = 0 =⇒ (v1 , v2 ), (v1 , v2 ) = 0 =⇒ v12 − v22 = 0 =⇒ v12 = v22 =⇒ |v1 | = |v2 |. Thus, in this space,
null vectors are given by {(v1 , v2 ) ∈ R2 : |v1 | = |v2 |} or equivalently, v = r(1, 1) or v = s(1, −1) where
r, s ∈ R.
24. v, v < 0 =⇒ (v1 , v2 ), (v1 , v2 ) < 0 =⇒ v12 − v22 < 0 =⇒ v12 < v22 . In this space, timelike vectors are
given by {(v1 , v2 ) ∈ R2 : v12 < v22 }.
25. v, v > 0 =⇒ (v1 , v2 ), (v1 , v2 ) > 0 =⇒ v12 − v22 > 0 =⇒ v12 > v22 . In this space, spacelike vectors are
given by {(v1 , v2 ) ∈ R2 : v12 > v22 }.
26. This is illustrated in the figure.
27. We verify the four properties in Definition 5.1.3. For property (1), note that
v, v = v12 − v1 v2 − v2 v1 + 4v22 = v12 − 2v1 v2 + 4v22 = (v1 − v2 )2 + 3v22 ≥ 0,
and in order for v, v to be zero, we must have v1 − v2 = 0 and v2 = 0. That is v1 = v2 = 0, which means
v = 0. This verifies property (1).
For property (2), we have
v, w = v1 w1 − v1 w2 − v2 w1 + 4v2 w2 = w1 v1 − w1 v2 − w2 v1 + 4w2 v2 = w, v.
For property (3), we have
kv, w = (kv1 )w1 − (kv1 )w2 − (kv2 )w1 + 4(kv2 )w2 = k(v1 w1 − v1 w2 − v2 w1 + 4v2 w2 ) = kv, w.
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y
Null Vectors
Null Vectors
Timelike Vectors
Spacelike Vectors
Spacelike Vectors
x
Spacelike Vectors
Spacelike Vectors
Timelike Vectors
Null Vectors
Null Vectors
Figure 0.0.64: Figure for Problem 26
Finally, for property (4), if we write u = (u1 , u2 ), then
u + v, w = (u1 + v1 )w1 − (u1 + v1 )w2 − (u2 + v2 )w1 + 4(u2 + v2 )w2
= (u1 w1 − u1 w2 − u2 w1 + 4u2 w2 ) + (v1 w1 − v1 w2 − v2 w1 + 4v2 w2 )
= u, w + v, w,
as needed.
28. Suppose that some ki ≤ 0. If ei denotes the standard basis vector in Rn with 1 in the ith position and
zeros elsewhere, then the given formula yields ei , ei = ki ≤ 0, violating the first axiom of an inner product.
Therefore, if the given formula defines a valid inner product, then we must have that ki > 0 for all i.
Now we prove the converse. Suppose that each ki > 0. We verify the axioms of an inner product for the
given proposed inner product. We have
v, v = k1 v12 + k2 v22 + · · · + kn vn2 ≥ 0,
and we have equality if and only if v1 = v2 = · · · = vn = 0. For the second axiom, we have
w, v = k1 w1 v1 + k2 w2 v2 + · · · + kn wn vn = k1 v1 w1 + k2 v2 w2 + · · · + kn vn wn = v, w.
For the third axiom, we have
kv, w = k1 (kv1 )w1 + k2 (kv2 )w2 + · · · + kn (kvn )wn = k[k1 v1 w1 + k2 v2 w2 + · · · + kn vn wn ] = kv, w.
Finally, for the fourth axiom, we have
u + v, w = k1 (u1 + v1 )w1 + k2 (u2 + v2 )w2 + · · · + kn (un + vn )wn
= [k1 u1 w1 + k2 u2 w2 + · · · + kn un wn ] + [k1 v1 w1 + k2 v2 w2 + · · · + kn vn wn ]
= u, w + v, w.
29. We have
v, 0 = v, 0 + 0 = 0 + 0, v = 0, v + 0, v = v, 0 + v, 0 = 2v, 0,
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which implies that v, 0 = 0.
30. We have
u, v + w = v + w, u
= v, u + w, u
= v, u + w, u
= u, v + u, w,
as required.
31. We have
u + v, w + x = u + v, w + u + v, x
= u, w + v, w + u, x + v, x.
32. (a). We have
||v + w|| =
=
=
=
√
v + w, v + w
||v||2 + v, w + w, v + ||w||2
32 + (−2) + (−2) + 16
29.
(b). We have
3v + w, −2v + 3w = 3v, −2v + 3v, 3w + w, −2v + w, 3w
= −6||v||2 + 9v, w − 2w, v + 3||w||2
= −6 · 9 + 9 · (−2) − 2 · (−2) + 3 · 16
= −54 − 18 + 4 + 48
= −20.
(c). We have
−w, 5v + 2w = −w, 5v + −w, 2w
= −5w, v − 2||w||2
= −5 · (−2) − 2 · 16
= −22.
33. (a). We have
w − 2v, w − 2v
= w, w − 2w, v − 2v, w + 4v, v
= ||w||2 − 4v, w + 4||v||2
= 62 − 4 · 3 + 4 · 22
√
√
= 40 = 2 10.
||w − 2v|| =
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(b). We have
||2v + 5w|| =
2v + 5w, 2v + 5w
4||v||2 + 10v, w + 10w, v + 25||w||2
= 4 · 22 + 10 · 3 + 10 · 3 + 25 · 62
√
= 16 + 30 + 30 + 900
√
= 976.
=
(c). We have
2v − w, 4w = 8v, w − 4||w||2
= 8 · 3 − 4 · 62
= 24 − 144
= −120.
34. (a). We have
u + v + w, u + v + w
= ||u||2 + ||v||2 + ||w||2 + 2(u, v + u, w + v, w)
= 62 + 22 + 32 + 2(4 + 5 + 0)
√
= 36 + 4 + 9 + 18
√
= 47.
||u + v + w|| =
(b). We have
2u − 3v − w, 2u − 3v − w
= 4||u||2 + 9||v||2 + ||w||2 − 12u, v − 4u, w + 6v, w
= 4 · 62 + 9 · 22 + 32 − 12 · 4 − 4 · 5 + 6 · 0
√
= 144 + 36 + 9 − 48 − 20 + 0
√
= 121
= 11.
||2u − 3v − w|| =
(c). We have
u + 2w, 3u − 3v + w = 3||u||2 + 2||w||2 − 3u, v + u, w + 6w, u − 6w, v
= 3 · 62 + 2 · 32 − 3 · 4 + 5 + 6 · 5 − 6 · 0
= 108 + 18 − 12 + 5 + 30
= 149.
35. (a). For all v, w ∈ V .
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||v + w||2 = (v + w), (v + w)
= v, v + w + w, v + w by Property 4
= v + w, v + v + w, w by Property 2
= v, v + w, v + v, w + w, w by Property 4
= v, v + v, w + w, v + w, w by Property 2
= ||v||2 + 2v, w + ||w||2 .
(b). This follows immediately by substituting v, w = 0 in the formula given in part (a).
36.(a). From Problem 35(a), it follows that ||v + w||2 = ||v||2 + 2v, w + ||w||2 and ||v − w||2 = ||v +
2
(−w)||2 = ||v||2 + 2v, −w + ||− w||2 = ||v||2 − 2v, w + ||w||
.
2
2
2
2
Thus, ||v + w|| − ||v − w|| = ||v|| + 2v, w + ||w|| − ||v||2 − 2v, w + ||w||2 = 4v, w.
(b).
+ w||2 + ||v − w||2 = ||v||2 + 2v, w + ||w||2 + ||v||2 − 2v, w + ||w||2 = 2||v||2 + 2||w||2 =
||v
2 ||v||2 + ||w||2 .
37. For all v, w ∈ V and vi , wi ∈ C.
||v + w||2 = v + w, v + w
= v, v + w + w, v + w by Property 4
= v + w, v + v + w, w by Property 2
= v, v + w, v + v, w + w, w by Property 4
= v, v + w, v + v, w + w, w
= v, v + w, w + v, w + v, w
= ||v||2 + ||w||2 + 2Re{v, w}
= ||v||2 + 2Re{v, w} + ||v||2 .
Solutions to Section 5.2
(a): TRUE. An orthonormal basis is simply an orthogonal basis consisting of unit vectors.
(b): FALSE. The converse of this statement is true, but for the given statement, many counterexamples
exist. For instance, the set of vectors {(1, 1), (1, 0)} in R2 is linearly independent, but does not form an
orthogonal set.
(c): TRUE. We can verify easily that
π
cos t sin tdt =
0
sin2 t π
| = 0,
2 0
which means that {cos x, sin x} is an orthogonal set. Moreover, since they are non-proportional functions,
they are linearly independent. Therefore, they comprise an orthogonal basis for the 2-dimensional inner
product space span{cos x, sin x}.
(d): FALSE. The vector w − P(w, v) is orthogonal to v, not w.
(e): TRUE. This is the content of Theorem 5.2.7.
(f ): TRUE. The vector P(w, v) is a scalar multiple of the vector v, and since v is orthogonal to u, P(w, v)
is also orthogonal to u, so that its projection onto u must be 0.
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(g): TRUE. We have
P(w1 + w2 , v) =
w1 , v + w2 , v
w1 , v
w2 , v
w1 + w2 , v
v=
v=
v+
v = P(w1 , v) + P(w2 , v).
v2
v2
v2
v2
Problems:
1. YES. Observe that (1,
that √
(1, 2) is √
orthogonal to
= 0, thus showing
√ 2), (−4, 2)
√ = (1)(−4) + (2)(2)
√
(−4, 2). Since ||(1, 2)|| = 12 + 22 = 5 and ||(−4, 2)|| = (−4)2 + 22 = 16 + 4 = 20 = 2 5, we obtain
the orthonormal set
1
1
√ (1, 2), √ (−4, 2) .
5
2 5
2. YES. (2, −1, 1), (1, 1, −1) = 2 + (−1) + (−1) = 0;
(2, −1, 1), (0, 1, 1) = 0 + (−1) + 1 = 0;
(1, 1, −1), (0, 1, 1) = 0 + 1 + (−1) = 0.
Since each vector in the set is orthogonal to every other vector in the set, the vectors form an orthogonal
set. To generate
√ set, we divide each vector by its norm:
√ an orthonormal
||(2, −1, 1)|| = √4 + 1 + 1 = √6,
||(1, 1, −1)|| =√ 1 + 1 + 1 =√ 3, and
||(0, 1, 1)|| = 0 + 1 + 1 = 2. Thus, the corresponding orthonormal set is:
√
√
6
3
2
(2, −1, 1),
(1, 1, −1),
(0, 1, 1) .
6
3
2
√
3. YES. (1, 3, −1, 1), (−1, 1, 1, −1) = −1 + 3 + (−1) + (−1) = 0;
(1, 3, −1, 1), (1, 0, 2, 1) = 1 + 0 + (−2) + 1 = 0;
(−1, 1, 1, −1), (1, 0, 2, 1) = −1 + 0 + 2 + (−1) = 0.
Since all vectors in the set are orthogonal to each other, they form an orthogonal set. To generate an
orthonormal set, √
we divide each vector
√ by its
√ norm:
||(1, 3, −1, 1)|| = √
1+9+1+1= √
12 = 2 3,
||(−1, 1, 1, −1)||√= 1 + 1 + 1 + 1√= 4 = 2, and
||(1, 0, 2, 1)|| = 1 + 0 + 4 + 1 = 6. Thus, an orthonormal set is:
√
3
6
1
(1, 3, −1, 1), (−1, 1, 1, −1),
(1, 0, 2, 1) .
6
2
6
√
4. NO. (1, 2, −1, 0), (1, 0, 1, 2) = 1 + 0 + (−1) + 0 = 0;
(1, 2, −1, 0), (−1, 1, 1, 0) = −1 + 2 + (−1) + 0 = 0;
(1, 2, −1, 0), (1, −1, −1, 0) = 1 + (−2) + 1 + 0 = 0;
(1, 0, 1, 2), (−1, 1, 1, 0) = −1 + 0 + 1 + 0 = 0;
(1, 0, 1, 2), (1, −1, −1, 0) = 1 + 0 + (−1) + 0 = 0;
(−1, 1, 1, 0), (1, −1, −1, 0) = −1 + (−1) + (−1) = −3.
Hence, this is not an orthogonal set of vectors.
5. YES. (1, 2, −1, 0, 3), (1, 1, 0, 2, −1) = 1 + 2 + 0 + 0 + (−3) = 0;
(1, 2, −1, 0, 3), (4, 2, −4, −5, −4) = 4 + 4 + 4 + 0 + (−12) = 0;
(1, 1, 0, 2, −1), (4, 2, −4, −5, −4) = 4 + 2 + 0 + (−10) + 4 = 0.
Since all vectors in the set are orthogonal to each other, they form an orthogonal set. To generate an
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orthonormal set, we√divide each vector by√its norm:
||(1, 2, −1, 0, 3)|| = √1 + 4 + 1 + 0 + 9 = √15,
||(1, 1, 0, 2, −1)|| = 1 √
+ 1 + 0 + 4 + 1 = 7, and√
||(4, 2, −4, −5, −4)|| = 16 + 4 + 16 + 25 + 16 = 77. Thus, an orthonormal set is:
√
√
√
15
7
77
(1, 2, −1, 0, 3),
(1, 1, 0, 2, −1),
(4, 2, −4, −5, −4) .
15
7
77
6. Write the nonzero vector w as w = (a, b). In order for w to be orthogonal to v, it is necessary and
sufficient that v, w = 0; that is, 7a − 2b = 0, or b = 72 a. That is, any vector of the form w = (a, 72 a), where
a = 0, satisfies the given requirement. Therefore, any nonzero multiple w of the vector (1, 72 ) will create the
orthogonal set {v, w}.
7. The vectors w = (a, b, c) in R3 that are orthogonal to v must satisfy (−3, 6, 1), (a, b, c) = 0; that is,
−3a + 6b + c = 0. Therefore, all vectors lying on the plane −3a + 6b + c = 0 in R3 are orthogonal to v. To
obtain a basis for R3 , we need a set of three vectors. Along with v, we will add two additional vectors that
lie on the plane to form the basis. In order to ensure that we arrive at an orthogonal basis, the two vectors
we choose on the plane must be orthogonal to one another. There are many choices we can make. For
instance, we can take w1 = (1, 0, 3) and w2 = (−3, − 53 , 1). Thus, one orthogonal basis for R3 that includes
v is {(−3, 6, 1), (1, 0, 3), (−3, − 53 , 1)}.
8. We require that v1 , v2 = v1 , w = v2 , w = 0. Let w = (a, b, c) where a, b, c ∈ R.
v1 , v2 = (1, 2, 3), (1, 1, −1) = 0.
v1 , w = (1, 2, 3), (a, b, c) =⇒ a + 2b + 3c = 0.
v2 , w = (1, 1, −1), (a, b, c) =⇒ a + b − c = 0.
Letting the free variable c = t ∈ R, the system has the solution a = 5t, b = −4t, and c = t. Consequently,
{(1, 2, 3), (1, 1, −1), (5t, −4t, t)} will form an orthogonal set whenever t = 0. To determine the corresponding
orthonormal
√
√
√ divide each
√
√
√ set, we must
√ vector by its√norm:
||v1 || = 1 + 4 + 9 = 14, ||v2 || = 1 + 1 + 1 = 3, ||w|| = 25t2 + 16t2 + t2 = 42t2 = |t| 42 = t 42
if t ≥ 0. Setting t = 1, an orthonormal set is:
√
√
√
14
3
42
(1, 2, 3),
(1, 1, −1),
(5, −4, 1) .
14
3
42
9. The vectors w = (a, b, c, d) in R4 that are orthogonal to v1 and v2 must satisfy (−4, 0, 0, 1), (a, b, c, d) = 0
and (1, 2, 3, 4), (a, b, c, d) = 0; that is
−4a + d = 0
and
a + 2b + 3c + 4d = 0.
Solving this system of two equations by the usual process of Gaussian elimination, we find that d = t and c = s
1
are free variables. Then b = − 32 s− 17
8 t and a = 4 t. Thus, the set of vectors that are orthogonal to both v1 and
1
v2 are precisely those vectors of the form ( 4 t, − 32 s− 17
8 t, s, t), where s and t are real numbers. We must obtain
two vectors of this form that are orthogonal to one another as well. By choosing t = 0, we obtain one vector
as (0, − 32 , 1, 0). Taking the inner product of this vector with a general vector of the type we must choose, we
51
3
obtain 13
4 s + 16 t, so for the next vector to be orthogonal to (0, − 2 , 1, 0), we must choose s and t to satisfy
13
51
4 s+ 16 t = 0. One solution is s = −51 and t = 52. Hence, we take our second vector to be (13, −34, −51, 52).
The orthogonal basis we have therefore obtained is {v1 , v2 , (0, − 32 , 1, 0), (13, −34, −51, 52)}.
10. We have
(2 + i, −5i), (−2 − i, −i) = (2 + i)(−2 + i) + (−5i)(i)
= −4 − 1 + 5 = 0.
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Thus, the two given vectors form an orthogonal set in C2 . To obtain an orthonormal set, we must normalize
the given vectors. Note that
√
√
||(2 + i, −5i)|| = (2 + i, −5i), (2 + i, −5i) = (2 + i)(2 − i) + (−5i)(5i) = 5 + 25 = 30
and
||(−2 − i, −i)|| =
√
√
(−2 − i, −i), (−2 − i, −i) = (−2 − i)(−2 + i) + (−i)(i) = 5 + 1 = 6.
Thus, we obtain the orthonormal set
1
1
√ (2 + i, −5i), √ (−2 − i, −i) .
30
6
11. (1 − i, 3 + 2i), (2 + 3i, 1 − i) = (1 − i)(2 + 3i) + (3 + 2i)(1 − i) = (1 − i)(2 − 3i) + (3 + 2i)(1 + i) =
(2 − 3i − 2i − 3) + (3 + 3i + 2i − 2) = (−1 − 5i) + (1 + √
5i) = 0. The vectors
√ are orthogonal.
||(1 − i, 3 + 2i)|| = (1 − i)(1 + i) + (3 + 2i)(3 − 2i) = 1 + 1 + 9 + 4 = 15.
√
√
||(2 + 3i, 1 − i)|| = (2 + 3i)(2 − 3i) + (1 − i)(1 + i) = 4 + 9 + 1 + 1 = 15. Thus, the corresponding
orthonormal set is:
√
√
15
15
(1 − i, 3 + 2i),
(2 + 3i, 1 − i) .
15
15
12. (1 − i, 1 + i, i), (0, i, 1 − i) = (1 − i) · 0 + (1 + i)(−i) + i(1 + i) = 0.
(1 − i, 1 + i, i), (−3 + 3i, 2 + 2i, 2i) = (1 − i)(−3 − 3i) + (1 + i)(2 − 2i) + i(−2i) = 0.
(0, i, 1 − i), (−3 + 3i, 2 + 2i, 2i) = 0 + i(2 − 2i) + (1 − i)(−2i) = (2i + 2) + (−2i − 2) = 0.
Hence, the vectors are orthogonal. To obtain a corresponding orthonormal set, we divide each vector by its
norm.
√
√
||(1 − i, 1 + i, i)||= (1 − i)(1 + i) + (1 + i)(1 − i) + i(−i) = 1 + 1 + 1 + 1 + 1 = 5.
√
√
||(0, i, 1 − i)|| = 0 + i(−i)+ (1 − i)(1 + i) = 1 + 1 + 1 = 3.
√
√
||(−3 + 3i, 2 + 2i, 2i)|| = (−3 + 3i)(−3 − 3i) + (2 + 2i)(2 − 2i) + 2i(−2i) = 9 + 9 + 4 + 4 + 4 = 30.
Consequently, an orthonormal set is:
√
√
√
5
3
30
(1 − i, 1 + i, i),
(0, i, 1 − i),
(−3 + 3i, 2 + 2i, 2i) .
5
3
30
13. Let z = a + bi where a, b ∈ R. We require that v, w = 0.
(1 − i, 1 + 2i), (2 + i, a + bi) = 0 =⇒ (1 − i)(2 − i) + (1 + 2i)(a − bi) = 0 =⇒ 1 − 3i + a + 2b + (2a − b)i = 0.
a + 2b = −1
Equating real parts and imaginary parts from the last equality results in the system:
2a − b = 3.
This system has the solution a = 1 and b = −1; hence z = 1 − i. Our desired orthogonal set is given by
{(1 − i, 1 + 2i), (2 + i, 1 − i)}.
√
√
||(1 − i, 1 + 2i)|| = (1 − i)(1 + i) + (1 + 2i)(1 − 2i) = 1 + 1 + 1 + 4 = 7.
√
√
||(2 + i, 1 − i)|| = (2 + i)(2 − i) + (1 − i)(1 + i) = 4 + 1 + 1 + 1 = 7.
The corresponding orthonormal set is given by:
√
√
7
7
(1 − i, 1 + 2i),
(2 + i, 1 − i) .
7
7
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1
− cos πx
14. f1 , f2 = 1, sin πx =
sin πxdx =
= 0.
π
−1
−1
1
1
sin πx
cos πxdx =
= 0.
f1 , f3 = 1, cos πx =
π
−1
−1
1
1
1 1
−1
f2 , f3 = sin πx, cos πx =
sin πx cos πxdx =
sin 2πxdx =
= 0. Thus, the
cos 2πx
2 −1
4π
−1
−1
vectors are orthogonal.
1
√
1dx = [x]1−1 = 2.
||f1 || =
−1
1
1
1
x 1
1
−
cos
2πx
1
2
||f2 || =
sin πxdx =
−
= 1.
dx =
sin 2πx
2
2 −1
4π
−1
−1
−1
1
1
1
x 1
1
1 + cos 2πx
||f3 || =
cos2 πxdx =
+
= 1.
dx =
sin 2πx
2
2 −1
4π
−1
−1
−1
√
2
Consequently,
, sin πx, cos πx is an orthonormal set of functions on [−1, 1].
2
1
2 1
x
= 0.
2 −1
−1
1
3
1
3x2 − 1
3x2 − 1
x −x
f1 , f3 = 1,
= 0.
=
dx =
2
2
2
−1
−1
1
1
3x2 − 1
3x2 − 1
1 3x4
x2
x·
= 0. Thus, the vectors are orthogonal.
=
dx =
−
f2 , f3 = x,
2
2
2 4
2 −1
−1
1
√
||f1 || =
dx = [x]1−1 = 2.
−1
√
1
1
x3
6
2
x dx =
=
||f2 || =
.
3 −1
3
−1 √
1
2
1
3x2 − 1
1 1
1 9x5
10
4
2
3
dx =
(9x − 6x + 1)dx =
=
− 2x + x
.
||f3 || =
2
4 −1
4 5
5
−1
−1
To obtain a set of orthonormal vectors, we divide each vector by its norm:
15. f1 , f2 = 1, x =
1
1 · xdx =
√
√
√
2
6
10
f2
f3
f1
=
f1 ,
=
f2 ,
=
f3 .
||f1 ||
2
||f2 ||
2
||f3 ||
2
√
√
2
6
10
,
x,
(3x2 − 1)
2
2
4
√
Thus,
16. We have
f1 , f2 =
f1 , f3 =
1
1
is an orthonormal set of vectors.
(2x)(1 + 2x2 )dx =
1
−1
−1
3
(2x)(x3 − x)dx =
5
−1
(2x + 4x3 )dx = (x2 + x4 )|1−1 = 0,
1
6
2
2
(2x4 − x2 )dx = ( x5 − x3 )|1−1 = 0,
5
5
5
−1
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and
1
3
(1+2x )(x − x)dx =
f2 , f3 =
5
−1
2
3
1
3
6
(x − x+2x5 − x3 )dx =
5
5
−1
3
1
1
3
1
3
1
(− x3 − x+2x5 )dx = (− x4 − x2 + x6 )|1−1 = 0.
5
5
20
10
3
−1
Thus, {f1 , f2 , f3 } forms an orthogonal set of vectors. To create an orthonormal set, we must normalize
each of these vectors. Note that
1
4 31
8
2
||f1 || = f1 , f1 =
(2x) dx =
x |−1 =
,
3
3
−1
||f2 || = f2 , f2 =
1
(1 + 2x2 )2 dx =
1
−1
(1 + 4x2 + 4x4 )dx =
−1
4
4
(x + x3 + x5 )|1−1 =
3
5
94
,
15
and
||f3 || = f3 , f3 =
1
3
(x3 − x)2 dx =
5
−1
1
−1
(x6 −
6 4
9
x + x2 )dx =
5
25
1
6
3
( x7 − x5 + x3 )|1−1 =
7
25
25
Thus, we obtain the orthonormal set
3
15
175 3 3
2
(2x),
(1 + 2x ),
(x − x) .
8
94
8
5
1 1
17. f1 , f2 = cos πx, cos 2πx =
cos πx cos 2πxdx =
(cos 3πx + cos πx)dx = 0.
2
1 −1
1 −1
1
f1 , f3 = cos πx, cos 3πx =
cos πx cos 3πxdx =
(cos 4πx + cos 2πx)dx = 0.
2 −1
−1
1
1
1
f2 , f3 = cos 2πx, cos 3πx =
cos 2πx cos 3πxdx =
(cos 5πx + cos πx)dx = 0.
2 −1
−1
Therefore,
set.
{f1 , f2 , f3 } is anorthogonal
1
1
1
sin 2πx
1
1
2
x+
||f1 || =
cos πxdx =
(1 + cos 2πx)dx =
= 1.
2
2
2π
−1
−1
−1
1
1
1 1
1
sin 4πx
2
||f2 || =
cos 2πxdx =
(1 + cos 4πx)dx =
= 1.
x+
2
2
4π
−1
−1
−1
1
1
sin 6πx
1 1
1
2
x+
||f3 || =
cos 3πxdx =
(1 + cos 6πx)dx =
= 1.
2 −1
2
6π
−1
−1
Thus, it follows that{f1 , f2 , f3 } is an orthonormal set of vectors on [−1, 1].
1
1 1
18. f1 , f2 =
sin πx sin 2πxdx =
(cos 3πx − cos πx)dx = 0.
2
1 −1
1 −1
1
f1 , f3 =
sin πx sin 3πxdx =
(cos 4πx − cos 2πx)dx = 0.
2 −1
−1
1
1 1
f2 , f3 =
sin 2πx sin 3πxdx =
(cos 5πx − cos πx)dx = 0.
2 −1
−1
Therefore, {f1 , f2 , f3 } is an orthogonal set.
1
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.
175
363
1
1
sin 2πx
1
1
x−
||f1 || =
sin πxdx =
= 1.
(1 − cos 2πx)dx =
2
2
2π
−1
−1
−1
1
1
1
sin 4πx
1
1
2
x−
||f2 || =
sin 2πxdx =
= 1.
(1 − cos 4πx)dx =
2
2
4π
−1
−1
−1
1
1
1
sin 6πx
1
1
2
x−
||f3 || =
sin 3πxdx =
= 1.
(1 − cos 6πx)dx =
2
6π
−1
−1 2
−1
Thus, it follows that {f1 , f2 , f3 } is an orthonormal set of vectors on [−1, 1].
2
1
19. We must have 2a − b − c = 0 and a + b + c = 0. Sovling this system via Gaussian elimination,
we take c = t to be a free variable, from which we find that b = −t and a = 0. Thus, we must have
r(x) = −t + tx2 = t(−1 + x2 ). Hence, {p(x), q(x), r(x)} is an orthogonal set precisely when r(x) is a multiple
of −1 + x2 .
20. It is easily verified that A1 , A2 = 0, A1 , A3 = 0, A2 , A3 = 0. Thus we require a, b, c, d such that
A1 , A4 = 0 =⇒ a + b − c + 2d = 0,
A2 , A4 = 0 =⇒ −a + b + 2c + d = 0,
A3 , A4 = 0 =⇒ a − 3b + 2d = 0.
3
1
Solving this system for a, b, c, and d, we obtain: a = c, b = − c, d = 0. Thus,
2
2
3
1
3 −1
3 −1
c −2c
2
= 2c
=k
,
A4 =
2
0
2
0
c
0
where k is any nonzero real number.
21. (a). Let us determine a vector that is parallel to L. We can create a vector joining two points on L.
By choose x = 0 and x = 1, respectively, we will obtain two points on L. These are the points (0, b) and
(1, m + b). Hence, a parallel vector to L can be obtained by subtraction: (1, m + b) − (0, b) = (1, m) = v, as
required.
(b). We consider a point Q(x1 , y1 ) on the line L (so that y1 = mx1 + b) and form the vector w from Q to
P via vector subtraction: w = (x0 − x1 , y0 − y1 ). Now
P(w, v) =
w, v
(x0 − x1 ) + m(y0 − y1 )
v=
(1, m).
2
||v||
1 + m2
The distance from P (x0 , y0 ) to the line L is given by
m
1
||w − P(w, v)|| = ||
(mx0 − y0 + b),
(y0 − mx0 − b) ||
1 + m2
1 + m2
1 2
m (mx0 − y0 + b)2 + (y0 − mx0 − b)2
=
1 + m2
1 2
=
(m + 1)(y0 − mx0 − b)2
1 + m2
|y0 − mx0 − b|
√
=
.
1 + m2
22. Using the formula derived in Problem 21(b), here we take x0 = −8, y0 = 0, m = 3, and b = −4 to
obtain
| − 3 · (−8) − (−4)|
28
√
=√ .
10
1 + 32
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23. The equation of the given line can be written in the form y = − 45 x + 15 . Using the formula derived in
Problem 21(b), here we take x0 = 1, y0 = −1, m = − 45 , and b = 15 to obtain
| − 1 + 45 − 15 |
2
5 2 = √ − = √ .
5
41
41
1 + 16
25
24. The equation of the given line can be written in the form y = x − 3. Using the formula derived in
Problem 21(b), here we take x0 = −6, y0 = 4, m = 1, and b = −3 to obtain
13
|4 − (−6) − (−3)|
√
=√ .
2
2
1+1
25. Setting t = 0, we obtain a point on the line: Q(0, −4, 0). Therefore, we can form the vector pointing
from Q to P : w = (4, 1, −1) − (0, −4, 0) = (4, 5, −1). A parallel vector to L is v = (2, −1, 3). Thus, we have
P(w, v) =
(4, 5, −1), (2, −1, 3)
(2, −1, 3) = 0.
||(2, −1, 3)||2
Thus, the distance from P to L is given by
||w − P(w, v)|| = ||w|| =
42 + 52 + (−1)2 =
√
42.
26. Setting t = 0, we obtain a point on the line: Q(4, 6, 0). Therefore, we can form the vector pointing from
Q to P : w = (9, 0, 0) − (4, 6, 0) = (5, −6, 0). A parallel vector to L is v = (3, 0, −1). Thus, we have
P(w, v) =
(5, −6, 0), (3, 0, −1)
15
9
3
(3, 0, −1) =
(3, 0, −1) = ( , 0, − ).
||(3, 0, −1)||2
10
2
2
Thus, the distance from P to L is given by
9
3
1
3
||w−P(w, v)|| = ||(5, −6, 0)−( , 0, − )|| = ||( , −6, )|| =
2
2
2
2
2 2
1
1
77
3
9
+ (−6)2 +
=
+ 36 + =
.
2
2
4
4
2
27. Setting t = 0, we obtain a point on the line: Q(−3, 0, 1, 0, 2). Therefore, we can form the vector pointing
from Q to P : w = (1, 2, 3, 2, 1)−(−3, 0, 1, 0, 2) = (4, 2, 2, 2, −1). A parallel vector to L is v = (0, −2, 2, −1, 5).
Thus, we have
(4, 2, 2, 2, −1), (0, −2, 2, −1, 5)
7
7
7 7
35
.
P(w, v) =
(0, −2, 2, −1, 5) = − (0, −2, 2, −1, 5) = 0, , − , , −
||(0, −2, 2, −1, 5)||2
34
17 17 34 34
Thus, the distance from P to L is given by
7
7 7
35
27 41 61 1
||w − P(w, v)|| = ||(4, 2, 2, 2, −1) − 0, , − , , −
|| = ||(4, , , , )||
17 17 34 34
17 17 34 34
√
√
1362 + 542 + 822 + 612 + 12
31858
1
||(136, 54, 82, 61, 1)|| =
=
≈ 5.250.
=
34
34
34
28. (c). Note that w = (x0 − x1 , y0 − y1 , z0 − z1 ) and n = (a, b, c). Now, we have
P(w, n) =
w, n
n,
||n||2
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so that
||P(w, n)|| =
|w, n|
|ax0 + by0 + cz0 + d|
|a(x0 − x1 ) + b(y0 − y1 ) + c(z0 − z1 )|
√
√
=
.
=
||n||
a 2 + b2 + c 2
a 2 + b2 + c 2
29. We have a = 1, b = 2, c = −4, d = −2, x0 = −4, y0 = 7, and z0 = −2. Therefore, from the formula in
Problem 28(c), we obtain the requested distance:
|(1)(−4) + (2)(7) + (−4)(−2) + (−2)|
16
=√ .
2
2
2
21
1 + 2 + (−4)
30. We have a = 3, b = −1, c = −1,d = −5, x0 = 0, y0 = −1, and z0 = 3. Therefore, from the formula in
Problem 28(c), we obtain the requested distance:
|(3)(0) + (−1)(−1) + (−1)(3) + (−5)|
7
=√ .
2
2
2
11
3 + (−1) + (−1)
31. We have a = 2, b = 0, c = −1, d = 0, x0 = −1, y0 = 1, and z0 = −1. Therefore, from the formula in
Problem 28(c), we obtain the requested distance:
|(2)(−1) + (0)(1) + (−1)(−1) + 0|
1
=√ .
2
2
2
5
2 + 0 + (−1)
32. We have a = 0, b = 1, c = −4, d = −2, x0 = 8, y0 = 8, and z0 = −1. Therefore, from the formula in
Problem 28(c), we obtain the requested distance:
10
|(0)(8) + (1)(8) + (−4)(−1) + (−2)|
=√ .
2
2
2
17
0 + 1 + (−4)
33. {u1 , u2 , v} is a linearly independent set of vectors, and u1 , u2 = 0. If we let u3 = v + λu1 + μu2 , then
it must be the case that u3 , u1 = 0 and u3 , u2 = 0. u3 , u1 = 0 =⇒ v + λu1 + μu2 , u1 = 0
=⇒ v, u1 + λu1 , u1 + μu2 , u1 = 0 =⇒ v, u1 + λu1 , u1 + μ · 0 = 0
v, u1 .
=⇒ λ = −
||u1 ||2
u3 , u2 = 0 =⇒ v + λu1 + μu2 , u2 = 0
=⇒ v, u2 + λu1 , u2 + μu2 , u2 = 0 =⇒ v, u2 + λ · 0 + μu2 , u2 = 0
v, u2 .
=⇒ μ = −
||u2 ||2
v, u1 v, u2 Hence, if u3 = v −
u1 −
u2 , then {u1 , u2 , u3 } is an orthogonal basis for the subspace spanned
2
||u1 ||
||u2 ||2
by {u1 , u2 , v}.
34. It was shown in Remark 3 following Definition 5.2.1 that each vector ui is a unit vector. Moreover, for
i = j,
1
1
1
ui , uj = vi ,
vj =
vi , vj = 0,
vi vj vi vj since vi , vj = 0. Therefore {u1 , u2 , . . . , uk } is an orthonormal set of vectors.
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35. (a). Using technology we find that
π
π
π
sin nx dx = 0,
cos nx dx = 0,
sin nx cos mx dx = 0.
−π
Further, for m = n,
−π
−π
π
π
sin nx sin mx dx = 0 and
−π
cos nx cos mx dx = 0.
−π
Consequently the given set of vectors is orthogonal on [−π, π].
(b). Multiplying (5.2.5) by cos mx and integrating over [−π, π] yields
π
1
f (x) cos mx dx = a0
2
−π
π
cos mx dx +
−π
π ,
∞
−π n=1
(an cos nx + bn sin nx) cos mx dx.
Assuming that interchange of the integral and infinite summation is permissible, this can be written
π
π
∞ π
,
1
f (x) cos mx dx = a0
cos mx dx +
(an cos nx + bn sin nx) cos mx dx.
2
−π
−π
n=1 −π
which reduces to
π
f (x) cos mx dx =
−π
1
a0
2
π
−π
π
cos mx dx + am
cos2 mx dx
−π
where we have used the results from part (a). When m = 0, this gives
π
π
1
1 π
f (x) dx = a0
dx = πa0 =⇒ a0 =
f (x) dx,
2
π −π
−π
−π
whereas for m = 0,
π
π
1 π
f (x) cos mx dx = am
cos2 mx dx = πam =⇒ am =
f (x) cos mx dx.
π −π
−π
−π
(c). Multiplying (5.2.5) by sin(mx), integrating over [−π, π], and interchanging the integration and summation yields
π
π
∞ π
,
1
f (x) sin mx dx = a0
sin mx dx +
(an cos nx + bn sin nx) sin mx dx.
2
−π
−π
n=1 −π
Using the results from (a), this reduces to
π
π
1 π
f (x) sin mx dx = bn
sin2 mx dx = πbn =⇒ bn =
f (x) sin mx dx.
π −π
−π
−π
(d). The Fourier coefficients for f are
1 π
1 π
a0 =
xdx = 0, an =
x cos nx dx = 0,
π −π
π −π
1 π
2
2
x sin nx dx = − cos nπ = (−1)n+1 .
bn =
π −π
n
n
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The Fourier series for f is
∞
,
2
n
n=1
(−1)n+1 sin nx.
(e). The approximations using the first term, the first three terms, the first five terms, and the first ten
terms in the Fourier series for f are shown in the accompanying figures.
S3(x)
3
2
1
x
3
2
1
1
2
3
1
2
3
Figure 0.0.65: Figure for Problem 35(e) - 3 terms included
These figures suggest that the Fourier series is converging to the function f (x) at all points in the interval
(−π, π).
Solutions to Section 5.3
(a): TRUE. In this case, all of the inner products appearing in the formulas in Theorem 5.3.2 are 0, so
that we obtain vi = xi for each i. That is, the new set obtained via the Gram-Schmidt Process is identical
to the original set we are given.
(b): TRUE. We have
(x1 + x2 ), x1 x1
||x1 ||2
||x1 ||2 + x2 , x1 = (x1 + x2 ) −
x1
||x1 ||2
= (x1 + x2 ) − x1
= x2 .
v2 = (x1 + x2 ) −
(c): FALSE. For instance, in R3 we can take the vectors x1 = (1, 0, 0), x2 = (1, 1, 0), and x3 = (0, 0, 1). Applying the Gram-Schmidt process to the ordered set {x1 , x2 , x3 } yields the standard basis {e1 , e2 , e3 }. However, applying the Gram-Schmidt process to the ordered set {x3 , x2 , x1 } yields the basis {x3 , x2 , ( 12 , − 12 , 0)}
instead.
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S5(x)
3
2
1
3
2
1
1
2
3
x
1
2
3
Figure 0.0.66: Figure for Problem 35(e) - 5 terms included
S10(x)
3
2
1
3
2
1
1
2
3
x
1
2
3
Figure 0.0.67: Figure for Problem 35(e) - 10 terms included
(d): TRUE. This is assumed in Theorem 5.3.2. If the given vectors were not linearly independent, then
we could not expect to obtain a basis after carrying out the Gram-Schmidt Process.
(e): FALSE. Here is a counterexample (there are many such). In R2 , take x1 = y1 = v1 = (1, 0), and take
x2 = (1, 1) and y2 = (−1, 1). Then B1 = B2 , but applying the Gram-Schmidt Process to B1 and B2 , we
obtain the same orthogonal basis: {(1, 0), (0, 1)}.
(f ): TRUE. We have v1 = x1 , and thus, 2v1 = 2x1 . Now, to find the second vector when the Gram-Schmidt
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Process is applied to B2 , we must compute 2x2 − P(2x2 , 2v1 ). We are given already that
v2 = x2 − P(x2 , v1 ) = x2 −
Now,
2x2 − P(2x2 , 2v1 ) = 2x2 −
x2 , v1 v1 .
||v1 ||2
2x2 , 2v1 x2 , v1 (2v1 ) = 2x2 −
(2v1 ) = 2v2 .
||2v1 ||2
||v1 ||2
Thus, we have obtained the orthogonal set {2v1 , 2v2 }, as requested.
Problems:
1. The given set of vectors is already an orthogonal set, so we √
need not apply the
√ Gram-Schmidt pro2 + 22 + 32 =
1
14 and ||(6, −3, 0)|| =
cess.
Instead,
we
proceed
directly
to
normalization:
||(1,
2,
3)||
=
√
√
62 + (−3)2 + 02 = 45 = 3 5. Thus, we obtain the orthonormal basis
1
1
√ (1, 2, 3), √ (6, −3, 0) .
14
3 5
= (2, 1, −1).
2. Let x1 = (1, −1, −1) and let x2 √
v1 = x1 = (1, −1, −1), ||v1 || = 12 + (−1)2 + (−1)2 = 3.
x2 , v1 = (2, 1, −1), (1, −1, −1) = (2)(1) + (1)(−1) + (−1)(−1) = 2.
x2 , v1 2
1
v1 = (2, 1, −1) − (1, −1, −1) = (4, 5, −1).
v2 = x2 −
||v1 ||2
3
3
√
||v2 || = 13 42 + 52 + (−1)2 = 342 . Hence, an orthonormal basis is
1
1
√ (1, −1, −1), √ (4, 5, −1) .
3
42
3. Let x1 = (2, 1, −2) and let x2 = (1, 3, −1).
v1 = x1 = (2, 1, −2), ||v1 || = 22 + 12 + (−2)2 = 3.
x2 , v1 = (1, 3, −1), (2, 1, −2) = (1)(2) + (3)(1) + (−1)(−2) = 7.
x2 , v1 7
5
v1 = (1, 3, −1) − (2, 1, −2) = (−1, 4, 1).
v2 = x2 −
2
||v ||
9
9
√
1
||v2 || = 59 (−1)2 + 42 + 12 = 5 3 2 . Hence, an orthonormal basis is
1
1
(2, 1, −2), √ (−1, 4, 1) .
3
3 2
4. Note that this particular set of vectors can be shown to be linearly independent, and since we have
three vectors in R3 , they therefore span all of R3 . As such, we could merely replace this basis with the
orthonormal basis {(1, 0, 0), (0, 1, 0), (0, 0, 1)} for the same space. However, our aim here is to demonstrate
the Gram-Schmidt Process.
Let x1 = (1, −5, −3), let x2 = (0,
−1, 3), and let x3 = (−6,
√ 0, −2).
v1 = x1 = (1, −5, −3), ||v1 || = 12 + (−5)2 + (−3)2 = 35.
x2 , v1 = (0, −1, 3), (1, −5, −3) = (0)(1) + (−1)(−5) + (3)(−3) = −4.
x2 , v1 4
1
v1 = (0, −1, 3) + (1, −5, −3) =
v2 = x2 −
(4, −55, 93).
||v1 ||2
35
35
√
1
42 + (−55)2 + 932 = 11690
||v2 || = 35
35 .
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x3 , v1 = (−6, 0, −2), (1, −5, −3) = (−6)(1) + (0)(−5) + (−2)(−3) = 0.
1
1
x3 , v2 = 35
(−6, 0, −2), (4, −55, 93) = 35
(−210) = −6.
x3 , v1
x3 , v2
105 1
3
v1 −
v2 = (−6, 0, −2)−(0, 0, 0)+
v3 = x3 −
· (4, −55, 93) = (−6, 0, −2)+
(4, −55, 93) =
||v1 ||2
||v2 ||2
167 35
167
55
(−18, −3, −1).
167
√
√
55
55
√ 2 . Hence, an orthonormal basis is
(−18)2 + (−3)2 + (−1)2 = 167
· 334 = 55
||v3 || = 167
167
1
1
1
√ (1, −5, −3), √
(4, −55, 93), √
(−18, −3, −1) .
35
11690
334
5. Note that this particular set of vectors can be shown to be linearly independent, and since we have
three vectors in R3 , they therefore span all of R3 . As such, we could merely replace this basis with the
orthonormal basis {(1, 0, 0), (0, 1, 0), (0, 0, 1)} for the same space. However, our aim here is to demonstrate
the Gram-Schmidt Process.
Let x1 = (2, 0, 1), let x2 = (−3,
x3 = (1, −3, 8).
√ 1, 1), and let √
v1 = x1 = (2, 0, 1), ||v1 || = 22 + 02 + 12 = 5.
x2 , v1 = (−3, 1, 1), (2, 0, 1) = (−3)(2) + (1)(0) + (1)(1) = −5.
x2 , v1 v 2 = x2 −
v1 = (−3, 1, 1) + (2, 0, 1) = (−1, 1, 2).
2
||v1 ||
√
||v2 || = (−1)2 + 12 + 22 = 6.
x3 , v1 = (1, −3, 8), (2, 0, 1) = (1)(2) + (−3)(0) + (8)(1) = 10.
x3 , v2 = (1, −3, 8), (−1, 1, 2) = (1)(−1) + (−3)(1) + (8)(2) = 12.
x3 , v1
x3 , v2
v 3 = x3 −
v1 −
v2 = (1, −3, 8) − 2(2, 0, 1) − 2(−1, 1, 2) = (−1, −5, 2).
2
||v2 ||2 √
||v1 ||
||v3 || = (−1)2 + (−5)2 + 22 = 30. Hence, an orthonormal basis is
1
1
1
√ (2, 0, 1), √ (−1, 1, 2), √ (−1, −5, 2) .
5
6
30
= (1, 2, 1, 2).
6. Let x1 = (−1, 1, 1, 1) and let x2 v1 = x1 = (−1, 1, 1, 1), ||v1 || = (−1)2 + 12 + 12 + 12 = 2.
x2 , v1 = (1, 2, 1, 2), (−1, 1, 1, 1) = (1)(−1) + (2)(1) + (1)(1) + (2)(1) = 4.
x2 , v1 v2 = x2 −
v1 = (1, 2, 1, 2) − (−1, 1, 1, 1) = (2, 1, 0, 1).
2
√ ||v1 ||
√
||v2 || = 22 + 12 + 02 + 12 = 6. Hence, an orthonormal basis is
1
1
(−1, 1, 1, 1), √ (2, 1, 0, 1) .
2
6
7. Let x1 = (1, 0, −1, 0), let x2 = (1,
1, −1, 0), and let x3 = (−1,
√ 1, 0, 1).
v1 = x1 = (1, 0, −1, 0), ||v1 || = 12 + 02 + (−1)2 + 02 = 2.
x2 , v1 = (1, 1, −1, 0), (1, 0, −1, 0) = (1)(1) + (1)(0) + (−1)(−1) + (0)(0) = 2.
√
x2 , v1 v2 = x2 −
v1 = (1, 1, −1, 0) − (1, 0, −1, 0) = (0, 1, 0, 0). ||v2 || = 02 + 12 + 02 + 02 = 1.
2
||v1 ||
x3 , v1 = (−1, 1, 0, 1), (1, 0, −1, 0) = (−1)(1) + (1)(0) + (0)(−1) + (1)(0) = −1.
x3 , v2 = (−1, 1, 0, 1), (0, 1, 0, 0) = (−1)(0) + (1)(1) + (0)(0) + (1)(0) = 1.
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x3 , v1 x3 , v2 1
1
v1 −
v2 = (−1, 1, 0, 1) + (1, 0, −1, 0) − (0, 1, 0, 0) = (−1, 0, −1, 2). ||v3 || =
v3 = x3 −
2
2
||v1 ||
||v√
2
2
2 ||
6
1
2 + 02 + (−1)2 + 22 =
(−1)
.
Hence,
an
orthonormal
basis
is
2
2
1
1
√ (1, 0, −1, 0), (0, 1, 0, 0), √ (−1, 0, −1, 2) .
2
6
8. Let x1 = (1, 2, 0, 1), let x2 = (2,
√ 1, 1, 0), and let x3 =√(1, 0, 2, 1).
v1 = x1 = (1, 2, 0, 1), ||v1 || = 12 + 22 + 02 + 12 = 6.
x2 , v1 = (2, 1, 1, 0), (1, 2, 0, 1) = (2)(1) + (1)(2) + (1)(0) + (0)(1) = 4.
x2 , v1 4
1
v1 = (2, 1, 1, 0) − (1, 2, 0, 1) = (4, −1, 3, −2).
v2 = x2 −
2
||v
||
6
3
1
√
||v2 || = 13 42 + (−1)2 + 32 + (−2)2 = 13 30.
x3 , v1 = (1, 0, 2, 1), (1, 2, 0, 1) = (1)(1) + (0)(2) + (2)(0) + (1)(1) = 2.
x3 , v2 = (1, 0, 2, 1), 13 (4, −1, 3, −2) = (1)( 43 ) + (0)(− 13 ) + (2)(1) + (1)(− 23 ) = 83 .
x3 , v1 x3 , v2 2
4
2
v3 = x3 −
v1 −
v2 = (1, 0, 2, 1) − (1, 2, 0, 1) − (4, −1, 3, −2) = (−1, −1, 3, 3).
2
2
||v
||
||v
||
6
15
5
2
1
||v3 || = 25 (−1)2 + (−1)2 + 32 + 32 = √45 . Hence, an orthonormal basis is
1
1
1
√ (1, 2, 0, 1), √ (4, −1, 3, −2), √ (−1, −1, 3, 3) .
6
30
2 5
9. Let x1 = (1, 1, −1, 0), let x2 = (−1,
0, 1, 1), and let x3 = (2,
√−1, 2, 1).
v1 = x1 = (1, 1, −1, 0), ||v1 || = 12 + 12 + (−1)2 + 02 = 3.
x2 , v1 = (−1, 0, 1, 1), (1, 1, −1, 0) = (−1)(1) + (0)(1) + (1)(−1) + (1)(0) = −2.
x2 , v1 2
1
v1 = (−1, 0, 1, 1) + (1, 1, −1, 0) = (−1, 2, 1, 3).
v 2 = x2 −
2
||v
||
3
3
1
√
||v2 || = 13 (−1)2 + 22 + 12 + 32 = 13 15.
x3 , v1 = (2, −1, 2, 1), (1, 1, −1, 0) = (2)(1) + (−1)(1) + (2)(−1) + (1)(0) = −1.
x3 , v2 = (2, −1, 2, 1), 13 (−1, 2, 1, 3) = (2)(− 13 ) + (−1)( 23 ) + (2)( 13 ) + (1)(1) = 13 .
x3 , v1 x3 , v2 1
1
4
v3 = x3 −
v1 −
v2 = (2, −1, 2, 1) + (1, 1, −1, 0) − (−1, 2, 1, 3) = (3, −1, 2, 1).
2
2
||v
||
||v
||
3
15
5
1
2
√
||v3 || = 45 32 + (−1)2 + 22 + 12 = 45 15. Hence, an orthonormal basis is
1
1
1
√ (1, 1, −1, 0), √ (−1, 2, 1, 3), √ (3, −1, 2, 1) .
3
15
15
10. Let x1 = (1, 2, 3, 4, 5) and let x√
2 = (−7, 0, 1, −2, 0).
√
v1 = x1 = (1, 2, 3, 4, 5), ||v1 || = 12 + 22 + 32 + 42 + 52 = 55.
x2 , v1 = (1, 2, 3, 4, 5), (−7, 0, 1, −2, 0) = (1)(−7) + (2)(0) + (3)(1) + (4)(−2) + (5)(0) = −12.
x2 , v1 12
373 24 91 62 12
v1 = (−7, 0, 1, −2, 0) + (1, 2, 3, 4, 5) = (−
v 2 = x2 −
, , , − , ).
||v1 ||2
55
55 55 55 55 11
√
1
(−373)2 + 242 + 912 + (−62)2 + (60)2 = 3 17,270
. Hence, an orthonormal basis is
||v2 || = 55
55
1
1
√ (1, 2, 3, 4, 5), √
(−373, 24, 91, −62, 60) .
55
3 17270
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1 −3
2 0 −1
. Hence, a basis for rowspace(A) is {(1, −3, 2, 0, −1), (0, 1, −3, 13 , 2)}.
2
0
1 −3 13
The vectors in this basis are not orthogonal. We therefore use the Gram-Schmidt process to determine an
orthogonal basis. Let x1 = (1, −3, 2, 0, −1) and x2 = (0, 1, −3, 13 , 2).
√
Then v1 = x1 = (1, −3, 2, 0, −1). ||v1 || = 12 + (−3)2 + 22 + 02 + (−1)2 = 15.
x2 , v1 = (0, 1, −3, 13 , 2), (1, −3, 2, 0, −1) = (0)(1) + (1)(−3) + (−3)(2) + ( 13 )(0) + (2)(−1) = −11.
x2 , v1 1
11
1
v1 = (0, 1, −3, , 2)+ (1, −3, 2, 0, −1) =
(11, −18, −23, 5, 19). Thus, an orthogonal
v2 = x2 −
||v1 ||2
3
15
15
basis for rowspace(A) is given by {(1, −3, 2, 0, −1), (11, −18, −23, 5, 19)}.
Since colspace(A) = R2 , we can simply take {(1, 0), (0, 1)} as an orthogonal basis for colspace(A).
⎤
⎡
1 5
⎢ 0 1 ⎥
⎥
⎢
2
⎥
12. A row-echelon form of A is ⎢
⎢ 0 0 ⎥. Obviously, rowspace(A) = R , so we may simply take
⎣ 0 0 ⎦
0 0
{(1, 0), (0, 1)} as orthogonal basis for rowspace(A).
A basis for colspace(A) is simply composed of the two columns of A:
⎧⎡
⎤⎫
⎤ ⎡
5 ⎪
1
⎪
⎪
⎪
⎪
⎪
⎪⎢ 2 ⎥ ⎢ 4 ⎥⎪
⎨
⎥⎬
⎥ ⎢
⎢
⎢ 3 ⎥,⎢ 3 ⎥ .
⎥
⎥ ⎢
⎢
⎪
⎪
⎪⎣ 4 ⎦ ⎣ 2 ⎦⎪
⎪
⎪
⎪
⎪
⎭
⎩
1
5
11. A row-echelon form of A is
The vectors in this basis are not orthogonal. We therefore use the Gram-Schmidt process to determine
an orthogonal basis. (We will render the columns as rows while we perform the steps to save space.) Let
x1 = (1, 2, 3, 4, 5) and let x2 = (5, 4, 3, 2, 1).
√
√
Then v1 = x1 = (1, 2, 3, 4, 5). ||v1 || = 12 + 22 + 32 + 42 + 52 = 55.
x2 , v1 = (5, 4, 3, 2, 1), (1, 2, 3, 4, 5) = (5)(1) + (4)(2) + (3)(3) + (2)(4) + (1)(5) = 35.
x2 , v1 35
1
v 2 = x2 −
v1 = (5, 4, 3, 2, 1) −
(1, 2, 3, 4, 5) =
(48, 30, 12, −6, −24). Thus, an orthogonal
||v1 ||2
55
11
basis for colspace(A) is given by
⎧⎡
⎤⎫
⎤ ⎡
48 ⎪
1
⎪
⎪
⎪
⎪
⎪
⎥⎪
⎥ ⎢
⎪
⎨⎢
⎢ 2 ⎥ ⎢ 30 ⎥⎬
⎢ 3 ⎥ , ⎢ 12 ⎥ .
⎥⎪
⎥ ⎢
⎢
⎪
⎪
⎣ 4 ⎦ ⎣ −6 ⎦⎪
⎪
⎪
⎪
⎪
⎭
⎩
−24
5
⎡
⎤
1 −2 1
1 17 ⎦. Hence, a basis for rowspace(A) is {(1, −2, 1), (0, 7, 1)}.
13. A row-echelon form of A is ⎣ 0
0
0 0
The vectors in this basis are not orthogonal. We therefore use the Gram-Schmidt process to determine an
orthogonal basis. Let x1 = (1, −2, 1) and x2 = (0, 7, 1). We have v1 = x1 = (1, −2, 1). Now ||v1 || =
√
x2 , v1 12 + (−2)2 + 12 = 6 and x2 , v1 = (1)(0) + (−2)(7) + (1)(1) = −13. Thus, v2 = x2 −
v1 =
||v1 ||2
13
1
(0, 7, 1) + (1, −2, 1) = (13, 16, 19).
6
6
Then an orthogonal basis for rowspace(A) is {(1, −2, 1), (13, 16, 19)}.
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Next we consider colspace(A), and we will render the column vectors that arise as rows in order to save space. From the row-echelon form of A obtained above, a basis for colspace(A) is given by {(3, 1, 1), (1, −2, 5)}.
The vectors in this basis are not orthogonal. We therefore use the Gram-Schmidt process to determine an
orthogonal
basis. √
Let x1 = (3, 1, 1) and x2 = (1, −2, 5). We have v1 = x1 = (3, 1, 1). Now ||v1 || =
√
32 + 12 + 12 = 11 and x2 , v1 = (1, −2, 5), (3, 1, 1) = (1)(3) + (−2)(1) + (5)(1) = 6. Thus, ,
x2 , v1 6
7
v1 = (1, −2, 5) − (3, 1, 1) =
v2 = x2 −
(−1, −4, 7). Then, an orthogonal basis for colspace(A)
2
||v
||
11
11
1
⎧⎡
⎤ ⎡
⎤⎫
−1 ⎬
⎨ 3
is ⎣ 1 ⎦ , ⎣ −4 ⎦ .
⎩
⎭
1
7
⎡
⎤
1 −4
7
1 −3 ⎦. Hence, a basis for rowspace(A) is {(1, −4, 7), (0, 1, −3)}.
14. A row-echelon form of A is ⎣ 0
0
0
0
The vectors in this basis are not orthogonal. We therefore use the Gram-Schmidt process to determine
an orthogonal
basis. Let x1√ = (1, −4, 7) and x2 = (0, 1, −3). We have v1 = x1 = (1, −4, 7). Now
||v1 || = 12 + (−4)2 + 72 = 66 and x2 , v1 = (0, 1, −3), (1, −4, 7) = (0)(1) + (1)(−4) + (−3)(7) = −25.
x2 , v1 25
1
v1 = (0, 1, −3) + (1, −4, 7) =
Thus, v2 = x2 −
(25, −34, −23).
||v1 ||2
66
66
Then an orthogonal basis for rowspace(A) is {(1, −4, 7), (25, −34, −23)}.
Next we consider colspace(A), and we will render the column vectors that arise as rows in order to save space. From the row-echelon form of A obtained above, a basis for colspace(A) is given by {(1, −2, −1), (−4, 6, 0)}.
The vectors in this basis are not orthogonal. We therefore use the Gram-Schmidt process to determine an
orthogonal
basis. Let x1 √
= (1, −2, −1) and x2 = (−4, 6, 0). We have v1 = x1 = (1, −2, −1). Now ||v1 || =
12 + (−2)2 + (−1)2 = 6 and x2 , v1 = (−4, 6, 0), (1, −2, −1) = (−4)(1) + (6)(−2) + (0)(−1) = −16.
x2 , v1 16
1
v1 = (−4, 6, 0) +
Thus, v2 = x2 −
(1, −2, −1) = (−4, 2, −8). Then an orthogonal basis for
2
||v
||
6
3
1⎤ ⎡
⎧⎡
⎤⎫
1
−4
⎨
⎬
colspace(A) is ⎣ −2 ⎦ , ⎣ 2 ⎦ .
⎩
⎭
−1
−8
15. Let v1 = (1 + i, i, 2 − i) and v
2 = (1 + 2i, 1 − i, i).
√
u1 = v1 = (1 + i, i, 2 − i), ||u1 || = (1 + i)(1 − i) + i(−i) + (2 − i)(2 + i) = 2 2.
v2 , u1 = (1 + 2i)(1 − i) + (1 − i)(−i) + i(2 + i) = 1 + 2i.
v2 , u1 1
1
u1 = (1 + 2i, 1 − i, i) − (1 + 2i)(1 + i, i, 2 − i) = (9 + 13i, 10 − 9i, −4 + 5i).
u2 = v 2 −
||u1 ||2
8
8
√
118
1
(9 + 13i)(9 − 13i) + (10 − 9i)(10 + 9i) + (−4 + 5i)(−4 − 5i) =
. Hence, an orthonormal
||u2 || =
8
4
basis is:
√
√
2
118
(1 + i, i, 2 − i),
(9 + 13i, 10 − 9i, −4 + 5i) .
4
236
16. Let v1 = (1 − i, 0, i) and v
2 = (1, 1 + i, 0).
√
u1 = v1 = (1 − i, 0, i), ||u1 || = (1 − i)(1 + i) + 0 + i(−i) = 3.
v2 , u1 = (1, 1 + i, 0), (1 − i, 0, i) = 1(1 + i) + (1 + i)0 + 0(−i) = 1 + i.
v2 , u1 1+i
1
u1 = (1, 1 + i, 0) −
u2 = v2 −
(1 − i, 0, i) = (1, 3 + 3i, 1 − i).
||u1 ||2
3
3
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||u2 || =
√
1
21
1 + (3 + 3i)(3 − 3i) + (1 − i)(1 + i) =
. Hence, an orthonormal basis is:
3
3
√
√
3
21
(1 − i, 0, i),
(1, 3 + 3i, 1 − i) .
3
21
17. Let f1 = 1 + 2x and f2 = −2 − x + x2 . We will define the orthogonal basis as {g1 , g2 } as follows. Set
g1 = f1 = 1 + 2x. Then
1
1
13
2
2
||g1 || = g1 , g1 =
(1 + 2x) dx =
(1 + 4x + 4x )dx =
3
0
0
and
g1 , f2 =
1
(1 + 2x)(−2 + x + x2 )dx =
0
Thus,
g 2 = f2 −
1
0
(−2 − 3x + 3x2 + 2x3 )dx = −2.
g1 , f2 6
20
1
g1 = (−2 − x + x2 ) + (1 + 2x) = − − x + x2 .
2
||g1 ||
13
13 13
Thus, we have the orthogonal basis
1 + 2x, −
20
1
− x + x2 .
13 13
18. Let f1 = 1, f2 = x and f3 = x2 .
g1 = f1 = 1;
2 1
1
1
1
x
2
dx = 1; f2 , g1 =
xdx =
= .
||g1 || =
2
2
0
0
0
f2 , g1 1
1
g 2 = f2 −
g1 = x − = (2x − 1).
||g1 ||2
2
2
2
3
1
1
1
x
1
1
1
x2
x
2
2
dx =
x−
x −x+
dx =
=
||g2 || =
−
+
.
2
4
3
2
4
12
0
0
3 1 0
1
1
x
f3 , g1 =
x2 dx =
= .
3
3
0
0
4
1
1 1
1
x
x3
2
x x−
=
f3 , g2 =
dx =
−
.
2
4
6
12
0
0
1
f3 , g1 f3 , g2 1
1
2
= (6x2 − 6x + 1). Thus, an orthogonal basis is given
g1 −
g2 = x − − x −
g 3 = f3 −
2
2
||g1 ||
||g2 ||
3
2
6
by:
1
1
2
1, (2x − 1), (6x − 6x + 1) .
2
6
19. Let f1 = 1, f2 = x2 and f3 = x4 for all x in [−1, 1].
1
1
1
2
2
2
2
4
x dx = , f1 , f3 =
x dx = , and f2 , f3 =
x6 dx = .
f1 , f2 =
3
5
7
−1
−1
1−1
1
2
||f1 ||2 =
dx = 2, ||f2 ||2 =
x4 dx = .
5
−1
−1
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Let g1 = f1 = 1.
f2 , g1 x2 , 1
1
1
g 2 = f2 −
g 1 = x2 −
· 1 = x2 − = (3x − 1).
2
||g1 ||
||1||2
3
3
x4 , x2 − 13 f3 , g1 f3 , g2 x4 , 1
1
4
2
x −
g1 −
g2 = x −
·1−
g 3 = f3 −
||g1 ||2
||g2 ||2
||1||2
3
||x2 − 13 ||2
= x4 −
6x2
3
1
+
=
(35x4 − 30x2 + 3).
7
35
35
Thus, an orthogonal basis is given by:
1
1
2
4
2
1, (3x − 1),
(35x − 30x + 3) .
3
35
π π
20. Let f1 = 1, f2 = sin x and f3 = cos x for all x in − , .
2 2
π/2
π/2
sin xdx = [− cos 2x]−π/2 = 0. Therefore, f1 and f2 are orthogonal.
f1 , f2 =
−π/2
π/2
1
sin x cos xdx =
f2 , f3 =
2
−π/2
nal.
π/2
f1 , f3 =
π/2
1
sin 2xdx = − cos 2x
= 0. Therefore, f2 and f3 are orthogo4
−π/2
−π/2
π/2
π/2
cos xdx = [sin x]−π/2 = 2.
π/2
Let g1 = f1 = 1 so that ||g1 ||2 =
dx = π.
−π/2
π/2
π/2
1 − cos 2x
π
sin2 xdx =
g2 = f2 = sin x, and ||g2 ||2 =
dx = .
2
2
−π/2
−π/2
f3 , g1 f3 , g2 2
1
g1 −
g2 = cos x − · 1 − 0 · sin x = (π cos x − 2). Thus, an orthogonal basis for
g 3 = f3 −
||g1 ||2
||g2 ||2
π
π
the subspace of C 0 [−π/2, π/2] spanned by {1, sin x, cos x} is:
1
1, sin x, (π cos x − 2) .
π
−π/2
1 −1
2 −3
21. Given A1 =
and A2 =
. Using the Gram-Schmidt procedure:
1
4
1
2
1 −1
, A2 , B1 = 10 + 6 + 24 + 5 = 45, and ||B1 ||2 = 5 + 2 + 12 + 5 = 24.
B1 =
2
1
1
A2 , B1 15 1 −1
− 98
2 −3
8
=
. Thus, an orthogonal basis for the
B
=
−
B2 = A2 −
1
1
2
1
4
1
− 78
||B1 ||2
8
4
subspace of M2 (R) spanned by A1 and A2 is:
1
− 98
1 −1
8
, 1
.
2
1
− 78
4
0 1
0 1
1
, A2 =
and A3 =
22. Given A1 =
1
0
1
1
1
0 1
2
, A2 , B1 = 5, and ||B1 || = 5.
B1 =
1 0
1
0
. Using the Gram-Schmidt procedure:
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A2 , B1 0 1
0 1
0 0
B
=
−
=
.
1
1 1
1 0
0 1
||B1 ||2
2
Also, A3 , B1 = 5, A3 , B2 = 0, and ||B
2 || =5, so
that A3 , B2 A3 , B1 1 1
0 1
0 0
1
B
−
B
=
−
−
0
=
B3 = A3 −
1
2
1 0
1 0
0 1
0
||B1 ||2
||B2 ||2
basis for the subspace of M2 (R) spanned by A1 , A2 , and A3 is:
0 1
1 0
0 0
,
,
,
1 0
0 0
0 1
B2 = A2 −
0
0
. Thus, an orthogonal
which is the subspace of all symmetric matrices in M2 (R).
23. Given p1 (x) = 1 − 2x + 2x2 and p2 (x) = 2 − x − x2 . Using the Gram-Schmidt procedure:
q1 = 1 − 2x + 2x2 , p2 , q1 = 2 · 1 + (−1)(−2) + (−1)2 = 2, ||q1 ||2 = 12 + (−2)2 + 22 = 9. So,
p2 , q1 2
1
q1 = 2 − x − x2 − (1 − 2x + 2x2 ) = (16 − 5x − 13x2 ). Thus, an orthogonal basis for the
q 2 = p2 −
2
||q1 ||
9
9
subspace spanned by p1 and p2 is {1 − 2x + 2x2 , 16 − 5x − 13x2 }.
24. Given p1 (x) = 1 + x2 , p2 (x) = 2 − x + x3 , and p3 (x) = −x + 2x2 . Using the Gram-Schmidt procedure:
q1 = 1 + x2 , p2 , q1 = 2 · 1 + (−1)(0) + 0 · 1 + 1 · 2 = 2, and q1 2 = 12 + 12 = 2. So,
p2 , q1 q1 = 2 − x + x3 − (1 + x2 ) = 1 − x − x2 + x3 . Also,
q 2 = p2 −
||q1 ||2
p3 , q1 = 0 · 1 + (−1)0 + 2 · 1 + 02 = 2
p3 , q2 = 0 · 1 + (−1)2 + 2(−1) + 0 · 1 = −1, and
||q2 ||2 = 12 + (−1)2 + (−1)2 + 12 = 4 so that
p3 , q1 p3 , q2 1
1
q1 −
q2 = −x + 2x2 − (1 + x2 ) + (1 − x − x2 + x3 ) = (−3 − 5x + 3x2 + x3 ). Thus, an
q 3 = p3 −
||q1 ||2
||q2 ||2
4
4
orthogonal basis for the subspace spanned by p1 , p2 , and p3 is {1 + x2 , 1 − x − x2 + x3 , −3 − 5x + 3x2 + x3 }.
25. Set
z = x − P(x, v1 ) − P(x, v2 ) − · · · − P(x, vk ).
To verify that z is orthogonal to vi , we compute the inner product of z with vi :
z, vi = x − P(x, v1 ) − P(x, v2 ) − · · · − P(x, vk ), vi = x, vi − P(x, v1 ), vi − P(x, v2 ), vi − · · · − P(x, vk ), vi .
Since P(x, vj ) is a multiple of vj and the set {v1 , v2 , . . . , vk } is orthogonal, all of the subtracted terms on
the right-hand side of this expression are zero except
P(x, vi ), vi = x, vi x, vi vi , vi =
vi , vi = x, vi .
vi 2
vi 2
Therefore,
z, vi = x, vi − x, vi = 0,
which implies that z is orthogonal to vi .
Solutions to Section 5.4
(a): TRUE. The given conclusion is essentially the definition of a least squares solution.
(b): FALSE. The correct formula is x0 = (AT A)−1 AT b.
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(c): TRUE. Any least squares solution x0 to Ax = b must satisfy the normal equations in the system
AT Ax0 = AT b. Since A is invertible, so is AT , and thus we can solve this system for x0 which yields
precisely x0 = A−1 b.
(d): FALSE. Example 5.4.3 provides a counterexample to this statement.
(e): TRUE. In the text, we observed the projection matrix P projects b onto the column space of A as
follows:
P b = A(AT A)−1 AT b = Ax0 ∈ colspace(A).
(f ): TRUE. We have
P = A(AT A)−1 AT = AA−1 (AT )−1 AT = In .
Problems:
1. We construct
A=
6 1
−2 1
,
A quick computation shows that
40 4
AT A =
,
and thus,
4 2
x=
a
b
,
(AT A)−1 =
and
1
64
3
0
b=
.
2 −4
−4 40
=
1
32
1 −2
−2 20
Continuing with the computation, we now find
1
1 −2
6 −2
3
3/8
x0 = (AT A)−1 AT b =
=
.
1
1
0
3/4
32 −2 20
Therefore, the least squares line is
3
3
x+ .
8
4
In this case, this is precisely the line that joins the two given points in the xy-plane.
y(x) =
2. We construct
1
2
1
1
A quick computation shows that
5
T
A A=
3
3
2
A=
,
x=
a
b
,
and
,
T
and thus,
(A A)
Continuing with the computation, we now find
2 −3
1
T
−1 T
x0 = (A A) A b =
−3
5
1
10
20
b=
−1
2
1
.
2 −3
−3
5
=
10
20
=
10
0
.
Therefore, the least squares line is
y(x) = 10x.
In this case, this is precisely the line that joins the two given points in the xy-plane.
3. We construct
⎡
1
A=⎣ 2
3
⎤
1
1 ⎦,
1
x=
a
b
⎤
10
b = ⎣ 20 ⎦ .
10
⎡
,
and
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.
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A quick computation shows that
14 6
,
AT A =
6 3
1
6
(AT A)−1 =
and thus,
3 −6
−6 14
.
Continuing with the computation, we now find
1
x0 = (AT A)−1 AT b =
6
3 −6
−6 14
1
1
2
1
3
1
⎡
⎤
10
0
⎣ 20 ⎦ =
.
40/3
10
Therefore, the least squares line is
y(x) =
4. We construct
⎡
⎤
1
1 ⎦,
1
2
A=⎣ 1
0
A quick computation shows that
5
AT A =
3
x=
a
b
40
.
3
⎡
,
3
3
,
⎤
−2
b = ⎣ −3 ⎦ .
0
and
(AT A)−1 =
and thus,
1
6
3 −3
−3
5
.
Continuing with the computation, we now find
1
x0 = (AT A)−1 AT b =
6
3 −3
−3
5
2
1
1
1
0
1
⎡
⎤
−2
−1
⎣ −3 ⎦ =
.
−2/3
0
Therefore, the least squares line is
2
y(x) = −x − .
3
5. We construct
⎡
2
⎢ 0
A=⎢
⎣ 5
1
⎤
1
1 ⎥
⎥,
1 ⎦
1
x=
A quick computation shows that
30 8
T
A A=
,
and thus,
8 4
a
b
T
(A A)
⎤
5
⎢ −1 ⎥
⎥
b=⎢
⎣ 3 ⎦.
−3
⎡
,
−1
and
1
=
56
4 −8
−8 30
1
=
28
2 −4
−4 15
Continuing with the computation, we now find
x0 = (AT A)−1 AT b =
1
28
2 −4
−4 15
2
1
0
1
5
1
Therefore, the least squares line is
y(x) = x − 1.
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1
⎤
5
⎢ −1 ⎥
1
⎥=
⎢
.
⎣ 3 ⎦
−1
−3
⎡
.
379
⎤
1
1 ⎥
⎥,
1 ⎦
1
⎡
6. We construct
0
⎢ 1
A=⎢
⎣ 2
4
A quick computation shows that
21
T
A A=
7
7
4
a
b
x=
,
and
,
⎤
3
⎢ −1 ⎥
⎥
b=⎢
⎣ 6 ⎦.
6
⎡
T
and thus,
(A A)
−1
1
=
35
Continuing with the computation, we now find
x0 = (AT A)−1 AT b =
1
35
4 −7
−7 21
0
1
1
1
2
1
4
1
4 −7
−7 21
.
⎤
3
⎢ −1 ⎥
⎥ = 6/5 .
⎢
⎣ 6 ⎦
7/5
6
⎡
Therefore, the least squares line is
6
7
x+ .
5
5
y(x) =
−7
⎢ −4
⎢
A=⎢
⎢ 2
⎣ 3
6
A quick computation shows that
114
T
A A=
0
0
5
⎤
3
⎢ 0 ⎥
⎥
⎢
⎥
b=⎢
⎢ −1 ⎥ .
⎣ 6 ⎦
−1
⎡
⎤
1
1 ⎥
⎥
1 ⎥
⎥,
1 ⎦
1
⎡
7. We construct
x=
a
b
,
and
,
and thus,
T
(A A)
−1
Continuing with the computation, we now find
T
x0 = (A A)
−1
T
A b=
1
114
0
=
1
114
0
0
1
5
.
⎤
3
⎢ 0 ⎥ ⎥
−11/114
0
−7 −4 2 3 6 ⎢
⎥
⎢
−1 ⎥ =
.
1
7/5
1
1 1 1 1 ⎢
5
⎣ 6 ⎦
−1
⎡
Therefore, the least squares line is
y(x) = −
8. We construct
⎡
1
A=⎣ 4
9
1
2
3
⎤
1
1 ⎦,
1
A quick computation shows that
⎤
⎡
98 36 14
AT A = ⎣ 36 14 6 ⎦ ,
14 6
3
11
7
x+ .
114
5
⎡
⎤
a
x = ⎣ b ⎦,
c
and thus,
⎡
and
⎤
10
b = ⎣ 20 ⎦ .
10
⎡
⎤
3 −12
10
1
49 −42 ⎦ .
(AT A)−1 = ⎣ −12
2
10 −42
38
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Continuing with the computation, we now find
⎡
⎤⎡
3 −12
10
1
1
49 −42 ⎦ ⎣ 1
x0 = (AT A)−1 AT b = ⎣ −12
2
10 −42
38
1
4
2
1
⎤⎡
⎤ ⎡
⎤
9
10
−10
3 ⎦ ⎣ 20 ⎦ = ⎣ 40 ⎦ .
1
10
−20
Therefore, the least squares parabola is
y(x) = −10x2 + 40x − 20.
In this case, this is precisely the unique parabola that joins the three given points in the xy-plane.
9. We construct
⎡
4
A=⎣ 1
0
2
1
0
⎤
1
1 ⎦,
1
A quick computation shows that
⎡
⎤
17 9 5
AT A = ⎣ 9 5 3 ⎦ ,
5 3 3
⎡
⎤
a
x = ⎣ b ⎦,
c
⎡
and
⎡
⎤
3 −6
1
1
(AT A)−1 = ⎣ −6 13 −3 ⎦ .
2
1 −3
2
and thus,
Continuing with the computation, we now find
⎡
⎤⎡
3 −6
1
4
1
x0 = (AT A)−1 AT b = ⎣ −6 13 −3 ⎦ ⎣ 2
2
1 −3
2
1
Therefore, the least squares parabola is
⎤
−2
b = ⎣ −3 ⎦ .
0
1
1
1
⎤⎡
⎤ ⎡
⎤
−2
0
2
0 ⎦ ⎣ −3 ⎦ = ⎣ −5 ⎦ .
1
0
0
y(x) = 2x2 − 5x.
In this case, this is precisely the unique parabola that joins the three given points in the xy-plane.
13. We construct
⎤
0
⎢ 3 ⎥
⎥
A=⎢
⎣ 5 ⎦,
6
⎤
0
⎢ −10 ⎥
⎥
b=⎢
⎣ −14 ⎦ .
−19
⎡
⎡
x=
a
,
and
A quick computation shows that
AT A =
70
,
and thus,
(AT A)−1 =
1
70
.
Continuing with the computation, we now find
⎤
0
⎢ −10 ⎥
⎥
⎢
⎣ −14 ⎦ =
−19
⎡
x0 = (AT A)−1 AT b =
1
70
0
3
5
6
− 107
35
.
Thus, we can estimate the spring constant k for the spring with k = −a = 107
35 pounds per inch.
14. Since the population growth of a bacteria culture is known to follow an exponential model, we will seek
to write P (t) in the form P (t) = Cekt , for some constants C and k. Using the notation y(t) = ln P (t), we
compute the following table from the given data (all values are approximated to three decimal places):
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t
y(t)
0
1.253
4
2.833
5
3.239
8
4.449
10
5.244
We can now formulate this least squares problem via the vector equation Ax = y, where
⎤
⎡
⎤
⎡
1.253
0 1
⎢ 2.833 ⎥
⎢ 4 1 ⎥
⎥
⎢
⎥
⎢
⎢ 3.239 ⎥ .
⎥, x = k ,
5
1
and
y
=
A=⎢
⎥
⎢
⎥
⎢
b
⎣ 4.449 ⎦
⎣ 8 1 ⎦
5.244
10 1
We compute
205
27
AT A =
27
5
,
(AT A)−1 =
and thus,
1
296
5 −27
−27 205
.
Continuing the computation, we now find
⎤
1.253
⎢ 2.833 ⎥ ⎥
.400
10 ⎢
⎢ 3.239 ⎥ =
.
⎥
1.275
1 ⎢
⎣ 4.449 ⎦
5.244
⎡
x0 = (AT A)−1 AT y =
1
296
5 −27
−27 205
0
1
4
1
5
1
8
1
Thus, we approximate k0 = 0.400 and b0 = 1.275. Hence, C0 = eb0 ≈ 3.579. Thus, our best fit exponential
function describing the growth of this bacteria culture is P (t) = 3.579e0.400t thousands of bacteria after t
hours.
15. Since the population growth of a bacteria culture is known to follow an exponential model, we will seek
to write P (t) in the form P (t) = Cekt , for some constants C and k. Using the notation y(t) = ln P (t), we
compute the following table from the given data (all values are approximated to three decimal places):
t
y(t)
0
.405
12
1.609
20
2.398
24
2.773
We can now formulate this least squares problem via the vector equation Ax = y, where
⎤
⎤
⎡
⎡
.405
0 1
⎢ 1.609 ⎥
⎢ 12 1 ⎥
k
⎥
⎥
and y = ⎢
A=⎢
⎣ 2.398 ⎦ .
⎣ 20 1 ⎦ , x = b ,
2.773
24 1
We compute
T
A A=
1120
56
56
4
,
and thus,
T
(A A)
−1
1
=
1344
4
−56
−56
1120
1
=
336
1 −14
−14 280
Continuing the computation, we now find
x0 = (AT A)−1 AT y =
1
336
1 −14
−14 280
0
1
12
1
20
1
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1
⎤
.405
⎢ 1.609 ⎥
⎥ = .099 .
⎢
⎣ 2.398 ⎦
.412
2.773
⎡
.
382
Thus, we approximate k0 = 0.099 and b0 = .412. Hence, C0 = eb0 ≈ 1.510. Thus, our best fit exponential
function describing the growth of this bacteria culture is P (t) = 1.510e0.099t thousands of bacteria after t
minutes.
16. We will use the notation y(t) = ln A(t). We compute the following table from the given data (all values
are approximated to three decimal places):
t
y(t)
0
4.605
1
3.199
3
.405
4
-.916
6
-2.303
We can now formulate this least squares problem via the vector equation Ax = y, where
⎤
⎡
⎡
⎤
4.605
0 1
⎢ 3.199 ⎥
⎢ 1 1 ⎥
⎥
⎢
⎢
⎥
k
⎢
⎥
.405 ⎥
,
and y = ⎢
A = ⎢ 3 1 ⎥, x =
⎥.
⎢
b
⎣ −.916 ⎦
⎣ 4 1 ⎦
−2.303
6 1
We compute
T
A A=
62
14
14
5
,
T
and thus,
(A A)
−1
Continuing the computation, we now find
T
x0 = (A A)
−1
1
A y=
114
T
5 −14
−14
62
1
=
114
5 −14
−14
62
.
⎤
4.605
⎢ 3.199 ⎥ ⎥
−1.186
6 ⎢
⎥
⎢
.405 ⎥ =
.
4.319
1 ⎢
⎣ −.916 ⎦
−2.303
⎡
0
1
1
1
3
1
4
1
Thus, we approximate k0 = −1.186 and b0 = 4.319. Hence, C0 = eb0 ≈ 75.113. Thus, our best fit exponential
function describing the decay of this radioactive sample is A(t) = 75.113e−1.186t grams after t hours.
17. (a). Since A is m × n, AT A is n × n and AT is n × m, we conclude from matrix multiplication that P
is m × m.
(b). We have
P A = [A(AT A)−1 AT ]A = A(AT A)−1 (AT A) = AI = A
and
P 2 = [A(AT A)−1 AT ][A(AT A)−1 AT ] = A(AT A)−1 (AT A)(AT A)−1 AT = A(AT A)−1 AT = P,
as needed.
(c). We will show that P T = P :
P T = [A(AT A)−1 AT ]T = (AT )T [(AT A)−1 ]T AT = A[(AT A)T ]−1 AT = A[AT (AT )T ]−1 AT = A[AT A]−1 AT = P,
as needed.
18. According to Theorem 5.4.2, the least squares solution to Ax = b must satisfy AT Ax0 = AT b. In this
case, A is invertible, so we can solve explicitly for x0 :
x0 = (AT A)−1 AT b = A−1 (AT )−1 AT b = A−1 b.
This is the only least squares solution for this system.
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Solutions to Section 5.5
Problems:
1. We have
u, v = (2, 3), (4, −1) = 2 · 4 + 3 · (−1) = 5, u =
and so
θ = cos
−1
u, v
uv
= cos
−1
5
√ √
13 17
√
13, v =
√
17,
≈ 1.23 radians.
2. We have
u, v = (−2, −1, 2, 4), (−3, 5, 1, 1) = (−2) · (−3) + (−1) · 5 + 2 · 1 + 4 · 1 = 7, u = 5, v = 6,
and so
θ = cos
−1
u, v
uv
= cos
−1
7
5·6
= cos−1 (7/30) ≈ 1.34 radians.
3. For Problem 1, we have
u, v = (2, 3), (4, −1) = 2 · 2 · 4 + 3 · (−1) = 13, u =
and so
θ = cos−1
u, v
uv
= cos−1
√
13
√
17 33
√
17, v =
√
33,
≈ 0.99 radians.
For Problem 2, we have
u, v = (−2, −1, 2, 4), (−3, 5, 1, 1) = 2 · (−2) · (−3) + (−1) · 5 + 2 · 1 + 4 · 1 = 13, u =
and so
θ = cos−1
u, v
uv
= cos−1
√
13
√
29 · 45
√
29, v =
√
45,
≈ 1.20 radians.
4. We will obtain bases for the
⎡ rowspace,⎤ columnspace, and nullspace and orthonormalize them. A row1 2 6
⎢ 0 1 2 ⎥
⎥
echelon form of A is given by ⎢
⎣ 0 0 0 ⎦. We see that a basis for the rowspace of A is given by
0 0 0
{(1, 2, 6), (0, 1, 2)}.
We apply Gram-Schmidt to this set, and thus we need to replace (0, 1, 2) by
14 13
2
14
.
(0, 1, 2) − (1, 2, 6) = − , , −
41
41 41 41
So an orthogonal basis for the rowspace of A is given by
14 13
2
(1, 2, 6) , − , , −
.
41 41 41
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To replace this with an orthonormal basis, we must normalize each vector. The first one has norm
the second one has norm √341 . Hence, an orthonormal basis for the rowspace of A is
√
41 and
√
√ 14 13
41
41
2
− , ,−
.
(1, 2, 6),
41
3
41 41 41
Returning to the row-echelon form of A obtained above, we see that a basis for the columnspace of A is
⎧⎡
⎤⎫
⎤ ⎡
2 ⎪
1
⎪
⎪
⎪
⎨⎢
⎥⎬
⎥ ⎢
⎢ 2 ⎥,⎢ 1 ⎥ .
⎣ 0 ⎦ ⎣ 1 ⎦⎪
⎪
⎪
⎪
⎭
⎩
0
1
⎤
⎡
2
⎢ 1 ⎥
⎥
We apply Gram-Schmidt to this set, and thus we need to replace ⎢
⎣ 1 ⎦ by
0
⎤
⎡
⎤
⎤
⎡
⎡
2
1
4
⎢ 1 ⎥ 4 ⎢ 2 ⎥ 1 ⎢ −1 ⎥
⎥
⎢
⎥
⎥
⎢
⎢
⎣ 1 ⎦ − 6 ⎣ 0 ⎦ = 3 ⎣ 3 ⎦.
0
1
−2
So an orthogonal basis for the columnspace of A is given by
⎧⎡
⎤
⎤⎫
⎡
1
4 ⎪
⎪
⎪
⎪
⎨⎢
⎥
⎥⎬
⎢
⎢ 2 ⎥ , 1 ⎢ −1 ⎥ .
⎣ 0 ⎦ 3 ⎣ 3 ⎦⎪
⎪
⎪
⎪
⎩
⎭
1
−2
√
√
The norms of these vectors are, respectively, 6 and 30/3. Hence, we normalize the above orthogonal
basis to obtain the orthonormal basis for the columnspace:
⎧
⎤⎫
⎡
⎡
⎤
1
4 ⎪
⎪
√
√
⎪
⎪
⎨ 6⎢
⎥⎬
⎢
⎥
⎢ 2 ⎥ , 30 ⎢ −1 ⎥ .
⎪
6 ⎣ 0 ⎦ 30 ⎣ 3 ⎦⎪
⎪
⎪
⎭
⎩
−2
1
Returning once more to the row-echelon form of A obtained above, we see that, in order to find the
nullspace of A, we must solve the equations x + 2y + 6z = 0 and y + 2z = 0. Setting z = t as a free variable,
we find that y = −2t and x = −2t. Thus, a basis for the nullspace of A is {(−2, −2, 1)}, which can be
normalized to
2 2 1
− ,− ,
.
3 3 3
⎤
⎡
1 3 5
⎢ 0 1 3/2 ⎥
⎥
⎢
⎥
5. A row-echelon form for A is ⎢
⎢ 0 0 1 ⎥. Note that since rank(A) = 3, nullity(A) = 0, and so
⎣ 0 0 0 ⎦
0 0 0
there is no basis for nullspace(A). Moreover, rowspace(A) is a 3-dimensional subspace of R3 , and therefore,
rowspace(A) = R3 . An orthonormal basis for this is {(1, 0, 0), (0, 1, 0), (0, 0, 1)}.
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Finally, consider the columnspace of A. We must apply the Gram-Schmidt process to the three columns
of A. Thus, we replace the second column vector by
⎤
⎤ ⎡
⎤
⎡
⎡
−1
1
3
⎢ −1 ⎥ ⎢ 1 ⎥
⎢ −3 ⎥
⎥
⎥ ⎢
⎥
⎢
⎢
⎥ − 4⎢ 0 ⎥ = ⎢ 2 ⎥.
2
v2 = ⎢
⎥
⎥ ⎢
⎥
⎢
⎢
⎣ 1 ⎦ ⎣ 1 ⎦
⎣ 5 ⎦
1
1
5
Next, we replace the third column vector by
⎤
⎤
⎡
⎡
⎡
⎤
⎤ ⎡
5
1
−1
3
⎥
⎥
⎢ 1 ⎥
⎢
⎢
⎥ ⎢
⎢
⎥ 7 ⎢ −1 ⎥ 3 ⎢ 1 ⎥ ⎢ 3 ⎥
⎥
⎢
⎢
⎢
⎢
⎥
⎥
v3 = ⎢ 3 ⎥ − ⎢ 0 ⎥ − ⎢ 2 ⎥ = ⎢ 0 ⎥
⎥.
⎣ 2 ⎦ 2 ⎣ 1 ⎦ 2 ⎣ 1 ⎦ ⎣ −3 ⎦
3
1
8
1
Hence, an orthogonal basis for the columnspace of A is
⎧⎡
⎤⎫
⎤ ⎡
⎤ ⎡
3 ⎪
−1
1
⎪
⎪
⎪
⎪
⎪
⎥⎪
⎥ ⎢
⎥ ⎢
⎪
⎨⎢
⎢ −1 ⎥ ⎢ 1 ⎥ ⎢ 3 ⎥⎬
⎢ 0 ⎥,⎢ 2 ⎥,⎢ 0 ⎥ .
⎥
⎥ ⎢
⎥ ⎢
⎢
⎪
⎪
⎪
⎣ 1 ⎦ ⎣ 1 ⎦ ⎣ −3 ⎦⎪
⎪
⎪
⎪
⎪
⎭
⎩
3
1
1
Normalizing each vector yields the orthonormal basis
⎧ ⎡
⎤
⎤⎫
⎤
⎡
⎡
1
3 ⎪
−1
⎪
⎪
⎪
⎪
⎪ ⎢ −1 ⎥
⎪
⎢ 3 ⎥⎪
⎢ 1 ⎥
⎨
⎬
⎥
⎥
⎥
⎢
⎢
⎢
1⎢
⎥ , √1 ⎢ 2 ⎥ , 1 ⎢ 0 ⎥ .
0
⎥
⎥
⎥
⎢
⎪
2⎢
8⎢
⎪
⎪
⎣ 1 ⎦
⎣ 1 ⎦ 6 ⎣ −3 ⎦⎪
⎪
⎪
⎪
⎪
⎩
⎭
1
3
1
6. Let x1 = (5, −1, 2) and let x2 = (7, 1, 1). Using the Gram-Schmidt process, we have
v1 = x1 = (5, −1, 2)
and
v2 = x2 −
x2 , v1 36
6 12
11 7
v1 = (7, 1, 1) − (5, −1, 2) = (7, 1, 1) − (6, − , ) = (1, , − ).
2
|v1 |
30
5 5
5
5
Hence, an orthogonal basis is given by
{(5, −1, 2), (1,
11 7
, − )}.
5
5
7. We can readily verify that S spans R3 , so therefore an obvious orthogonal basis for span(S) is {(1, 0, 0), (0, 1, 0), (0, 0, 1)}.
Alternatively, for practice with Gram-Schmidt, we would proceed as follows: Let x1 = (6, −3, 2), x2 =
(1, 1, 1), and x3 = (1, −8, −1). Using the Gram-Schmidt process, we have
v1 = x1 = (6, −3, 2),
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v2 = x2 −
x2 , v1 5
v1 = (1, 1, 1) − (6, −3, 2) =
2
v1 49
19 64 39
, ,
49 49 49
and
x3 , v1 x3 , v2 4
38
v3 = x3 −
v1 −
v2 = (1, −8, −1) − (6, −3, 2) +
(19, 64, 39) =
2
2
v1 v2 7
427
,
45 36 81
− ,− ,
61 61 61
.
Hence, an orthogonal basis is given by
19 64 39
45 36 81
(6, −3, 2),
, − ,− ,
.
, ,
49 49 49
61 61 61
8. We can check that S spans P3 (R), so therefore we can apply Gram-Schmidt to the basis {1, x, x2 , x3 }
for P3 (R), instead of the given set of polynomials. Let x1 = 1, x2 = x, x3 = x2 , and x4 = x3 . Using the
Gram-Schmidt process, we have
v1 = x1 = 1,
v2 = x2 −
v3 = x3 −
x2 , v1 1
v1 = x − ,
2
v1 2
x3 , v1 x3 , v2 1
1
1
v1 −
v2 = x2 − (x − ) − = x2 − x + ,
v1 2
v2 2
2
3
6
and
v4 = x4 −
x4 , v1 x4 , v2 x4 , v3 1
1
3
3
9
1
3
1
v1 −
v2 −
v3 = x3 − − (x − ) − (x2 − x + ) = x3 − x2 + x − .
v1 2
v2 2
v3 2
4 10
2
2
6
2
5
20
Hence, an orthogonal basis is given by
1
1
3
3
1
.
1, x − , x2 − x + , x3 − x2 + x −
2
6
2
5
20
9. It is easy to see that the span of the set of vectors in this problem is the set of all 2 × 2 symmetric
matrices. Therefore, we can simply give the orthogonal basis
1 0
0 1
0 0
,
,
0 0
1 0
0 1
for the set of all 2 × 2 symmetric matrices.
10.
(a). We must verify the four axioms for an inner product given in Definition 5.1.3.
Axiom 1: We have
p · p = p(t0 )p(t0 ) + p(t1 )p(t1 ) + · · · + p(tn )p(tn ) = p(t0 )2 + p(t1 )2 + · · · + p(tn )2 ≥ 0.
Moreover,
p(t0 )2 + p(t1 )2 + · · · + p(tn )2 = 0 ⇐⇒ p(t0 ) = p(t1 ) = · · · = p(tn ) = 0.
But the only polynomial of degree ≤ n which has more than n roots is the zero polynomial. Thus,
p · p = 0 ⇐⇒ p = 0.
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Axiom 2: We have
p · q = p(t0 )q(t0 ) + p(t1 )q(t1 ) + · · · + p(tn )q(tn ) = q(t0 )p(t0 ) + q(t1 )p(t1 ) + · · · + q(tn )p(tn ) = q · p
for all p, q ∈ Pn (R).
Axiom 3: Let k be a scalar, and let p, q ∈ Pn (R). Then
(kp) · q = (kp)(t0 )q(t0 ) + (kp)(t1 )q(t1 ) + · · · + (kp)(tn )q(tn )
= kp(t0 )q(t0 ) + kp(t1 )q(t1 ) + · · · + kp(tn )q(tn )
= k[p(t0 )q(t0 ) + p(t1 )q(t1 ) + · · · + p(tn )q(tn )]
= k[p · q],
as required.
Axiom 4: Let p1 , p2 , q ∈ Pn (R). Then we have
(p1 + p2 ) · q = (p1 + p2 )(t0 )q(t0 ) + (p1 + p2 )(t1 )q(t1 ) + · · · + (p1 + p2 )(tn )q(tn )
= [p1 (t0 ) + p2 (t0 )]q(t0 ) + [p1 (t1 ) + p2 (t1 )]q(t1 ) + · · · + [p1 (tn ) + p2 (tn )]q(tn )
= [p1 (t0 )q(t0 ) + p1 (t1 )q(t1 ) + · · · + p1 (tn )q(tn )] + [p2 (t0 )q(t0 ) + p2 (t1 )q(t1 ) + · · · + p2 (tn )q(tn )]
= (p1 · q) + (p2 · q),
as required.
(b). We have
p0 , p1 = p0 (−3)p1 (−3) + p0 (−1)p1 (−1) + p0 (1)p1 (1) + p0 (3)p1 (3) = (1)(−3) + (1)(−1) + (1)(1) + (1)(3) = 0,
which proves that p0 and p1 are orthogonal.
(c). The projection of p2 onto span{p0 , p1 } is given by
p2 , p0 p2 , p1 20
0
p0 +
p1 =
p0 + p1 = 5.
p0 2
p1 2
4
11
Therefore, we can take q = t2 − 5.
11. We are given the point P (2, 3, 4), and we can write the equation of the given line L parametrically via
x(t) = 6t, −t, −4t. A parallel vector to L is given by v = (6, −1, −4). Since the point Q(0, 0, 0) lies on L,
we form the vector pointing from Q to P : w = (2, 3, 4) − (0, 0, 0) = (2, 3, 4). Thus, we have
P(w, v) =
(2, 3, 4), (6, −1, −4)
7
(2, −1, 3) = − (6, −1, −4).
2
||(6, −1, −4)||
53
Thus, the distance from P to L is given by
||w − P(w, v)|| = ||(2, 3, 4) − (−
1
42 7 28
, , )|| =
||(148, 152, 184)|| =
53 53 53
53
√
78864
.
53
12. We will use the formula obtained in Problem 28(c) in Section 5.2 with a = 2, b = −1, c = 3, d = −6,
and x0 = y0 = z0 = 0. The requested distance is:
|ax0 + by0 + cz0 + d|
6
| − 6|
√
=√ .
=
2
2
2
2
2
2
14
a +b +c
2 + (−1) + 3
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13. We will use the formula obtained in Problem 28(c) in Section 5.2 with a = −1, b = 3, c = 3, d = −8,
and x0 = −1, y0 = 3, and z0 = 5. The requested distance is:
|(−1)(−1) + (3)(3) + (3)(5) + (−8)|
17
|ax0 + by0 + cz0 + d|
√
=
=√ .
19
a 2 + b2 + c 2
(−1)2 + 32 + 32
14. We must start by determining the equation of the given plane. We can obtain a normal vector to this
plane by computing the cross product of v and w:
v × w = −15, 14, −3.
Thus, we can write the equation of the plane as
−15(x − x0 ) + 14(y − y0 ) − 3(z − z0 ) = 0,
or using the point (−1, 0, 5),
−15(x + 1) + 14y − 3(z − 5) = −15x + 14y − 3z = 0.
Next, we will use the formula obtained in Problem 28(c) in Section 5.2 with a = −15, b = 14, c = −3, d = 0,
and x0 = −2, y0 = 8, and z0 = 0. The requested distance is:
|(−15)(−2) + (14)(8) + (−3)(0) + 0|
142
|ax0 + by0 + cz0 + d|
√
=
=√
.
2
2
2
2
2
2
150
a +b +c
(−15) + 14 + (−3)
⎡
15. We construct
0
⎢ 1
⎢
A=⎢
⎢ 2
⎣ 3
4
⎤
1
1 ⎥
⎥
1 ⎥
⎥,
1 ⎦
1
A quick computation shows that
30 10
AT A =
,
and thus,
10 5
⎤
−2
⎢ −1 ⎥
⎥
⎢
⎥
b=⎢
⎢ 1 ⎥.
⎣ 2 ⎦
2
⎡
x=
a
b
,
and
(AT A)−1 =
1
50
5 −10
−10
30
=
1
10
1 −2
−2
6
Continuing with the computation, we now find
⎤
−2
⎢ −1 ⎥ ⎥
4 ⎢
⎢ 1 ⎥ = 11/10 .
⎥
−9/5
1 ⎢
⎣ 2 ⎦
2
⎡
T
x0 = (A A)
−1
1
A b=
10
T
1 −2
−2
6
0
1
1
1
2
1
3
1
Therefore, the least squares line is
y(x) =
16. We construct
⎡
−1
⎢ 1
A=⎢
⎣ 2
3
⎤
1
1 ⎥
⎥,
1 ⎦
1
x=
11
9
x− .
10
5
a
b
⎤
5
⎢ 1 ⎥
⎥
b=⎢
⎣ 1 ⎦.
−3
⎡
,
and
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A quick computation shows that
15
T
A A=
5
5
4
,
T
and thus,
(A A)
−1
1
=
35
4 −5
−5 15
.
Continuing with the computation, we now find
1
1
4 −5
−64
−64/35
T
−1 T
x0 = (A A) A b =
=
=
.
135
27/7
35 −5 15
35
Therefore, the least squares line is
y(x) = −
⎡
17. We construct
−4
⎢ −3
A=⎢
⎣ −2
0
⎤
1
1 ⎥
⎥,
1 ⎦
1
A quick computation shows that
29 −9
T
A A=
,
−9
4
x=
64
23
x+ .
35
7
a
b
⎤
−1
⎢ 1 ⎥
⎥
b=⎢
⎣ 3 ⎦.
7
⎡
,
and
T
and thus,
(A A)
−1
1
=
35
4
9
9
29
.
Continuing with the computation, we now find
x0 = (AT A)−1 AT b =
1
35
4
9
9
29
−4 −3 −2 0
1
1
1 1
⎤
−1
⎢ 1 ⎥
⎥= 2 .
⎢
⎣ 3 ⎦
7
7
⎡
Therefore, the least squares line is
y(x) = 2x + 7.
⎡
18. We construct
−3
⎢ −2
⎢
A=⎢
⎢ −1
⎣ 0
2
⎤
1
1 ⎥
⎥
1 ⎥
⎥,
1 ⎦
1
A quick computation shows that
18 −4
AT A =
,
−4
5
⎤
1
⎢ 0 ⎥
⎥
⎢
⎥
b=⎢
⎢ 1 ⎥.
⎣ −1 ⎦
−1
⎡
x=
a
b
and thus,
,
and
(AT A)−1 =
1
74
5
4
4
18
.
Continuing with the computation, we now find
⎤
1
⎢ 0 ⎥ ⎥
4
−3 −2 −1 0 2 ⎢
⎢ 1 ⎥ = −15/37 .
⎥
−12/37
18
1
1
1 1 1 ⎢
⎣ −1 ⎦
−1
⎡
T
x0 = (A A)
−1
1
A b=
74
T
5
4
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Therefore, the least squares line is
y(x) = −
15
12
x− .
37
37
20. Note that
x − y, vi = x, vi − y, vi = 0,
by assumption. Let v be an arbitrary vector in V , and write
v = a1 v 1 + a2 v 2 + · · · + a n v n ,
for some scalars a1 , a2 , . . . , an . Observe that
x − y, v = x − y, a1 v1 + a2 v2 + · · · + an vn = a1 x − y, v1 + a2 x − y, v2 + · · · + an x − y, vn = a1 · 0 + a2 · 0 + · · · + an · 0 = 0.
Thus, x − y is orthogonal to every vector in V . In particular
x − y, x − y = 0,
and hence, x − y = 0. Therefore x = y.
Chapter 6 Solutions
Solutions to Section 6.1
True-False Review:
(a): FALSE. The conditions T (u + v) = T (u) + T (v) and T (c · v) = c · T (v) must hold for all vectors u, v
in V and for all scalars c, not just “for some”.
(b): FALSE. The dimensions of the matrix A should be m × n, not n × m, as stated in the question.
(c): FALSE. This will only necessarily hold for a linear transformation, not for more general mappings.
(d): TRUE. This is precisely the definition of the matrix associated with a linear transformation, as given
in the text.
(e): TRUE. Since
0 = T (0) = T (v + (−v)) = T (v) + T (−v),
we conclude that T (−v) = −T (v).
(f ): TRUE. Using the properties of a linear transformation, we have
T ((c + d)v) = T (cv + dv) = T (cv) + T (dv) = cT (v) + dT (v),
as stated.
Problems:
1. Let (x1 , x2 , x3 ), (y1 , y2 , y3 ) ∈ R3 and c ∈ R.
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T ((x1 , x2 , x3 ) + (y1 , y2 , y3 )) = T (x1 + y1 , x2 + y2 , x3 + y3 )
= (x1 + y1 + 3x2 + 3y2 + x3 + y3 , x1 + y1 − x2 − y2 )
= (x1 + 3x2 + x3 , x1 − x2 ) + (y1 + 3y2 + y3 , y1 − y2 )
= T (x1 , x2 , x3 ) + T (y1 , y2 , y3 ).
T (c(x1 , x2 , x3 )) = T (cx1 , cx2 , cx3 ) = (cx1 +3cx2 +cx3 , cx1 −cx2 ) = c(x1 +3x2 +x3 , x1 −x2 ) = cT (x1 , x2 , x3 ).
Thus, T is a linear transformation.
2. Let (x1 , x2 ), (y1 , y2 ) ∈ R2 and c ∈ R.
T ((x1 , x2 ) + (y1 , y2 )) = T (x1 + y1 , x2 + y2 )
= (x1 + y1 + 2x2 + 2y2 , 2x1 + 2y1 − x2 − y2 )
= (x1 + 2x2 , 2x1 − x2 ) + (y1 + 2y2 , 2y1 − y2 )
= T (x1 , x2 ) + T (y1 , y2 ).
T (c(x1 , x2 )) = T (cx1 , cx2 ) = (cx1 + 2cx2 , 2cx1 − cx2 ) = c(x1 + 2x2 , 2x1 − x2 ) = cT (x1 , x2 ).
Thus, T is a linear transformation.
3. Let y1 , y2 ∈ C 2 (I) and c ∈ R. Then,
T (y1 + y2 ) = (y1 + y2 ) + a1 (y1 + y2 ) + a2 (y1 + y2 )
= (y1 + a1 y1 + a2 y1 ) + (y2 + a1 y2 + a2 y2 ) = T (y1 ) + T (y2 ).
T (cy1 ) = (cy1 ) + a1 (cy1 ) + a2 (cy1 ) = c(y1 + a1 y1 + a2 y1 ) = cT (y1 ).
Consequently, T is a linear transformation.
4. Let y1 , y2 ∈ C 2 (I) and c ∈ R. Then,
T (y1 + y2 ) = (y1 + y2 ) − 16(y1 + y2 ) = (y1 − 16y1 ) + (y2 − 16y2 ) = T (y1 ) + T (y2 ).
T (cy1 ) = (cy1 ) − 16(cy1 ) = c(y1 − 16y1 ) = cT (y1 ).
Consequently, T is a linear transformation.
5. Let A1 , A2 , B ∈ Mn (R) and c ∈ R.
T (A1 + A2 ) = (A1 + A2 )B − B(A1 + A2 ) = A1 B + A2 B − BA1 − BA2
= (A1 B − BA1 ) + (A2 B − BA2 ) = T (A1 ) + T (A2 ).
T (cA1 ) = (cA1 B) − B(cA1 ) = c(A1 B − BA1 ) = cT (A1 ).
Consequently, T is a linear transformation.
b
b
b
b
6. Let f, g ∈ V and c ∈ R. Then T (f +g) =
(f +g)(x)dx =
[f (x)+g(x)]dx =
f (x)dx+
g(x)dx =
a
a
a
T (f ) + T (g).
b
b
T (cf ) =
[cf (x)]dx = c
f (x)dx = cT (f ). Therefore, T is a linear transformation.
a
a
7. Let A, B ∈ Mn (R) and c ∈ R. Then,
n
n
n
,
,
,
T (A + B) = tr(A + B) =
(akk + bkk ) =
akk +
bkk = tr(A) + tr(B) = T (A) + T (B).
k=1
k=1
k=1
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T (cA) = tr(cA) =
n
,
k=1
cakk = c
n
,
akk = ctr(A) = cT (A).
k=1
Consequently, T is a linear transformation.
8. Let A, B ∈ Mn (R) and c ∈ R. Then,
S(A + B) = (A + B) + (A + B)T = A + AT + B + B T = S(A) + S(B).
S(cA) = (cA) + (cA)T = c(A + AT ) = cS(A).
Consequently, S is a linear transformation.
9. According to the given mapping, we have T (0) = 1, which violates part (1) of Theorem 6.1.11. The
conclusion is that T must not be linear.
10. We show with a specific example
addition
(the first property
in the
that T fails to respect
definition of
0 1
0 0
0 1
a linear transformation). Let A =
and B =
. We have A + B =
, so that
0 0
1 0
1 0
2
0 1
T (A + B) = (A + B)2 =
= I2 ,
1 0
whereas
T (A) + T (B) = A2 + B 2 = 02 + 02 = 02 .
Therefore, in this example, we have T (A + B) = T (A) + T (B). (Many other examples can be given as well.)
11. According to the given mapping, we have T (0) = x, which violates part (1) of Theorem 6.1.11. The
conclusion is that T must not be linear.
12. Let x = (x1 , x2 ), y = (y1 , y2 ) be in R2 . Then
T (x + y) = T (x1 + y1 , x2 + y2 ) = (x1 + x2 + y1 + y2 , 2),
whereas
T (x) + T (y) = (x1 + x2 , 2) + (y1 + y2 , 2) = (x1 + x2 + y1 + y2 , 4).
We see that T (x + y) = T (x) + T (y), hence T is not a linear transformation.
13. Let A ∈ M2 (R) and c ∈ R. Then
T (cA) = det(cA) = c2 det(A) = c2 T (A).
Since T (cA) = cT (A) in general, it follows that T is not a linear transformation.
3 −2
14. If T (x1 , x2 ) = (3x1 − 2x2 , x1 + 5x2 ), then A = [T (e1 ), T (e2 )] =
.
1
5
⎡
⎤
1
3
15. If T (x1 , x2 ) = (x1 + 3x2 , 2x1 − 7x2 , x1 ), then A = [T (e1 ), T (e2 )] = ⎣ 2 −7 ⎦.
1
0
1 −1 1
.
16. If T (x1 , x2 , x3 ) = (x1 − x2 + x3 , x3 − x1 ), then A = [T (e1 ), T (e2 ), T (e3 )] =
−1
0 1
17. If T (x1 , x2 , x3 ) = x1 + 5x2 − 3x3 , then A = [T (e1 ), T (e2 ), T (e3 )] =
1
5 −3 .
⎡
−1
⎢ −1
18. If T (x1 , x2 , x3 ) = (x3 − x1 , −x1 , 3x1 + 2x3 , 0), then A = [T (e1 ), T (e2 ), T (e3 )] = ⎢
⎣ 3
0
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0
0
0
⎤
1
0 ⎥
⎥.
2 ⎦
0
393
19. T (x) = Ax =
1 3
−4 7
x1
x2
=
x1 + 3x2
−4x1 + 7x2
, which we write as
T (x1 , x2 ) = (x1 + 3x2 , −4x1 + 7x2 ).
20. T (x) = Ax =
2 −1
5
3
1 −2
⎤
x1
⎣ x2 ⎦ = 2x1 − x2 + 5x3 , which we write as
3x1 + x2 − 2x3
x3
⎡
T (x1 , x2 , x3 ) = (2x1 − x2 + 5x3 , 3x1 + x2 − 2x3 ).
⎤ ⎡
⎤
⎤⎡
2
2 −3
x1
2x1 + 2x2 − 3x3
2 ⎦ ⎣ x2 ⎦ = ⎣ 4x1 − x2 + 2x3 ⎦ , which we write as
21. T (x) = Ax = ⎣ 4 −1
5
7 −8
x3
5x1 + 7x2 − 8x3
⎡
T (x1 , x2 , x3 ) = (2x1 + 2x2 − 3x3 , 4x1 − x2 + 2x3 , 5x1 + 7x2 − 8x3 ).
⎤
⎤
⎡
−3x
−3
⎢ −2x ⎥
⎢ −2 ⎥
⎥ , which we write as
⎥
⎢
22. T (x) = Ax = ⎢
⎣ 0 ⎦ [x] = ⎣
0 ⎦
x
1
⎡
T (x) = (−3x, −2x, 0, x).
⎡
23. T (x) = Ax =
1
−4 −6
0
2
⎤
x1
⎢ x2 ⎥
⎢
⎥
⎢ x3 ⎥ = [x1 − 4x2 − 6x3 + 2x5 ], which we write as
⎢
⎥
⎣ x4 ⎦
x5
T (x1 , x2 , x3 , x4 , x5 ) = x1 − 4x2 − 6x3 + 2x5 .
24. Let u be a fixed vector in V , v1 , v2 ∈ V , and c ∈ R. Then
T (v1 + v2 ) = u, v1 + v2 = u, v1 + u, v2 = T (v1 ) + T (v2 ).
T (cv1 ) = u, cv1 = cu, v1 = cT (v1 ). Thus, T is a linear transformation.
25. We must show that the linear transformation T respects addition and scalar multiplication:
T respects addition: Let v1 and v2 be vectors in V . Then we have
T (v1 + v2 ) = (u1 , v1 + v2 , u2 , v1 + v2 )
= (v1 + v2 , u1 , v1 + v2 , u2 )
= (v1 , u1 + v2 , u1 , v1 , u2 + v2 , u2 )
= (v1 , u1 , v1 , u2 ) + (v2 , u1 , v2 , u2 )
= (u1 , v1 , u2 , v1 ) + (u1 , v2 , u2 , v2 )
= T (v1 ) + T (v2 ).
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T respects scalar multiplication: Let v be a vector in V and let c be a scalar. Then we have
T (cv) = (u1 , cv, u2 , cv)
= (cv, u1 , cv, u2 )
= (cv, u1 , cv, u2 )
= c(v, u1 , v, u2 )
= c(u1 , v, u2 , v)
= cT (v).
1
1 then det(D) = −2 = 0, so by Corollary 4.5.17 the vectors v1 = (1, 1)
26. (a). If D = [v1 , v2 ] = 1 −1 and v2 = (1, −1) are linearly independent. Since dim[R2 ] = 2, it follows from Theorem 4.6.10 that {v1 , v2 }
is a basis for R2 .
(b). Let x = (x1 , x2 ) be an arbitrary vector in R2 . Since {v1 , v2 } forms a basis for R2 , there exist c1 and
c2 such that
(x1 , x2 ) = c1 (1, 1) + c2 (1, −1),
that is, such that
c 1 + c 2 = x1 , c 1 − c 2 = x2 .
Solving this system yields c1 =
1
1
(x1 + x2 ), c2 = (x1 − x2 ). Thus,
2
2
(x1 , x2 ) =
1
1
(x1 + x2 )v1 + (x1 − x2 )v2 ,
2
2
1
1
T [(x1 , x2 )] = T
(x1 + x2 )v1 + (x1 − x2 )v2
2
2
1
1
= (x1 + x2 )T (v1 ) + (x1 − x2 )T (v2 )
2
2
1
1
= (x1 + x2 )(2, 3) + (x1 − x2 )(−1, 1)
2
2 x1
3x2
=
+
, 2x1 + x2 .
2
2
In particular, when (4, −2) is substituted for (x1 , x2 ), it follows that T (4, −2) = (−1, 6).
so that
27. The matrix of T is the 4 × 2 matrix [T (e1 ), T (e2 )]. Therefore, we must determine T (1, 0) and T (0, 1),
which we can determine from the given information by using the linear transformation properties. A quick
calculation shows that (1, 0) = − 23 (−1, 1) + 13 (1, 2), so
2
1
2
1
2
1
5 1 5
T (1, 0) = T (− (−1, 1)+ (1, 2)) = − T (−1, 1)+ T (1, 2) = − (1, 0, −2, 2)+ (−3, 1, 1, 1) = (− , , , −1).
3
3
3
3
3
3
3 3 3
Similarly, we have (0, 1) = 13 (−1, 1) + 13 (1, 2), so
1
1
1
1
1
1
2 1 1
T (0, 1) = T ( (−1, 1) + (1, 2)) = T (−1, 1) + T (1, 2) = (1, 0, −2, 2) + (−3, 1, 1, 1) = (− , , − , 1).
3
3
3
3
3
3
3 3 3
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Therefore, we have the matrix of T :
⎤
−5/3 −2/3
⎢ 1/3
1/3 ⎥
⎥
⎢
⎣ 5/3 −1/3 ⎦ .
−1
1
⎡
28. The matrix of T is the 2 × 4 matrix [T (e1 ), T (e2 ), T (e3 ), T (e4 )]. Therefore, we must determine
T (1, 0, 0, 0), T (0, 1, 0, 0), T (0, 0, 1, 0), and T (0, 0, 0, 1), which we can determine from the given information
by using the linear transformation properties. We are given that
T (1, 0, 0, 0) = (3, −2).
Next,
T (0, 1, 0, 0) = T (1, 1, 0, 0) − T (1, 0, 0, 0) = (5, 1) − (3, −2) = (2, 3),
T (0, 0, 1, 0) = T (1, 1, 1, 0) − T (1, 1, 0, 0) = (−1, 0) − (5, 1) = (−6, −1),
and
T (0, 0, 0, 1) = T (1, 1, 1, 1) − T (1, 1, 1, 0) = (2, 2) − (−1, 0) = (3, 2).
Therefore, we have the matrix of T :
3
−2
2 −6
3 −1
3
2
.
29. The matrix of T is the 3 × 3 matrix [T (e1 ), T (e2 ), T (e3 )]. Therefore, we must determine T (1, 0, 0),
T (0, 1, 0), and T (0, 0, 1), which we can determine from the given information by using the linear transformation properties. A quick calculation shows that (1, 0, 0) = (1, 2, 0) − 6(0, 1, 1) + 2(0, 2, 3), so
T (1, 0, 0) = T (1, 2, 0) − 6T (0, 1, 1) + 2T (0, 2, 3) = (2, −1, 1) − 6(3, −1, −1) + 2(6, −5, 4) = (32, −5, 4).
Similarly, (0, 1, 0) = 3(0, 1, 1) − (0, 2, 3), so
T (0, 1, 0) = 3T (0, 1, 1) − T (0, 2, 3) = 3(3, −1, −1) − (6, −5, 4) = (3, 2, −7).
Finally, (0, 0, 1) = −2(0, 1, 1) + (0, 2, 3), so
T (0, 0, 1) = −2T (0, 1, 1) + T (0, 2, 3) = −2(3, −1, −1) + (6, −5, 4) = (0, −3, 6).
Therefore, we have the matrix of T :
⎡
⎤
32
3
0
⎣ −5
2 −3 ⎦ .
4 −7
6
30. The matrix of T is the 4 × 3 matrix [T (e1 ), T (e2 ), T (e3 )]. Therefore, we must determine T (1, 0, 0),
T (0, 1, 0), and T (0, 0, 1), which we can determine from the given information by using the linear transformation properties. A quick calculation shows that (1, 0, 0) = 41 (0, −1, 4) − 14 (0, 3, 3) + 14 (4, 4, −1), so
T (1, 0, 0) =
1
1
1
1
1
1
3 1 1
T (0, −1, 4)− T (0, 3, 3)+ T (4, 4, −1) = (2, 5, −2, 1)− (−1, 0, 0, 5)+ (−3, 1, 1, 3) = (0, , − , − ).
4
4
4
4
4
4
2 4 4
4
(0, 3, 3), so
Similarly, (0, 1, 0) = − 15 (0, −1, 4) + 15
1
4
2
2 17
T (0, 1, 0) = − (2, 5, −2, 1) + (−1, 0, 0, 5) = (− , −1, , ).
5
15
3
5 15
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1
(0, 3, 3), so
Finally, (0, 0, 1) = 15 (0, −1, 4) + 15
T (0, 0, 1) =
1
1
1
1
1
2 8
T (0, −1, 4) + T (0, 3, 3) = (2, 5, −2, 1) + (−1, 0, 0, 5) = ( , 1, − , ).
5
15
5
15
3
5 15
Therefore, we have the matrix of T :
⎤
0 −2/3
1/3
⎢ 3/2
−1
1 ⎥
⎥.
⎢
⎣ −1/4
2/5 −2/5 ⎦
−1/4 17/15 8/15
⎡
31. T (ax2 +bx+c) = aT (x2 )+bT (x)+cT (1) = a(3x+2)+b(x2 −1)+c(x+1) = bx2 +(3a+c)x+(2a−b+c).
32. Using the linearity of T , we have T (2v1 + 3v2 ) = v1 + v2 and T (v1 + v2 ) = 3v1 − v2 . That is,
2T (v1 ) + 3T (v2 ) = v1 + v2 and T (v1 ) + T (v2 ) = 3v1 − v2 . Solving this system for the unknowns T (v1 ) and
T (v2 ), we obtain T (v2 ) = 3v2 − 5v1 and T (v1 ) = 8v1 − 4v2 .
33. Since T is a linear transformation we obtain:
T (x2 ) − T (1) = x2 + x − 3,
2T (x) = 4x,
(33.1)
(33.2)
3T (x) + 2T (1) = 2(x + 3) = 2x + 6.
(33.3)
From Equation (33.2) it follows that T (x) = 2x, so upon substitution into Equation (33.3) we have
3(2x) + 2T (1) = 2(x + 3) or T (1) = −2x + 3. Substituting this last result into Equation (33.1) yields
T (x2 ) − (−2x + 3) = x2 + x − 3 so T (x2 ) = x2 − x. Now if a, b and c are arbitrary real numbers, then
T (ax2 + bx + c) = aT (x2 ) + bT (x) + cT (1) = a(x2 − x) + b(2x) + c(−2x + 3)
= ax2 − ax + 2bx − 2cx + 3c = ax2 + (−a + 2b − 2c)x + 3c.
34. Let v ∈ V . Since {v1 , v2 } is a basis for V , there exists a, b ∈ R such that v = av1 + bv2 . Hence
T (v) = T (av1 + bv2 ) = aT (v1 ) + bT (v2 ) = a(3v1 − v2 ) + b(v1 + 2v2 )
= 3av1 − av2 + bv1 + 2bv2 = (3a + b)v1 + (2b − a)v2 .
35. Let v be any vector in V . Since {v1 , v2 , . . . , vk } spans V , we can write v = c1 v1 + c2 v2 + · · · + ck vk for
suitable scalars c1 , c2 , . . . , ck . Then
T (v) = T (c1 v1 + c2 v2 + · · · + ck vk ) = c1 T (v1 ) + c2 T (v2 ) + · · · + ck T (vk )
= c1 S(v1 ) + c2 S(v2 ) + · · · + ck S(vk )
= S(c1 v1 + c2 v2 + · · · + ck vk ) = S(v),
as required.
36. Let v be any vector in V . Since {v1 , v2 , . . . , vk } is a basis for V , we can write v = c1 v1 +c2 v2 +· · ·+ck vk
for suitable scalars c1 , c2 , . . . , ck . Then
T (v) = T (c1 v1 + c2 v2 + · · · + ck vk ) = c1 T (v1 ) + c2 T (v2 ) + · · · + ck T (vk ) = c1 0 + c2 0 + . . . ck 0 = 0,
as required.
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37. Let v1 and v2 be arbitrary vectors in V . Then,
(T1 + T2 )(v1 + v2 ) = T1 (v1 + v2 ) + T2 (v1 + v2 )
= T1 (v1 ) + T1 (v2 ) + T2 (v1 ) + T2 (v2 )
= T1 (v1 ) + T2 (v1 ) + T1 (v2 ) + T2 (v2 )
= (T1 + T2 )(v1 ) + (T1 + T2 )(v2 ).
Further, if k is any scalar, then
(T1 + T2 )(kv) = T1 (kv) + T2 (kv) = kT1 (v) + kT2 (v) = k[T1 (v) + T2 (v)] = k(T1 + T2 )(v).
It follows that T1 + T2 is a linear transformation.
Now consider the transformation cT , where c is an arbitrary scalar.
(cT )(v1 + v2 ) = cT (v1 + v2 ) = c[T (v1 ) + T (v2 )] = cT (v1 ) + cT (v2 ) = (cT )(v1 ) + (cT )(v2 ).
(cT )(kv1 ) = cT (kv1 ) = c[kT (v1 )] = (ck)T (v1 ) = (kc)T (v1 ) = k[cT (v1 )].
Thus, cT is a linear transformation.
38. (T1 + T2 )(x) = T1 (x) + T2 (x) = Ax + Bx = (A + B)x =
Hence, (T1 + T2 )(x1 , x2 ) = (5x1 + 6x2 , 2x1 − 2x2 ).
(cT1 )(x) = cT1 (x) = c(Ax) = (cA)x =
3c
−c
c
2c
5
6
2 −2
x1
x2
x1
x2
=
=
5x1 + 6x2
2x1 − 2x2
3cx1 + cx2
−cx1 + 2cx2
.
.
Hence, (cT1 )(x1 , x2 ) = (3cx1 + cx2 , −cx1 + 2cx2 ).
39. (T1 + T2 )(x) = T1 (x) + T2 (x) = Ax + Bx = (A + B)x.
(cT1 )(x) = cT1 (x) = c(Ax) = (cA)x.
40. Problem 37 establishes that if T1 and T2 are in L(V, W ) and c is any scalar, then T1 + T2 and cT1 are
in L(V, W ). Consequently, Axioms (A1) and (A2) are satisfied.
A3: Let v be any vector in L(V, W ). Then
(T1 + T2 )(v) = T1 (v) + T2 (v) = T2 (v) + T1 (v) = (T2 + T1 )(v).
Hence T1 + T2 = T2 + T1 , therefore the addition operation is commutative.
A4: Let T3 ∈ L(V, W ). Then
[(T1 + T2 ) + T3 ] (v) = (T1 + T2 )(v) + T3 (v) = [T1 (v) + T2 (v)] + T3 (v)
= T1 (v) + [T2 (v) + T3 (v)] = T1 (v) + (T2 + T3 )(v) = [T1 + (T2 + T3 )](v).
Hence (T1 + T2 ) + T3 = T1 + (T2 + T3 ), therefore the addition operation is associative.
A5: The zero vector in L(V, W ) is the zero transformation, O : V → W , defined by
O(v) = 0, for all v in V,
where 0 denotes the zero vector in V . To show that O is indeed the zero vector in L(V, W ), let T be any
transformation in L(V, W ). Then
(T + O)(v) = T (v) + O(v) = T (v) + 0 = T (v) for all v ∈ V,
so that T + O = T .
A6: The additive inverse of the transformation T ∈ L(V, W ) is the linear transformation −T defined by
−T = (−1)T , since
[T + (−T )](v) = T (v) + (−T )(v) = T (v) + (−1)T (v) = T (v) − T (v) = 0,
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for all v ∈ V , so that T + (−T ) = O.
A7-A10 are all straightforward verifications.
Solutions to Section 6.2
True-False Review:
(a): FALSE. For example, T (x1 , x2 ) = (0, 0) is a linear transformation that maps every line to the origin.
(b): TRUE. All of the matrices Rx , Ry , Rxy , LSx , LSy , Sx , and Sy discussed in this section are elementary
matrices.
(c): FALSE.
A shear
parallel to the
x-axis composed with a shear parallel to the y-axis is given by matrix
1 k
1 0
1 + kl k
=
, which is not a shear.
0 1
l 1
l
1
(d): TRUE. This is explained prior to Example 6.2.1.
(e): FALSE. For example,
Rxy · Rx =
0
1
1
0
1
0
0
−1
=
0 −1
1
0
,
and this matrix is not in the form of a stretch.
(f ): FALSE. For example,
k
0
0
1
1
0
0
l
=
k
0
0
l
is not a stretch.
Problems:
1. T (1, 1) = (0, 3), T (2, 1) = (1, 4), T (2, 2) = (0, 6), T (1, 2) = (−1, 5).
2. T (1, 1) = (1, −1), T (2, 1) = (1, −2), T (2, 2) = (2, −2), T (1, 2) = (2, −1).
3. T (1, 1) = (−4, −2), T (2, 1) = (−6, −4), T (2, 2) = (−8, −4), T (1, 2) = (−6, −2).
4. T (1, 1) = (2, 0), T (2, 1) = (3, −1), T (2, 2) = (4, 0), T (1, 2) = (3, 1).
0 2 1 2 0 2 1 0 3 1 0
5.
∼
∼
∼
2 0
0 0
0 2
0 1
1. P12
2. M1 (1/2)
3. M2 (1/2).
So,
T (x) = Ax = P12 M1 (2)M2 (2)x
which corresponds to a stretch in the y-direction, followed by a stretch in the x-direction, followed by a
reflection in y = x.
1 2
6. A =
=⇒ T (x) = Ax corresponds to a shear parallel to the x-axis.
0 1
−1
0 1 1
0 2 1 0
1. M1 (−1) 2. M2 (−1) So,
7.
∼
∼
0 −1
0 −1
0 1
T (x) = Ax = M1 (−1)M2 (−1)x
which corresponds to a reflection in the x-axis, followed by a reflection in the y-axis.
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y
6
5
4
3
2
1
x
-2
-1
1
2
Figure 0.0.68: Figure for Problem 1
y
2
1
1
2
x
-1
-2
Figure 0.0.69: Figure for Problem 2
8. A =
9.
1
3
1 0
=⇒ T (x) = Ax corresponds to a shear parallel to the y-axis.
3 1
2 1 1
2 2 1
0 3 1 0
1. A12 (−3) 2. A21 (1)
∼
∼
∼
4
0 −2
0 −2
0 1
T (x) = Ax = A12 (3)A21 (−1)M2 (−2)x
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3. M2 (−1/2).
So,
400
y
2
1
x
-8
-6
-4
-2
1
2
-2
-4
Figure 0.0.70: Figure for Problem 3
y
2
1
1
2
3
4
x
-1
Figure 0.0.71: Figure for Problem 4
which corresponds to a reflection in the x-axis followed by a stretch in they y-direction, followed by a shear
parallel to the x-axis, followed by a shear parallel to the y-axis.
1 −3 1 1 −3 2 1 −3 3 1 0
1. A12 (2) 2. M2 (1/2) 3. A21 (3). So,
10.
∼
∼
∼
−2
8
0
2
0
1
0 1
T (x) = Ax = A12 (−2)M2 (2)A21 (−3)x
which corresponds to a shear parallel to the x-axis, followed by a stretch in the y-direction, followed by a
shear parallel to the y-axis.
−1 −1
−1 −1
1 1
1 0
1
2
3
1. A12 (−1) 2. M1 (−1) 3. A21 (−1).
11.
∼
∼
∼
−1
0
0
1
0 1
0 1
So,
T (x) = Ax = A12 (1)M1 (−1)A21 (1)x
which corresponds to a shear parallel to the x-axis, followed by a reflection in the y-axis, followed by a shear
parallel to the y-axis.
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1
0
1 0
1
0
=
. So T (x) = Ax corresponds to a reflection in the x-axis followed by
0 −2
0 2
0 −1
a stretch in the y-direction.
π
0 −1
13. The matrix for a counter-clockwise rotation through an angle θ = is
. Now,
1
0
2
12.
0 −1
1
0
1
∼
1
0
0 −1
2
∼
1
0
0
1
1. P12
2. M2 (−1)
So,
T (x) = Ax = P12 M2 (−1)x
which corresponds to a reflection in the x-axis followed by a reflection in y = x.
14.
R(θ) =
cos θ
0
0
1
cos θ
0
0
1
1 − tan θ
0
1
0
1 − tan θ
cos θ 0
1
=
1
0
sec θ
0
1
sin θ
0
1
1
0
0
sec θ
1
− tan θ
=
sin θ
cos θ
cos θ − sin θ
=
sin θ
cos θ
which coincides with the matrix of the transformation of R2 corresponding to a rotation through an angle θ
in the counter-clockwise direction.
1
sin θ
Solutions to Section 6.3
True-False Review:
(a): FALSE. The statement should read
dim[Ker(T )] + dim[Rng(T )] = dim[V ],
not dim[W ] on the right-hand side.
(b): FALSE. As a specific illustration, we could take T : P4 (R) → R7 defined by
T (a0 + a1 x + a2 x2 + a3 x3 + a4 x4 ) = (a0 , a1 , a2 , a3 , a4 , 0, 0),
and it is easy to see that T is a linear transformation with Ker(T ) = {0}. Therefore, Ker(T ) is 0-dimensional.
(c): FALSE. The solution set to the homogeneous linear system Ax = 0 is Ker(T ), not Rng(T ).
(d): FALSE. Rng(T ) is a subspace of W , not V , since it consists of vectors of the form T (v), and these
belong to W .
(e): TRUE. From the given information, we see that Ker(T ) is at least 2-dimensional, and therefore, since
M23 is 6-dimensional, the Rank-Nullity Theorem requires that Rng(T ) have dimension at most 6 − 2 = 4.
(f ): TRUE. Any vector of the form T (v) where v belongs to Rn can be written as Av, and this in turn
can be expressed as a linear combination of the columns of A. Therefore, T (v) belongs to colspace(A).
Problems:
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⎤
⎤
⎡
0
1 2
⎢ 0 ⎥
⎢ 2 4 ⎥ −10
⎥
⎥
=⎢
1. (a). T (−10, 5) = ⎢
⎣ 0 ⎦. Thus, x does belong to Ker(T ).
⎣ 4 8 ⎦
5
0
8 16
⎤
⎡
⎤
⎡
−1
1 2
⎢ −2 ⎥
⎢ 2 4 ⎥
1
⎥
⎢
⎥
(b). T (1, −1) = ⎢
⎣ 4 8 ⎦ −1 = ⎣ −4 ⎦. Thus, x does not belong to Ker(T ).
−8
8 16
⎤
⎡
⎤
⎡
1 2
0
⎢ 2 4 ⎥
⎢ 0 ⎥
2
⎥
⎥
⎢
(c). T (2, −1) = ⎢
⎣ 4 8 ⎦ −1 = ⎣ 0 ⎦. Thus, x does belong to Ker(T ).
8 16
0
⎤
⎡
7
1 −1
2 ⎣
0
⎦
5 =
2. (a). T (7, 5, −1) =
=⇒ (7, 5, −1) ∈ Ker(T ).
1 −2 −3
0
−1
⎡
⎤
−21
1 −1
2 ⎣
−2
−15 ⎦ =
(b). T (−21, −15, 2) =
=⇒ (−21, −15, 2) ∈
/ Ker(T ).
1 −2 −3
3
2
⎡
⎤
35
1 −1
2 ⎣
0
⎦
25 =
(c). T (35, 25, −5) =
=⇒ (35, 25, −5) ∈ Ker(T ).
1 −2 −3
0
−5
⎡
2
2
3.
{x ∈ R : T (x) = 0} = {x ∈ R : Ax
= 0}. The
augmented matrix of the system Ax = 0 is:
Ker(T ) =
3 6 0
1 2 0
, with reduced row-echelon form of
. It follows that
1 2 0
0 0 0
Ker(T ) = {x ∈ R2 : x = (−2t, t), t ∈ R} = {x ∈ R2 : x = t(−2, 1), t ∈ R}.
Geometrically, this is a line in R2 . It is the subspace of R2 spanned by the vector (−2, 1).
dim[Ker(T )] = 1.
For the given transformation, Rng(T ) = colspace(A). From the preceding reduced row-echelon form of A,
we see that colspace(A) is generated by the first column vector of A. Consequently,
Rng(T ) = {y ∈ R2 : y = r(3, 1), r ∈ R}.
Geometrically, this is a line in R2 . It is the subspace of R2 spanned by the vector (3, 1).
dim[Rng(T )] = 1.
Since dim[Ker(T )]+ dim[Rng(T )] = 2 = dim[R2 ], Theorem 6.3.8 is satisfied.
matrix of the system Ax = 0
4. Ker(T
) = {x ∈ R⎤3 : T (x) = 0} = {x ∈ R3 : Ax = 0}.
⎤
⎡
⎡ The augmented
1 −1 0 0
1 0 0 0
1 2 0 ⎦, with reduced row-echelon form ⎣ 0 1 0 0 ⎦. Thus x1 = x2 = x3 = 0, so
is: ⎣ 0
2 −1 1 0
0 0 1 0
Ker(T ) = {0}. Geometrically, this describes a point (the origin) in R3 .
dim[Ker(T )] = 0.
For the given transformation, Rng(T ) = colspace(A). From the preceding reduced row-echelon form of A,
we see that colspace(A) is generated by the first three column vectors of A. Consequently,
Rng(T ) = R3 , dim[Rng(T )] = dim[R3 ] = 3, and Theorem 6.3.8 is satisfied since
dim[Ker(T )]+ dim[Rng(T )] = 0 + 3 = dim[R3 ].
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5.
R3 : T (x) = 0} = {x ∈ R3 : Ax = 0}.
⎡ Ker(T ) = {x ∈ ⎤
⎤ matrix of the system Ax = 0 is:
⎡ The augmented
1 −2
1 0
1 0 −5 0
⎣ 2 −3 −1 0 ⎦, with reduced row-echelon form of ⎣ 0 1 −3 0 ⎦. Thus
5 −8 −1 0
0 0 0 0
Ker(T ) = {x ∈ R3 : x = t(5, 3, 1), t ∈ R}.
Geometrically, this describes the line in R3 through the origin, spanned by (5, 3, 1).
dim[Ker(T )] = 1.
For the given transformation, Rng(T ) = colspace(A). From the preceding reduced row-echelon form of A,
we see that a basis for colspace(A) is given by the first two column vectors of A. Consequently,
Rng(T ) = {y ∈ R3 : y = r(1, 2, 5) + s(−2, −3, −8), r, s ∈ R}.
Geometrically, this is a plane through the origin in R3 .
dim[Rng(T )] = 2 and Theorem 6.3.8 is satisfied since dim[Ker(T )]+ dim[Rng(T )] = 1 + 2 = 3 = dim[R3 ].
3
3
6.
Ker(T ) = {x ∈ R : T (x) = 0} = {x ∈ R : Ax = 0}. The augmentedmatrix of the system Ax = 0 is:
1 −1
2 0
1 −1 2 0
, with reduced row-echelon form of
. Thus
−3
3 −6 0
0
0 0 0
Ker(T ) = {x ∈ R3 : x = r(1, 1, 0) + s(−2, 0, 1), r, s ∈ R}.
Geometrically, this describes the plane through the origin in R3 , which is spanned by the linearly independent set {(1, 1, 0), (−2, 0, 1)}.
dim[Ker(T )] = 2.
For the given transformation, Rng(T ) = colspace(A). From the preceding reduced row-echelon form of A,
we see that a basis for colspace(A) is given by the first column vector of A. Consequently,
Rng(T ) = {y ∈ R2 : y = t(1, −3), t ∈ R}.
Geometrically, this is the line through the origin in R2 spanned by (1, −3).
dim[Rng(T )] = 1 and Theorem 6.3.8 is satisfied since dim[Ker(T )]+ dim[Rng(T )] = 2 + 1 = 3 = dim[R3 ].
3
3
= 0}. The augmented
matrix of the system Ax = 0 is:
7.
Ker(T ) = {x
∈ R : T (x) = 0} = {x ∈ R : Ax 1 3 2 0
1 3 0 0
, with reduced row-echelon form of
. Thus
2 6 5 0
0 0 1 0
Ker(T ) = {x ∈ R3 : x = r(−3, 1, 0), r ∈ R}.
Geometrically, this describes the line through the origin in R3 , which is spanned by (−3, 1, 0).
dim[Ker(T )] = 1.
For the given transformation, Rng(T ) = colspace(A). From the preceding reduced row-echelon form of A,
we see that a basis for colspace(A) is given by the first and third column vectors of A. Consequently,
Rng(T ) = span{(1, 2), (2, 5)} = R2 , so that dim[Rng(T )] = 2. Geometrically, Rng(T ) is the xy-plane, and
Theorem 6.3.8 is satisfied since dim[Ker(T )]+ dim[Rng(T )] = 1 + 2 = 3 = dim[R3 ].
⎤
⎡
−5/3 −2/3
⎢ 1/3
1/3 ⎥
⎥
8. The matrix of T in Problem 27 of Section 6.1 is A = ⎢
⎣ 5/3 −1/3 ⎦. Thus,
−1
1
Ker(T ) = nullspace(A) = {0}
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⎧⎡
⎤⎫
⎤ ⎡
−2/3 ⎪
−5/3
⎪
⎪
⎪
⎨⎢
⎢ 1/3 ⎥⎬
1/3 ⎥
⎥
⎢
⎥
⎢
.
,
Rng(T ) = colspace(A) = span ⎣
5/3 ⎦ ⎣ −1/3 ⎦⎪
⎪
⎪
⎪
⎭
⎩
1
−1
and
9. The matrix of T in Problem 28 of Section 6.1 is A =
3 2 −6
−2 3 −1
3
2
. Thus,
Ker(T ) = nullspace(A) = span{(16/13, 15/13, 1, 0), (−5/13, −12/13, 0, 1)}
and
Rng(T ) = colspace(A) = R2 .
⎡
⎤
32
3
0
2 −3 ⎦. Thus,
10. The matrix of T in Problem 29 of Section 6.1 is A = ⎣ −5
4 −7
6
Ker(T ) = nullspace(A) = {0}
and
Rng(T ) = colspace(A) = R3 .
⎡
0
⎢ 3/2
11. The matrix of T in Problem 30 of Section 6.1 is A = ⎢
⎣
0
−1/4
⎤
−2/3
1/3
−1
1 ⎥
⎥. Thus,
2/5 −2/5 ⎦
17/15 8/15
Ker(T ) = nullspace(A) = {0}
and
⎧⎡
⎪
⎪
⎨⎢
Rng(T ) = colspace(A) = span ⎢
⎣
⎪
⎪
⎩
⎤ ⎡
⎤ ⎡
⎤⎫
−2/3
0
1/3 ⎪
⎪
⎬
⎢
⎢
−1 ⎥
3/2 ⎥
1 ⎥
⎥,⎢
⎥,⎢
⎥ .
2/5 ⎦ ⎣ −2/5 ⎦⎪
0 ⎦ ⎣
⎪
⎭
17/15
−1/4
8/15
12. (a). Ker(T ) = {v ∈ R3 : u, v = 0}. For v to be in the kernel of T , u and v must be orthogonal. Since
u is any fixed vector in R3 , then v must lie in the plane orthogonal to u. Hence dim[Ker(T )] = 2.
(b). Rng(T ) = {y ∈ R : y = u, v, v ∈ R3 }, and dim[Rng(T )] = 1.
13. (a). Ker(S) = {A ∈ Mn (R) : A − AT = 0} = {A ∈ Mn (R) : A = AT }. Hence, any matrix in Ker(S) is
symmetric by definition.
(b). Since any matrix in Ker(S) is symmetric, it has been shown that {A1 , A2 , A3 } is a spanning set for the
set of all symmetric matrices in M2 (R), where
1 0
0 1
0 0
, A2 =
, A3 =
.
A1 =
0 0
1 0
0 1
Thus, dim[Ker(S)] = 3.
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14. Ker(T ) = {A ∈ Mn (R) : AB − BA = 0} = {A ∈ M2 (R) : AB = BA}. This is the set of matrices that
commute with B.
15. (a).
Ker(T ) = {p ∈ P2 (R) : T (p) = 0} = {ax2 + bx + c ∈ P2 (R) : ax2 + (a + 2b + c)x + (3a − 2b − c) = 0, for all
x}. Thus, for p(x) = ax2 + bx + c to be in Ker(T ), a, b, and c must satisfy the system:
a
a + 2b + c
=0
=0
3a − 2b − c
=0
Solving this system, we obtain that a = 0 and c = −2b. Consequently, all polynomials of the form
0x2 + bx + (−2b) are in Ker(T ), so
Ker(T ) = {b(x − 2) : b ∈ R}.
Since Ker(T ) is spanned by the nonzero vector x − 2, it follows that dim[Rng(T )] = 1.
(b). In this case,
Rng(T ) = {T (ax2 + bx + c) : a, b, c ∈ R}
= {ax2 + (a + 2b + c)x + (3a − 2b − c) : a, b, c ∈ R}
= {a(x2 + x + 3) + b(2x − 2) + c(x − 1) : a, b, c ∈ R}
= {a(x2 + x + 3) + (2b + c)(x − 1) : a, b, c ∈ R}
= span{x2 + x + 3, x − 1}.
Since the vectors in this spanning set are linearly independent on any interval, it follows that the spanning
set is a basis for Rng(T ). Hence, dim[Rng(T )] = 2.
16. Ker(T ) = {p ∈ P2 (R) : T (p) = 0} = {ax2 + bx + c ∈ P2 (R) : (a + b) + (b − c)x = 0, for all x}. Thus,
a, b, and c must satisfy: a + b = 0 and b − c = 0 =⇒ a = −b and b = c. Letting c = r ∈ R, we have
ax2 + bx + c = r(−x2 + x + 1). Thus, Ker(T ) = {r(−x2 + x + 1) : r ∈ R} and dim[Ker(T )] = 1.
Rng(T ) = {T (ax2 + bx + c) : a, b, c ∈ R} = {(a + b) + (b − c)x : a, b, c ∈ R} = {c1 + c2 x : c1 , c2 ∈ R}.
Consequently, a basis for Rng(T ) is {1, x}, so that Rng(T ) = P1 (R), and dim[Rng(T )] = 2.
17. Ker(T ) = {p ∈ P1 (R) : T (p) = 0} = {ax + b ∈ P1 (R) : (b − a) + (2b − 3a)x + bx2 = 0, for all x}. Thus, a
and b must satisfy: b−a = 0, 2b−3a = 0, and b = 0 =⇒ a = b = 0. Thus, Ker(T ) = {0} and dim[Ker(T )] = 0.
Rng(T ) = {T (ax + b) : a, b ∈ R} = {(b − a) + (2b − 3a)x + bx2 : a, b ∈ R}
= {−a(1 + 3x) + b(1 + 2x + x2 ) : a, b ∈ R}
= span{1 + 3x, 1 + 2x + x2 }.
Since the vectors in this spanning set are linearly independent on any interval, it follows that the spanning
set is a basis for Rng(T ), and dim[Rng(T )] = 2.
18. To determine Ker(T ), we set
T
a
c
b
d
= 0.
Thus,
(a − b + d) + (−a + b − d)x2 = 0,
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thus leading to the system of linear equations
a−b+d=0
− a + b − d = 0.
and
We quickly find that b = r, c = s, and d = t are free variables, and thus, a = r − t. Hence,
r−t r
Ker(T ) =
: r, s, t ∈ R .
s
t
By extracting the free variables, a basis for Ker(T ) can be given as
1 1
0 0
−1 0
,
,
.
0 0
1 0
0 1
Hence, dim[Ker(T )] = 3.
Next, Rng(T ) consists of all polynomials of the form (a − b + d) + (−a + b − d)x2 . The constant coefficient
and x2 are additive inverses of one another, so we can describe Rng(T ) more simply as
Rng(T ) = span{1 − x2 },
and thus, dim[Rng(T )] = 1.
19. To determine Ker(T ) we set
T (x, y) =
0
0
0
0
0
0
.
Thus,
−x − y = 0,
2x + 2y = 0,
3x + 3y = 0,
−9x − 9y = 0.
All of these equations are equivalent to the single equation x + y = 0. That is, y = −x. Hence,
Ker(T ) = {(x, y) : x, y ∈ R and y = −x} = {(x, −x) : x ∈ R}.
A basis for Ker(T ) can be obtained by eliminating the free variable x to get the basis {(1, −1)}. Hence,
dim[Ker(T ) = 1.
−x − y
0
2x + 2y
Next, Rng(T ) consists of all 2 × 3 matrices of the form
. That is,
0 3x + 3y −9x − 9y
−z 0
2z
Rng(T ) consists of all 2 × 3 matrices of the form
, a subspace of M23 (R) with basis
0
3z
−9z
−1 0
2
.
0 3 −9
20. To determine Ker(T ), we set T (A) = 0. That is, AT = 0, which implies that A = 0. Hence,
Ker(T ) = {02×4 }, the trivial subspace. Hence, its dimension is zero.
Next, Rng(T ) consists of all 4 × 2 matrices that arise as outputs from the transformation T . However,
given any 4 × 2 matrix B, note that T (B T ) = (B T )T = B. Thus, Rng(T ) = M42 (R), which is an eightdimensional space.
21.
T (v) = 0 ⇐⇒ T (av1 + bv2 + cv3 ) = 0
⇐⇒ aT (v1 ) + bT (v2 ) + cT (v3 ) = 0
⇐⇒ a(2w1 − w2 ) + b(w1 − w2 ) + c(w1 + 2w2 ) = 0
⇐⇒ (2a + b + c)w1 + (−a − b + 2c)w2 = 0
⇐⇒ 2a + b + c = 0 and a − b + 2c = 0.
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Reducing the augmented matrix of the system yields:
2
1 1 0
1 −1 2 0
1 0
1 0
∼
∼
.
1 −1 2 0
2
1 1 0
0 1 −1 0
Setting c = r =⇒ b = r, a = −r. Thus,
Ker(T ) = {v ∈ V : v = r(−v1 + v2 + v3 ), r ∈ R} and dim[Ker(T )] = 1.
Rng(T ) = {T (v) : v ∈ V } = {(2a + b + c)w1 + (−a − b + 2c)w2 : a, b, c ∈ V } = span{w1 , w2 } = W .
Consequently, dim[Rng(T )] = 2.
22. (a). If w ∈ Rng(T ), then T (v) = w for some v ∈ V , and since {v1 , v2 , . . . , vn } is a basis for V , there
exist c1 , c2 , . . . , cn ∈ R for which v = c1 v1 + c2 v2 + · · · + cn vn .
Accordingly, w = T (v) = a1 T (v1 ) + a2 T (v2 ) + · · · + an T (vn ). Thus,
Rng(T ) = span{T (v1 ), T (v2 ), . . . , T (vn )}.
We must show that {T (v1 ), T (v2 ), . . . , T (vn )} is also linearly independent. Suppose that
b1 T (v1 ) + b2 T (v2 ) + · · · + bn T (vn ) = 0.
Then T (b1 v1 + b2 v2 + · · · + bn vn ) = 0, so since Ker(T ) = {0}, b1 v1 + b2 v2 + · · · + bn vn = 0. Since
{v1 , v2 , . . . , vn } is a linearly independent set, the preceding equation implies that b1 = b2 = · · · = bn = 0.
Consequently, {T (v1 ), T (v2 ), . . . , T (vn )} is a linearly independent set in W . Therefore, since we have already
shown that it is a spanning set for Rng(T ), {T (v1 ), T (v2 ), . . . , T (vn )} is a basis for Rng(T ).
(b). As an example, let T : R3 → R2 be defined by T ((a, b, c)) = (a, b) for all (a, b, c) in R3 . Then for the
basis {e1 , e2 , e3 } of R3 , we have {T (e1 ), T (e2 ), T (e3 )} = {(1, 0), (0, 1), (0, 0)}, which is clearly not a basis for
R2 .
23. Let v ∈ V . Since {v1 , v2 , . . . , vk } spans V , we can write
v = c1 v 1 + c2 v2 + · · · + ck v k
for some constants c1 , c2 , . . . , ck . Thus,
T (v) = T (c1 v1 + c2 v2 + · · · + ck vk )
= c1 T (v1 ) + c2 T (v2 ) + · · · + ck T (vk )
= c1 S(v1 ) + c2 S(v2 ) + · · · + ck S(vk )
= S(c1 v1 + c2 v2 + · · · + ck vk )
= S(v),
as required.
24. Let v ∈ V . Since {v1 , v2 , . . . , vk } spans V , we can write
v = c1 v 1 + c2 v2 + · · · + ck v k
for some constants c1 , c2 , . . . , ck . Thus,
T (v) = T (c1 v1 + c2 v2 + · · · + ck vk )
= c1 T (v1 ) + c2 T (v2 ) + · · · + ck T (vk )
= c1 0 + c2 0 + · · · + ck 0
= 0,
as required.
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Solutions to Section 6.4
True-False Review:
(a): FALSE. Many one-to-one linear transformations T : P3 (R) → M32 can be constructed. One possible
example would be to define
⎡
⎤
a0 a1
T (a0 + a1 x + a2 x2 + a3 x3 ) = ⎣ a2 a3 ⎦ .
0 0
It is easy to check that with T so defined, T is a one-to-one linear transformation.
(b): TRUE. We can define an isomorphism T : V → M32 via
⎛⎡
⎤⎞ ⎡
a b c
a
T ⎝⎣ 0 d e ⎦⎠ = ⎣ c
0 0 f
e
⎤
b
d ⎦.
f
With T so defined, it is easy to check that T is an isomorphism.
(c): TRUE. Both Ker(T1 ) and Ker(T2 T1 ) are subspaces of V1 , and since if T1 (v1 ) = 0, then (T2 T1 )v1 =
T2 (T1 (v1 )) = T2 (0) = 0, we see that every vector in Ker(T1 ) belongs to Ker(T2 T1 ). Therefore, Ker(T1 ) is a
subspace of Ker(T2 T1 ).
(d): TRUE. Observe that T is not one-to-one since Ker(T ) = {0} (because Ker(T ) is 1-dimensional).
Moreover, since M22 is 4-dimensional, then by the Rank-Nullity Theorem, Rng(T ) is 3-dimensional. Since
P2 (R) is 3-dimensional, we conclude that Rng(T ) = P2 (R); that is, T is onto.
(e): TRUE. Since M2 (R) is 4-dimensional, Rng(T ) can be at most 4-dimensional. However, P4 (R) is
5-dimensional. Therefore, any such linear transformation T cannot be onto.
(f ): TRUE. If we assume that (T2 T1 )v = (T2 T1 )w, then T2 (T1 (v)) = T2 (T1 (w)). Since T2 is one-to-one,
then T1 (v) = T1 (w). Next, since T1 is one-to-one, we conclude that v = w. Therefore, T2 T1 is one-to-one.
(g): FALSE. This linear transformation is onto one-to-one. The reason is essentially because the derivative
of any constant is zero. Therefore, Ker(T ) consists of all constant functions, and therefore, Ker(T ) = {0}.
(h): TRUE. Since M23 is 6-dimensional and Rng(T ) is only 4-dimensional, T is not onto. Moreover,
since P3 (R) is 4-dimensional, the Rank-Nullity Theorem implies that Ker(T ) is 0-dimensional. Therefore,
Ker(T ) = {0}, and this means that T is one-to-one.
(i): TRUE. Recall that dim[Rn ] = n and dim[Rm ] = m. In order for such an isomorphism to exist, Rn and
Rm must have the same dimension; that is, m = n.
(j): FALSE. For example, the vector space of all polynomials with real coefficients is an infinite-dimensional
real vector space, and since Rn is finite-dimensional for all positive integers n, this statement is false.
(k): FALSE. In order for this to be true, it would also have to be assumed that T1 is onto. For example,
suppose V1 = V2 = V3 = R2 . If we define T2 (x, y) = (x, y) for all (x, y) in R2 , then T2 is onto. However,
if we define T1 (x, y) = (0, 0) for all (x, y) in R2 , then (T2 T1 )(x, y) = T2 (T1 (x, y)) = T2 (0, 0) = (0, 0) for all
(x, y) in R2 . Therefore T2 T1 is not onto, since Rng(T2 T1 ) = {(0, 0)}, even though T2 itself is onto.
(l): TRUE. This is a direct application of the Rank-Nullity Theorem. Since T is assumed to be onto,
Rng(T ) = R3 , which is three-dimensional. Therefore the dimension of Ker(T ) is 8 − 3 = 5.
Problems:
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1. We have
T2 T1 (x) = T2 (T1 (x)) = T2 (Ax) = (BA)x.
Now,
BA =
−1
1
1 −1
3
2
Thus,
T2 T1 (x) = (BA)x =
2
3
x1
x2
=
2
3
.
=
2x1 + 3x2
.
In order for T1 T2 to exist, the outputs from T2 must be received as inputs for T1 for the composition of
functions T1 T2 to work. However, the outputs from T2 belong to R, whereas the domain of T1 is R2 .
Therefore, T1 T2 does not exist.
T2 T1 = BA.
2. T1 T2 (x) = T1 (T
2 (x)) = T1 (Bx)
= (AB)x, so T1 T2 = AB. Similarly,
−1 2
1 5
−5 −5
1 5
−1 2
14
7
AB =
=
. BA =
=
.
3 1
−2
0
−2 0
3 1
2 −4
1 15 −5x1 − 5x2
−5 −5
x1
=
= (−5(x1 + x2 ), x1 + 15x2 ) and
T1 T2 (x) = (AB)x =
x2
x1 + 15x2
1 15
14x1 + 7x2
14
7
x1
=
= (7(2x1 + x2 ), 2(x1 − 2x2 )).
T2 T1 (x) = (BA)x =
x2
2x1 − 4x2
2 −4
Clearly, T1 T2 = T2 T1 .
3. We have
T1 T2 (x) = T1 (T2 (x)) = T1 (Bx) = (AB)x.
Now,
⎤
⎡
⎤
2 −1 −1 −8 5
0 −4 3
1 ⎦
0 1 ⎦.
AB = ⎣ 0
=⎣ 1
1
0 1
1
1
1 −4 4
Thus,
⎤ ⎡
⎤
⎤⎡
−1 −8 5
x1
−x1 − 8x2 + 5x3
⎦.
x1 + x3
0 1 ⎦ ⎣ x2 ⎦ = ⎣
T1 T2 (x) = (AB)x = ⎣ 1
1 −4 4
x3
x1 − 4x2 + 4x3
⎡
⎡
Now, Ker(T1 T2 ) = nullspace(AB), which we ⎡compute by first
⎤ finding a row-echelon form for the matrix AB.
1 0
1
One row-echelon form for AB is the matrix ⎣ 0 1 − 34 ⎦. Let z = t be a free variable. Then y = 34 t, and
0 0
0
then x = −t. Thus,
3
3
Ker(T1 T2 ) = nullspace(AB) = {(−t, t, t) : t ∈ R} = span{(−1, , 1)}.
4
4
Next, Rng(T1 T2 ) = colspace(AB). Using our calculations above, we see that the column space of AB is
spanned by the first two columns of AB, so that
⎧⎡
⎤ ⎡
⎤⎫
−8 ⎬
⎨ −1
Rng(T1 T2 ) = colspace(AB) = span ⎣ 1 ⎦ , ⎣ 0 ⎦ .
⎩
⎭
1
−4
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Next, we turn our attention to T2 T1 . We have
T2 T1 (x) = T2 (Ax) = (BA)x.
Now
BA =
0 −4 3
1
0 1
Thus,
T2 T1 (x) = (BA)x =
⎡
⎤
2 −1
3 −1
⎣ 0
⎦
1 =
.
3
0
1
1
3 −1
3
0
x1
x2
=
3x1 − x2
3x1
.
Now, Ker(T1 T2 ) = nullspace(BA), and since BA is an invertible matrix, its reduced row echelon form is
I2 . Thus,
Ker(T2 T1 ) = nullspace(AB) = {(0, 0)}.
Also,
Rng(T2 T1 ) = colspace(AB) = R2 .
0
1 −1
x1
=
. This matrix equation results in
4. Ker(T1 ) = {x ∈ R : Ax = 0}. Ax = 0 =⇒
0
x2
2 −2
the system: x1 − x2 = 0 and 2x1 − 2x2 = 0, or equivalently, x1 = x2 . Thus,
2
Ker(T1 ) = {x ∈ R2 : x = r(1, 1) where r ∈ R}.
Geometrically, this is the line through the origin
in R2 spanned
by
(1,1).
0
2
1
x
1
=
. This matrix equation results in the
Ker(T2 ) = {x ∈ R2 : Bx = 0}. Bx = 0 =⇒
x2
0
3 −1
system: 2x1 + x2 = 0 and 3x1 − x2 = 0, or equivalently, x1 = x2 = 0. Thus, Ker(T2 ) = {0}. Geometrically,
this is a point (the origin).
0
1 −1
2
1
x1
2
=
Ker(T1 T2 ) = {x ∈ R : (AB)x = 0}. (AB)x = 0 =⇒
x2
0
2 −2
3 −1
0
−1 2
x1
=
. This matrix equation results in the system:
=⇒
0
x2
−2 4
−x1 + 2x2 = 0 and − 2x1 + 4x2 = 0,
or equivalently, x1 = 2x2 .
Thus, Ker(T1 T2 ) = {x ∈ R2 : x = s(2, 1) where s ∈ R}. Geometrically, this is the line through the origin in
R2 spanned by the vector (2, 1).
0
2
1
1 −1
x1
2
=
Ker(T2 T1 ) = {x ∈ R : (BA)x = 0}. (BA)x = 0 =⇒
0
x2
3 −1
2 −2
0
4 −4
x1
=
. This matrix equation results in the system:
=⇒
x2
0
1 −1
4x1 − 4x2 = 0 and x1 − x2 = 0,
or equivalently, x1 = x2 .
Thus, Ker(T2 T1 ) = {x ∈ R2 : x = t(1, 1) where t ∈ R}. Geometrically, this is the line through the origin in
R2 spanned by the vector (1, 1).
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5.
(T2 T1 )(A) = T2 (T1 (A)) = T2 (A − AT ) = (A − AT ) + (A − AT )T = (A − AT ) + (AT − A) = 0n .
d
6. (a). (T1 (f ))(x) =
[sin(x − a)] = cos(x − a).
dx
x
x
(T2 (f ))(x) =
sin(t − a)dt = [− cos(t − a)]a = 1 − cos(x − a).
a
(T1 T2 )(f )(x) =
d
[1 − cos(x − a)] = sin(x − a) = f (x).
dxx
x
(T2 T1 )(f )(x) =
a
cos(t − a)dt = [sin(t − a)]a = sin(x − a) = f (x). Consequently,
(T1 T2 )(f ) = (T2 T1 )(f ) = f .
x
d
f (x)dx =
f (x)dx = f (x).
(b). (T1 T2 )(f ) = T1 (T2 (f )) = T1
dx
a
a x
dg(x)
dg(x)
=
dx = g(x) − g(a).
(T2 T1 )(g) = T2 (T1 (g)) = T2
dx
dx
a
x
7. Let v ∈ V . There exists a, b ∈ R such that v = av1 + bv2 . Then
T2 T1 (v) = T2 [aT1 (v1 ) + bT1 (v2 )] = T2 [a(v1 − v2 ) + b(2v1 + v2 )]
= T2 [(a + 2b)v1 + (b − a)v2 ] = (a + 2b)T2 (v1 ) + (b − a)T (v2 )
= (a + 2b)(v1 + 2v2 ) + (b − a)(3v1 − v2 ) = (5b − 2a)v1 + 3(a + b)v2 .
8. Let v ∈ V . There exists a, b ∈ R such that v = av1 + bv2 . Then
T2 T1 (v) = T2 [aT1 (v1 ) + bT1 (v2 )] = T2 [a(3v1 + v2 )]
= 3a(−5v2 ) + a(−v1 + 6v2 ) = −av1 − 9av2 .
9. ONTO only. The transformation T2 T1 : Mn (R) → R is given by (T2 T1 )(A) = 2 tr(A). Given any real
number r, the matrix 12 r ·E11 (which has (1, 1)-entry 12 r and all other entries zero) satisfies (T2 T1 )( 12 r ·E11 ) =
r. Thus, Rng(T2 T1 ) = R, which implies that T2 T1 is onto. However, for n > 1, T2 T1 cannot be one-to-one,
since dim[Mn (R)] > 1 = dim(R).
10. ONE-TO-ONE
only.
By inspection, we see that Ker(T ) = {0}. Moreover, Rng(T ) = {Ax : x ∈
⎧⎡
⎫
⎤
x
⎨
⎬
R} = ⎣ 2x ⎦ : x ∈ R . Since Ker(T ) is trivial, T is one-to-one. However, Rng(T ) is only one-dimensional
⎩
⎭
3x
⎡
⎤
1
(spanned by ⎣ 2 ⎦), so that T is not onto. Since T fails to be onto, we know that T −1 does not exist.
3
11. BOTH. Ker(T ) = {x ∈ R2 : T (x) = 0} = {x ∈ R2 : Ax= 0}. Theaugmented matrix of the system
1 0 0
4 2 0
, with reduced row-echelon form of
. It follows that Ker(T ) = {0}.
Ax = 0 is:
1 3 0
0 1 0
Hence T is one-to-one by Theorem 6.4.8.
For the given transformation,
Rng(T
) = {y ∈ R2 : Ax = y is consistent}.
The augmented
matrix of the
1 0 (3y1 − y2 )/5
4 2 y1
with reduced row-echelon form of
. The system is,
system Ax = y is
0 1 (2y2 − y1 )/5
1 3 y2
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therefore, consistent for all (y1 , y2 ), so that Rng(T ) = R2 . Consequently, T is onto. T −1 exists since T has
been shown to be both one-to-one and onto. Using the Gauss-Jordan method for computing A−1 , we have
1
3
0
− 15
1 12
1 0
4 2 1 0
4
10
.
∼
∼
1
2
1 3 0 1
0 1 − 10
0 52 − 14 1
5
3
− 15
−1
10
Thus, A =
, so T −1 (y) = A−1 y.
2
1
− 10
5
matrix of the
12. NEITHER. Ker(T ) = {x ∈ R2 : T (x) = 0} = {x ∈ R2 : Ax = 0}. The augmented
1 2 0
1
2 0
, with reduced row-echelon form of
. It follows that
system Ax = 0 is:
−2 −4 0
0 0 0
Ker(T ) = {x ∈ R2 : x = t(−2, 1), t ∈ R}.
By Theorem 6.4.8, since Ker(T ) = {0}, T is not one-to-one. This also implies that T −1 does not exist.
2
For the given transformation,
Rng(T
The augmented
matrix of the
) = {y ∈ R : Ax = y is consistent}.
y1
1 2
1
2 y1
with reduced row-echelon form of
. The last row of
system Ax = y is
−2 −4 y2
0 0 2y1 + y2
this matrix implies that 2y1 + y2 = 0 is required for consistency. Therefore, it follows that
Rng(T ) = {(y1 , y2 ) ∈ R2 : 2y1 + y2 = 0} = {y ∈ R2 : y = s(1, −2), s ∈ R}.
T is not onto because dim[Rng(T )] = 1 = 2 = dim[R2 ].
matrix of the
13. ONTO only. Ker(T ) = {x ∈R3 : T (x) = 0} = {x ∈ R3 : Ax = 0}. The augmented
1 0 −7 0
1 2 −1 0
, with reduced row-echelon form
. It follows that
system Ax = 0 is:
2 5
1 0
0 1
3 0
Ker(T ) = {(x1 , x2 , x3 ) ∈ R3 : x1 = 7t, x2 = −3t, x3 = t, t ∈ R} = {x ∈ R3 : x = t(7, −3, 1), t ∈ R}.
By Theorem 6.4.8, since Ker(T ) = {0}, T is not one-to-one. This also implies that T −1 does not exist.
For the given transformation, Rng(T ) = {y ∈ R2 : Ax = y is consistent}. The augmented matrix of the
system Ax = y is
1 2 −1 y1
1 0 −7 5y1 − 2y2
∼
.
2 5
1 y2
0 1
3 y2 − 2y1
We clearly have a consistent system for all y = (y1 , y2 ) ∈ R2 , thus Rng(T ) = R2 . Therefore, T is onto.
14. NEITHER. Reducing A to row-echelon form, we obtain
⎤
⎡
1 3 5
⎢ 0 1 2 ⎥
⎥
REF(A) = ⎢
⎣ 0 0 0 ⎦.
0 0 0
We quickly find that
Ker(T ) = nullspace(A) = span{(1, −2, 1)}.
Moreover,
⎧⎡
⎤⎫
⎤ ⎡
1 ⎪
0
⎪
⎪
⎪
⎨⎢
⎥⎬
⎢
3 ⎥
⎥,⎢ 4 ⎥ .
Rng(T ) = colspace(A) = span ⎢
⎦ ⎣ 4 ⎦⎪
⎣
⎪
⎪
⎪ 5
⎭
⎩
1
2
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Based on these calculations, we see that T is neither one-to-one nor onto.
15. We have
and so
1
2
1
(0, 0, 1) = − (2, 1, −3) + (1, 0, 0) + (0, 1, 0),
3
3
3
2
1
1
T (0, 0, 1) = − T (2, 1, −3) + T (1, 0, 0) + T (0, 1, 0)
3
3
3
1
2
1
= − (7, −1) + (4, 5) + (−1, 1)
3
3
3
= (0, 4).
(a). From the given information and the above calculation, we find that the matrix of T is A =
4 −1 0
5
1 4
.
(b). Because A has more columns than rows, REF(A) must have an unpivoted column, which implies that
nullspace(A) = {0}. Hence, T is not one-to-one. On the other hand, colspace(A) = R2 since the first two
columns, for example, are linearly independent vectors in R2 . Thus, T is onto.
16. Show T is a linear transformation:
Let x, y ∈ V and c, λ ∈ R where λ = 0.
T (x + y) = λ(x + y) = λx + λy = T (x) + T (y), and
T (cx) = λ(cx) = c(λx) = cT (x). Thus, T is a linear transformation.
Show T is one-to-one:
x ∈ Ker(T ) ⇐⇒ T (x) = 0 ⇐⇒ λx = 0 ⇐⇒ x = 0, since λ = 0. Thus, Ker(T ) = {0}. By Theorem 6.4.8, T
is one-to-one.
Show T is onto:
dim[Ker(T )]+ dim[Rng(T )] = dim[V ] =⇒ dim[{0}]+ dim[Rng(T )] = dim[V ]
=⇒ 0 + dim[Rng(T )] = dim[V ] =⇒ dim[Rng(T )] = dim[V ] =⇒ Rng(T ) = V . Thus, T is onto.
Find T −1 :
1
1
1
T −1 (x) = x since T (T −1 (x)) = T
x =λ
x = x.
λ
λ
λ
17. T : P1 (R) → P1 (R) where T (ax + b) = (2b − a)x + (b + a).
Show T is one-to-one:
Ker(T ) = {p ∈ P1 (R) : T (p) = 0} = {ax + b ∈ P1 (R) : (2b − a)x + (b + a) = 0}. Thus, a and b must satisfy
the system 2b − a = 0, b + a = 0. The only solution to this system is a = b = 0. Consequently, Ker(T ) = {0},
so T is one-to-one.
Show T is onto:
Since Ker(T ) = {0}, dim[Ker(T )] = 0. Thus, dim[Ker(T )]+ dim[Rng(T )] = dim[P1 (R)]
=⇒ dim[Rng(T )] = dim[P1 (R)], and since Rng(T ) is a subspace of P1 (R), it follows that Rng(T ) = P1 (R).
Consequently, T is onto by Theorem 6.4.8.
Determine T −1 :
Since T is one-to-one and onto, T −1 exists. T (ax + b) = (2b − a)x + (b + a)
A+B
2B − A
and a =
=⇒ T −1 [(2b − a)x + (b + a)] = ax + b. If we let A = 2b − a and B = a + b, then b =
3
3
1
1
−1
so that T [Ax + B] = (2B − A)x + (A + B).
3
3
18. T is not one-to-one:
Ker(T ) = {p ∈ P2 (R) : T (p) = 0} = {ax2 + bx + c ∈ P2 (R) : c + (a − b)x = 0}. Thus, a, b, and c must satisfy
the system c = 0 and a − b = 0. Consequently, Ker(T ) = {r(x2 + x) : r ∈ R} = {0}. Thus, by Theorem
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6.4.8, T is not one-to-one. T −1 does not exist because T is not one-to-one.
T is onto:
Since Ker(T ) = {r(x2 + x) : r ∈ R}, we see that dim[Ker(T )] = 1.
Thus, dim[Ker(T )]+ dim[Rng(T )] = dim[P2 (R)] =⇒ 1+ dim[Rng(T )] = 3 =⇒ dim[Rng(T )] = 2. Since
Rng(T ) is a subspace of P1 (R),
2 = dim[Rng(T )] ≤ dim[P1 (R)] = 2,
and so equality holds: Rng(T ) = P1 (R). Thus, T is onto by Theorem 6.4.8.
19: ONTO only. Since dim[P2 (R)] = 3 > 2 = dim[R2 ], we know by inspection that T cannot be one-to-one.
To see that T is onto, note that T (x2 ) = (1, 0) and T (x) = (−3, 1). Therefore, any linear combination of
(1, 0) and (−3, 1) belongs to Rng(T ). Since {(1, 0), (−3, 1)} spans R2 , we conclude that Rng(T ) = R2 , which
implies that T is onto.
Since T fails to be one-to-one, T −1 does not exist.
a b
20. ONE-TO-ONE and ONTO. Note that
belongs to Ker(T ) if and only if ax2 + bx + c = 0;
b c
0 0
that is, if and only if a = b = c = 0. Thus, Ker(T ) =
. Therefore, T is one-to-one. Since
0 0
−1
that
dim[V ] = 3 = dim[P2 (R)], Proposition 6.4.14 immediately implies
T is also onto. Therefore, T exists,
a b
.
and we have T −1 : P2 (R) → V given by T (ax2 + bx + c) =
b c
21. ONE-TO-ONE only. We can see by inspection that Rng(T ) = M2 (R), since the outputs from T all
have a zero in the lower right corner. Therefore, T is not onto. (Alternatively, this conclusion can bereached
0 0
by comparing dimensions for R3 and M2 (R).) To check if T is one-to-one, set T (a, b, c) =
. Then
0 0
we have
−a + 3c = 0, a − b − c = 0, 2a + b = 0.
⎡
⎤
−1
0
3 0
This leads to the linear system of equations with augmented matrix ⎣ 1 −1 −1 0 ⎦. Since the coeffi2
1
0 0
cient matrix for this system is invertible (check this, for example, by using the determinant), we know that
the system has only the trivial solution a = b = c = 0. Therefore, T is one-to-one.
Since T fails to be onto, T −1 does not exist.
a b
22. NEITHER. To check if T is one-to-one, set T
= 0. This implies that
c d
(a − b + d) + (2a + b)x + cx2 + (4a − b + 2d)x3 = 0.
This leads to the linear system of equations
a − b + d = 0,
2a + b = 0,
c = 0,
The augmented matrix for this linear system is
⎡
1 −1 0 1
⎢ 2
1 0 0
⎢
⎣ 0
0 1 0
4 −1 0 2
4a − b + 2d = 0.
⎤
0
0 ⎥
⎥,
0 ⎦
0
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⎤
1 −1 0 1
⎢ 2
1 0 0 ⎥
⎥, we can use cofactor expansion along
and considering the 4 × 4 coefficient matrix A = ⎢
⎣ 0
0 1 0 ⎦
4 −1 0 2
the third row, followed by the arrow method for 3 × 3 determinants, to verify that the coefficient matrix is
not invertible. Therefore, the linear system has nontrivial solutions, so that Ker(T ) is nontrivial. Hence, T
is not one-to-one.
Moreover, since dim[M2 (R)] = 4 = dim[P3 (R)], we conclude that T is also not onto by Proposition 6.4.14.
Since T is not one-to-one and not onto, we know that T −1 does not exist.
⎡
23. {v1 , v2 } is a basis for V and T : V → V is a linear transformation.
Show T is one-to-one:
Any v ∈ V can be expressed as v = av1 + bv2 where a, b ∈ R.
T (v) = 0 ⇐⇒ T (av1 + bv2 ) = 0 ⇐⇒ aT (v1 ) + bT (v2 ) = 0 ⇐⇒ a(v1 + 2v2 ) + b(2v1 − 3v2 ) = 0
⇐⇒ (a + 2b)v1 + (2a − 3b)v2 = 0 ⇐⇒ a + 2b = 0 and 2a − 3b = 0 ⇐⇒ a = b = 0 ⇐⇒ v = 0. Hence
Ker(T ) = {0}. Therefore T is one-to-one.
Show T is onto:
dim[Ker(T )] = dim[{0}] = 0 and dim[Ker(T )]+ dim[Rng(T )] = dim[V ] implies that 0+ dim[Rng(T )] = 2
or dim[Rng(T )] = 2. Since Rng(T ) is a subspace of V , it follows that Rng(T ) = V . Thus, T is onto by
Theorem 6.4.8.
Determine T −1 :
Since T is one-to-one and onto, T −1 exists.
T (av1 + bv2 ) = (a + 2b)v1 + (2a − 3b)v2
1
=⇒ T −1 [(a + 2b)v1 + (2a − 3b)v2 ] = (av1 + bv2 ). If we let A = a + 2b and B = 2a − 3b, then a = (3A + 2B)
7
1
1
1
and b = (2A − B). Hence, T −1 (Av1 + Bv2 ) = (3A + 2B)v1 + (2A − B)v2 .
7
7
7
24. Let v ∈ V . Then there exists a, b ∈ R such that v = av1 + bv2 .
a
b
(T1 T2 )v = T1 [aT2 (v1 ) + bT2 (v2 )] = T1
(v1 + v2 ) + (v1 − v2 )
2
2
a−b
a+b
a−b
a+b
=
T1 (v1 ) +
T1 (v2 ) =
(v1 + v2 ) +
(v1 − v2 )
2
2
2
2
a+b a−b
a+b a−b
v1 +
v2 = av1 + bv2 = v.
=
+
−
2
2
2
2
(T2 T1 )v = T2 [aT1 (v1 ) + bT1 (v2 )] = T2 [a(v1 + v2 ) + b(v1 − v2 )]
= T2 [(a + b)v1 + (a − b)v2 ] = (a + b)T2 (v1 ) + (a − b)T2 (v2 )
a−b
a+b
=
(v1 + v2 ) +
(v1 − v2 ) = av1 + bv2 = v.
2
2
Since (T1 T2 )v = v and (T2 T1 )v = v for all v ∈ V , it follows that T2 is the inverse of T1 , thus T2 = T1−1 .
25. An arbitrary vector in P1 (R) can be written as
p(x) = ap0 (x) + bp1 (x),
where p0 (x) = 1, and p1 (x) = x denote the standard basis vectors in P1 (R). Hence, we can define an
isomorphism T : R2 → P1 (R) by
T (a, b) = a + bx.
26. Let S denote the subspace of M2 (R) consisting of all upper triangular matrices. An arbitrary vector in
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S can be written as
a
0
b
c
=a
1
0
0
0
0
0
1
0
a
0
b
c
+b
+c
0
0
0
1
.
Therefore, we define an isomorphism T : R3 → S by
T (a, b, c) =
.
27. Let S denote the subspace of M2 (R) consisting of all skew-symmetric matrices. An arbitrary vector in
S can be written as
0 a
0 1
=a
.
−a 0
−1 0
Therefore, we can define an isomorphism T : R → S by
0
T (a) =
−a
a
0
.
28. Let S denote the subspace of M2 (R) consisting of all symmetric matrices. An arbitrary vector in S can
be written as
a b
1 0
0 1
0 0
=a
+b
+c
.
b c
0 0
1 0
0 1
Therefore, we can define an isomorphism T : R3 → S by
a
T (a, b, c) =
b
⎡
a
⎢ 0
29. A typical vector in V takes the form A = ⎢
⎣ 0
0
b
e
0
0
c
f
h
0
b
c
.
⎤
d
g ⎥
⎥. Therefore, we can define T : V → R10 via
i ⎦
j
T (A) = (a, b, c, d, e, f, g, h, i, j).
It is routine to verify that T is an invertible linear transformation. Therefore, we have n = 10.
30. A typical vector in V takes the form p = a0 + a2 x2 + a4 x4 + a6 x6 + a8 x8 . Therefore, we can define
T : V → R5 via
T (p) = (a0 , a2 , a4 , a6 , a8 ).
It is routine to verify that T is an invertible linear transformation. Therefore, we have n = 5.
31. We can see that V is isomorphic to R3 via the isomorphism T : V → R3 given by
⎛⎡
⎤⎞
0
a b
0 c ⎦⎠ = (a, b, c).
T ⎝⎣ −a
−b −c 0
It is easy to verify that T is one-to-one and onto, and thus, T is an isomorphism. Therefore, we conclude
that n = 3.
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32. We have
T1−1 (x) = A−1 x =
33. We have
T2−1 (x) = A−1 x =
−1/6
2/3
3 −1
−2
1
x.
x.
⎡
⎤
8 −29
3
19 −2 ⎦ x.
T3−1 (x) = A−1 x = ⎣ −5
2 −8
1
34. We have
⎡
⎤
11/14 −8/7
1/14
5/7
1/7 ⎦ x.
T4−1 (x) = A−1 x = ⎣ −3/7
3/14
1/7 −1/14
35. We have
36. The matrix of T2 T1 is
−2 1
.
−2 4
−1/3
1/3
−4 −1
2
2
1
2
1
3
=
−6 −7
6
8
. The matrix of T1 T2 is
⎤
⎤⎡
⎡
1 1
3
6
3 5 1
⎦
⎦
⎣
⎣
⎣
0
1
2
4
1
2
1
=
37. The matrix of T4 T3 is
3
5
−1
2
6
7
23
⎤ ⎡
⎤⎡
⎤
⎡
3 5 1
10 25 23
1 1
3
⎣ 0 1
2 ⎦ ⎣ 1 2 1 ⎦ = ⎣ 5 14 15 ⎦.
2 6 7
3 5 −1
12 19 1
⎡
⎤
13 18
8
6 ⎦.
43 11
1
2
1
3
−4 −1
2
2
The matrix of T3 T4 is
38.
(a). To show that T2 T1 : V1 → V3 is one-to-one, we show that Ker(T2 T1 ) = {0}. Suppose that v1 ∈
Ker(T2 T1 ). This means that (T2 T1 )(v1 ) = 0. Hence, T2 (T1 (v1 )) = 0. However, since T2 is one-to-one, we
conclude that T1 (v1 ) = 0. Next, since T1 is one-to-one, we conclude that v1 = 0, which shows that the only
vector in Ker(T2 T1 ) is 0, as expected.
(b). To show that T2 T1 : V1 → V3 is onto, we begin with an arbitrary vector v3 in V3 . Since T2 : V2 → V3
is onto, there exists v2 in V2 such that T2 (v2 ) = v3 . Moreover, since T1 : V1 → V2 is onto, there exists v1 in
V1 such that T1 (v1 ) = v2 . Therefore,
(T2 T1 )(v1 ) = T2 (T1 (v1 )) = T2 (v2 ) = v3 ,
and therefore, we have found a vector, namely v1 , in V1 that is mapped under T2 T1 to v3 . Hence, T2 T1 is
onto.
(c). This follows immediately from parts (a) and (b).
39. We have
(T2 T1 )(cu1 ) = T2 (T1 (cu1 ))
= T2 (cT1 (u1 ))
= c(T2 (T1 (u1 )))
= c(T2 T1 )(u1 ),
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418
as needed.
40. We first prove that if T is one-to-one, then T is onto. The assumption that T is one-to-one implies that
dim[Ker(T )] = 0. Hence, by Theorem 6.3.8,
dim[W ] = dim[V ] = dim[Rng(T )],
which implies that Rng(T ) = W . That is, T is onto.
Next we show that if T is onto, then T is one-to-one. We have Rng(T ) = W , so that dim[Rng(T )] =
dim[W ] = dim[V ]. Hence, by Theorem 6.3.8, we have dim[Ker(T )] = 0. Hence, Ker(T ) = {0}, which implies
that T is one-to-one.
41. If T : V → W is linear, then show that T −1 : W → V is also linear.
Let y, z ∈ W, c ∈ R, and T (u) = y and T (v) = z where u, v ∈ V . Then T −1 (y) = u and T −1 (z) = v. Thus,
T −1 (y + z) = T −1 (T (u) + T (v)) = T −1 (T (u + v)) = u + v = T −1 (y) + T −1 (z),
and
T −1 (cy) = T −1 (cT (u)) = T −1 (T (cu)) = cu = cT −1 (y).
Hence, T −1 is a linear transformation.
42. Let T : V → V be a one-to-one linear transformation. Since T is one-to-one, it follows from Theorem
6.4.8 that Ker(T ) = {0}, so dim[Ker(T )] = 0. By Theorem 6.3.8 and substitution,
dim[Ker(T )]+ dim[Rng(T )] = dim[V ] =⇒ 0 + dim[Rng(T )] = dim[V ] =⇒ dim[Rng(T )] = dim[V ],
and since Rng(T ) is a subspace of V , it follows that Rng(T ) = V , thus T is onto. T −1 exists because T is
both one-to-one and onto.
43. To show that {T (v1 ), T (v2 ), . . . , T (vk )} is linearly independent, assume that
c1 T (v1 ) + c2 T (v2 ) + · · · + ck T (vk ) = 0.
We must show that c1 = c2 = · · · = ck = 0. Using the linearity properties of T , we can write
T (c1 v1 + c2 v2 + · · · + ck vk ) = 0.
Now, since T is one-to-one, we can conclude that c1 v1 + c2 v2 + · · · + ck vk = 0, and since {v1 , v2 , . . . , vk } is
linearly independent, we conclude that c1 = c2 = · · · = ck = 0 as desired.
44. To prove that T is onto, let w be an arbitrary vector in W . We must find a vector v in V such that
T (v) = w. Since {w1 , w2 , . . . , wm } spans W , we can write
w = c 1 w1 + c 2 w2 + · · · + c m wm
for some scalars c1 , c2 , . . . , cm . Therefore
T (c1 v1 + c2 v2 + · · · + cm vm ) = c1 T (v1 ) + c2 T (v2 ) + · · · + cm T (vm ) = c1 w1 + c2 w2 + · · · + cm wm = w,
which shows that v = c1 v1 + c2 v2 + · · · + cm vm maps under T to w, as desired.
45. Since T is a linear transformation, Rng(T ) is a subspace of W , but
dim[W ] = dim[Rng(T )] = n, so W = Rng(T ), which means, by definition, that T is onto.
46. Let v ∈ V be arbitrary. Since T2 is onto, there exists w ∈ V such that T2 (w) = v. Applying T1 to both
sides of this, we get
T1 (T2 (w)) = T1 (v).
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However, by assumption, (T1 T2 )(w) = w. We conclude that T1 (v) = w. Applying T2 to both sides of this,
we get
(T2 T1 )(v) = T2 (w) = v,
as required.
47. Suppose that x belongs to Rng(T ). This means that there exists a vector v in V such that T (v) = x.
Applying T to both sides of this equation, we have T (T (v)) = T (x), or T 2 (v) = T (x). However, since
T 2 = 0, we conclude that T 2 (v) = 0, and hence, T (x) = 0. By definition, this means that x belongs to
Ker(T ). Thus, since every vector in Rng(T ) belongs to Ker(T ), Rng(T ) is a subset of Ker(T ). In addition,
Rng(T ) is closed under addition and scalar multiplication, by Theorem 6.3.5, and therefore, Rng(T ) forms
a subspace of Ker(T ).
Solutions to Section 6.5
True-False Review:
(a): FALSE. The matrix representation is an m × n matrix, not an n × m matrix.
(b): FALSE. The matrix [T ]B
C would only make sense if C was a basis for V and B was a basis for W ,
and this would only be true if V and W were the same vector space. Of course, in general V and W can be
different, and so this statement does not hold.
(c): FALSE. The correct equation is given in (6.5.2).
−1
(d): FALSE. The correct statement is [T ]C
= [T −1 ]B
B
C.
(e): TRUE. Many examples are possible. A fairly simple one is the following. Let T1 : R2 → R2 be given
by T1 (x, y) = (x, y), and let T2 : R2 → R2 be given by T2 (x, y) = (y,x). Clearly,
T1 and T2 are different
1
0
1
. If B2 = {(1, 0), (0, 1)} and
linear transformations. Now if B1 = {(1, 0), (0, 1)} = C1 , then [T1 ]C
B1 =
0 1
1 0
2
. Thus, although T1 = T2 , we found suitable bases B1 , C1 , B2 , C2
C2 = {(0, 1), (1, 0)}, then [T2 ]C
B2 =
0 1
C2
1
such that [T1 ]C
B1 = [T2 ]B2 .
(f ): TRUE. This is the content of part (b) of Corollary 6.5.10.
Problems:
1.
(a). We must determine T (E11 ), T (E12 ), T (E21 ), and T (E22 ), and find the components of the resulting
vectors in P3 (R) relative to the basis C. We have
T (E11 ) = 1 − x3 ,
T (E12 ) = 3x2 ,
T (E21 ) = x3 ,
T (E22 ) = −1.
Therefore, relative to the standard basis C on P3 (R), we have
⎤
1
⎢ 0 ⎥
⎥
[T (E11 )]C = ⎢
⎣ 0 ⎦,
−1
⎡
⎤
0
⎢ 0 ⎥
⎥
[T (E12 )]C = ⎢
⎣ 3 ⎦,
0
⎡
⎤
0
⎢ 0 ⎥
⎥
[T (E21 )]C = ⎢
⎣ 0 ⎦,
1
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⎡
⎤
−1
⎢ 0 ⎥
⎥
[T (E22 )]C = ⎢
⎣ 0 ⎦.
0
⎡
420
Putting these results into the columns of [T ]C
B , we obtain
⎤
⎡
1 0 0 −1
⎢ 0 0 0
0 ⎥
⎥.
⎢
[T ]C
B =⎣
0 3 0
0 ⎦
−1 0 1
0
(b). The values of T (E21 ), T (E11 ), T (E22 ), and T (E12 ) were determined in part (a). We must express those
results in terms of the ordered basis C given in (b). We have
⎤
⎤
⎤
⎤
⎡
⎡
⎡
⎡
0
0
0
0
⎥
⎥
⎥
⎢ 0 ⎥
⎢
⎢
⎢
⎥ , [T (E11 )]C = ⎢ 1 ⎥ , [T (E22 )]C = ⎢ −1 ⎥ , [T (E12 )]C = ⎢ 0 ⎥ .
[T (E21 )]C = ⎢
⎣ 1 ⎦
⎣ −1 ⎦
⎣ 0 ⎦
⎣ 0 ⎦
0
0
0
3
Putting these results into the columns of [T ]C
B , we obtain
⎤
⎡
0
0
0 0
⎢ 0
1 −1 0 ⎥
⎥.
⎢
[T ]C
B = ⎣ 1 −1
0 0 ⎦
0
0
0 3
2.
(a). We must determine T (1), T (x), and T (x2 ), and find the components of the resulting vectors in R2
relative to the basis C. We have
T (1) = (1, 2),
T (x) = (0, 1),
T (x2 ) = (−3, −2).
Therefore, relative to the standard basis C on R2 , we have
1
0
[T (1)]C =
, [T (x)]C =
,
2
1
Therefore,
[T ]C
B =
1
2
0 −3
1 −2
2
[T (x )]C =
−3
−2
.
.
(b). We must determine T (1), T (1 + x), and T (1 + x + x2 ), and find the components of the resulting vectors
in R2 relative to the basis C. We have
T (1) = (1, 2),
T (1 + x) = (1, 3),
T (1 + x + x2 ) = (−2, 1).
Setting
T (1) = (1, 2) = c1 (1, −1) + c2 (2, 1)
−1
. Next, setting
and solving, we find c1 = −1 and c2 = 1. Thus, [T (1)]C =
1
T (1 + x) = (1, 3) = c1 (1, −1) + c2 (2, 1)
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and solving, we find c1 = −5/3 and c2 = 4/3. Thus, [T (1 + x)]C =
−5/3
4/3
. Finally, setting
T (1 + x + x2 ) = (−2, 1) = c1 (1, −1) + c2 (2, 1)
−4/3
and solving, we find c1 = −4/3 and c2 = −1/3. Thus, [T (1 + x + x2 )]C =
. Putting the results for
−1/3
[T (1)]C , [T (1 + x)]C , and [T (1 + x + x2 )]C into the columns of [T ]C
B , we obtain
−1 − 53 − 43
[T ]C
.
4
B =
1
− 13
3
3.
(a). We must determine T (1), T (x), and T (x2 ), and find the components of the resulting vectors relative
to the standard basis C on P3 (R). We have
T (1) = 1 + x,
Therefore,
⎤
1
⎢ 1 ⎥
⎥
[T (1)]C = ⎢
⎣ 0 ⎦,
0
⎡
T (x) = x + x2 ,
T (x2 ) = x2 + x3 .
⎤
0
⎢ 1 ⎥
⎥
[T (x)]C = ⎢
⎣ 1 ⎦,
0
⎡
Putting these results into the columns of [T ]C
B , we obtain
⎡
1 0
⎢
⎢ 1 1
[T ]C
B =⎣ 0 1
0 0
⎤
0
⎢ 0 ⎥
⎥
[T (x2 )]C = ⎢
⎣ 1 ⎦.
1
⎡
⎤
0
0 ⎥
⎥.
1 ⎦
1
(b). We must determine T (1), T (x − 1), and T ((x − 1)2 ), and find the components of the resulting vectors
relative to the given basis C on P3 (R). We have
T (1) = 1 + x,
Setting
T (x − 1) = −1 + x2 ,
T ((x − 1)2 ) = 1 − 2x − x2 + x3 .
1 + x = c1 (1) + c2 (x − 1) + c3 (x − 1)2 + c4 (x − 1)3
⎡
⎤
2
⎢ 1 ⎥
⎥
and solving, we find c1 = 2, c2 = 1, c3 = 0, and c4 = 0. Therefore [T (1)]C = ⎢
⎣ 0 ⎦. Next, setting
0
−1 + x2 = c1 (1) + c2 (x − 1) + c3 (x − 1)2 + c4 (x − 1)3
⎤
0
⎢ 2 ⎥
⎥
and solving, we find c1 = 0, c2 = 2, c3 = 1, and c4 = 0. Therefore, [T (x − 1)]C = ⎢
⎣ 1 ⎦. Finally, setting
0
⎡
1 − 2x − x2 + x3 = c1 (1) + c2 (x − 1) + c3 (x − 1)2 + c4 (x − 1)3
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⎤
−1
⎢ −1 ⎥
⎥
and solving, we find c1 = −1, c2 = −1, c3 = 2, and c4 = 1. Therefore, [T ((x − 1)2 )]C = ⎢
⎣ 2 ⎦. Putting
1
these results into the columns of [T ]C
B , we obtain
⎤
⎡
2 0 −1
⎢ 1 2 −1 ⎥
⎥.
⎢
[T ]C
B =⎣ 0 1
2 ⎦
0 0
1
⎡
4.
(a). We must determine T (1, 0, 0), T (0, 1, 0), and T (0, 0, 1), and find the components of the resulting vectors
relative to the basis C. We have
T (1, 0, 0) = cos x,
T (0, 1, 0) = 3 sin x,
T (0, 0, 1) = −2 cos x + sin x.
Therefore, relative to the basis C, we have
1
0
[T (1, 0, 0)]C =
, [T (0, 1, 0)]C =
,
0
3
[T (0, 0, 1)]C =
−2
1
.
Putting these results into the columns of [T ]C
B , we obtain
1 0 −2
=
.
[T ]C
B
0 3
1
(b). We must determine T (2, −1, −1), T (1, 3, 5), and T (0, 4, −1), and find the components of the resulting
vectors relative to the basis C. We have
T (2, −1, −1) = 4 cos x − 4 sin x,
T (1, 3, 5) = −9 cos x + 14 sin x,
T (0, 4, −1) = 2 cos x + 11 sin x.
Setting
4 cos x − 4 sin x = c1 (cos x − sin x) + c2 (cos x + sin x)
4
. Next, setting
and solving, we find c1 = 4 and c2 = 0. Therefore [T (2, −1, −1)]C =
0
−9 cos x + 14 sin x = c1 (cos x − sin x) + c2 (cos x + sin x)
−23/2
. Finally, setting
and solving, we find c1 = −23/2 and c2 = 5/2. Therefore, [T (1, 3, 5)]C =
5/2
2 cos x + 11 sin x = c1 (cos x − sin x) + c2 (cos x + sin x)
−9/2
. Putting these results
and solving, we find c1 = −9/2 and c2 = 13/2. Therefore [T (0, 4, −1)]C =
13/2
into the columns of [T ]C
B , we obtain
− 92
4 − 23
2
.
=
[T ]C
13
5
B
0
2
2
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5.
(a). We must determine T (E11 ), T (E12 ), T (E21 ), and T (E22 ), and find the components of the resulting
vectors relative to the standard basis C on R2 . We have
T (E11 ) = (1, 1),
T (E12 ) = (0, 0),
T (E21 ) = (0, 0),
T (E22 ) = (1, 1).
Therefore, since C is the standard basis on R2 , we have
1
0
0
, [T (E12 )]C =
, [T (E21 )]C =
,
[T (E11 )]C =
1
0
0
Putting these results into the columns of [T ]C
B , we obtain
1 0 0
C
[T ]B =
1 0 0
1
1
[T (E22 )]C =
1
1
.
.
(b). Let us denote the four matrices in the ordered basis for B as A1 , A2 , A3 , and A4 , respectively. We
must determine T (A1 ), T (A2 ), T (A3 ), and T (A4 ), and find the components of the resulting vectors relative
to the standard basis C on R2 . We have
T (A1 ) = (−4, −4),
T (A2 ) = (3, 3),
T (A3 ) = (−2, −2),
Therefore, since C is the standard basis on R2 , we have
−4
3
−2
, [T (A2 )]C =
, [T (A3 )]C =
,
[T (A1 )]C =
−4
3
−2
Putting these results into the columns of [T ]C
B , we obtain
−4 3 −2
=
[T ]C
B
−4 3 −2
0
0
T (A4 ) = (0, 0).
[T (A4 )]C =
0
0
.
.
6.
(a). We must determine T (1), T (x), T (x2 ), and T (x3 ), and find the components of the resulting vectors
relative to the standard basis C on P2 (R). We have
T (1) = 0,
T (x) = 1,
T (x2 ) = 2x,
T (x3 ) = 3x2 .
Therefore, if C is the standard basis on P2 (R), then we have
⎡
⎤
⎡
⎤
⎡
⎤
0
0
1
[T (1)]C = ⎣ 0 ⎦ , [T (x)]C = ⎣ 0 ⎦ , [T (x2 )]C = ⎣ 2 ⎦ ,
0
0
0
Putting these results into the columns of [T ]C
B , we obtain
⎡
0 1 0
⎣ 0 0 2
[T ]C
B =
0 0 0
⎤
0
0 ⎦.
3
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⎡
⎤
0
[T (x3 )]C = ⎣ 0 ⎦ .
3
424
(b). We must determine T (x3 ), T (x3 + 1), T (x3 + x), and T (x3 + x2 ), and find the components of the
resulting vectors relative to the given basis C on P2 (R). We have
T (x3 ) = 3x2 ,
Setting
T (x3 + 1) = 3x2 ,
T (x3 + x) = 3x2 + 1,
T (x3 + x2 ) = 3x2 + 2x.
3x2 = c1 (1) + c2 (1 + x) + c3 (1 + x + x2 )
⎡
⎤
0
and solving, we find c1 = 0, c2 = −3, and c3 = 3. Therefore [T (x3 )]C = ⎣ −3 ⎦. Likewise, [T (x3 + 1)]C =
3
⎡
⎤
0
⎣ −3 ⎦. Next, setting
3
3x2 + 1 = c1 (1) + c2 (1 + x) + c3 (1 + x + x2 )
⎡
⎤
1
and solving, we find c1 = 1, c2 = −3, and c3 = 3. Therefore, [T (x3 + x)]C = ⎣ −3 ⎦. Finally, setting
3
3x2 + 2x = c1 (1) + c2 (1 + x) + c3 (1 + x + x2 )
⎡
⎤
−2
and solving, we find c1 = −2, c2 = −1, and c3 = 3. Therefore, [T (x3 + 2x)]C = ⎣ −1 ⎦. Putting these
3
,
we
obtain
results into the columns of [T ]C
B
⎡
⎤
0
0
1 −2
⎣ −3 −3 −3 −1 ⎦ .
[T ]C
B =
3
3
3
3
7.
(a). We must determine T (e2x ) and T (e−3x ), and find the components of the resulting vectors relative to
the given basis C. We have
T (e2x ) = 2e2x and T (e−3x ) = −3e−3x .
Therefore, we have
[T ]C
B =
2
0
0
−3
.
(b). We must determine T (e2x − 3e−3x ) and T (2e−3x ), and find the components of the resulting vectors
relative to the given basis C. We have
T (e2x − 3e−3x ) = 2e2x + 9e−3x
and
T (2e−3x ) = −6e−3x .
Now, setting
2e2x + 9e−3x = c1 (e2x + e−3x ) + c2 (−e2x )
9
2x
−3x
)]C =
. Finally, setting
and solving, we find c1 = 9 and c2 = 7. Therefore, [T (e − 3e
7
−6e−3x = c1 (e2x + e−3x ) + c2 (−e2x )
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and solving, we find c1 = c2 = −6. Therefore, [T (2e−3x )]C =
of [T ]C
B , we obtain
[T ]C
B =
9 −6
7 −6
−6
−6
. Putting these results into the columns
.
8.
(a). We must determine T (E11 ), T (E12 ), T (E21 ), and T (E22 ), and find the components of the resulting
vectors relative to the standard basis C on M2 (R). We have
T (E11 ) = E11 ,
T (E12 ) = 2E12 − E21 ,
T (E21 ) = 2E21 − E12 ,
T (E22 ) = E22 .
⎤
0
⎢ 2 ⎥
⎥
[T (E12 )]C = ⎢
⎣ −1 ⎦ ,
0
⎤
0
⎢ −1 ⎥
⎥
[T (E21 )]C = ⎢
⎣ 2 ⎦,
0
⎤
0
⎢ 0 ⎥
⎥
[T (E22 )]C = ⎢
⎣ 0 ⎦.
1
Therefore, we have
⎤
1
⎢ 0 ⎥
⎥
[T (E11 )]C = ⎢
⎣ 0 ⎦,
0
⎡
⎡
⎡
⎡
Putting these results into the columns of [T ]C
B , we obtain
⎤
1
0
0 0
⎢ 0
2 −1 0 ⎥
⎥.
⎢
[T ]C
B = ⎣ 0 −1
2 0 ⎦
0
0
0 1
⎡
(b). Let us denote the four matrices in the ordered basis for B as A1 , A2 , A3 , and A4 , respectively. We
must determine T (A1 ), T (A2 ), T (A3 ), and T (A4 ), and find the components of the resulting vectors relative
to the standard basis C on M2 (R). We have
−1 −2
1 0
0 −8
0 7
T (A1 ) =
, T (A2 ) =
, T (A3 ) =
, T (A4 ) =
.
−2 −3
3 2
7 −2
−2 0
Therefore, we have
⎤
−1
⎢ −2 ⎥
⎥
[T (A1 )]C = ⎢
⎣ −2 ⎦ ,
−3
⎡
⎤
1
⎢ 0 ⎥
⎥
[T (A2 )]C = ⎢
⎣ 3 ⎦,
2
⎡
⎤
0
⎢ −8 ⎥
⎥
[T (A3 )]C = ⎢
⎣ 7 ⎦,
−2
⎡
Putting these results into the columns of [T ]C
B , we obtain
⎡
−1
⎢ −2
C
[T ]B = ⎢
⎣ −2
−3
⎤
1
0
0
0 −8
7 ⎥
⎥.
3
7 −2 ⎦
2 −2
0
9.
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⎤
0
⎢ 7 ⎥
⎥
[T (A4 )]C = ⎢
⎣ −2 ⎦ .
0
⎡
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(a). Let us first compute [T ]C
B . We must determine T (1, 0, 0), T (0, 1, 0), and T (0, 0, 1), and find the
components of the resulting vectors relative to the standard basis C = {1, x, x2 , x3 } on P3 (R). We have
T (1, 0, 0) = 2 − x − x3 ,
Therefore,
⎤
2
⎢ −1 ⎥
⎥
[T (1, 0, 0)]C = ⎢
⎣ 0 ⎦,
−1
T (0, 1, 0) = −x,
T (0, 0, 1) = x + 2x3 .
⎤
0
⎢ −1 ⎥
⎥
[T (0, 1, 0)]C = ⎢
⎣ 0 ⎦,
0
⎡
⎤
0
⎢ 1 ⎥
⎥
[T (0, 0, 1)]C = ⎢
⎣ 0 ⎦.
2
⎡
⎡
Putting these results into the columns of [T ]C
B , we obtain
⎤
⎡
2
0 0
⎢ −1 −1 1 ⎥
⎥.
⎢
[T ]C
B =⎣
0
0 0 ⎦
−1
0 2
Now,
and therefore,
⎡
⎤
2
[v]B = [(2, −1, 5)]B = ⎣ −1 ⎦ ,
5
⎤
⎤
⎡
⎤
4
2
0 0 ⎡
2
⎥
⎢
⎢ −1 −1 1 ⎥
⎥ ⎣ −1 ⎦ = ⎢ 4 ⎥ .
⎢
[T (v)]C = [T ]C
B [v]B = ⎣
⎣ 0 ⎦
0
0 0 ⎦
5
8
−1
0 2
⎡
Therefore,
(b). We have
T (v) = 4 + 4x + 8x3 .
T (v) = T (2, −1, 5) = 4 + 4x + 8x3 .
10.
(a). Let us first compute [T ]C
B . We must determine T (1) and T (x), and find the components of the resulting
vectors relative to the standard basis C = {E11 , E12 , E21 , E22 } on M2 (R). We have
1
0
−1 0
T (1) =
and T (x) =
.
0 −1
−2 1
Therefore
Hence,
⎤
1
⎢ 0 ⎥
⎥
[T (1)]C = ⎢
⎣ 0 ⎦
−1
⎤
−1
⎢ 0 ⎥
⎥
and [T (x)]C = ⎢
⎣ −2 ⎦ .
1
⎡
⎡
⎤
1 −1
⎢ 0
0 ⎥
⎥.
⎢
[T ]C
B =⎣
0 −2 ⎦
−1
1
⎡
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Now,
[p(x)]B = [−2 + 3x]B =
−2
3
.
Therefore, we have
⎤
⎡
⎤
−5
1 −1 ⎥
⎢
⎢ 0
0 ⎥
⎥ −2 = ⎢ 0 ⎥ .
⎢
[T (p(x))]C = [T ]C
B [p(x)]B = ⎣
⎣
⎦
−6 ⎦
3
0 −2
5
−1
1
⎡
Thus,
T (p(x)) =
−5
−6
(b). We have
0
5
T (p(x)) = T (−2 + 3x) =
.
−5
−6
0
5
.
11.
2
(a). Let us first compute [T ]C
B . We must determine T (1), T (x), and T (x ), and find the components of the
2
3
4
resulting vectors relative to the standard basis C = {1, x, x , x , x }. We have
T (1) = x2 ,
Therefore,
⎤
0
⎢ 0 ⎥
⎥
⎢
⎥
[T (1)]C = ⎢
⎢ 1 ⎥,
⎣ 0 ⎦
0
⎡
T (x) = x3 ,
⎤
0
⎢ 0 ⎥
⎥
⎢
⎥
[T (x)]C = ⎢
⎢ 0 ⎥,
⎣ 1 ⎦
0
and therefore,
⎤
0
⎢ 0 ⎥
⎥
⎢
⎥
[T (x2 )]C = ⎢
⎢ 0 ⎥.
⎣ 0 ⎦
1
⎡
Putting these results into the columns of [T ]C
B , we obtain
⎡
0 0
⎢ 0 0
⎢
⎢
[T ]C
B =⎢ 1 0
⎣ 0 1
0 0
Now,
T (x2 ) = x4 .
⎡
⎤
0
0 ⎥
⎥
0 ⎥
⎥.
0 ⎦
1
⎡
⎤
−1
[p(x)]B = [−1 + 5x − 6x2 ]B = ⎣ 5 ⎦ ,
−6
⎡
0
⎢ 0
⎢
⎢
[T (p(x))]C = [T ]C
B [p(x)]B = ⎢ 1
⎣ 0
0
0
0
0
1
0
⎤
⎡
⎤
0
0 ⎡
⎤
⎢ 0 ⎥
0 ⎥
⎥
⎢
⎥ −1
⎥
⎥
⎣
⎦
5 =⎢
0 ⎥
⎢ −1 ⎥ .
⎣
⎦
5 ⎦
−6
0
−6
1
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Hence,
T (p(x)) = −x2 + 5x3 − 6x4 .
(b). We have
T (p(x)) = −x2 + 5x3 − 6x4 .
12.
(a). Let us first compute [T ]C
B . We must determine T (E11 ), T (E12 ), T (E21 ), and T (E22 ), and find the
components of the resulting vectors relative to the standard basis C = {E11 , E12 , E21 , E22 }. We have
2 −1
−1
0
0 0
1 3
, T (E12 ) =
, T (E21 ) =
, T (E22 ) =
.
T (E11 ) =
0 −1
0 −1
0 3
0 0
Therefore,
⎤
2
⎢ −1 ⎥
⎥
[T (E11 )]C = ⎢
⎣ 0 ⎦,
−1
⎡
⎤
−1
⎢ 0 ⎥
⎥
[T (E12 )]C = ⎢
⎣ 0 ⎦,
−1
⎡
⎤
0
⎢ 0 ⎥
⎥
[T (E21 )]C = ⎢
⎣ 0 ⎦,
3
⎡
⎤
1
⎢ 3 ⎥
⎥
[T (E22 )]C = ⎢
⎣ 0 ⎦.
0
⎡
Putting these results into the columns of [T ]C
B , we obtain
⎤
⎡
2 −1 0 1
⎢ −1
0 0 3 ⎥
⎥.
⎢
[T ]C
B =⎣
0
0 0 0 ⎦
−1 −1 3 0
Now,
and therefore,
⎤
−7
⎢ 2 ⎥
⎥
[A]B = ⎢
⎣ 1 ⎦,
−3
⎡
⎤ ⎡
⎤
⎤⎡
−19
−7
2 −1 0 1
⎥ ⎢
⎥
⎢
⎢ −1
0 0 3 ⎥
⎥ ⎢ 2 ⎥ = ⎢ −2 ⎥ .
⎢
[T (A)]C = [T ]C
B [A]B = ⎣
⎦
⎣
⎦
⎣
0 ⎦
1
0
0 0 0
−3
8
−1 −1 3 0
⎡
Hence,
T (A) =
(b). We have
T (A) =
−19 −2
0
8
−19 −2
0
8
.
.
13.
2
3
4
(a). Let us first compute [T ]C
B . We must determine T (1), T (x), T (x ), T (x ), and T (x ), and find the
components of the resulting vectors relative to the standard basis C = {1, x, x2 , x3 }. We have
T (1) = 0,
T (x) = 1,
T (x2 ) = 2x,
T (x3 ) = 3x2 ,
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T (x4 ) = 4x3 .
429
Therefore,
⎤
0
⎢ 0 ⎥
⎥
[T (1)]C = ⎢
⎣ 0 ⎦,
0
⎡
⎤
1
⎢ 0 ⎥
⎥
[T (x)]C = ⎢
⎣ 0 ⎦,
0
⎡
⎤
0
⎢ 2 ⎥
⎥
[T (x2 )]C = ⎢
⎣ 0 ⎦,
0
⎤
0
⎢ 0 ⎥
⎥
[T (x3 )]C = ⎢
⎣ 3 ⎦,
0
⎡
⎡
⎤
0
⎢ 0 ⎥
⎥
[T (x4 )]C = ⎢
⎣ 0 ⎦.
4
⎡
Putting these results into the columns of [T ]C
B , we obtain
⎡
0
⎢ 0
C
[T ]B = ⎢
⎣ 0
0
Now,
1
0
0
0
0
2
0
0
0
0
3
0
⎤
0
0 ⎥
⎥.
0 ⎦
4
⎤
3
⎢ −4 ⎥
⎥
⎢
⎥
[p(x)]B = [3 − 4x + 6x2 + 6x3 − 2x4 ]B = ⎢
⎢ 6 ⎥,
⎣ 6 ⎦
−2
⎡
and therefore,
⎡
0
⎢ 0
0
0
⎢
[T (p(x))]C = [T ]C
B [p(x)]B = ⎣
1
0
0
0
0
2
0
0
0
0
3
0
⎤
⎤
⎡
3
−4
0 ⎢
⎥
−4
⎥
⎥ ⎢
⎢
0 ⎥
⎥ ⎢ 6 ⎥ = ⎢ 12 ⎥ .
⎥
⎢
⎣
⎦
18 ⎦
0 ⎣
6 ⎦
−8
4
−2
⎤
⎡
Therefore,
T (p(x)) = T (3 − 4x + 6x2 + 6x3 − 2x4 ) = −4 + 12x + 18x2 − 8x3 .
(b). We have
T (p(x)) = p (x) = −4 + 12x + 18x2 − 8x3 .
14.
(a). Let us first compute [T ]C
B . We must determine T (Eij ) for 1 ≤ i, j ≤ 3, and find components of the
resulting vectors relative to the standard basis C = {1}. We have
T (E11 ) = T (E22 ) = T (E33 ) = 1
and T (E12 ) = T (E13 ) = T (E23 ) = T (E21 ) = T (E31 ) = T (E32 ) = 0.
Therefore
[T (E11 )]C = [T (E22 ]C = [T (E33 ]C = [1]
and all other component vectors are [0].
Putting these results into the columns of [T ]C
B , we obtain
[T ]C
B =
1
0
0
0
1
0
0
0
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Now,
⎤
2
⎢ −6 ⎥
⎥
⎢
⎢ 0 ⎥
⎥
⎢
⎢ 1 ⎥
⎥
⎢
⎥
[A]B = ⎢
⎢ 4 ⎥.
⎢ −4 ⎥
⎥
⎢
⎢ 0 ⎥
⎥
⎢
⎣ 0 ⎦
−3
⎡
and therefore,
⎤
2
⎢ −6 ⎥
⎥
⎢
⎢ 0 ⎥
⎥
⎢
⎢ 1 ⎥
⎥
⎢
⎢ 4 ⎥ = [3].
⎥
⎢
⎢ −4 ⎥
⎥
⎢
⎢ 0 ⎥
⎥
⎢
⎣ 0 ⎦
−3
⎡
[T (A)]C = [T ]C
B [A]B =
1
0
0
0
1
0
0
0
1
Hence, T (A) = 3.
(b). We have T (A) = 3.
15.
2
3
(a). Let us first compute [T ]C
B . We must determine T (1), T (x), T (x ), and T (x ), and find the components
of the resulting vectors relative to the standard basis C = {1}. We have
T (1) = 1,
T (x) = 2, T (x2 ) = 4,
Therefore,
[T (1)]C = [1],
[T (x)]C = [2],
T (x3 ) = 8.
[T (x2 )]C = [4],
[T (x3 )]C = [8].
Putting these results into the columns of [T ]C
B , we obtain
[T ]C
B =
Now,
1
2
4
8
.
⎤
0
⎢ 2 ⎥
⎥
[p(x)]B = [2x − 3x2 ]B = ⎢
⎣ −3 ⎦ ,
0
⎡
and therefore,
⎤
0
⎢ 2 ⎥
⎥
⎢
⎣ −3 ⎦ = [−8].
0
⎡
[T (p(x))]C = [T ]C
B [p(x)]B =
1
2
4
8
Therefore, T (p(x)) = −8.
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(b). We have p(2) = 2 · 2 − 3 · 22 = −8.
16. The linear transformation T2 T1 : P4 (R) → R is given by
(T2 T1 )(p(x)) = p (2).
Let A denote the standard basis on P4 (R), let B denote the standard basis on P3 (R), and let C denote the
standard basis on R.
(a). To determine [T2 T1 ]C
A , we compute
(T2 T1 )(1) = 0,
(T2 T1 )(x2 ) = 4,
(T2 T1 )(x) = 1,
(T2 T1 )(x3 ) = 12,
(T2 T1 )(x4 ) = 32.
Therefore
[(T2 T1 )(1)]C = [0],
[(T2 T1 )(x2 )]C = [4],
[(T2 T1 )(x)]C = [1],
[(T2 T1 )(x3 )]C = [12],
[(T2 T1 )(x4 )]C = [32].
Putting these results into the columns of [T2 T1 ]C
A , we obtain
[T2 T1 ]C
A =
0
1
4
12
0
2
0
0
0
0
3
0
⎤
0
0 ⎥
⎥=
0 ⎦
4
32
.
0
1
(b). We have
⎡
B
[T2 ]C
B [T1 ]A =
1
2
4
8
0
⎢ 0
⎢
⎣ 0
0
1
0
0
0
4
12
32
= [T2 T1 ]C
A.
2
4
(c). Let p(x)
⎡ = 2⎤ + 5x − x + 3x . Then the component vector of p(x) relative to the standard basis A is
2
⎢ 5 ⎥
⎥
⎢
⎥
[p(x)]A = ⎢
⎢ −1 ⎥. Thus,
⎣ 0 ⎦
3
⎤
2
⎢ 5 ⎥
⎥
⎢
⎢ −1 ⎥ = [97].
⎥
⎢
⎣ 0 ⎦
3
⎡
[(T2 T1 )(p(x))]C = [T2 T1 ]C
A [p(x)]A =
Therefore,
Of course,
0
1
4
12
(T2 T1 )(2 + 5x − x2 + 3x4 ) = 97.
p (2) = 5 − 2 · 2 + 12 · 23 = 97
by direct calculation as well.
17. The linear transformation T2 T1 : P1 (R) → R2 is given by
(T2 T1 )(a + bx) = (0, 0).
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Let A denote the standard basis on P1 (R), let B denote the standard basis on M2 (R), and let C denote the
standard basis on R2 .
(a). To determine [T2 T1 ]C
A , we compute
(T2 T1 )(1) = (0, 0)
and
(T2 T1 )(x) = (0, 0).
Therefore, we obtain [T2 T1 ]C
A = 02 .
(b). We have
B
[T2 ]C
B [T1 ]A =
1
1
0
0
0
0
1
1
⎤
1 −1
⎢ 0
0 ⎥
C
⎥
⎢
⎣ 0 −2 ⎦ = 02 = [T2 T1 ]A .
−1
1
⎡
(c). The component vector of p(x) = −3 + 8x relative to the standard basis A is [p(x)]A =
[(T2 T1 )(p(x))]C = [T2 T1 ]C
A [p(x)]A = 02 [p(x)]A =
0
0
−3
8
. Thus,
.
Therefore,
(T2 T1 )(−3 + 8x) = (0, 0).
Of course
(T2 T1 )(−3 + 8x) = T1
−11
−16
0
11
= (0, 0)
by direct calculation as well.
18. The linear transformation T2 T1 : P2 (R) → P2 (R) is given by
(T2 T1 )(p(x)) = [(x + 1)p(x)] .
Let A denote the standard basis on P2 (R), let B denote the standard basis on P3 (R), and let C denote the
standard basis on P2 (R).
(a). To determine [T2 T1 ]C
A , we compute as follows:
(T2 T1 )(1) = 1,
Therefore,
⎡
⎤
1
[(T2 T1 )(1)]C = ⎣ 0 ⎦ ,
0
(T2 T1 )(x) = 1 + 2x,
(T2 T1 )(x2 ) = 2x + 3x2 .
⎡
⎡
⎤
1
[(T2 T1 )(x)]C = ⎣ 2 ⎦ ,
0
⎤
0
[(T2 T1 )(x2 )]C = ⎣ 2 ⎦ .
3
Putting these results into the columns of [T2 T1 ]C
A , we obtain
⎡
⎤
1 1 0
C
[T2 T1 ]A = ⎣ 0 2 2 ⎦ .
0 0 3
(b). We have
⎡
0
B
⎣
0
[T
]
=
[T2 ]C
B 1 A
0
1
0
0
0
2
0
⎡
⎤ 1
0 ⎢
1
0 ⎦⎢
⎣ 0
3
0
0
1
1
0
⎤
⎡
0
1
⎥
0 ⎥ ⎣
0
=
1 ⎦
0
1
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2
0
⎤
0
2 ⎦ = [T2 T1 ]C
A.
3
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⎡
⎤
7
(c). The component vector of p(x) = 7 − x + 2x2 relative to the standard basis A is [p(x)]A = ⎣ −1 ⎦.
2
Thus,
⎡
⎤⎡
⎤ ⎡
⎤
1 1 0
7
6
⎣
⎦
⎣
⎦
⎣
0
2
2
−1
2 ⎦.
[p(x)]
=
[(T2 T1 )(p(x))]C = [T2 T1 ]C
=
A
A
0 0 3
2
6
Therefore,
(T2 T1 )(7 − x + 2x2 ) = 6 + 2x + 6x2 .
Of course
(T2 T1 )(7 − x + 2x2 ) = T2 ((x + 1)(7 − x + 2x2 )) = T2 (7 + 6x + x2 + 2x3 ) = 6 + 2x + 6x2
by direct calculation as well.
(d). YES. Since the matrix [T2 T1 ]C
A computed in part (a) is invertible, T2 T1 is invertible.
19. NO. The matrices [T ]C
B obtained in Problem 1 are not invertible (they contain rows of zeros), and
therefore, the corresponding linear transformation T is not invertible.
20. YES. We can explain this answer by using a matrix representation of T . Let B = {1, x, x2 } and let
C = {(1, 0, 0), (0, 1, 0), (0, 0, 1)}. Then T (1) = (1, 1, 1), T (x) = (0, 1, 2), and T (x2 ) = (0, 1, 4), and so
⎡
⎤
1 0 0
⎣ 1 1 1 ⎦.
[T ]C
B =
1 2 4
Since this matrix is invertible, the corresponding linear transformation T is invertible.
21. Note that
w belongs to Rng(T ) ⇐⇒ w = T (v) for some v in V
⇐⇒ [w]C = [T (v)]C for some v in V
⇐⇒ [w]C = [T ]C
B [v]B for some v in V .
The right-hand side of this last expression can be expressed as a linear combination of the columns of [T ]C
B,
and therefore, w belongs to Rng(T ) if and only if [w]C can be expressed as a linear combination of the
C
columns of [T ]C
B . That is, if and only if [w]C belongs to colspace([T ]B ).
Solutions to Section 6.6
Additional Problems:
1. NO. Note that T (1, 1) = (2, 0, 0, 1) and T (2, 2) = (4, 0, 0, 4) = 2T (1, 1). Thus, T is not a linear
transformation.
2. YES. The function T can be represented by the matrix function
T (x) = Ax,
where
A=
2 −3 0
−1
0 0
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and x is a vector in R3 . Every matrix transformation of the form T (x) = Ax is linear. Since the domain of
T has larger dimension than the codomain of T , T cannot be one-to-one. However, since
T (0, −1/3, 0) = (1, 0)
and
T (−1, −2/3, 0) = (0, 1),
we see that T is onto. Thus, Rng(T ) = R2 (2-dimensional), and so a basis for Rng(T ) is {(1, 0), (0, 1)}. The
kernel of T consists of vectors of the form (0, 0, z), and hence, a basis for Ker(T ) is {(0, 0, 1)} and Ker(T ) is
1-dimensional.
3. YES. The function T can be represented by the matrix function
T (x) = Ax,
where
A=
0 −3
−1
5
0
2
and x is a vector in R3 . Every matrix transformation of the form T (x) = Ax is linear. Since the domain of
T has larger dimension than the codomain of T , T cannot be one-to-one. However, since
T (1, 1/3, −1/3) = (1, 0)
and T (1, 1, 0) = (0, 1),
we see that T is onto. Thus, Rng(T ) = R2 , and so a basis for Rng(T ) is {(1, 0), (0, 1)}, and Rng(T ) is
2-dimensional. The kernel of T consists of vectors of the form (t, 2t, 0), where t ∈ R, and hence, a basis for
Ker(T ) is {(1, 2, 0)}. We have that Ker(T ) is 1-dimensional.
4. YES. The function T is a linear transformation, because if g, h ∈ C[0, 1], then
T (g + h) = ((g + h)(0), (g + h)(1)) = (g(0) + h(0), g(1) + h(1)) = (g(0), g(1)) + (h(0), h(1)) = T (g) + T (h),
and if c is a scalar,
T (cg) = ((cg)(0), (cg)(1)) = (cg(0), cg(1)) = c(g(0), g(1)) = cT (g).
Note that any function g ∈ C[0, 1] for which g(0) = g(1) = 0 (such as g(x) = x2 − x) belongs to Ker(T ), and
hence, T is not one-to-one. However, given (a, b) ∈ R2 , note that g defined by
g(x) = a + (b − a)x
satisfies
T (g) = (g(0), g(1)) = (a, b),
so T is onto. Thus, Rng(T ) = R2 , with basis {(1, 0), (0, 1)} (2-dimensional). Now, Ker(T ) is infinitedimensional. We cannot list a basis for this subspace of C[0, 1].
5. YES. The function T can be represented by the matrix function
T (x) = Ax,
where
A=
1/5
1/5
and x is a vector in R2 . Every such matrix transformation is linear. Since the domain of T has larger
dimension than the codomain of T , T cannot be one-to-one. However, T (5, 0) = 1, so we see that T is onto.
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Thus, Rng(T ) = R, a 1-dimensional space with basis {1}. The kernel of T consists of vectors of the form
t(1, −1), and hence, a basis for Ker(T ) is {(1, −1)}. We have that Ker(T ) is 1-dimensional.
6. NO. For instance, note that
1
T
1
and so
T
However,
1
1
T
0
1
0
1
1
1
= (1, 0)
+T
0
1
+
and
1
0
1
1
1
0
1
1
T
1
0
1
1
= (0, 1),
= (1, 0) + (0, 1) = (1, 1).
=T
2
1
1
2
= (2, 2).
Thus, T does not respect addition, and hence, T is not a linear transformation. Similar work could be given
to show that T also fails to respect scalar multiplication.
7. YES. We can verify that T respects addition and scalar multiplication as follows:
T respects addition: Let a1 + b1 x + c1 x2 and a2 + b2 x + c2 x2 belong to P2 (R). Then
T ((a1 + b1 x + c1 x2 ) + (a2 + b2 x + c2 x2 )) = T ((a1 + a2 ) + (b1 + b2 )x + (c1 + c2 )x2 )
0
−(a1 + a2 ) − (b1 + b2 )
=
3(c1 + c2 ) − (a1 + a2 ) −2(b1 + b2 )
0
0
−a2 − b2
−a1 − b1
+
=
3c1 − a1 −2b1
3c2 − a2 −2b2
= T (a1 + b1 x + c1 x2 ) + T (a2 + b2 x + c2 x2 ).
T respects scalar multiplication: Let a + bx + cx2 belong to P2 (R) and let k be a scalar. Then we have
−ka − kb
0
T (k(a + bx + cx2 )) = T ((ka) + (kb)x + (kc)x2 ) =
3(kc) − (ka) −2(kb)
k(−a − b)
0
=
k(3c − a) k(−2b)
−a − b
0
=k
3c − a −2b
= kT (a + bx + cx2 ).
Next, observe that a + bx + cx2 belongs to Ker(T ) if and only if −a − b = 0, 3c − a = 0, and −2b = 0.
These equations require that b = 0, a = 0, and c = 0. Thus, Ker(T ) = {0}, which implies that Ker(T ) is
0-dimensional (with basis ∅), and that T is one-to-one. However, since M2 (R) is 4-dimensional and P2 (R)
is only 3-dimensional, we see immediately that T cannot be onto. By the Rank-Nullity Theorem, in fact,
Rng(T ) must be 3-dimensional, and a basis is given by
−1 0
−1
0
0 0
Basis for Rng(T ) = {T (1), T (x), T (x2 )} =
,
,
.
−1 0
0 −2
3 0
8. YES. We can verify that T respects addition and scalar multiplication as follows:
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T respects addition: Let A, B belong to M2 (R). Then
T (A + B) = (A + B) + (A + B)T = A + B + AT + B T = (A + AT ) + (B + B T ) = T (A) + T (B).
T respects scalar multiplication: Let A belong to M2 (R), and let k be a scalar. Then
T (kA) = (kA) + (kA)T = kA + kAT = k(A + AT ) = kT (A).
Thus, T is a linear transformation. Note that if A is any skew-symmetric matrix, then T (A) = A + AT =
A + (−A)
= 0, so Ker(T ) consists precisely of the 2 × 2 skew-symmetric matrices.
matrices take the
These
0 −a
0 −1
form
, for a constant a, and thus a basis for Ker(T ) is given by
, and Ker(T ) is
a
0
1
0
1-dimensional. Consequently, T is not one-to-one.
Therefore, by Proposition 6.4.14, T also fails to be onto. In fact, by the Rank-Nullity Theorem, Rng(T )
must be 3-dimensional. A typical element of the range of T takes the form
a b
a b
a c
2a
b+c
T
=
+
=
.
c d
c d
b d
c+b
2d
The characterizing feature of this matrix is that it is symmetric. So Rng(T ) consists of all 2 × 2 symmetric
matrices, and hence a basis for Rng(T ) is
1 0
0 1
0 0
,
,
.
0 0
1 0
0 1
9. YES. We can verify that T respects addition and scalar multiplication as follows:
T respects addition: Let (a1 , b1 , c1 ) and (a2 , b2 , c2 ) belong to R3 . Then
T ((a1 , b1 , c1 ) + (a2 , b2 , c2 )) = T (a1 + a2 , b1 + b2 , c1 + c2 )
= (a1 + a2 )x2 + (2(b1 + b2 ) − (c1 + c2 ))x + (a1 + a2 − 2(b1 + b2 ) + (c1 + c2 ))
= [a1 x2 + (2b1 − c1 )x + (a1 − 2b1 + c1 )] + [a2 x2 + (2b2 − c2 )x + (a2 − 2b2 + c2 )]
= T ((a1 , b1 , c1 )) + T ((a2 , b2 , c2 )).
T respects scalar multiplication: Let (a, b, c) belong to R3 and let k be a scalar. Then
T (k(a, b, c)) = T (ka, kb, kc) = (ka)x2 + (2kb − kc)x + (ka − 2kb + kc)
= k(ax2 + (2b − c)x + (a − 2b + c))
= kT ((a, b, c)).
Thus, T is a linear transformation. Now, (a, b, c) belongs to Ker(T ) if and only if a = 0, 2b − c = 0, and
a − 2b + c = 0. These equations collectively require that a = 0 and 2b = c. Setting c = 2t, we find that b = t.
Hence, (a, b, c) belongs to Ker(T ) if and only if (a, b, c) has the form (0, t, 2t) = t(0, 1, 2). Hence, {(0, 1, 2)}
is a basis for Ker(T ), which is therefore 1-dimensional. Hence, T is not one-to-one.
By Proposition 6.4.14, T is also not onto. In fact, the Rank-Nullity Theorem implies that Rng(T ) must
be 2-dimensional. It is spanned by
{T (1, 0, 0), T (0, 1, 0), T (0, 0, 1)} = {x2 + 1, 2x − 2, −x + 1},
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but the last two polynomials are proportional to each other. Omitting the polynomial 2x − 2 (this is an
arbitrary choice; we could have omitted −x + 1 instead), we arrive at a basis for Rng(T ): {x2 + 1, −x + 1}.
10. YES. We can verify that T respects addition and scalar multiplication as follows:
T respects addition: Let (x1 , x2 , x3 ) and (y1 , y2 , y3 ) be vectors in R3 . Then
T ((x1 , x2 , x3 ) + (y1 , y2 , y3 )) = T (x1 + y1 , x2 + y2 , x3 + y3 )
0
(x1 + y1 ) − (x2 + y2 ) + (x3 + y3 )
=
−(x1 + y1 ) + (x2 + y2 ) − (x3 + y3 )
0
0
y1 − y2 + y3
0
x1 − x2 + x3
+
=
−x1 + x2 − x3
0
−y1 + y2 − y3
0
= T ((x1 , x2 , x3 )) + T ((y1 , y2 , y3 )).
T respects scalar multiplication: Let (x1 , x2 , x3 ) belong to R3 and let k be a scalar. Then
0
(kx1 ) − (kx2 ) + (kx3 )
T (k(x1 , x2 , x3 )) = T (kx1 , kx2 , kx3 ) =
−(kx1 ) + (kx2 ) − (kx3 )
0
0
x1 − x2 + x3
=k
−x1 + x2 − x3
0
= kT ((x1 , x2 , x3 )).
Thus, T is a linear transformation. Now, (x1 , x2 , x3 ) belongs to Ker(T ) if and only if x1 − x2 + x3 = 0
and −x1 + x2 − x3 = 0. Of course, the latter equation is equivalent to the former, so the kernel of T consists
simply of ordered triples (x1 , x2 , x3 ) with x1 − x2 + x3 = 0. Setting x3 = t and x2 = s, we have x1 = s − t,
so a typical element of Ker(T ) takes the form (s − t, s, t), where s, t ∈ R. Extracting the free variables, we
find a basis for Ker(T ): {(1, 1, 0), (−1, 0, 1)}. Hence, Ker(T ) is 2-dimensional.
By the Rank-Nullity Theorem, Rng(T ) must be 1-dimensional.
In fact, Rng(T ) consists precisely of the
0 −1
set of 2 × 2 skew-symmetric matrices, with basis
. Since M2 (R) is 4-dimensional, T fails to be
1
0
onto.
11. We have
T (x, y, z) = (−x + 8y, 2x − 2y − 5z).
12. We have
T (x, y) = (−x + 4y, 2y, 3x − 3y, 3x − 3y, 2x − 6y).
x
x
x 5x
T (2) = (−1, 5, 0, −2) = − , , 0, −x .
2
2
2 2
a b
14. For an arbitrary 2 × 2 matrix
, if we write
c d
a b
1 0
0 1
1 0
1 1
=r
+s
+t
+u
,
c d
0 1
1 0
0 0
0 0
13. We have
T (x) =
we can solve for r, s, t, u to find
r = d,
s = c,
t = a − b + c − d,
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Thus,
T
a
c
b
d
1 0
0 1
1 0
1 1
+ (a − b + c − d)
+ (b − c)
=T d
+c
0 1
1 0
0 0
0 0
1 0
0 1
1 0
1 1
= dT
+ cT
+ (a − b + c − d)T
+ (b − c)T
0 1
1 0
0 0
0 0
= d(2, −5) + c(0, −3) + (a − b + c − d)(1, 1) + (b − c)(−6, 2)
= (a − 7b + 7c + d, a + b − 4c − 6d).
15. For an arbitrary element ax2 + bx + c in P2 (R), if we write
ax2 + bx + c = r(x2 − x − 3) + s(2x + 5) + 6t = rx2 + (−r + 2s)x + (−3r + 5s + 6t),
1
5
a − 12
b + 16 c. Thus,
we can solve for r, s, t to find r = a, s = 12 (a + b), and t = 12
1
1
5
2
2
T (ax + bx + c) = T a(x − x − 3) + (a + b)(2x + 5) + a − b + c
2
2
2
1
1
5
c
2
T (6)
a− b+
= aT (x − x − 3) + (a + b)T (2x + 5) +
2
12
12
6
1
1
5
c
12 6
−2
1
0
1
=a
+ (a + b)
+
a− b+
6 18
−4 −1
2 −2
2
12
12
6
−a − 5b + 2c
2a − 2b + c
=
.
− 52 a − 32 b + c − 12 a − 17
2 b + 3c
16. Since dim[P5 (R)] = 6 and dim[M2 (R)] = 4 = dim[Rng(T )] (since T is onto), the Rank-Nullity Theorem
gives
dim[Ker(T )] = 6 − 4 = 2.
17. Since T is one-to-one, dim[Ker(T )] = 0, so the Rank-Nullity Theorem gives
dim[Rng(T )] = dim[M2×3 (R)] = 6.
18. We must have n = 8, since dim[M42 (R)] = 8.
19. We know that dim[Ker(T )] + dim[Rng(T )] = dim[V ]. In this case, dim[V ] = 5. Since T is not
onto, we must have dim[Rng(T )] < 4. Therefore, dim[Rng(T )] ∈ {0, 1, 2, 3}, from which it follows that
dim[Ker(T )] ∈ {5, 4, 3, 2}.
20. Note that
T (1) = (1, 1) = 1 · (1, 0) + 1 · (0, 1),
T (x) = (0, 1) = 0 · (1, 0) + 1 · (0, 1),
T (x2 ) = (0, 1) = 0 · (1, 0) + 1 · (0, 1),
T (x3 ) = (0, 1) = 0 · (1, 0) + 1 · (0, 1).
Thus,
[T ]C
B =
1
1
0
1
0
1
0
1
.
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⎤
2
⎢ 0 ⎥
⎥
Since [v]B = ⎢
⎣ −1 ⎦, by Theorem 6.5.4, we have
2
⎡
[T (v)]C =
1
1
0
1
0
1
0
1
⎤
2
⎢ 0 ⎥
2
⎥=
⎢
.
⎣ −1 ⎦
−1
2
⎡
Therefore, T (v) = 2 · (1, 0) + (−1) · (0, 1) = (2, −1).
21. Note that
T (0, 1, 0) =
T (0, 0, 1) =
T (1, 0, 0) =
Thus,
0
0
1
−3
0
−1
3
0
0 −1
5
0
= 0 · E21 + (−3) · E22 + 0 · E11 + 1 · E12 ,
= (−1) · E21 + 0 · E22 + 0 · E11 + 3 · E12 ,
= 5 · E21 + 0 · E22 + 0 · E11 + (−1) · E12 .
⎤
0 −1
5
⎢ −3
0
0 ⎥
⎥.
⎢
[T ]C
B =⎣
0
0
0 ⎦
1
3 −1
⎡
⎡
⎤
1
Since [v]B = ⎣ −2 ⎦, by Theorem 6.5.4, we have
−2
⎤
⎤
⎡
⎤
−8
0 −1
5 ⎡
1
⎥
⎢
⎢ −3
0
0 ⎥
⎥ ⎣ −2 ⎦ = ⎢ −3 ⎥ .
[T (v)]C = ⎢
⎣ 0 ⎦
⎣ 0
0
0 ⎦
−2
−3
1
3 −1
⎡
Thus,
T (v) = −8 · E21 + (−3) · E22 + 0 · E11 + (−3) · E12 =
0 −3
−8 −3
.
22. Note that
⎤
4
⎢ 0 ⎥
⎥
T (1, 1) = ⎢
⎣ −2 ⎦ = 7 · (0, 0, 0, 1) + (−2) · (0, 0, 1, 0) + 0 · (0, 1, 0, 0) + 4 · (1, 0, 0, 0),
7
⎡
⎤
1
⎢ −1 ⎥
⎥
T (1, 0) = ⎢
⎣ −2 ⎦ = 5 · (0, 0, 0, 1) + (−2) · (0, 0, 1, 0) + (−1) · (0, 1, 0, 0) + 1 · (1, 0, 0, 0).
5
⎡
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Thus,
⎤
7
5
⎢ −2 −2 ⎥
⎥.
⎢
[T ]C
B =⎣
0 −1 ⎦
4
1
⎡
Next we compute [v]B by writing
(−2,
4) = c1 (1, 1) + c2 (1, 0). A short calculation shows that c1 = 4 and
4
, by Theorem 6.5.4, we have
c2 = −6. Thus, since [v]B =
−6
⎡
⎤
⎤
⎡
−2
7
5 ⎢ 4 ⎥
⎢ −2 −2 ⎥
4
⎥
⎢
⎥
[T (v)]C = ⎢
⎣ 0 −1 ⎦ −6 = ⎣ 6 ⎦ .
10
4
1
Thus,
T (v) = −2 · (0, 0, 0, 1) + 4 · (0, 0, 1, 0) + 6 · (0, 1, 0, 0) + 10 · (1, 0, 0, 0) = (10, 6, 4, −2).
23. Note that
T (x2 + x) = 2x + 1 = (−1) · 1 + 2 · (1 + x),
T (1) = 0 = 0 · 1 + 0 · (1 + x),
T (x) = 1 = 1 · 1 + 0 · 1 + x.
Thus,
[T ]C
B =
−1
2
0
0
1
0
.
2
Next we compute [v]B by writing −3 + x2 = c1 · (x
⎡ + x)
⎤ + c2 · 1 + c3 · x. A short calculation shows that
1
c1 = 1, c2 = −3, and c3 = −1. Thus, since [v]B = ⎣ −3 ⎦, by Theorem 6.5.4, we have
−1
⎡
⎤
1
−1 0 1 ⎣
−2
−3 ⎦ =
[T (v)]C =
.
2 0 0
2
−1
Thus,
T (v) = −2 · 1 + 2 · (1 + x) = 2x.
24. To see that T1 + T2 is a linear transformation, we must verify that it respects addition and scalar
multiplication:
T1 + T2 respects addition: Let v1 and v2 belong to V . Then we have
(T1 + T2 )(v1 + v2 ) = T1 (v1 + v2 ) + T2 (v1 + v2 )
= [T1 (v1 ) + T1 (v2 )] + [T2 (v1 ) + T2 (v2 )]
= [T1 (v1 ) + T2 (v1 )] + [T1 (v2 ) + T2 (v2 )]
= (T1 + T2 )(v1 ) + (T1 + T2 )(v2 ),
where we have used the linearity of T1 and T2 individually in the second step.
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T1 + T2 respects scalar multiplication: Let v belong to V and let k be a scalar. Then we have
(T1 + T2 )(kv) = T1 (kv) + T2 (kv)
= kT1 (v) + kT2 (v)
= k[T1 (v) + T2 (v)]
= k(T1 + T2 )(v),
as required.
There is no particular relationship between Ker(T1 ), Ker(T2 ), and Ker(T1 + T2 ).
25. FALSE. For instance, consider T1 : R → R defined by T1 (x) = x, and consider T2 : R → R defined by
T2 (x) = −x. Both T1 and T2 are linear transformations, and both of them are onto. However, (T1 + T2 )(x) =
T1 (x) + T2 (x) = x + (−x) = 0, so Rng(T1 + T2 ) = {0}, which implies that T1 + T2 is not onto.
26. FALSE. For instance, consider T1 : R → R defined by T1 (x) = x, and consider T2 : R → R defined
by T2 (x) = −x. Both T1 and T2 are linear transformations, and both of them are one-to-one. However,
(T1 + T2 )(x) = T1 (x) + T2 (x) = x + (−x) = 0, so Ker(T1 + T2 ) = R, which implies that T1 + T2 is not
one-to-one.
27. Assume that
c1 T (v1 + c2 T (v2 ) + · · · + cn T (vn ) = 0.
We wish to show that
c1 = c2 = · · · = cn = 0.
To do this, use the linearity of T to rewrite the above equation as
T (c1 v1 + c2 v2 + · · · + cn vn ) = 0.
Now, since Ker(T ) = {0}, we conclude that
c1 v1 + c2 v2 + · · · + cn vn = 0.
Since {v1 , v2 , . . . , vn } is a linearly independent set, we conclude that c1 = c2 = · · · = cn = 0, as required.
28. Assume that V1 ∼
= V2 and V2 ∼
= V3 . Then there exist isomorphisms T1 : V1 → V2 and T2 : V2 → V3 .
Since the composition of two linear transformations is a linear transformation (Theorem 6.4.2), we have a
linear transformation T2 T1 : V1 → V3 . Moreover, since both T1 and T2 are one-to-one and onto, T2 T1 is
also one-to-one and onto (see Problem 38 in Section 6.4). Thus, T2 T1 : V1 → V3 is an isomorphism. Hence,
V1 ∼
= V3 , as required.
29. We first show that T is a linear transformation:
T respects addition: Let A and B belong to Mn (R). Then
T (A + B) = S −1 (A + B)S = S −1 AS + S −1 BS = T (A) + T (B),
and so T respects addition.
T respects scalar multiplication: Let A belong to Mn (R), and let k be a scalar. Then
T (kA) = S −1 (kA)S = k(S −1 AS) = kT (A),
and so T respects scalar multiplication.
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Next, we verify that T is both one-to-one and onto (of course, in view of Proposition 6.4.14, it is only
necessary to confirm one of these two properties, but we will nonetheless verify them both):
T is one-to-one: Assume that T (A) = 0n . That is, S −1 AS = 0n . Left multiplying by S and right multiplying
by S −1 on both sides of this equation yields A = S0n S −1 = 0n . Hence, Ker(T ) = {0n }, and so T is one-toone.
T is onto: Let B be an arbitrary matrix in Mn (R). Then
T (SBS −1 ) = S −1 (SBS −1 )S = (S −1 S)B(S −1 S) = In BIn = B,
and hence, B belongs to Rng(T ). Since B was an arbitrary element of Mn (R), we conclude that Rng(T ) =
Mn (R). That is, T is onto.
Chapter 7 Solutions
Solutions to Section 7.1
True-False Review:
(a): FALSE. If v = 0, then Av = λv = 0, but by definition, an eigenvector must be a nonzero vector.
(b): TRUE. When we compute det(A − λI) for an upper or lower triangular matrix A, the determinant is
the product of the entries lying along the main diagonal of A − λI:
det(A − λI) = (a11 − λ)(a22 − λ) . . . (ann − λ).
The roots of this characteristic equation are precisely the values a11 , a22 , . . . , ann along the main diagonal of
the matrix A.
(c): TRUE. The eigenvalues of a matrix are precisely the set of roots of its characteristic equation. Therefore, two matrices A and B that have the same characteristic equation will have the same eigenvalues.
(d): TRUE. The characteristic equation of an n × n matrix A, det(A − λI) is a polynomial of degree n i
the indeterminate λ. Since such a polynomial always possesses n roots (with possible repeated or complex
roots) by the Fundamental Theorem of Algebra, the statement is true.
(e): TRUE. Geometrically, all nonzero points v = (x, y) in R2 are oriented in a different direction from
the origin after a 90◦ rotation than they are initially. Therefore, the vectors v and Av are not parallel.
0 0
(f ): FALSE. Many examples of this can be found. As a simple one, consider A =
and B =
0 0
0 1
. We have det(A − λI) = (−λ)2 = λ2 = det(B − λI). Note that every nonzero vector in R2 is an
0 0
a
eigenvector of A corresponding to λ = 0. However, only vectors of the form
with a = 0 are eigenvectors
0
of B. Therefore, A and B do not have precisely the same set of eigenvectors. In this case, every eigenvector
of B is also an eigenvector of A, but not conversely.
(g): FALSE. This is not true, in general, when the linear combination
formed involves eigenvectors cor
1 0
responding to different eigenvalues. For example, let A =
, with eigenvalues λ = 1 and λ = 2.
0 2
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It is easy to see that corresponding eigenvectors to these eigenvalues are, respectively, v1 =
0
. However, note that
v2 =
1
1
A(v1 + bf v 2 ) =
,
2
1
0
and
which is not of the form λ(v1 + v2 ), and therefore v1 + v2 is not an eigenvector of A. As a more trivial
illustration, note that if v is an eigenvector of A, then 0v is a linear combination of {v} that is no longer an
eigenvector of A.
(h): TRUE. This is basically a fact about roots of polynomials. Complex roots of real polynomials always
occur in complex conjugate pairs. Therefore, if λ = a + ib (b = 0) is an eigenvalue of A, then so is λ = a − ib.
(i): TRUE. We are given that Av = λv for a nonzero eigenvector v. Then
A3 v = A2 (Av) = A2 (λv) = λ(A2 v) = λA(Av) = λA(λv) = λ2 (Av) = λ2 (λv) = λ3 v,
which shows that λ3 is an eigenvalue of A3 .
Problems:
1. We have
−7
−9
Av =
as required.
1
2. Av =
2
⎡
3
2
1
1
=
4
4
=4
1
1
⎡
2
4
2
9
=
4
18
= 2v,
= λv.
⎤⎡
⎤
⎤
⎡
⎤
1 −2 −6
2
6
2
2 −5 ⎦ ⎣ 1 ⎦ = ⎣ 3 ⎦ = 3 ⎣ 1 ⎦ = λv.
3. Av = ⎣ −2
2
1
8
−1
−3
−1
⎡
⎤
⎡
⎤ ⎡
⎤
1
4
c1 + 4c2
4. Since v = c1 ⎣ 0 ⎦ + c2 ⎣ −3 ⎦ = ⎣ −3c2 ⎦, it follows that
−3
0
−3c1
⎤ ⎡
⎤
⎤
⎡
⎡
⎤⎡
−2c1 − 8c2
c1 + 4c2
1 4
1
c1 + 4c2
⎦ = −2 ⎣ −3c2 ⎦ = λv.
6c2
1 ⎦ ⎣ −3c2 ⎦ = ⎣
Av = ⎣ 3 2
−3c1
6c1
−3c1
3 4 −1
5. We have
⎤ ⎡
⎤
⎤
⎡
⎤⎡
10c1
c1
6 −1 0
c1
6 0 ⎦ ⎣ −4c1 ⎦ = ⎣ −40c1 ⎦ = 10 ⎣ −4c1 ⎦ = 10v,
Av = ⎣ −16
c2
10c2
c2
−4 −1 10
⎡
as required.
6. We have
Av1 =
−5
1
2
−4
−2
1
=
12
−6
= −6
−2
1
= −6v1 ,
which shows that λ1 = −6 is an eigenvalue of A. Furthermore,
−5
2
1
−3
1
=
= −3
= −3v2 ,
Av2 =
1 −4
1
−3
1
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which shows that λ2 = −3 is an eigenvalue of A. Thus, the eigenvalues of A are λ1 = −6 and λ2 = −3.
7. We have
Av1 =
4
2
1
3
1
−2
=
2
−4
=2
1
−2
= 2v1 ,
which shows that λ1 = 2 is an eigenvalue of A. Furthermore,
4 1
1
5
1
Av2 =
=
=5
= 5v2 ,
2 3
1
5
1
which shows that λ2 = 5 is an eigenvalue of A. Thus, the eigenvalues of A are λ1 = 2 and λ2 = 5.
8. The only vectors that are mapped into a scalar multiple of themselves under a reflection in the x-axis
are those vectors that either point along the x-axis, or that point along the y-axis. Hence, the eigenvectors
are of the form (a, 0) or (0, b) where a and b are arbitrary nonzero real numbers. A vector that points along
the x-axis will have neither its magnitude nor its direction altered by a reflection in the x-axis. Hence, the
eigenvectors of the form (a, 0) correspond to the eigenvalue λ = 1. A vector of the form (0, b) will be mapped
into the vector (0, −b) = −1(0, b) under a reflection in the x-axis. Consequently, the eigenvectors of the form
(0, b) correspond to the eigenvalue λ = −1.
9. Any vectors lying along the line y = x are unmoved by the action of T , and hence, any vector (t, t) with
t = 0 is an eigenvector with corresponding eigenvalue λ = 1. On the other hand, any vector lying along
the line y = −x will be reflected across the line y = x, thereby experiencing a 180◦ change of direction.
Therefore, any vector (t, −t) with t = 0 is an eigenvector with corresponding eigenvalue λ = −1. All vectors
that do not lie on the line y = x or the line y = −x are not eigenvectors of this linear transformation.
10. If θ = 0, π, there are no vectors that are mapped into scalar multiples of themselves under the rotation,
and consequently, there are no real eigenvalues and eigenvectors in this case. If θ = 0, then every vector is
mapped onto itself under the rotation, therefore λ = 1, and every nonzero vector in R2 is an eigenvector. If
θ = π, then every vector is mapped onto its negative under the rotation, therefore λ = −1, and once again,
every nonzero vector in R2 is an eigenvector.
11. Any vectors lying on the y-axis are unmoved by the action of T , and hence, any vector (0, y, 0) with y
not zero is an eigenvector of T with corresponding eigenvalue λ = 1. On the other hand, any vector lying
in the xz-plane, say (x, 0, z) is transformed under T to (0, 0, 0). Thus, any vector (x, 0, z) with x and z not
both zero is an eigenvector with corresponding eigenvalue λ = 0.
5−λ
−4 = 0 ⇐⇒ λ2 + 2λ − 3 = 0 ⇐⇒ (λ + 3)(λ − 1) = 0
12. det(A − λI) = 0 ⇐⇒ 8
−7 − λ ⇐⇒ λ = −3 or λ = 1.
8 −4
v1
0
=
If λ = −3 then (A − λI)v = 0 assumes the form
v2
0
8 −4
1
=⇒ 8v1 − 4v2 = 0 =⇒ v1 = 2 v2 . If we let v2 = r ∈ R, then the solution set of this equation is
{( 12 r, r) : r ∈ R}, so the eigenvectors corresponding to λ = −3 are v = r( 12 , 1), where r ∈ R (and r = 0).
0
4 −4
v1
If λ = 1 then (A − λI)v = 0 assumes the form
=
v2
0
8 −8
=⇒ v1 = v2 . If we let v2 = t ∈ R, then v1 = t and the solution set of this system is {(t, t) : t ∈ R}, so the
eigenvectors corresponding to λ = 1 are v = t(1, 1), where t ∈ R (and t = 0).
1−λ
6
= 0 ⇐⇒ λ2 + 2λ − 15 = 0
13. det(A − λI) = 0 ⇐⇒ 2
−3 − λ ⇐⇒ (λ − 3)(λ + 5) = 0 ⇐⇒ λ = 3 or λ = −5.
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0
−2
6
v1
=
0
v2
2 −6
=⇒ v1 − 3v2 = 0. If we let v2 = r ∈ R, then the solution set of this system is {(3r, r) : r ∈ R} so the
eigenvectors corresponding to λ = 3 are v = r(3, 1)
where r ∈ R. 0
6 6
v1
=
If λ = −5 then (A − λI)v = 0 assumes the form
0
v2
2 2
=⇒ v1 + v2 = 0. If we let v2 = s ∈ R, then the solution set of this system is {(−s, s) : s ∈ R} so the
eigenvectors corresponding to λ = −5 are v = s(−1, 1) where s ∈ R.
7−λ
4 14. det(A − λI) = 0 ⇐⇒ = 0 ⇐⇒ λ2 − 10λ + 25 = 0
−1
3−λ ⇐⇒ (λ − 5)2 = 0 ⇐⇒ λ = 5 of multiplicity two.
0
2
4
v1
=
If λ = 5 then (A − λI)v = 0 assumes the form
v2
0
−1 −2
=⇒ v1 + 2v2 = 0 =⇒ v1 = −2v2 . If we let v2 = t ∈ R, then the solution set of this system is {(−2t, t) : t ∈ R}
so the eigenvectors corresponding to λ = 5 are v = t(−2, 1) where t ∈ R.
2−λ
0 = 0 ⇐⇒ (2 − λ)2 = 0 ⇐⇒ λ = 2 of multiplicity two.
15. det(A − λI) = 0 ⇐⇒ 0
2−λ 0
0 0
v1
If λ = 2 then (A − λI)v = 0 assumes the form
=
0
v2
0 0
Thus, if we let v1 = s and v2 = t where s, t ∈ R, then the solution set of this system is {(s, t) : s, t ∈ R} so
the eigenvectors corresponding to λ = 2 are v = s(1, 0) + t(0, 1) where s, t ∈ R.
7−λ
3 = 0 ⇐⇒ λ2 − 8λ + 25 = 0, from which the quadratic formula
16. det(A − λI) = 0 ⇐⇒ −6
1−λ supplies the eigenvalues: λ = 4 ± 3i.
0
3 − 3i
3
v1
=
. The given
If λ = 4 + 3i then (A − λI)v = 0 assumes the form
0
v2
−6
−3 − 3i
system of equations reduces to (3−3i)v1 +3v2 = 0, or after dividing by 3, (1−i)v1 +v2 = 0. Setting v1 = t ∈ R,
we have v2 = −t(1 − i), so the eigenvectors corresponding to λ = 4 + 3i are v = (t, −t(1 − i)) = t(1, −(1 − i)),
where t ∈ R, with t = 0.
If λ = 4 − 3i then we can apply Theorem 7.1.8 and use our calculations above to conclude that the
corresponding eigenvectors in this case have the form v = t(1, −(1 + i)), where t ∈ R, with t = 0.
−2 − λ
−6 = 0 ⇐⇒ λ2 − 2λ + 10 = 0, from which the quadratic formula
17. det(A − λI) = 0 ⇐⇒ 3
4−λ supplies the eigenvalues: λ = 1 ± 3i.
0
−3 − 3i
−6
v1
=
. The given
If λ = 1 + 3i then (A − λI)v = 0 assumes the form
v2
0
3
3 − 3i
system of equations reduces to 3v1 +(3−3i)v2 = 0, or after dividing by 3, v1 +(1−i)v2 = 0. Setting v2 = t ∈ R,
we have v1 = −t(1 − i), so the eigenvectors corresponding to λ = 1 + 3i are v = (−t(1 − i), t) = t(−(1 − i), 1),
where t ∈ R, with t = 0.
If λ = 1 − 3i then we can apply Theorem 7.1.8 and use our calculations above to conclude that the
corresponding eigenvectors in this case have the form v = t(−(1 + i), 1), where t ∈ R, with t = 0.
3−λ
−2 = 0 ⇐⇒ λ2 − 2λ + 5 = 0 ⇐⇒ λ = 1 ± 2i.
18. det(A − λI) = 0 ⇐⇒ 4
−1 − λ 0
2 + 2i
−2
v1
If λ = 1 − 2i then (A − λI)v = 0 assumes the form
=
v2
0
4
−2 + 2i
=⇒ (1 + i)v1 − v2 = 0. If we let v1 = s ∈ C, then the solution set of this system is {(s, (1 + i)s) : s ∈ C} so
If λ = 3 then (A − λI)v = 0 assumes the form
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the eigenvectors corresponding to λ = 1 − 2i are v = s(1, 1 + i) where s ∈ C. By Theorem 7.1.8, since the
entries of A are real, λ = 1 + 2i has corresponding eigenvectors of the form v = t(1, 1 − i) where t ∈ C.
2−λ
3 = 0 ⇐⇒ (2 − λ)2 = −9 ⇐⇒ λ = 2 ± 3i.
19. det(A − λI) = 0 ⇐⇒ −3
2−λ 0
3i 3
v1
=
If λ = 2 − 3i then (A − λI)v = 0 assumes the form
v2
0
−3 3i
=⇒ v2 = −iv1 . If we let v1 = t ∈ C, then the solution set of this system is {(t, −it) : t ∈ C} so the
eigenvectors corresponding to λ = 2 − 3i are v = t(1, −i) where t ∈ C. By Theorem 7.1.8, since the entries
of A are real, λ = 2 + 3i has corresponding eigenvectors of the form v = r(1, i) where r ∈ C.
10 − λ −12
8
= 0 ⇐⇒ (λ − 2)3 = 0 ⇐⇒ λ = 2 of multiplicity three.
0
2−λ
0
20. det(A − λI) = 0 ⇐⇒ −8
12
−6 − λ ⎤ ⎡
⎡
⎤⎡
⎤
8 −12
8
v1
0
0
0 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
If λ = 2 then (A − λI)v = 0 assumes the form ⎣ 0
−8
12 −8
v3
0
=⇒ 2v1 − 3v2 + 2v3 = 0. Thus, if we let v2 = 2s and v3 = t where s, t ∈ R, then the solution set of this
system is {(3s − t, 2s, t) : s, t ∈ R} so the eigenvectors corresponding to λ = 2 are v = s(3, 2, 0) + t(−1, 0, 1)
where s, t ∈ R.
3−λ
0
0 2−λ
−1 = 0 ⇐⇒ (λ − 1)(λ − 3)2 = 0 ⇐⇒ λ = 1 or λ = 3 of
21. det(A − λI) = 0 ⇐⇒ 0
1
−1
2−λ multiplicity two.
⎤ ⎡
⎡
⎤⎡
⎤
2
0
0
v1
0
1 −1 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
If λ = 1 then (A − λI)v = 0 assumes the form ⎣ 0
1 −1
1
v3
0
=⇒ v1 = 0 and v2 − v3 = 0. Thus, if we let v3 = s where s ∈ R, then the solution set of this system is
{(0, s, s) : s ∈ R} so the eigenvectors corresponding
v = s(0,
1) where
⎤ 1,⎡
⎤ s ∈ R.
⎡ to λ = 1 are
⎤⎡
0
0
0
v1
0
If λ = 3 then (A − λI)v = 0 assumes the form ⎣ 0 −1 −1 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
0
1 −1 −1
v3
=⇒ v1 = 0 and v2 + v3 = 0. Thus, if we let v3 = t where t ∈ R, then the solution set of this system is
{(0, −t, t) : t ∈ R} so the eigenvectors corresponding to λ = 3 are v = t(0, −1, 1) where t ∈ R.
1−λ
0
0
= 0 ⇐⇒ (λ − 1)3 = 0 ⇐⇒ λ = 1 of multiplicity three.
3−λ
2
22. det(A − λI) = 0 ⇐⇒ 0
2
−2
−1 − λ ⎤⎡
⎤ ⎡
⎤
⎡
v1
0
0
0
0
2
2 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
If λ = 1 then (A − λI)v = 0 assumes the form ⎣ 0
2 −2 −2
0
v3
=⇒ v1 = 0 and v2 + v3 = 0. Thus, if we let v3 = s where s ∈ R, then the solution set of this system is
{(0, −s, s) : s ∈ R} so the eigenvectors corresponding to λ = 1 are v = s(0, −1, 1) where s ∈ R.
6−λ
3
−4 = 0 ⇐⇒ (λ − 1)(λ + 1)(λ − 3) = 0 ⇐⇒ λ = −1, λ = 1,
−2 − λ
2
23. det(A − λI) = 0 ⇐⇒ −5
0
0
−1 − λ or λ = 3.
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⎤ ⎡
⎤⎡
⎤
7
3 −4
v1
0
2 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
If λ = −1 then (A − λI)v = 0 assumes the form ⎣ −5 −1
0
0
0
v3
0
=⇒ v1 = v3 − v2 and 4v2 − 3v3 = 0. Thus, if we let v3 = 4r where r ∈ R, then the solution set of this system
is {(r, 3r, 4r) : r ∈ R} so the eigenvectors corresponding
to λ = −1
⎡
⎤ ⎡are v ⎤= r(1,
⎡ 3, 4)
⎤ where r ∈ R.
5
3 −4
v1
0
2 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
If λ = 1 then (A − λI)v = 0 assumes the form ⎣ −5 −3
0
0 −2
v3
0
=⇒ 5v1 + 3v2 = 0 and v3 = 0. Thus, if we let v2 = −5s where s ∈ R, then the solution set of this system is
{(3s, −5s, 0) : s ∈ R} so the eigenvectors corresponding
to λ = 1 ⎤are
⎡
⎡ v =⎤s(3,⎡−5, 0)
⎤ where s ∈ R.
3
3 −4
v1
0
2 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
If λ = 3 then (A − λI)v = 0 assumes the form ⎣ −5 −5
0
0 −4
v3
0
=⇒ v1 + v2 = 0 and v3 = 0. Thus, if we let v2 = t where t ∈ R, then the solution set of this system is
{(−t, t, 0) : t ∈ R} so the eigenvectors corresponding to λ = 3 are v = t(−1, 1, 0) where t ∈ R.
7−λ
−8
6
= 0 ⇐⇒ (λ + 1)3 = 0 ⇐⇒ λ = −1 of multiplicity
−9 − λ
6
24. det(A − λI) = 0 ⇐⇒ 8
0
0
−1 − λ three.
⎤ ⎡
⎡
⎤⎡
⎤
8 −8 6
v1
0
If λ = −1 then (A − λI)v = 0 assumes the form ⎣ 8 −8 6 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
0
0 0
v3
0
=⇒ 4v1 − 4v2 + 3v3 = 0. Thus, if we let v2 = r and v3 = 4s where r, s ∈ R, then the solution set of this
system is {(r −3s, r, 4s) : r, s ∈ R} so the eigenvectors corresponding to λ = −1 are v = r(1, 1, 0)+s(−3, 0, 4)
where r, s ∈ R.
−λ
1
−1 0 = 0 ⇐⇒ (λ − 1)(λ − 2)2 = 0 ⇐⇒ λ = 1 or λ = 2 of
25. det(A − λI) = 0 ⇐⇒ 0 2 − λ
2
−1
3−λ multiplicity two.
⎤ ⎡
⎡
⎤⎡
⎤
−1
1 −1
v1
0
1
0 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
If λ = 1 then (A − λI)v = 0 assumes the form ⎣ 0
2 −1
2
v3
0
=⇒ v2 = 0 and v1 + v3 = 0. Thus, if we let v3 = r where r ∈ R, then the solution set of this system is
{(−r, 0, r) : r ∈ R} so the eigenvectors corresponding
to λ = 1 are
v = r(−1,
r ∈ R.
⎤ 0,
⎡
⎤⎡
⎡ 1) where
⎤
−2
1 −1
v1
0
0
0 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
If λ = 2 then (A − λI)v = 0 assumes the form ⎣ 0
2 −1
1
v3
0
=⇒ 2v1 − v2 + v3 = 0. Thus, if we let v1 = s and v3 = t where s, t ∈ R, then the solution set of this system
is {(s, 2s + t, t) : s, t ∈ R} so the eigenvectors corresponding to λ = 2 are v = s(1, 2, 0) + t(0, 1, 1) where
s, t ∈ R.
1−λ 0
0 −λ 1 = 0 ⇐⇒ (1 − λ)(1 + λ2 ) = 0 ⇐⇒ λ = 1 or λ = ±i.
26. det(A − λI) = 0 ⇐⇒ 0
0
−1 −λ ⎤ ⎡
⎡
⎤⎡
⎤
0
0
0
v1
0
1 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
If λ = 1 then (A − λI)v = 0 assumes the form ⎣ 0 −1
0 −1 −1
v3
0
=⇒ −v2 + v3 = 0 and −v2 − v3 = 0. The solution set of this system is {(r, 0, 0) : r ∈ C} so the eigenvectors
⎡
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corresponding to λ = 1 are v = r(1, 0, 0) where r⎡∈ C.
⎤ ⎡
⎤
⎤⎡
1+i
0 0
v1
0
i 1 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
If λ = −i then (A − λI)v = 0 assumes the form ⎣ 0
0
0
−1 i
v3
=⇒ v1 = 0 and −v2 + iv3 = 0. The solution set of this system is {(0, si, s) : s ∈ C} so the eigenvectors
corresponding to λ = −i are v = s(0, i, 1) where s ∈ C. By Theorem 7.1.8, since the entries of A are real,
λ = i has corresponding eigenvectors of the form v = t(0, −i, 1) where t ∈ C.
0
−1 = 0 ⇐⇒ (λ + 2)(λ2 + 4λ + 5) = 0 ⇐⇒ λ = −2 or
−3 − λ ⎤ ⎡
⎡
⎤⎡
⎤
0 1
0
v1
0
If λ = −2 then (A − λI)v = 0 assumes the form ⎣ 1 1 −1 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
1 3 −1
v3
0
=⇒ v2 = 0 and v1 − v3 = 0. Thus, if we let v3 = r where r ∈ C, then the solution set of this system is
{(r, 0, r) : r ∈ C} so the eigenvectors corresponding to⎡ λ = −2 are v = r(1, ⎤
0,⎡1) where
⎤ r⎡∈ C.⎤
−i
1
0
v1
0
−1 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
If λ = −2 + i then (A − λI)v = 0 assumes the form ⎣ 1 1 − i
1
3
−1 − i
v3
0
−2 + i
−1 − 2i
v3 = 0 and v2 +
v3 = 0. Thus, if we let v3 = 5s where s ∈ C, then the solution
=⇒ v1 +
5
5
set of this system is {((2 − i)s, (1 + 2i)s, 5s) : s ∈ C} so the eigenvectors corresponding to λ = −2 + i are
v = s(2 − i, 1 + 2i, 5) where s ∈ C. By Theorem 7.1.8, since the entries of A are real, λ = −2 − i has
corresponding eigenvectors of the form v = t(2 + i, 1 − 2i, 5) where t ∈ C.
−2 − λ
1
27. det(A − λI) = 0 ⇐⇒ 1
λ = −2 ± i.
1
−1 − λ
3
3 0 = 0 ⇐⇒ λ(λ − 2)(λ − 4) = 0 ⇐⇒ λ = 0, λ = 2, or λ = 4.
3−λ ⎤⎡
⎤ ⎡
⎤
⎡
v1
2 −1 3
0
1 0 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
If λ = 0 then (A − λI)v = 0 assumes the form ⎣ 3
2 −1 3
0
v3
=⇒ v1 + 2v2 − 3v3 = 0 and −5v2 + 9v3 = 0. Thus, if we let v3 = 5r where r ∈ R, then the solution set of this
system is {(−3r, 9r, 5r) : r ∈ R} so the eigenvectors
0 are⎤v = r(−3, 9, 5) where r ∈ R.
⎡ corresponding
⎤ ⎡ to ⎤λ = ⎡
0 −1 3
v1
0
If λ = 2 then (A − λI)v = 0 assumes the form ⎣ 3 −1 0 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
2 −1 1
v3
0
=⇒ v1 − v3 = 0 and v2 − 3v3 = 0. Thus, if we let v3 = s where s ∈ R, then the solution set of this system is
{(s, 3s, s) : s ∈ R} so the eigenvectors corresponding
to λ = 2 are⎤v⎡= s(1,
⎤ 3, 1)
⎡
⎡ where
⎤ s ∈ R.
−2 −1
3
v1
0
0 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
If λ = 4 then (A − λI)v = 0 assumes the form ⎣ 3 −3
2 −1 −1
v3
0
=⇒ v1 − v3 = 0 and v2 − v3 = 0. Thus, if we let v3 = t where t ∈ R, then the solution set of this system is
{(t, t, t) : t ∈ R} so the eigenvectors corresponding to λ = 4 are v = t(1, 1, 1) where t ∈ R.
2−λ
28. det(A − λI) = 0 ⇐⇒ 3
2
−1
1−λ
−1
5−λ
29. det(A − λI) = 0 ⇐⇒ 0
0
0
5−λ
0
0 0 = 0 ⇐⇒ (λ − 5)3 = 0 ⇐⇒ λ = 5 of multiplicity three.
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⎤ ⎡
⎤⎡
⎤
0 0 0
v1
0
If λ = 5 then (A − λI)v = 0 assumes the form ⎣ 0 0 0 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
0 0 0
v3
0
Thus, if we let v1 = r, v2 = s, and v3 = t where r, s, t ∈ R, then the solution set of this system is
{(r, s, t) : r, s, t ∈ R} so the eigenvectors corresponding to λ = 5 are v = r(1, 0, 0) + s(0, 1, 0) + t(0, 0, 1) where
r, s, t ∈ R. That is, every nonzero vector in R3 is an eigenvector of A corresponding to λ = 5.
⎡
−λ
2
2 2 = 0 ⇐⇒ (λ − 4)(λ + 2)2 = 0 ⇐⇒ λ = 4 or λ = −2 of multiplicity
30. det(A − λI) = 0 ⇐⇒ 2 −λ
2
2 −λ two.
⎤ ⎡
⎡
⎤⎡
⎤
−4
2
2
v1
0
2 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
If λ = 4 then (A − λI)v = 0 assumes the form ⎣ 2 −4
2
2 −4
v3
0
=⇒ v1 − v3 = 0 and v2 − v3 = 0. Thus, if we let v3 = r where r ∈ R, then the solution set of this system is
{(r, r, r) : r ∈ R} so the eigenvectors corresponding
r(1, 1,
r ∈ R.
⎡ to λ = 4 are
⎡ 1) where
⎤ ⎡v = ⎤
⎤
2 2 2
0
v1
If λ = −2 then (A − λI)v = 0 assumes the form ⎣ 2 2 2 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
v3
2 2 2
0
=⇒ v1 + v2 + v3 = 0. Thus, if we let v2 = s and v3 = t where s, t ∈ R, then the solution set of this system is
{(−s − t, s, t) : s, t ∈ R} so the eigenvectors corresponding to λ = −2 are v = s(−1, 1, 0) + t(−1, 0, 1) where
s, t ∈ R.
1−λ
2
3
4 4
3−λ
2
1 4
3
2
31. det(A − λI) = 0 ⇐⇒ = 0 ⇐⇒ λ − 14λ − 32λ = 0
4
5
6
−
λ
7
7
6
5
4−λ 2
⇐⇒ λ (λ − 16)(λ + 2) = 0 ⇐⇒ λ = 16, λ = −2, ⎡or λ = 0 of multiplicity two.
⎤
⎤⎡
⎤ ⎡
v1
−15
2
3
4
0
⎥
⎢
⎢
⎥ ⎢
4 −13
2
1 ⎥
⎥ ⎢ v2 ⎥ = ⎢ 0 ⎥
If λ = 16 then (A − λI)v = 0 assumes the form ⎢
⎦
⎣
⎣
⎦
⎣
0 ⎦
4
5 −10
7
v3
0
7
6
5 −12
v4
=⇒ v1 − 1841v3 + 2078v4 = 0, v2 + 82v3 − 93v4 = 0, and 31v3 − 35v4 = 0. Thus, if we let v4 = 31r where
r ∈ R, then the solution set of this system is {(17r, 13r, 35r, 31r) : r ∈ R} so the eigenvectors corresponding
to λ = 16 are v = r(17, 13, 35, 31) where r ∈ R. ⎡
⎤⎡
⎤
⎤ ⎡
v1
3 2 3 4
0
⎢ 4 5 2 1 ⎥ ⎢ v2 ⎥ ⎢ 0 ⎥
⎥⎢
⎥
⎥ ⎢
If λ = −2 then (A − λI)v = 0 assumes the form ⎢
⎣ 4 5 8 7 ⎦ ⎣ v3 ⎦ = ⎣ 0 ⎦
7 6 5 6
0
v4
=⇒ v1 + v4 = 0, v2 − v4 = 0, and v3 + v4 = 0. Thus, if we let v4 = s where s ∈ R, then the solution set of
this system is {(−s, s, −s, s) : s ∈ R} so the eigenvectors corresponding to λ = −2 are v = s(−1, 1, −1, 1)
where s ∈ R.
⎤⎡
⎤
⎡
⎤ ⎡
v1
1 2 3 4
0
⎢ 4 3 2 1 ⎥ ⎢ v2 ⎥ ⎢ 0 ⎥
⎥⎢
⎥
⎥ ⎢
If λ = 0 then (A − λI)v = 0 assumes the form ⎢
⎣ 4 5 6 7 ⎦ ⎣ v3 ⎦ = ⎣ 0 ⎦
7 6 5 4
0
v4
=⇒ v1 − v3 − 2v4 = 0 and v2 + 2v3 + 3v4 = 0. Thus, if we let v3 = a and v4 = b where a, b ∈ R, then the
solution set of this system is {(a + 2b, −2a − 3b, a, b) : a, b ∈ R} so the eigenvectors corresponding to λ = 0
are v = a(1, −2, 1, 0) + b(2, −3, 0, 1) where a, b ∈ R.
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−λ
1
0
0 −1 −λ
0
0 32. det(A − λI) = 0 ⇐⇒ = 0 ⇐⇒ (λ2 + 1)2 = 0 ⇐⇒ λ = ±i, where each root is of
0 −λ −1 0
0
0
1 −λ multiplicity two.
⎤
⎤⎡
⎡
⎤ ⎡
v1
i 1 0
0
0
⎥
⎢
⎢ −1 i 0
⎥ ⎢
0 ⎥
⎥ ⎢ v2 ⎥ ⎢ 0 ⎥
If λ = −i then (A − λI)v = 0 assumes the form ⎢
⎣ 0 0 i −1 ⎦ ⎣ v3 ⎦ = ⎣ 0 ⎦
0
0 0 1
i
v4
=⇒ v1 − iv2 = 0 and v3 + iv4 = 0. Thus, if we let v2 = r and v4 = s where r, s ∈ C, then the solution set
of this system is {(ir, r, −is, s) : r, s ∈ C} so the eigenvectors corresponding to λ = −i are v = r(i, 1, 0, 0) +
s(0, 0, −i, 1) where r, s ∈ C. By Theorem 7.1.8, since the entries of A are real, λ = i has corresponding
eigenvectors v = a(−i, 1, 0, 0) + b(0, 0, i, 1) where a, b ∈ C.
33. This matrix is lower triangular, and therefore, the eigenvalues appear along the main diagonal of the
matrix: λ = 1 + i, 1 − 3i, 1. Note that the eigenvalues do not occur in complex conjugate pairs, but this does
not contradict Theorem 7.1.8 because the matrix does not consist entirely of real elements.
1−λ
−1 34. (a). p(λ) = det(A − λI2 ) = = λ2 − 5λ + 6.
2
4−λ (b).
1 −1
1 −1
1 0
−5
+6
2
4
2
4
0 1
−1 −5
−5
5
6 0
0 0
=
+
+
=
= 02 .
10 14
−10 −20
0 6
0 0
A2 − 5A + 6I2 =
1 −1
2
4
(c). Using part (b) of this problem:
A2 − 5A + 6I2 = 02 ⇐⇒ A−1 (A2 − 5A + 6I2 ) = A−1 · 02
⇐⇒ A − 5I2 + 6A−1 = 02
⇐⇒ 6A−1 = 5I2 − A
1
1 0
1 −1
5
−
⇐⇒ A−1 =
0 1
2
4
6
2 1 1
4 1
3
6
, or A−1 =
⇐⇒ A−1 =
.
− 13 16
6 −2 1
1−λ
2
= 0 ⇐⇒ λ2 + λ − 6 = 0 ⇐⇒ (λ − 2)(λ + 3) = 0 ⇐⇒ λ = 2 or
35. (a). det(A − λI2 ) = 2
−2 − λ λ = −3.
1
2
1
2
1 0
(b).
∼
∼
= B.
2 −2
0 −6
0 1
1−λ
0 = 0 ⇐⇒ (λ − 1)2 = 0 ⇐⇒ λ = 1 of multiplicity two. Matrices A
det(B − λI2 ) = 0 ⇐⇒ 0
1−λ and B do not have the same eigenvalues.
36.
A(3v1 − v2 ) = 3Av1 − Av2 = 3(2v1 ) − (−3v2 ) = 6v1 + 3v2
1
2
6
6
12
=6
+3
=
+
=
.
−1
1
−6
3
−3
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37. (a). Let a, b, c ∈ R. If v = av1 + bv2 + cv3 , then 5, 0, 3) = a(1, −1, 1) + b(2, 1, 3)
⎧ + c(−1, −1, 2), or
⎪
⎨ a + 2b − c = 5
(5, 0, 3) = (a + 2b − c, −a + b − c, a + 3b + 2c). The last equality results in the system:
⎪
⎩
−a + b − c = 0
a + 3b + 2c = 3.
This system has the solution a = 2, b = 1, and c = −1. Consequently, v = 2v1 + v2 − v3 .
(b). Using part (a):
Av = A(2v1 + v2 − v3 ) = 2Av1 + Av2 − Av3 = 2(2v1 ) + (−2v2 ) − (3v3 )
= 4v1 − 2v2 − 3v3 = 4(1, −1, 1) − 2(2, 1, 3) − 3(−1, −1, 2) = (3, −3, −8).
38.
A(c1 v1 + c2 v2 + c3 v3 ) = A(c1 v1 ) + A(c2 v2 ) + A(c3 v3 )
= c1 (Av1 ) + c2 (Av2 ) + c3 (Av3 )
= c1 (λv1 ) + c2 (λv2 ) + c3 (λv3 )
= λ(c1 v1 + c2 v2 + c3 v3 ).
Thus, c1 v1 + c2 v2 + c3 v3 is an eigenvector of A corresponding to the eigenvalue λ.
39. Recall that the determinant of an upper (lower) triangular matrix is just the product of its main diagonal
elements. Let A be an n × n upper (lower) triangular matrix. It follows that A − λIn is an upper (lower)
triangular matrix with main diagonal element aii − λ, i = 1, 2, . . . , n. Consequently,
det(A − λIn ) = 0 ⇐⇒
n
-
(aii − λ) = 0.
i=1
This implies that λ = a11 , a22 , . . . , ann .
40. Any scalar λ such that det(A − λI) = 0 is an eigenvalue of A. Therefore, if 0 is an eigenvalue of A, then
det(A − 0 · I) = 0, or det(A) = 0, which implies that A is not invertible. On the other hand, if 0 is not an
eigenvalue of A, then det(A − 0 · I) = 0, or det(A) = 0, which implies that A is invertible.
41. A is invertible, so A−1 exists. Also, λ is an eigenvalue of A so that Av = λv. Thus,
1
A−1 (Av) = A−1 (λv) =⇒ (A−1 A)v = λA−1 v =⇒ In v = λA−1 v =⇒ v = λA−1 v =⇒ v = A−1 v.
λ
1
Therefore is an eigenvalue of A−1 provided that λ is an eigenvalue of A.
λ
42. By assumption, we have Av = λv and Bv = μv.
(a). Therefore,
(AB)v = A(Bv) = A(μv) = μ(Av) = μ(λv) = (λμ)v,
which shows that v is an eigenvector of AB with corresponding eigenvalue λμ.
(b). Also,
(A + B)v = Av + Bv = λv + μv = (λ + μ)v,
which shows that v is an eigenvector of A + B with corresponding eigenvalue λ + μ.
43. Recall that a matrix and its transpose have the same determinant. Thus,
det(A − λIn ) = det([A − λIn ]T ) = det(AT − λIn ).
Since A and AT have the same characteristic polynomial, it follows that both matrices also have the same
eigenvalues.
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44. (a). v = r + is is an eigenvector with eigenvalue λ = a + bi, b = 0 =⇒ Av = λv
=⇒ A(r + is) = (a + bi)(r + is) = (ar − bs) + i(as + br) =⇒ Ar = ar − bs and As = as + br.
Now if r = 0, then A0 = a0 − bs =⇒ 0 = 0 − bs =⇒ 0 = bs =⇒ s = 0 since b = 0. This would mean that
v = 0 so v could not be an eigenvector. Thus, it must be that r = 0. Similarly, if s = 0, then r = 0, and
again, this would contradict the fact that v is an eigenvector. Hence, it must be the case that r = 0 and
s = 0.
(b). As in part (a), Ar = ar − bs and As = as + br.
Let c1 , c2 ∈ R. Then if
c1 r + c2 s = 0,
(44.1)
we have A(c1 r + c2 s) = 0 =⇒ c1 Ar + c2 As = 0 =⇒ c1 (ar − bs) + c2 (as + br) = 0.
Hence, (c1 a + c2 b)r + (c2 a − c1 b)s = 0 =⇒ a(c1 r + c2 s) + b(c2 r − c1 s) = 0 =⇒ b(c2 r − c1 s) = 0 where we
have used (39.1). Since b = 0, we must have c2 r − c1 s = 0. Combining this with (44.1) yields c1 = c2 = 0.
Therefore, it follows that r and s are linearly independent vectors.
45. λ1 = 2, v = r(−1, 1). λ2 = 5, v = s(1, 2).
46. λ1 = −2 (multiplicity two), v = r(1, 1, 1). λ2 = −5, v = s(20, 11, 14).
47. λ1 = 3 (multiplicity two), v = r(1, 0, −1) + s(0, 1, −1). λ2 = 6, v = t(1, 1, 1).
√
√
√
√
√
√
√
√
48. λ1 = 3 − 6, v = r( 6, −1 + 6, −5 + 6). λ2 = 3 + 6, v = s( 6, 1 + 6, 5 + 6), λ3 = −2, v =
t(−1, 3, 0).
49. λ1 = 0, v = r(2, 2, 1). λ2 = 3i, v = s(−4 − 3i, 5, −2 + 6i), λ3 = −3i, v = t(−4 + 3i, 5, −2 − 6i).
50. λ1 = −1 (multiplicity four), v = a(−1, 0, 0, 1, 0) + b(−1, 0, 1, 0, 0) + c(−1, 0, 0, 0, 1) + d(−1, 1, 0, 0, 0).
Solutions to Section 7.2
True-False Review:
(a): FALSE. There have to be n linearly independent eigenvectors.
(b): TRUE. The eigenspace Eλ is equal to the null space of the n × n matrix A − λI, and this null space
is a subspace of Rn .
(c): TRUE. The dimension of an eigenspace never exceeds the algebraic multiplicity of the corresponding
eigenvalue.
(d): TRUE. Eigenvectors corresponding to distinct eigenspaces are linearly independent. Therefore if we
choose one (nonzero) vector from each distinct eigenspace, the chosen vectors will form a linearly independent
set.
(e): TRUE. Since each eigenvalue of the matrix A occurs with algebraic multiplicity 1, we can simply
choose one eigenvector from each eigenspace to obtain a basis of eigenvectors for A. Thus, A is nondefective.
(f ): FALSE. Many examples will show that this statement is false, including the n × n identity matrix In
for n ≥ 2. The matrix In is not defective, and yet, has λ = 1 occurring with algebraic multiplicity n.
Problems:
−7 − λ
1. det(A − λI) = 0 ⇐⇒ −3
λ = −7 (with multiplicity 2).
0
= 0 ⇐⇒ (−7 − λ)2 = 0, so taht we obtain the eigenvalue
−7 − λ (c)2017 Pearson Education. Inc.
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0
0 0
v1
=
. The solution set of this
0
v2
−3 0
system is {(0, r) : r ∈ R}, so the eigenspace corresponding to λ = −7 is E1 = {v ∈ R2 : v = r(0, 1), r ∈ R}.
A basis for E1 is {(0, 1)}, and dim[E1 ] = 1. We conclude that A is defective since it does not have a complete
set of eigenvectors.
1−λ
4 2. det(A − λI) = 0 ⇐⇒ = 0 ⇐⇒ λ2 − 4λ − 5 = 0 ⇐⇒ (λ − 5)(λ + 1) = 0 ⇐⇒ λ = 5 or
2
3−λ λ = −1.
0
−4
4
v1
=
=⇒ v1 − v2 = 0. The solution
If λ1 = 5 then (A − λI)v = 0 assumes the form
0
v2
2 −2
set of this system is {(r, r) : r ∈ R}, so the eigenspace corresponding to λ1 = 5 is E1 = {v ∈ R2 : v =
r(1, 1), r ∈ R}. A basis for E1 is {(1, 1)}, and dim[E
1 ] = 1. 0
2 4
v1
=
=⇒ v1 + 2v2 = 0. The solution
If λ2 = −1 then (A − λI)v = 0 assumes the form
0
v2
2 4
set of this system is {(−2s, s) : s ∈ R}, so the eigenspace corresponding to λ2 = −1 is E2 = {v ∈ R2 : v =
s(−2, 1), s ∈ R}. A basis for E2 is {(−2, 1)}, and dim[E2 ] = 1.
A complete set of eigenvectors for A is given by {(1, 1), (−2, 1)}, so A is nondefective.
3−λ
0 3. det(A − λI) = 0 ⇐⇒ = 0 ⇐⇒ (3 − λ)2 = 0 ⇐⇒ λ = 3 of multiplicity two.
0
3−λ 0
0 0
v1
=
.
If λ1 = 3 then (A − λI)v = 0 assumes the form
v2
0
0 0
The solution set of this system is {(r, s) : r, s ∈ R}, so the eigenspace corresponding to λ1 = 3 is
E1 = {v ∈ R2 : v = r(1, 0) + s(0, 1), r, s ∈ R}. A basis for E1 is {(1, 0), (0, 1)}, and dim[E1 ] = 2.
A is nondefective.
1−λ
2 = 0 ⇐⇒ (λ − 3)2 = 0 ⇐⇒ λ = 3 of multiplicity two.
4. det(A − λI) = 0 ⇐⇒ −2
5−λ 0
−2 2
v1
=
=⇒ v1 − v2 = 0. The solution
If λ1 = 3 then (A − λI)v = 0 assumes the form
0
v2
−2 2
set of this system is {(r, r) : r ∈ R}, so the eigenspace corresponding to λ1 = 3 is
E1 = {v ∈ R2 : v = r(1, 1), r ∈ R}. A basis for E1 is {(1, 1)}, and dim[E1 ] = 1.
A is defective since it does not have a complete set of eigenvectors.
5−λ
5
= 0 ⇐⇒ λ2 − 4λ + 5 = 0 ⇐⇒ λ = 2 ± i.
5. det(A − λI) = 0 ⇐⇒ −2
−1 − λ 0
3+i
5
v1
=
=⇒ −2v1 +(−3+i)v2 =
If λ1 = 2 − i then (A−λI)v = 0 assumes the form
0
v2
−2 −3 + i
0. The solution set of this system is {((−3 + i)r, 2r) : r ∈ C}, so the eigenspace corresponding to λ1 = 2 − i
is E1 = {v ∈ C2 : v = r(−3 + i, 2), r ∈ C}. A basis for E1 is {(−3 + i, 2)}, and dim[E1 ] = 1.
If λ2 = 2 + i then from Theorem 7.1.8, the eigenvectors corresponding to λ2 = 2 + i are v = s(−3 − i, 2)
where s ∈ C, so the eigenspace corresponding to λ2 = 2 + i is E2 = {v ∈ C2 : v = s(−3 − i, 2), s ∈ C}. A
basis for E2 is {(−3 − i, 2)}, and dim[E2 ] = 1. A is nondefective since it has a complete set of eigenvectors,
namely {(−3 + i, 2), (−3 − i, 2)}.
3−λ
−4
−1 −1 − λ
−1 = 0 ⇐⇒ (λ + 2)(λ − 3)2 = 0 ⇐⇒ λ = −2 or λ = 3 of
6. det(A − λI) = 0 ⇐⇒ 0
0
−4
2−λ multiplicity two.
If λ = −7 then (A − λI)v = 0 assumes the form
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⎤ ⎡
⎤⎡
⎤
5 −4 −1
v1
0
1 −1 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦.
If λ1 = −2 then (A − λI)v = 0 assumes the form ⎣ 0
0 −4
4
v3
0
The solution set of this system is {(r, r, r) : r ∈ R}, so the eigenspace corresponding to λ1 = −2 is E1 =
{v ∈ R3 : v = r(1, 1, 1), r ∈ R}. A basis for E1 is⎡{(1, 1, 1)}, and⎤dim[E
] = 1.⎡
1⎤
⎡
⎤
0 −4 −1
v1
0
If λ2 = 3 then (A − λI)v = 0 assumes the form ⎣ 0 −4 −1 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ 4v2 + v3 = 0. The
0 −4 −1
v3
0
solution set of this system is {(s, t, −4t) : s, t ∈ R}, so the eigenspace corresponding to λ2 = 3 is
E2 = {v ∈ R3 : v = s(1, 0, 0) + t(0, 1, −4), s, t ∈ R}. A basis for E2 is {(1, 0, 0), (0, 1, −4)}, and dim[E2 ] = 2.
A complete set of eigenvectors for A is given by {(1, 1, 1), (1, 0, 0), (0, 1, −4)}, so A is nondefective.
4−λ
0
0 2−λ
−3 = 0 ⇐⇒ (λ + 1)(λ − 4)2 = 0 ⇐⇒ λ = −1 or λ = 4 of
7. det(A − λI) = 0 ⇐⇒ 0
0
−2
1−λ multiplicity two.
⎤ ⎡
⎡
⎤⎡
⎤
5
0
0
v1
0
3 −3 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
If λ1 = −1 then (A − λI)v = 0 assumes the form ⎣ 0
0 −2
2
v3
0
=⇒ v1 = 0 and v2 − v3 = 0. The solution set of this system is {(0, r, r) : r ∈ R}, so the eigenspace
corresponding to λ1 = −1 is E1 = {v ∈ R3 : v = r(0, 1, 1), r ∈ R}. A basis for E1 is {(0, 1, 1)}, and
dim[E1 ] = 1.
⎤
⎡
⎤⎡
⎡
⎤
0
0
0
v1
0
If λ2 = 4 then (A − λI)v = 0 assumes the form ⎣ 0 −2 −3 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ 2v2 + 3v3 = 0.
0 −2 −3
v3
0
The solution set of this system is {(s, 3t, −2t) : s, t ∈ R}, so the eigenspace corresponding to λ2 = 4 is
E2 = {v ∈ R3 : v = s(1, 0, 0) + t(0, 3, −2), s, t ∈ R}. A basis for E2 is {(1, 0, 0), (0, −1, 1)}, and dim[E2 ] = 2.
A complete set of eigenvectors for A is given by {(1, 0, 0), (0, 3, −2), (0, 1, 1)}, so A is nondefective.
3−λ
1
0 5−λ
0 = 0 ⇐⇒ (λ − 4)3 = 0 ⇐⇒ λ = 4 of multiplicity three.
8. det(A − λI) = 0 ⇐⇒ −1
0
0
4−λ ⎤ ⎡
⎡
⎤⎡
⎤
−1 1 0
v1
0
If λ1 = 4 then (A − λI)v = 0 assumes the form ⎣ −1 1 0 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
0 0 0
v3
0
=⇒ v1 − v2 = 0 and v3 ∈ R. The solution set of this system is {(r, r, s) : r, s ∈ R}, so the eigenspace corresponding to λ1 = 4 is E1 = {v ∈ R3 : v = r(1, 1, 0) + s(0, 0, 1), r, s ∈ R}. A basis for E1 is {(1, 1, 0), (0, 0, 1)},
and dim[E1 ] = 2.
A is defective since it does not have a complete set of eigenvectors.
3−λ
0
0 −λ −4 = 0 ⇐⇒ (λ − 3)(λ2 + 16) = 0 ⇐⇒ λ = 3 or λ = ±4i.
9. det(A − λI) = 0 ⇐⇒ 2
1
4 −λ ⎤ ⎡
⎤
⎡
⎤⎡
0
0
0
v1
0
If λ1 = 3 then (A − λI)v = 0 assumes the form ⎣ 2 −3 −4 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
0
1
4 −3
v3
=⇒ 11v1 − 25v3 = 0 and 11v2 − 2v3 . The solution set of this system is {(25r, 2r, 11r) : r ∈ C}, so the
eigenspace corresponding to λ1 = 3 is E1 = {v ∈ C3 : v = r(25, 2, 11), r ∈ C}. A basis for E1 is {(25, 2, 11)},
and dim[E1 ] = 1.
⎡
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⎤ ⎡
⎤⎡
⎤
3 + 4i 0
0
v1
0
2
4i −4 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
If λ2 = −4i then (A − λI)v = 0 assumes the form ⎣
1
4
4i
v3
0
=⇒ v1 = 0 and iv2 − v3 = 0. The solution set of this system is {(0, s, is) : s ∈ C}, so the eigenspace
corresponding to λ2 = −4i is E2 = {v ∈ C3 : v = s(0, 1, i), s ∈ C}. A basis for E2 is {(0, 1, i)}, and
dim[E2 ] = 1.
If λ3 = 4i then from Theorem 7.1.8, the eigenvectors corresponding to λ3 = 4i are v = t(0, 1, −i) where
t ∈ C, so the eigenspace corresponding to λ3 = 4i is E3 = {v ∈ C3 : v = t(0, 1, −i), t ∈ C}. A basis for E3 is
{(0, 1, −i)}, and dim[E3 ] = 1. A complete set of eigenvectors for A is given by {(25, 2, 11), (0, 1, i), (0, 1, −i)},
so A is nondefective.
4−λ
1
6
−λ
−7 = 0 ⇐⇒ (λ + 3)(λ − 2)2 = 0 ⇐⇒ λ = −3 or λ = 2 of
10. det(A − λI) = 0 ⇐⇒ −4
0
0 −3 − λ multiplicity two.
⎤ ⎡
⎡
⎤⎡
⎤
7 1
6
v1
0
If λ1 = −3 then (A − λI)v = 0 assumes the form ⎣ −4 3 −7 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
0 0
0
v3
0
=⇒ v1 + v3 = 0 and v2 − v3 = 0. The solution set of this system is {(−r, r, r) : r ∈ R}, so the eigenspace
corresponding to λ1 = −3 is E1 = {v ∈ R3 : v = r(−1, 1, 1), r ∈ R}. A basis for E1 is {(−1, 1, 1)}, and
dim[E1 ] = 1.
⎤ ⎡
⎡
⎤⎡
⎤
2
1
6
v1
0
If λ2 = 2 then (A − λI)v = 0 assumes the form ⎣ −4 −2 −7 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
0
0
5
v3
0
=⇒ 2v1 + v2 = 0 and v3 = 0. The solution set of this system is {(−s, 2s, 0) : s ∈ R}, so the eigenspace
corresponding to λ2 = 2 is E2 = {v ∈ R3 : v = s(−1, 2, 0), s ∈ R}. A basis for E2 is {(−1, 2, 0)}, and
dim[E2 ] = 1.
A is defective because it does not have a complete set of eigenvectors.
2−λ
0
0 2−λ
0 = 0 ⇐⇒ (λ − 2)3 = 0 ⇐⇒ λ = 2 of multiplicity three.
11. det(A − λI) = 0 ⇐⇒ 0
0
0
2−λ ⎤
⎡
⎤⎡
⎡
⎤
0 0 0
v1
0
If λ1 = 2 then (A − λI)v = 0 assumes the form ⎣ 0 0 0 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦. The solution set of this
0 0 0
v3
0
system is {(r, s, t) : r, s, t ∈ R}, so the eigenspace corresponding to λ1 = 2 is
E1 = {v ∈ R3 : v = r(1, 0, 0) + s(0, 1, 0) + t(0, 0, 1), r, s, t ∈ R}. A basis for E1 is {(1, 0, 0), (0, 1, 0), (0, 0, 1)},
and dim[E1 ] = 3.
A is nondefective since it has a complete set of eigenvectors.
7−λ
−8
6
= 0 ⇐⇒ (λ + 1)3 = 0 ⇐⇒ λ = −1 of multiplicity
−9 − λ
6
12. det(A − λI) = 0 ⇐⇒ 8
0
0
−1 − λ three.
⎤ ⎡
⎤
⎡
⎤⎡
8 −8 6
v1
0
If λ1 = −1 then (A − λI)v = 0 assumes the form ⎣ 8 −8 6 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ 4v1 − 4v2 + 3v3 = 0.
0
0
0 0
v3
The solution set of this system is {(r − 3s, r, 4s) : r, s ∈ R}, so the eigenspace corresponding to λ1 = −1 is
E1 = {v ∈ R3 : v = r(1, 1, 0) + s(−3, 0, 4), r, s ∈ R}. A basis for E1 is {(1, 1, 0), (−3, 0, 4)}, and dim[E1 ] = 2.
A is defective since it does not have a complete set of eigenvectors.
⎡
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2−λ
13. det(A − λI) = 0 ⇐⇒ 2
2
multiplicity two.
−1 −1 = 0 ⇐⇒ (λ − 2)λ2 = 0 ⇐⇒ λ = 2 or λ = 0 of
−1 − λ ⎤ ⎡
⎡
⎤⎡
⎤
0
2 −1
v1
0
If λ1 = 2 then (A − λI)v = 0 assumes the form ⎣ 2 −1 −1 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
2
3 −3
v3
0
=⇒ 4v1 − 3v3 = 0 and 2v2 − v3 = 0. The solution set of this system is {(3r, 2r, 4r) : r ∈ R}, so the
eigenspace corresponding to λ1 = 2 is E1 = {v ∈ R3 : v = r(3, 2, 4), r ∈ R}. A basis for E1 is {(3, 2, 4)},
and dim[E1 ] = 1.
⎤ ⎡
⎤
⎡
⎤⎡
2 2 −1
v1
0
If λ2 = 0 then (A − λI)v = 0 assumes the form ⎣ 2 1 −1 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
2 3 −1
v3
0
=⇒ 2v1 − v3 = 0 and v2 = 0. The solution set of this system is {(s, 0, 2s) : s ∈ R}, so the eigenspace
corresponding to λ2 = 0 is E2 = {v ∈ R3 : v = s(1, 0, 2), s ∈ R}. A basis for E2 is {(1, 0, 2)}, and
dim[E2 ] = 1.
A is defective because it does not have a complete set of eigenvectors.
1−λ
−1
2 −1 − λ
2 = 0 ⇐⇒ (λ − 2)λ2 = 0 ⇐⇒ λ = 2 or λ = 0 of
14. det(A − λI) = 0 ⇐⇒ 1
1
−1
2−λ multiplicity two.
⎤ ⎡
⎤
⎡
⎤⎡
−1 −1 2
v1
0
If λ1 = 2 then (A − λI)v = 0 assumes the form ⎣ 1 −3 2 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
0
1 −1 0
v3
=⇒ v1 − v3 = 0 and v2 − v3 = 0. The solution set of this system is {(r, r, r) : r ∈ R}, so the eigenspace
corresponding to λ1 = 2 is E1 = {v ∈ R3 : v = r(1, 1, 1), r ∈ R}. A basis for E1 is {(1, 1, 1)}, and
dim[E1 ] = 1.
⎤
⎡
⎤⎡
⎡
⎤
1 −1 2
v1
0
If λ2 = 0 then (A − λI)v = 0 assumes the form ⎣ 1 −1 2 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ v1 − v2 + 2v3 = 0.
1 −1 2
v3
0
The solution set of this system is {(s − 2t, s, t) : s, t ∈ R}, so the eigenspace corresponding to λ2 = 0 is
E2 = {v ∈ R3 : v = s(1, 1, 0) + t(−2, 0, 1), s, t ∈ R}. A basis for E2 is {(1, 1, 0), (−2, 0, 1)}, and dim[E2 ] = 2.
A is nondefective because it has a complete set of eigenvectors.
2−λ
3
0 −λ
1 = 0 ⇐⇒ (λ − 2)3 = 0 ⇐⇒ λ = 2 of multiplicity three.
15. det(A − λI) = 0 ⇐⇒ −1
−2
−1 4 − λ ⎤ ⎡
⎡
⎤⎡
⎤
0
3 0
v1
0
If λ1 = 2 then (A − λI)v = 0 assumes the form ⎣ −1 −2 1 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
−2 −1 2
v3
0
=⇒ v1 − v3 = 0 and v2 = 0. The solution set of this system is {(r, 0, r) : r ∈ R}, so the eigenspace
corresponding to λ1 = 2 is E1 = {v ∈ R3 : v = r(1, 0, 1), r ∈ R}. A basis for E1 is {(1, 0, 1)}, and
dim[E1 ] = 1.
A is defective since it does not have a complete set of eigenvectors.
−λ −1 −1 16. det(A − λI) = 0 ⇐⇒ −1 −λ −1 = 0 ⇐⇒ (λ + 2)(λ − 1)2 = 0 ⇐⇒ λ = −2 or λ = 1 of multiplicity
−1 −1 −λ two.
2
1−λ
3
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⎤ ⎡
⎤⎡
⎤
2 −1 −1
v1
0
2 −1 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
If λ1 = −2 then (A − λI)v = 0 assumes the form ⎣ −1
−1 −1
2
v3
0
=⇒ v1 − v3 = 0 and v2 − v3 = 0. The solution set of this system is {(r, r, r) : r ∈ R}, so the eigenspace
corresponding to λ1 = −2 is E1 = {v ∈ R3 : v = r(1, 1, 1), r ∈ R}. A basis for E1 is {(1, 1, 1)}, and
dim[E1 ] = 1.
⎤ ⎡
⎡
⎤⎡
⎤
−1 −1 −1
v1
0
If λ2 = 1 then (A − λI)v = 0 assumes the form ⎣ −1 −1 −1 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ v1 + v2 + v3 = 0.
−1 −1 −1
v3
0
The solution set of this system is {(−s − t, s, t) : s, t ∈ R}, so the eigenspace corresponding to λ2 = 1
is E2 = {v ∈ R3 : v = s(−1, 1, 0) + t(−1, 0, 1), s, t ∈ R}. A basis for E2 is {(−1, 1, 0), (−1, 0, 1)}, and
dim[E2 ] = 2.
A is nondefective because it has a complete set of eigenvectors.
⎡
17. (λ − 4)(λ + 1) = 0 ⇐⇒ λ = 4 or λ = −1. Since A has two distinct eigenvalues, it has two linearly
independent eigenvectors and is, therefore, nondefective.
18. (λ − 1)2 = 0 ⇐⇒ λ = 1 of multiplicity two. 0
5
5
v1
=
=⇒ v1 + v2 = 0. The solution
v2
0
−5 −5
set of this system is {(−r, r) : r ∈ R}, so the eigenspace corresponding to λ1 = 1
is E1 = {v ∈ R2 : v = r(−1, 1), r ∈ R}. A basis for E1 is {(1, 1)}, and dim[E1 ] = 1.
A is defective since it does not have a complete set of eigenvectors.
If λ1 = 1 then (A − λI)v = 0 assumes the form
19. λ2 − 4λ + 13 = 0 ⇐⇒ λ = 2 ± 3i. Since A has two distinct eigenvalues, it has two linearly independent
eigenvectors and is, therefore, nondefective.
20. (λ − 2)2 (λ + 1) = 0 ⇐⇒ λ = −1 or λ = 2 of multiplicity two. To determine whether A is nondefective,
all we require is the dimension of the eigenspace⎡corresponding ⎤to⎡λ = 2.
⎤ ⎡
⎤
−1 −3 1
v1
0
If λ = 2 then (A − λI)v = 0 assumes the form ⎣ −1 −3 1 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ −v1 − 3v2 + v3 = 0.
−1 −3 1
v3
0
The solution set of this system is {(−3s + t, s, t) : s, t ∈ R}, so the eigenspace corresponding to λ = 2 is
E = {v ∈ R3 : v = s(−3, 1, 0) + t(1, 0, 1), s, t ∈ R}. Since dim[E] = 2, A is nondefective.
21. (λ − 3)3 = 0 ⇐⇒ λ = 3 of multiplicity three.
⎡
⎤ ⎡
⎤
⎤⎡
−4 2 2
v1
0
If λ = 3 then (A − λI)v = 0 assumes the form ⎣ −4 2 2 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ −2v1 + v2 + v3 = 0 and
0
−4 2 2
v3
v2 = 0. The solution set of this system is {(r, 2r − s, s) : r, s ∈ R}, so the eigenspace corresponding to λ = 3
is E = {v ∈ R3 : v = r(1, 2, 0) + s(0, −1, 1), r, s ∈ R}.
A is defective since it does not have a complete set of eigenvectors.
22. Since A is upper triangular, we note that
det(A − λI) = (4 − λ)3 (−λ)2 ,
3) and λ2 = 0 ⎤(of
from which we obtain the eigenvalues λ1 = 4 (of multiplicity
⎤
⎡ multiplicity
⎡ 2).
⎡
⎤
v1
0
0 1 0
0
0
⎢
⎢ 0 ⎥
⎢ 0 0 1
⎥
0
0 ⎥
⎥
⎥ ⎢ v2 ⎥
⎢
⎢
⎥
⎥
⎢
⎢
⎥
0
0 ⎥ ⎢ v3 ⎥ = ⎢
If λ1 = 4 then (A − λI)v = 0 assumes the form ⎢ 0 0 0
⎢ 0 ⎥, from which
⎦
⎣
⎣
⎣ 0 0 0 −4
⎦
0 ⎦
1
v4
0
0 0 0
0 −4
v5
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we compute quickly that v2 = v3 = v4 = v5 = 0, leaving only v1 as a free variable. Thus, a solution set
for this system is {(r, 0, 0, 0, 0) : r ∈ R}, so the eigenspace corresponding to λ1 = 4 is E1 = {v ∈ R5 : v =
r(1, 0, 0, 0, 0), r ∈ R}. A basis for E1 is {(1, 0, 0, 0, 0)} and dim[E1 ] = 1. We conclude that A is defective
already since the eigenvalue λ1 occurred with multiplicity 3, but only generated one linearly independent
eigenvector. It is thus impossible to form a complete set of eigenvectors.
Note: We could also consider the eigenvalue λ2 = 0 and go through a similar analysis to that done above,
but since we already know that A is defective, it is unnecessary to carry out this consideration.
23. In Problem 1 in this section, we saw that there is only one eigenspace, E1 , corresponding to λ = −7. A
basis for this eigenspace was determined to be {(0, 1)} in R2 , so that this eigenspace consists of the y-axis
in the xy-plane R2 .
2−λ
1 = 0 ⇐⇒ (λ − 1)(λ − 5) = 0 ⇐⇒ λ = 1 or λ = 5.
24. det(A − λI) = 0 ⇐⇒ 3
4−λ 0
1 1
v1
If λ1 = 1 then (A − λI)v = 0 assumes the form
=
=⇒ v1 + v2 = 0. The eigenspace
v2
0
3 3
2
corresponding to λ1 = 1 is E1 = {v ∈ R : v = r(−1,
forE1 is
{(−1, 1)}.
1), r ∈ R}. A basis
0
−3
1
v1
=
=⇒ 3v1 − v2 = 0. The
If λ2 = 5 then (A − λI)v = 0 assumes the form
0
v2
3 −1
2
eigenspace corresponding to λ2 = 5 is E2 = {v ∈ R : v = s(1, 3), s ∈ R}. A basis for E2 is {(1, 3)}.
v2
3
2
1
E1
-2
-1
-1
E2
1
2
v1
Figure 0.0.72: Figure for Problem 24
2−λ
25. det(A − λI) = 0 ⇐⇒ 0
3 = 0 ⇐⇒ (2 − λ)2 = 0 ⇐⇒ λ = 2 of multiplicity two.
2−λ 0
0 3
v1
If λ1 = 2 then (A − λI)v = 0 assumes the form
=
=⇒ v1 ∈ R and v2 = 0. The
v2
0
0 0
eigenspace corresponding to λ1 = 2 is E1 = {v ∈ R2 : v = r(1, 0), r ∈ R}. A basis for E1 is {(1, 0)}.
0 = 0 ⇐⇒ (5 − λ)2 = 0 ⇐⇒ λ = 5 of multiplicity two.
5−λ 0
0 0
v1
If λ1 = 5 then (A − λI)v = 0 assumes the form
=
=⇒ v1 , v2 ∈ R. The eigenspace
v2
0
0 0
corresponding to λ1 = 5 is E1 = {v ∈ R2 : v = r(1, 0) + s(0, 1), r, s ∈ R}. A basis for E1 is {(1, 0), (0, 1)}.
5−λ
26. det(A − λI) = 0 ⇐⇒ 0
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v2
E1
v1
Figure 0.0.73: Figure for Problem 25
v2
E1 is the whole of R2
v1
Figure 0.0.74: Figure for Problem 26
3−λ
27. det(A − λI) = 0 ⇐⇒ 1
−1
multiplicity two.
−1 −1 = 0 ⇐⇒ (λ − 5)(λ − 2)2 = 0 ⇐⇒ λ = 5 or λ = 2 of
3−λ ⎤ ⎡
⎡
⎤⎡
⎤
−2
1 −1
v1
0
If λ1 = 5 then (A − λI)v = 0 assumes the form ⎣ 1 −2 −1 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
−1 −1 −2
v3
0
=⇒ v1 + v3 = 0 and v2 + v3 = 0. The eigenspace corresponding to λ1 = 5 is
E1 = {v ∈ R3 : v = r(1, 1, −1), r ∈ R}. A basis for
⎤ ⎡
⎡ E1 is {(1, 1, −1)}.
⎤⎡
⎤
1
1 −1
v1
0
1 −1 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ v1 + v2 − v3 = 0.
If λ2 = 2 then (A − λI)v = 0 assumes the form ⎣ 1
−1 −1
1
v3
0
The eigenspace corresponding to λ2 = 2 is E2 = {v ∈ R3 : v = s(−1, 1, 0) + t(1, 0, 1), s, t ∈ R}. A basis for
E2 is {(−1, 1, 0), (1, 0, 1)}.
1
3−λ
−1
0
= 0 ⇐⇒ (λ + 2)3 = 0 ⇐⇒ λ = −2 of multiplicity
2
−2 − λ ⎤
⎡
⎤⎡
⎡
⎤
−1 1 0
v1
0
If λ1 = −2 then (A − λI)v = 0 assumes the form ⎣ −1 1 2 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ v1 − v2 = 0 and
0 0 0
v3
0
v3 = 0. The eigenspace corresponding to λ1 = −2 is E1 = {v ∈ R3 : v = r(1, 1, 0), r ∈ R}. A basis for E1 is
{(1, 1, 0)}.
−3 − λ
28. det(A − λI) = 0 ⇐⇒ −1
0
three.
1
−1 − λ
0
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v3
E2
(1, 0, 1)
(-1, 1, 0)
v2
E1
V1
(1, 1, -1)
Figure 0.0.75: Figure for Problem 27
v3
E1
v2
V1
(1, 1, 0)
Figure 0.0.76: Figure for Problem 28
⎤ ⎡
⎤
⎤⎡
1 −2 3
v1
0
29. (a). If λ1 = 1 then (A − λI)v = 0 assumes the form ⎣ 1 −2 3 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
0
1 −2 3
v3
=⇒ v1 − 2v2 + 3v3 = 0. The eigenspace corresponding to λ1 = 1 is
E1 = {v ∈ R3 : v = r(2, 1, 0) + s(−3, 0, 1), r, s ∈ R}. A basis for E1 is {(2, 1, 0), (−3, 0, 1)}.
Now apply the Gram-Schmidt process where v1 = (−3, 0, 1), and v2 = (2, 1, 0). Let u1 = v1 so that
v2 , u1 = (2, 1, 0), (−3, 0, 1) = 2(−3) + 1 · 0 + 0 · 1 = −6 and ||u1 ||2 = (−3)2 + 02 + 12 = 10.
v2 , u1 6
1
u1 = (2, 1, 0) + (−3, 0, 1) = (1, 5, 3).
u2 = v2 −
2
||u1 ||
10
5
Thus, {(−3, 0, 1), (1, 5, 3)} is an orthogonal basis for E1 .
⎤ ⎡
⎡
⎤⎡
⎤
−1 −2 3
v1
0
(b). If λ2 = 3 then (A − λI)v = 0 assumes the form ⎣ 1 −4 3 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ v1 − v2 = 0 and
1 −2 1
v3
0
v2 − v3 = 0. The eigenspace corresponding to λ2 = 3 is E2 = {v ∈ R3 : v = r(1, 1, 1), r ∈ R}. A basis for
E2 is {(1, 1, 1)}.
To determine the orthogonality of the vectors, consider the following inner products:
(−3, 0, 1), (1, 1, 1) = −3 + 0 + 1 = −2 = 0 and (1, 5, 3), (1, 1, 1) = 1 + 5 + 3 = 9 = 0.
Thus, the vectors in E1 are not orthogonal to the vectors in E2 .
⎤ ⎡
⎡
⎤⎡
⎤
−1 −1
1
v1
0
1 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
30. (a). If λ1 = 2 then (A − λI)v = 0 assumes the form ⎣ −1 −1
1
1 −1
v3
0
=⇒ v1 + v2 − v3 = 0. The eigenspace corresponding to λ1 = 2 is
E1 = {v ∈ R3 : v = r(−1, 1, 0) + s(1, 0, 1), r, s ∈ R}. A basis for E1 is {(−1, 1, 0), (1, 0, 1)}.
⎡
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Now apply the Gram-Schmidt process where v1 = (1, 0, 1), and v2 = (−1, 1, 0). Let u1 = v1 so that
v2 , u1 = (−1, 1, 0), (1, 0, 1) = −1 · 1 + 1 · 0 + 0 · 1 = −1 and ||u1 ||2 = 12 + 02 + 12 = 2.
1
v2 , u1 1
u1 = (−1, 1, 0) + (1, 0, 1) = (−1, 2, 1).
u2 = v2 −
||u1 ||2
2
2
Thus, {(1, 0, 1), (−1, 2, 1)} is an orthogonal basis for E1 .
⎤ ⎡
⎡
⎤⎡
⎤
2 −1 1
v1
0
2 1 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ . The eigenspace
(b). If λ2 = −1 then (A − λI)v = 0 assumes the form ⎣ −1
1
1 2
v3
0
corresponding to λ2 = −1 is E2 = {v ∈ R3 : v = r(−1, −1, 1), r ∈ R}. A basis for E2 is {(−1, −1, 1)}.
To determine the orthogonality of the vectors, consider the following inner products:
(1, 0, 1), (−1, −1, 1) = −1 + 0 + 1 = 0 and (−1, 2, 1), (−1, −1, 1) = 1 − 2 + 1 = 0.
Thus, the vectors in E1 are orthogonal to the vectors in E2 .
31. We are given that the eigenvalues of A are λ1 = 0 (multiplicity two), and λ2 = a + b + c. There are two
cases to consider: λ1 = λ2 or λ1 = λ2 .
⎤
⎡
⎤⎡
a b c
v1
If λ1 = λ2 then λ1 = 0 is of multiplicity three, and (A − λI)v = 0 assumes the form ⎣ a b c ⎦ ⎣ v2 ⎦ =
a b c
v3
⎤
⎡
0
⎣ 0 ⎦, or equivalently,
0
av1 + bv2 + cv3 = 0.
(31.1)
The only way to have three linearly independent eigenvectors for A is if a = b = c = 0.
If λ1 = λ2 then λ1 = 0 is of multiplicity two, and λ2 = a + b + c = 0 are distinct eigenvalues. By Theorem
7.2.11, E1 must have dimension two for A to possess a complete set of eigenvectors. The system for determining the eigenvectors corresponding to λ1 = 0 is once more given by (31.1). Since we can choose two
variables freely in (31.1), it follows that there are indeed two corresponding linearly independent eigenvectors.
Consequently, A is nondefective in this case.
32. (a). Setting λ = 0 in (7.2.4), we have p(λ) = det(A − λI) = det(A), and in (7.2.5), we have p(λ) =
p(0) = bn . Thus, bn = det(A).
The value of det(A − λI) is the sum of products of its elements, one taken from each row and each column.
Expanding det(A − λI) yields equation (7.2.5). The expression involving λn in p(λ) comes from the product,
n
(aii − λ), of the diagonal elements. All the remaining products of the determinant have degree not higher
i=1
than n − 2, since, if one of the factors of the product is aij , where i = j, then this product cannot contain
the factors λ − aii and λ − ajj . Hence,
p(λ) =
n
-
(aii − λ) + (terms of degree not higher than n − 2)
i=1
so,
p(λ) = (−1)n λn + (−1)n−1 (a11 + a22 + · · · + ann )λn−1 + · · · + an .
Equating like coefficients from (5.7.5), it follows that b1 = (−1)n−1 (a11 + a22 + · · · + ann ).
(b). Letting λ = 0, we have from (7.2.6) that p(0) =
n
-
(λi − 0) or p(0) =
i=1
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i=1
λi , but from (7.2.5), p(0) = bn ,
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therefore bn =
n
-
λi . Letting λ = 1, we have from (7.2.6) that
i=1
p(λ) =
n
i=1
(λi − λ) = (−1)n
n
-
(λ − λi ) = (−1)n [λn − (λ1 + λ2 + · · · + λn )λn−1 + · · · + bn ].
i=1
Equating like coefficients with (7.2.5), it follows that b1 = (−1)n−1 (λ1 + λ2 + · · · + λn ).
(c). From (a), bn = det(A), and from (b), det(A) = λ1 λ2 · · · λn , so det(A) is the product of the eigenvalues
of A.
From (a), b1 = (−1)n−1 (a11 + a22 + · · · + ann ) and from (b), b1 = (−1)n−1 (λ1 + λ2 + · · · + λn ), thus
a11 + a22 + · · · + ann = λ1 + λ2 + · · · + λn . That is, tr(A) is the sum of the eigenvalues of A.
33. The sum of the eigenvalues of A is
tr(A) = (−1) + (−3) + 1 = −3
and the product of the eigenvalues of A (obtained using the arrow method for a 3 × 3 determinant) is
det(A) = 3 + 32 + 0 − 0 − 16 + 12 = 31.
34. We have det(A) = 19 and tr(A) = 3, so the product of the eigenvalues of A is 19, and the sum of the
eigenvalues of A is 3.
35. We have det(A) = −69 and tr(A) = 1, so the product of the eigenvalues of A is -69, and the sum of the
eigenvalues of A is 1.
36. We have det(A) = −607 and tr(A) = 24, so the product of the eigenvalues of A is -607, and the sum of
the eigenvalues of A is 24.
37. Note that Ei = ∅ since 0 belongs to Ei .
Closure under Addition: Let v1 , v2 ∈ Ei . Then A(v1 + v2 ) = Av1 + Av2 = λi v1 + λi v2 = λi (v1 + v2 ) =⇒
v 1 + v 2 ∈ Ei .
Closure under Scalar Multiplication: Let c ∈ C and v1 ∈ Ei . Then A(cv1 ) = c(Av1 ) = c(λi v1 ) = λi (cv1 ) =⇒
cv1 ∈ Ei .
Thus, by Theorem 4.3.2, Ei is a subspace of C n .
38. The condition
c1 v 1 + c 2 v 2 = 0
(38.1)
=⇒ A(c1 v1 ) + A(c2 v2 ) = 0 =⇒ c1 Av1 + c2 Av2 = 0
=⇒ c1 (λ1 v1 ) + c2 (λ2 v2 ) = 0.
(38.2)
Substituting c2 v2 from (38.1) into (38.2) yields (λ1 − λ2 )c1 v1 = 0.
Since λ2 = λ1 , we must have c1 v1 = 0, but v1 = 0, so c1 = 0. Substituting into (38.1) yields c2 = 0 also.
Consequently, v1 and v2 are linearly independent.
39. Consider
c1 v1 + c2 v2 + c3 v3 = 0.
(39.1)
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If c1 = 0, then the preceding equation can be written as
w1 + w2 = 0,
where w1 = c1 v1 and w2 = c2 v2 + c3 v3 . But this would imply that {w1 , w2 } is linearly dependent, which
would contradict Theorem 7.2.5 since w1 and w2 are eigenvectors corresponding to different eigenvalues.
Consequently, we must have c1 = 0. But then (39.1) implies that c2 = c3 = 0 since {v1 , v2 } is a linearly
independent set by assumption. Hence {v1 , v2 , v3 } is linearly independent.
40. λ1 = 1 (multiplicity 3), basis: {(0, 1, 1)}.
41. λ1 = 0 (multiplicity 2), basis: {(−1, 1, 0), (−1, 0, 1)}. λ2 = 3, basis: {(1, 1, 1)}.
√
√
√
42. λ1 = 2, basis: {(1, −2 2, 1)}. λ2 = 0, basis: {(1, 0, −1)}. λ3 = 7, basis: {( 2, 1, 2)}.
43. λ1 = −2, basis: {(2, 1, −4)}. λ2 = 3 (multiplicity 2), basis: {(0, 2, 1), (3, 11, 0)}.
44. λ1 = 0 (multiplicity 2), basis: {(0, 1, 0, −1), (1, 0, −1, 0)}. λ2 = 6, basis: {(1, 1, 1, 1)}. λ3 = −2, basis:
{1, −1, 1, −1)}.
45. A has eigenvalues:
3
1 2
a + 8b2 ,
a+
2
2
3
1 2
λ2 = a −
a + 8b2 ,
2
2
λ3 = 0.
λ1 =
Provided a = ±b, these eigenvalues are distinct, and therefore the matrix is nondefective.
If a = b = 0, then the eigenvalue λ = 0 has multiplicity two. A basis for the corresponding eigenspace is
{(−1, 0, 1), (−1, 1, 0)}. Since this is two-dimensional, the matrix is nondefective in this case.
If a = −b = 0, then the eigenvalue λ = 0 once more has multiplicity two. A basis for the corresponding
eigenspace is {(0, 1, 1), (1, 1, 0)}, therefore the matrix is nondefective in this case also.
If a = b = 0, then A = 02 , so that λ = 0 (multiplicity three), and the corresponding eigenspace is all of R3 .
Hence A is nondefective.
46. If a = b = 0, then A = 03 , which is nondefective. We now assume that at least one of either a or b is
nonzero.
A has eigenvalues: λ1 = b, λ2 = a − b, and λ3 = 3a + b.
Provided a = 0, 2b, −b, the eigenvalues are distinct, and therefore A is nondefective.
If a = 0, then the eigenvalue λ = b has multiplicity two. In this case, a basis for the corresponding eigenspace
is {(1, 0, 1), (0, 1, 0)}, so that A is nondefective.
If a = 2b, then the eigenvalue λ = b has multiplicity two. In this case, a basis for the corresponding eigenspace
is {(−2, 1, 0), (−1, 0, 1)}, so that A is nondefective.
If a = −b, then the eigenvalue λ = −2b has multiplicity two. In this case, a basis for the corresponding
eigenspace is {(0, 1, 1), (1, 0, −1)}, so that A is nondefective.
Solutions to Section 7.3
True-False Review:
(a): TRUE. The terms “diagonalizable” and “nondefective” are synonymous. The diagonalizability of
a matrix A hinges on the ability to form an invertible matrix S with a full set of linearly independent
eigenvectors of the matrix as its columns. This, in turn, requires the original matrix to be nondefective.
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(b): TRUE. If we assume that A is diagonalizable, then there exists an invertible matrix S and a diagonal
matrix D such that S −1 AS = D. Since A is invertible, we can take the inverse of each side of this equation
to obtain
D−1 = (S −1 AS)−1 = S −1 A−1 S,
and since D−1 is still a diagonal matrix, this equation shows that A−1 is diagonalizable.
1 1
(c): FALSE. For instance, the matrices A = I2 and B =
both have eigenvalue λ = 1 (with
0 1
multiplicity 2). However, A and B are not similar. [Reason: If A and B were similar, then S −1 AS = B
for some invertible matrix S, but since A = I2 , this would imply that B = I2 , contrary to our choice of B
above.]
(d): FALSE. An n × n matrix is diagonalizable if and only if it has n linearly independent eigenvectors.
Besides, every matrix actually has infinitely many eigenvectors, obtained by taking scalar multiples of a
single eigenvector v.
(e): TRUE. Assume A is an n × n matrix such that p(λ) = det(A − λI) has no repeated roots. This implies
that A has n distinct eigenvalues. Corresponding to each eigenvalue, we can select an eigenvector. Since
eigenvectors corresponding to distinct eigenvalues are linearly independent, this yields n linearly independent
eigenvectors for A. Therefore, A is nondefective, and hence, diagonalizable.
(f ): TRUE. Assuming that A is diagonalizable, then there exists an invertible matrix S and a diagonal
matrix D such that S −1 AS = D. Therefore,
D2 = (S −1 AS)2 = (S −1 AS)(S −1 AS) = S −1 ASS −1 AS = S −1 A2 S.
Since D2 is still a diagonalizable matrix, this equation shows that A2 is diagonalizable.
(g): TRUE. Since In−1 AIn = A, A is similar to itself.
(h): TRUE. The sum of the dimensions of the eigenspaces of such a matrix is even, and therefore not equal
to n. This means we cannot obtain n linearly independent eigenvectors for A, and therefore, A is defective
(and not diagonalizable).
(i): TRUE. Assume that
c1 v1 + c2 v2 = 0.
Left-multiplying the matrix A on both sides, we obtain
c1 Av1 + c2 Av2 = A0 = 0.
Using Av1 = λ1 v1 and Av2 = λ2 v2 , this becomes
c1 λ1 v1 + c2 λ2 v2 = 0.
On the other hand, multiplying the first equation through by λ1 , we obtain
c1 λ1 v1 + c2 λ1 v2 = 0.
Subtracting the last two equations, we obtain
c2 (λ2 − λ1 )v2 = 0.
Since v2 = 0 and λ2 = λ1 , we conclude that c2 = 0. Now the original equation implies that c1 = 0. Since
c1 = c2 = 0, we conclude that {v1 , v2 } is linearly independent.
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Problems:
0
= 0 ⇐⇒ (−9 − λ)2 = 0 ⇐⇒ λ = −9 of multiplicity two.
−9 − λ 0
0 0
v1
=
=⇒ v1 = 0. If we let
If λ = −9 then (A − λI)v = 0 assumes the form
0
v2
4 0
v2 = r ∈ R, then the solution set of the linear system is {(0, r) : r ∈ R}, so the eigenvectors corresponding
to λ = −9 are v = r(0, 1), where r ∈ R.
−9 − λ
1. det(A − λI) = 0 ⇐⇒ 4
A has only one linearly independent eigenvector, so by Theorem 7.3.4, A is not diagonalizable.
−1 − λ
−2 = 0 ⇐⇒ λ2 − λ − 6 = 0 ⇐⇒ (λ − 3)(λ + 2) = 0 ⇐⇒ λ = 3 or
2. det(A − λI) = 0 ⇐⇒ −2
2−λ λ = −2. A is diagonalizable because it has two distinct
eigenvalues.
0
−4 −2
v1
=
=⇒ 2v1 + v2 = 0. If we let
If λ1 = 3 then (A − λI)v = 0 assumes the form
0
v2
−2 −1
v1 = r ∈ R, then the solution set of this system is {(−r, 2r) : r ∈ R}, so the eigenvectors corresponding to
λ1 = 3 are v1 = r(−1, 2) where r ∈ R.
0
1 −2
v1
=
=⇒ v1 − 2v2 = 0. If we let
If λ2 = −2 then (A − λI)v = 0 assumes the form
v2
0
−2
4
v2 = s ∈ R, then the solution set of this system is {(2s, s) : s ∈ R}, so the eigenvectors corresponding to
λ2 = −2 are v2 = s(2, 1) where s ∈ R.
−1 2
Thus, the matrix S =
satisfies S −1 AS = diag(3, −2).
2 1
−7 − λ
4 3. det(A − λI) = 0 ⇐⇒ = 0 ⇐⇒ λ2 + 6λ + 9 = 0 ⇐⇒ (λ + 3)2 = 0 ⇐⇒ λ = −3 of
−4
1−λ multiplicity two.
0
−4 4
v1
=
=⇒ v1 − v2 = 0. If we let
If λ = −3 then (A − λI)v = 0 assumes the form
0
v2
−4 4
v1 = r ∈ R, then the solution set of this system is {(r, r) : r ∈ R}, so the eigenvectors corresponding to
λ = −3 are v = r(1, 1) where r ∈ R.
A has only one linearly independent eigenvector, so by Theorem 7.3.4, A is not diagonalizable.
1−λ
−8 = 0 ⇐⇒ λ2 + 6λ + 9 = 0 ⇐⇒ (λ + 3)2 = 0 ⇐⇒ λ = −3 of
4. det(A − λI) = 0 ⇐⇒ 2
−7 − λ multiplicity two.
0
4 −8
v1
=
=⇒ v1 − 2v2 = 0. If we let
If λ = −3 then (A − λI)v = 0 assumes the form
v2
0
2 −4
v2 = r ∈ R, then the solution set of this system is {(2r, r) : r ∈ R}, so the eigenvectors corresponding to
λ = −3 are v = r(2, 1) where r ∈ R.
A has only one linearly independent eigenvector, so by Theorem 7.3.4, A is not diagonalizable.
−λ
4 = 0 ⇐⇒ λ2 + 16 = 0 ⇐⇒ λ = ±4i. A is diagonalizable because it has
5. det(A − λI) = 0 ⇐⇒ −4 −λ two distinct eigenvalues.
0
4i 4
v1
=
=⇒ v1 − iv2 = 0. If we let
If λ = −4i then (A − λI)v = 0 assumes the form
0
v2
−4 4i
v2 = r ∈ C, then the solution set of this system is {(ir, r) : r ∈ C}, so the eigenvectors corresponding to
λ = −4i are v = r(i, 1) where r ∈ C. Since the entries of A are real, it follows from Theorem 7.1.8 that
v2 = (−i, 1) is an eigenvector corresponding to λ = 4i.
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i −i
satisfies S −1 AS = diag(−4i, 4i).
1
1
1−λ
0
0
= 0 ⇐⇒ (1 − λ)(λ + 4)(λ − 4) = 0 ⇐⇒ λ = 1, λ = −4 or
3−λ
7
6. det(A − λI) = 0 ⇐⇒ 0
1
1
−3 − λ λ = 4.
⎤ ⎡
⎡
⎤⎡
⎤
0 0
0
v1
0
7 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
If λ = 1 then (A − λI)v = 0 assumes the form ⎣ 0 2
1 1 −4
v3
0
=⇒ 2v1 − 15v3 = 0 and 2v2 + 7v3 = 0. If we let v3 = 2r where r ∈ R, then the solution set of this system is
{(15r, −7r, 2r) : r ∈ R} so the eigenvectors corresponding
to λ
⎤ v1 ⎡= r(15,
⎡
⎤=
⎡ 1 are
⎤ −7, 2) where r ∈ R.
5 0 0
v1
0
If λ = −4 then (A − λI)v = 0 assumes the form ⎣ 0 7 7 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
1 1 1
v3
0
=⇒ v1 = 0 and v2 + v3 = 0. If we let v2 = s ∈ R, then the solution set of this system is {(0, s, −s) : s ∈ R}
so the eigenvectors corresponding to λ = −4 are⎡ v2 = s(0, 1, −1)⎤where
s⎤∈ R.⎡
⎡
⎤
−3
0
0
v1
0
7 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
If λ = 4 then (A − λI)v = 0 assumes the form ⎣ 0 −1
1
1 −7
v3
0
=⇒ v1 = 0 and v2 − 7v3 = 0. If we let v3 = t ∈ R, then the solution set of this system is {(0, 7t, t) : t ∈ R}
so the eigenvectors corresponding
to λ⎤= 4 are v3 = t(0, 7, 1) where t ∈ R.
⎡
15 0
0
1 ⎦ satisfies S −1 AS = diag(1, 4, −4).
Thus, the matrix S = ⎣ −7 7
2 1 −1
1−λ
−2
0
= 0 ⇐⇒ (λ + 1)3 = 0 ⇐⇒ λ = −1 of multiplicity three.
−3 − λ
0
7. det(A − λI) = 0 ⇐⇒ 2
2
−2
−2 − λ ⎤
⎡
⎤⎡
⎡
⎤
2 −2 0
v1
0
If λ = −1 then (A − λI)v = 0 assumes the form ⎣ 2 −2 0 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ v1 − v2 = 0 and
2 −2 0
v3
0
v3 ∈ R. If we let v2 = r ∈ R and v3 = s ∈ R, then the solution set of this system is {(r, r, s) : r, s ∈ R}, so
the eigenvectors corresponding to λ = −1 are v1 = r(1, 1, 0) and v2 = s(0, 0, 1) where r, s ∈ R.
A has only two linearly independent eigenvectors, so by Theorem 7.3.4, A is not diagonalizable.
−λ −2 −2 8. det(A − λI) = 0 ⇐⇒ −2 −λ −2 = 0 ⇐⇒ (λ − 2)2 (λ + 4) = 0 ⇐⇒ λ = −4 or λ = 2 of multiplicity
−2 −2 −λ two.
⎤ ⎡
⎤
⎡
⎤⎡
4 −2 −2
v1
0
4 −2 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
If λ = −4 then (A − λI)v = 0 assumes the form ⎣ −2
0
−2 −2
4
v3
=⇒ v1 − v3 = 0 and v2 − v3 = 0. If we let v3 = r ∈ R, then the solution set of this system is {(r, r, r) : r ∈ R}
so the eigenvectors corresponding to λ = −4 are⎡ v1 = r(1, 1, 1) where
⎤ R. ⎡
⎤⎡ r ∈
⎤
−2 −2 −2
v1
0
If λ = 2 then (A − λI)v = 0 assumes the form ⎣ −2 −2 −2 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ v1 + v2 + v3 = 0. If
−2 −2 −2
v3
0
we let v2 = s ∈ R and v3 = t ∈ R, then the solution set of this system is {(−s − t, s, t) : s, t ∈ R}, so two
linearly independent eigenvectors corresponding to λ = 2 are v2 = s(−1, 1, 0) and v3 = t(−1, 0, 1).
Thus, the matrix S =
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⎡
⎤
1 −1 −1
0
1 ⎦ satisfies S −1 AS = diag(−4, 2, 2).
Thus, the matrix S = ⎣ 1
1
1
0
4 4 = 0 ⇐⇒ λ2 (λ−3) = 0 ⇐⇒ λ = 3, or λ = 0 of multiplicity
4−λ ⎤ ⎡
⎡
⎤⎡
⎤
−5
1 4
v1
0
If λ = 3 then (A − λI)v = 0 assumes the form ⎣ −2 −2 4 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
−2
1 1
v3
0
=⇒ v1 − v3 = 0 and v2 − v3 = 0. If we let v3 = r ∈ R, then the solution set of this system is {(r, r, r) : r ∈ R},
so the eigenvectors corresponding to λ = 3 are ⎡
v1 = r(1, 1, 1)⎤where
r⎤ ∈ R.
⎡
⎡
⎤
−2 1 4
v1
0
If λ = 0 then (A − λI)v = 0 assumes the form ⎣ −2 1 4 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ −2v1 + v2 + 4v3 = 0. If
−2 1 4
v3
0
we let v1 = s ∈ R and v3 = t ∈ R, then the solution set of this system is {(s, 2s − 4t, t) : s, t ∈ R}, so two
linearly independent eigenvectors
corresponding
to λ = 0 are v2 = s(1, 2, 0) and v3 = t(0, −4, 1).
⎡
⎤
1 1
0
Thus, the matrix S = ⎣ 1 2 −4 ⎦ satisfies S −1 AS = diag(3, 0, 0).
1 0
1
−2 − λ
9. det(A−λI) = 0 ⇐⇒ −2
−2
two.
2−λ
10. det(A − λI) = 0 ⇐⇒ 0
2
multiplicity two.
1
1−λ
1
0 0 = 0 ⇐⇒ (λ − 2)(λ − 1)2 = 0 ⇐⇒ λ = 2 or λ = 1 of
1−λ ⎤
⎡
⎤⎡
⎡
⎤
1
0 0
v1
0
0 0 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ v1 = v2 = 0 and
If λ = 1 then (A − λI)v = 0 assumes the form ⎣ 0
2 −1 0
v3
0
v3 ∈ R. If we let v3 = r ∈ R then the solution set of this system is {(0, 0, r) : r ∈ R}, so there is only one
corresponding linearly independent eigenvector. Hence, by Theorem 7.3.4, A is not diagonalizable.
4−λ
0
0 −1 − λ
−1 = 0 ⇐⇒ (λ2 + 1)(λ − 4) = 0 ⇐⇒ λ = 4, or λ = ±i.
11. det(A − λI) = 0 ⇐⇒ 3
0
2
1−λ ⎤ ⎡
⎡
⎤⎡
⎤
0
0
0
v1
0
If λ = 4 then (A − λI)v = 0 assumes the form ⎣ 3 −5 −1 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
0
2 −3
v3
0
=⇒ 6v1 − 17v3 = 0 and 2v2 − 3v3 = 0. If we let v3 = 6r ∈ C, then the solution set of this system is
{(17r, 9r, 6r) : r ∈ C}, so the eigenvectors corresponding
to λ = 4 are v1⎤=⎡r(17,⎤9, 6)⎡where
⎡
⎤ r ∈ C.
4−i
0
0
v1
0
−1 − i
−1 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ v1 = 0 and
If λ = i then (A − λI)v = 0 assumes the form ⎣ 3
0
2
1−i
v3
0
2v2 + (1 − i)v3 = 0. If we let v3 = −2s ∈ C, then the solution set of this system is {(0, (1 − i)s, 2s) : s ∈ C},
so the eigenvectors corresponding to λ = i are v2 = s(0, 1 − i, 2) where s ∈ C. Since the entries of A are real,
v3 = t(0, 1 + i, 2) where⎡ t ∈ C are the eigenvectors
corresponding to λ = −i by Theorem 7.1.8.
⎤
17
0
0
Thus, the matrix S = ⎣ 9 1 − i 1 + i ⎦ satisfies S −1 AS = diag(4, i, −i).
6
2
2
0
1−λ
−1
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−λ
2
12. det(A − λI) = 0 ⇐⇒ −2 −λ
1
2
−1 −2 = 0 ⇐⇒ λ(λ2 + 9) = 0 ⇐⇒ λ = 0, or λ = ±3i.
−λ ⎤ ⎡
⎡
⎤⎡
⎤
0 2 −1
v1
0
If λ = 0 then (A − λI)v = 0 assumes the form ⎣ −2 0 −2 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
1 2
0
v3
0
=⇒ v1 + v3 = 0 and 2v2 − v3 = 0. If we let v3 = 2r ∈ C, then the solution set of this system is
{(−2r, r, 2r) : r ∈ C}, so the eigenvectors corresponding
to λ = 0 are
⎡
⎤ ⎡v1 =⎤r(−2,
⎡ 1, 2)
⎤ where r ∈ C.
3i 2 −1
v1
0
If λ = −3i then (A − λI)v = 0 assumes the form ⎣ −2 3i −2 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
1 2
3i
v3
0
=⇒ 5v1 +(−4+3i)v3 = 0 and 5v2 +(2+6i)v3 = 0. If we let v3 = 5s ∈ C, then the solution set of this system is
{((4−3i)s, (−2−6i)s, 5s) : s ∈ C}, so the eigenvectors corresponding to λ = −3i are v2 = s(4−3i, −2−6i, 5)
where s ∈ C. Since the entries of A are real, v3 = t(4 + 3i, −2 + 6i, 5) where t ∈ C are the eigenvectors
corresponding to λ = 3i
⎡ by Theorem 7.1.8.
⎤
−2
4 + 3i
4 − 3i
Thus, the matrix S = ⎣ 1 −2 + 6i −2 − 6i ⎦ satisfies S −1 AS = diag(0, 3i, −3i).
2
5
5
1−λ
−2
0 1−λ
0 = 0 ⇐⇒ (λ − 3)2 (λ + 1) = 0 ⇐⇒ λ = −1 or λ = 3 of
13. det(A − λI) = 0 ⇐⇒ −2
0
0
3−λ multiplicity two.
⎤ ⎡
⎤
⎡
⎤⎡
2 −2 0
v1
0
2 0 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦
If λ = −1 then (A − λI)v = 0 assumes the form ⎣ −2
0
0
0 4
v3
=⇒ 2v1 − v2 = 0 and v3 = 0. If we let v1 = r ∈ R, then the solution set of this system is {(r, r, 0) : r ∈ R}
so the eigenvectors corresponding to λ = −1 are v⎡1 = r(1, 1, 0) where
⎤ ⎡ r ∈⎤R. ⎡
⎤
−2 −2 0
v1
0
If λ = 3 then (A − λI)v = 0 assumes the form ⎣ −2 −2 0 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ v1 + v2 = 0 and
0
0 0
v3
0
v3 ∈ R. If we let v2 = s ∈ R and v3 = t ∈ R, then the solution set of this system is {(−s, s, t) : s, t ∈ R}, so
the eigenvectors corresponding
to λ ⎤
= 3 are v2 = s(−1, 1, 0) and v3 = t(0, 0, 1).
⎡
1 −1 0
1 0 ⎦ satisfies S −1 AS = diag(−1, 3, 3).
Thus, the matrix S = ⎣ 1
0
0 1
−1 − λ
1
0
0 0
−1 − λ
0
0 = 0 ⇐⇒ (−1 − λ)3 (1 − λ) = 0 ⇐⇒ λ = −1
14. det(A − λI) = 0 ⇐⇒ 0
0
−1
−
λ
0 0
0
0
1−λ of multiplicity three and λ = 1 of multiplicity one.
⎤
⎤⎡
⎤
⎡
⎡
v1
0
0 1 0 0
⎢ 0 ⎥
⎢ 0 0 0 0 ⎥ ⎢ v2 ⎥
⎥
⎥⎢
⎥
⎢
If λ = −1 then (A − λI)v = 0 assumes the form ⎢
⎣ 0 0 0 0 ⎦ ⎣ v3 ⎦ = ⎣ 0 ⎦ =⇒ v2 = v4 = 0.
0
0 0 0 2
v4
If we let v1 = r and v3 = s be free variables (r, s ∈ R), then the solution set of the linear system is
{(r, 0, s, 0) : r, s ∈ R}, so the eigenvectors corresponding to λ = −1 are v = r(1, 0, 0, 0) + s(0, 0, 1, 0), where
r, s ∈ R.
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⎤⎡
⎤
⎤ ⎡
v1
−2
1
0 0
0
⎢
⎥
⎢ 0 −2
⎥ ⎢
0 0 ⎥
⎥ ⎢ v2 ⎥ = ⎢ 0 ⎥ =⇒ v1 = v2 =
If λ = 1 then (A − λI)v = 0 assumes the form ⎢
⎦
⎣
⎣ 0
⎦
⎣
0 −2 0
0 ⎦
v3
0
0
0 0
0
v4
v3 = 0. If we let v4 = t, (t ∈ R), then the solution set of the linear system is {(0, 0, 0, t) : t ∈ R}, so the
eigenvectors corresponding to λ = 1 are v = t(0, 0, 0, 1), where t ∈ R.
We have collected only three linearly independent eigenvectors for the 4 × 4 matrix A, so by Theorem
7.3.4, A is not diagonalizable.
⎡
15. Since all rows of A are multiples of the first row, this matrix is not invertible, and we see that λ = 0
must be an eigenvalue.
⎤⎡
⎤
⎡
⎤ ⎡
v1
1
2
3
4
0
⎢ −1 −2 −3 −4 ⎥ ⎢ v2 ⎥ ⎢ 0 ⎥
⎥⎢
⎥ =⇒ v1 + 2v2 +
⎥=⎢
If λ = 0 then (A − λI)v = 0 assumes the form ⎢
⎣ 2
4
6
8 ⎦ ⎣ v3 ⎦ ⎣ 0 ⎦
−2 −4 −6 −8
0
v4
3v3 + 4v4 = 0. If we let v2 = r, v3 = s, and v4 = t, where r, s, t ∈ R, then v1 = −2r − 3s − 4t. Thus, the
solution set of the linear system is {(−2r − 3s − 4t, r, s, t) : r, s, t ∈ R}, so the eigenvectors corresponding to
λ = 0 are v = r(−2, 1, 0, 0) + s(−3, 0, 1, 0) + t(−4, 0, 0, 1), where r, s, t ∈ R. Thus, we obtain three linearly
independent eigenvectors, v1 = (−2, 1, 0, 0), v2 = (−3, 0, 1, 0), and v3 = (−4, 0, 0, 1).
Since A is a 4 × 4 matrix, we know that A has four eigenvalues (including multiplicities). Moreover, by
Problem 32 in Section 7.2, we know that the sum of the four eigenvalues of A is tr(A) = 1+(−2)+6+(−8) =
−3. Since λ = 0 occurs with multiplicity three, we must⎡have the final eigenvalue
⎤ ⎡ λ =⎤−3. ⎡
⎤
v1
4
2
3
4
0
⎢
⎥
⎢ −1
⎥
⎢
1 −3 −4 ⎥
⎥ ⎢ v2 ⎥ = ⎢ 0 ⎥. A rowIf λ = −3 then (A − λI)v = 0 assumes the form ⎢
⎦
⎣
⎣ 2
⎦
⎣
4
9
8
0 ⎦
v3
−2 −4 −6 −5
0
v4
⎤
⎡
1 −1
3
4
⎢ 0
1 − 32 −2 ⎥
⎥. Setting v4 = t, by
echelon form for the coefficient matrix of this linear system is ⎢
⎣ 0
0
1
1 ⎦
0
0
0
0
back substitution we obtain v3 = −t, v2 = 12 t, and then v1 = − 12 t. Thus, the solution set of the linear system
is {(− 12 t, 12 t, −t, t) : t ∈ R}, so we obtain one linearly independent eigenvector corresponding to λ = −3,
namely v4 = (− 12 , 12 , −1, 1).
Altogether, we have obtained four linearly independent eigenvectors of A, which implies that A is diagonalizable. We construct S by placing the four eigenvectors we have obtained in the columns:
⎡
⎤
−2 −3 −4 − 21
1 ⎥
⎢ 1
0
0
2 ⎥.
S=⎢
⎣ 0
1
0 −1 ⎦
0
0
1
1
This matrix satisfies S −1 AS = diag(0, 0, 0, −3).
16. λ1 = 2 (multiplicity 2), basis for eigenspace: {(−3, 1, 0), (3, 0, 1)}.
λ2 = 1, basis
for eigenspace:
{(1, 2, 2)}.
⎡
⎤
−3 3 1
Set S = ⎣ 1 0 2 ⎦. Then S −1 AS = diag(2, 2, 1).
0 1 2
17. λ1 = 0 (multiplicity 2), basis for eigenspace: {(0, 1, 0, −1), (1, 0, −1, 0)}.
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for eigenspace: ⎤
{(1, 1, 1, 1)}. λ3 = 10, basis for eigenspace: {(−1, 1, −1, 1)}.
λ2 = 2, basis
⎡
0
1 1 −1
⎢ 1
0 1
1 ⎥
−1
⎥
Set S = ⎢
⎣ 0 −1 1 −1 ⎦. Then S AS = diag(0, 0, 2, 10).
−1
0 1
1
1 4
18. The given system can be written as x = Ax, where A =
.
2 3
A has eigenvalues λ1 = −1, λ2 = 5 with
corresponding linearly independent eigenvectors v1 = (−2, 1) and
−2 1
, then S −1 AS = diag(−1, 5), therefore, under the transformation
v2 = (1, 1). If we set S =
1 1
x = Sy, the given system of differential equations simplifies to
y1
y2
=
−1 0
0 5
y1
y2
.
Hence, y1 = −y1 and y2 = 5y2 . Integrating these equations, we obtain
y1 (t) = c1 e−t , y2 (t) = c2 e5t .
Returning to the original variables, we have
−2c1 e−t + c2 e5t
−2 1
c1 e−t
=
.
x = Sy =
c2 e5t
c1 e−t + c2 e5t
1 1
Consequently, x1 (t) = −2c1 e−t + c2 e5t and x2 (t) = c1 e−t + c2 e5t .
6 −2
.
19. The given system can be written as x = Ax, where A =
−2
6
A has eigenvalues λ1 = 4, λ2 = 8 withcorresponding linearly independent eigenvectors v1 = (1, 1) and
1 −1
, then S −1 AS = diag(4, 8), therefore, under the transformation
v2 = (−1, 1). If we set S =
1
1
x = Sy, the given system of differential equations simplifies to
y1
y2
=
4
0
0
8
y1
y2
.
Hence, y1 = 4y1 and y2 = 8y2 . Integrating these equations, we obtain
y1 (t) = c1 e4t , y2 (t) = c2 e8t .
Returning to the original variables, we have
c1 e4t − c2 e8t
1 −1
c1 e4t
.
=
x = Sy =
c2 e8t
c1 e4t + c2 e8t
1
1
Consequently, x1 (t) = c1 e4t − c2 e8t and x2 (t) = c1 e4t + c2 e8t .
20. The given system can be written as x = Ax, where A =
6 −1
−5
2
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A has eigenvalues λ1 = 1 and
λ2 = 7 with corresponding linearly independent eigenvectors v1 = (1, 5)
1 −1
, then S −1 AS = diag(1, 7). Therefore, under the transformation
and v2 = (−1, 1). If we set S =
5
1
x = Sy, the given system of differential equations simplifies to
y1
1 0
y1
=
.
0 7
y2
y2
Hence, y1 = y1 and y2 = 7y2 . Integrating these equations, we obtain
y1 (t) = c1 et ,
y2 (t) = c2 e7t .
Returning to the original variables, we have
1 −1
c1 et
c1 et − c2 e7t
x = Sy =
=
.
5
1
c2 e7t
5c1 et + c2 e7t
Consequently, x1 (t) = c1 et − c2 e7t and x2 (t) = 5c1 et + c2 e7t .
−12 −7
21. The given system can be written as x = Ax, where A =
.
16 10
A has eigenvalues λ1 = 2, λ2= −4 withcorresponding linearly independent eigenvectors v1 = (1, −2) and
1
7
, then S −1 AS = diag(2, −4), therefore, under the transformation
v2 = (7, −8). If we set S =
−2 −8
x = Sy, the given system of differential equations simplifies to
2
0
y1
y1
=
.
y2
0 −4
y2
Hence, y1 = 2y1 and y2 = −4y2 . Integrating these equations, we obtain
y1 (t) = c1 e2t , y2 (t) = c2 e−4t .
Returning to the original variables, we have
1
7
c1 e2t
c1 e2t + 7c2 e−4t
x = Sy =
=
.
−2 −8
c2 e−4t
−2c1 e2t − 8c2 e−4t
Consequently, x1 (t) = c1 e2t + 7c2 e−4t and x2 (t) = −2c1 e2t − 8c2 e−4t .
0 1
22. The given system can be written as x = Ax, where A =
.
−1 0
A has eigenvalues λ1 = i, λ2 = −i with
corresponding linearly independent eigenvectors v1 = (1, i) and
1
1
, then S −1 AS = diag(i, −i), therefore, under the transformation
v2 = (1, −i). If we set S =
i −i
x = Sy, the given system of differential equations simplifies to
i
0
y1
y1
=
.
0 −i
y2
y2
Hence, y1 = iy1 and y2 = −iy2 . Integrating these equations, we obtain
y1 (t) = c1 eit , y2 (t) = c2 e−it .
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Returning to the original variables, we have
c1 eit + c2 e−it
1
1
c1 eit
=
.
x = Sy =
c2 e−it
i(c1 eit − c2 e−it )
i −i
Consequently, x1 (t) = c1 eit + c2 e−it and x2 (t) = i(c1 eit − c2 e−it ). Using Euler’s formula, these expressions
can be written as
x1 (t) = (c1 + c2 ) cos t + i(c1 − c2 ) sin t,
x2 (t) = i(c1 − c2 ) cos t − (c1 + c2 ) sin t,
or equivalently,
x1 (t) = a cos t + b sin t, x2 (t) = b cos t − a sin t,
where a = c1 + c2 , and b = i(c1 − c2 ).
⎡
⎤
3 −4 −1
23. The given system can be written as x = Ax, where A = ⎣ 0 −1 −1 ⎦.
0 −4
2
= 3 with correA has eigenvalue λ1 = −2 with corresponding eigenvector v1 = (1, 1, 1), and eigenvalue λ2 ⎡
⎤
1 1
0
1 ⎦,
sponding linearly independent eigenvectors v2 = (1, 0, 0) and v3 = (0, 1, −4). If we set S = ⎣ 1 0
1 0 −4
then S −1 AS = diag(−2, 3, 3), therefore, under the transformation x = Sy, the given system of differential
equations simplifies to
⎡ ⎤ ⎡
⎤
⎤⎡
y1
−2 0 0
y1
⎣ y2 ⎦ = ⎣ 0 3 0 ⎦ ⎣ y2 ⎦ .
y3
0 0 3
y3
Hence, y1 = −2y1 , y2 = 3y2 , and y3 = 3y3 . Integrating these equations, we obtain
y1 (t) = c1 e−2t , y2 (t) = c2 e3t , y3 (t) = c3 e3t .
Returning to the original variables, we have
⎤ ⎡
⎤
⎡
⎤⎡
1 1
0
c1 e−2t
c1 e−2t + c2 e3t
1 ⎦ ⎣ c2 e3t ⎦ = ⎣ c1 e−2t + c3 e3t ⎦ .
x = Sy = ⎣ 1 0
c3 e3t
c1 e−2t − 4c3 e3t
1 0 −4
Consequently, x1 (t) = c1 e−2t + c2 e3t , x2 (t) = c1 e−2t + c3 e3t , and c1 e−2t − 4c3 e3t .
⎡
⎤
1 1 −1
1 ⎦.
24. The given system can be written as x = Ax, where A = ⎣ 1 1
−1 1
1
A has eigenvalue λ1 = −1 with corresponding eigenvector v1 = (−1, 1, −1), and eigenvalue λ⎡2 = 2 with corre⎤
−1 0
1
0 ⎦,
sponding linearly independent eigenvectors v2 = (0, 1, 1) and v3 = (1, 0, −1). If we set S = ⎣ 1 1
−1 1 −1
then S −1 AS = diag(−1, 2, 2), therefore, under the transformation x = Sy, the given system of differential
equations simplifies to
⎡ ⎤ ⎡
⎤
⎤⎡
y1
−1 0 0
y1
⎣ y2 ⎦ = ⎣ 0 2 0 ⎦ ⎣ y2 ⎦ .
y3
0 0 2
y3
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Hence, y1 = −y1 , y2 = 2y2 , and y3 = 2y3 . Integrating these equations, we obtain
y1 (t) = c1 e−t , y2 (t) = c2 e2t , y3 (t) = c3 e2t .
Returning to the original variables, we have
⎤ ⎡
⎤
⎡
⎤⎡
−1 0
1
c1 e−t
−c1 e−t + c3 e2t
⎦.
c1 e−t + c2 e2t
0 ⎦ ⎣ c2 e2t ⎦ = ⎣
x = Sy = ⎣ 1 1
2t
−1 1 −1
c3 e
−c1 e−t + c2 e2t − c3 e2t
Consequently, x1 (t) = −c1 e−t + c3 e2t , x2 (t) = c1 e−t + c2 e2t , and −c1 e−t + (c2 − c3 )e2t .
25. A2 = (SDS −1 )(SDS −1 ) = SD(S −1 S)DS −1 = SDIn DS −1 = SD2 S −1 .
We now use mathematical induction to establish the general result. Suppose that for k = m > 2 that
Am = SDm S −1 . Then
Am+1 = AAm = (SDS −1 )(SDm S −1 = SD(S −1 S)Dm S −1 ) = SDm+1 S −1 .
It follows by mathematical induction that Ak = SDk S −1 for k = 1, 2, . . .
26. Let A = diag(a1 , a2 , . . . , an ) and let B = diag(b1 , b2 , . . . , bn ). Then from the index form of the matrix
product,
n
,
0,
if i = j,
(AB)ij =
aik bkj = aii bij =
if i = j.
a i bi ,
k=1
Consequently, AB = diag(a1 b1 , a2 b2 , . . . , an bn ). Applying this result to the matrix D = diag(λ1 , λ2 , . . . , λk ),
it follows directly that Dk = diag(λk1 , λk2 , . . . , λkn ).
27. The matrix A has eigenvalues
λ1 = 5, λ
2 = −1, with corresponding eigenvectors v1 = (1, −3) and
1
2
, then S −1 AS = D, where D = diag(5, −1).
v2 = (2, −3). Thus, if we set S =
−3 −3
Equivalently, A = SDS −1 . It follows from the results of the previous two examples that
−127 −84
1
2
125
0
−1 − 23
3
3 −1
A = SD S =
=
,
1
378 251
−3 −3
0 −1
1
3
whereas
5
5
A = SD S
−1
=
1
2
−3 −3
3125
0
0
−1
−1
1
− 23
1
3
=
−3127 −2084
9378
6251
.
28.
(a). This is self-evident from matrix multiplication. Another perspective on this is that when we multiply
a matrix B on the left by a diagonal matrix√D, the
√ ith√diagonal element
√ ith row of B gets multiplied by the
of D. Thus, if we multiply the
ith
row
of
D,
λ
,
by
the
ith
diagonal
element
of
D,
i
√ √
√ λi , the result in
√ √
the ith row of the product is λi λi = λi . Therefore, D D = D, which means that D is a square root
of D.
(b). We have
√
√
√
√ √
(S DS −1 )2 = (S DS −1 )(S DS −1 ) = S( DI D)S −1 = SDS −1 = A,
as required.
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(c). We begin by diagonalizing A. We have
det(A − λI) = det
6−λ
−3
−2
7−λ
= (6 − λ)(7 − λ) − 6 = λ2 − 13λ + 36 = (λ − 4)(λ − 9),
so the eigenvalues of A are λ = 4 and λ = 9. An eigenvector of A corresponding to λ = 4 is
2
. Thus, we can form
eigenvector of A corresponding to λ = 9 is
−3
S=
√
We take D =
2 0
0 3
root of A is given by
1
1
2
−3
and
D=
4
0
. A fast computation shows that S
√
√
A = S DS −1 =
−1
12/5
−3/5
=
0
9
, and an
.
3/5
2/5
1/5 −1/5
−2/5
13/5
1
1
. By part (b), one square
.
Directly squaring this result confirms this matrix as a square root of A.
29. (a). Show: A ∼ A.
The identity matrix, I, is invertible, and I = I −1 . Since IA = AI =⇒ A = I −1 AI, it follows that A ∼ A.
(b). Show: A ∼ B ⇐⇒ B ∼ A.
A ∼ B =⇒ there exists an invertible matrix S such that B = S −1 AS =⇒ A = SBS −1 = (S −1 )−1 BS −1 .
But S −1 is invertible since S is invertible. Consequently, B ∼ A.
(c). Show A ∼ B and B ∼ C ⇐⇒ A ∼ C.
A ∼ B =⇒ there exists an invertible matrix S such that B = S −1 AS; moreover, B ∼ C =⇒ there exists an
invertible matrix P such that C = P −1 BP . Thus,
C = P −1 BP = P −1 (S −1 AS)P = (P −1 S −1 )A(SP ) = (SP )−1 A(SP ) where SP is invertible. Therefore
A ∼ C.
30. Let A ∼ B mean that A is similar to B. Therefore, we must show: A ∼ B ⇐⇒ AT ∼ B T .
A ∼ B =⇒ there exists an invertible matrix S such that B = S −1 AS
=⇒ B T = (S −1 AS)T = S T AT (S −1 )T = S T AT (S T )−1 . S T is invertible because S is invertible, [since
det(S) = det(S T )]. Thus, AT ∼ B T .
31. We are given that Av = λv and B = S −1 AS.
B(S −1 v) = (S −1 AS)(S −1 v) = S −1 A(SS −1 )v = S −1 AIv = S −1 Av = S −1 (λv) = λ(S −1 v).
Hence, S −1 v is an eigenvector of B corresponding to the eigenvalue λ.
32. (a). S −1 AS = diag(λ1 , λ2 , . . . , λn ) =⇒ det(S −1 AS) = λ1 λ2 · · · λn
=⇒ det(A) det(S −1 ) det(S) = λ1 λ2 · · · λn =⇒ det(A) = λ1 λ2 · · · λn . Since all eigenvalues are nonzero, it
follows that det(A) = 0. Consequently, A is invertible.
−1
(b). S −1 AS = diag(λ1 , λ2 , .
. . , λn ) =⇒ [S −1 AS]
= [diag(λ1 , λ2 , . . . , λn )]−1
1
1
1
=⇒ S −1 A−1 (S −1 )−1 = diag
, ,...,
λ1 λ 2 λ n
1 1
1
.
, ,...,
=⇒ S −1 A−1 S = diag
λ1 λ2
λn
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33. (a). S −1 AS = diag(λ1 , λ2 , . . . , λn ) =⇒ (S −1 AS)T = [diag(λ1 , λ2 , . . . , λn )]T
=⇒ S T AT (S −1 )T = diag(λ1 , λ2 , . . . , λn ) =⇒ S T AT (S T )−1 = diag(λ1 , λ2 , . . . , λn ). Since we have that
Q = (ST )−1 , this implies that Q−1 AT Q = diag(λ1 , λ2 , . . . , λn ).
(b). Let MC = [v1 , v2 , v3 , . . . , vn ] where MC denotes the matrix of cofactors of S. We see from part (a)
that AT is nondefective, which means it possesses a complete set of eigenvectors. Also from part (a),
Q−1 AT Q = diag(λ1 , λ2 , . . . , λn ) where Q = (S T )−1 , so
AT Q = Q diag(λ1 , λ2 , . . . , λn )
(33.1)
If we let MC denote the matrix of cofactors of S, then
T
MC
adj(S)
MCT
MC
MCT
=⇒ Q =
S −1 =
=⇒ (S T )−1 =
=
=⇒ (S −1 )T =
.
det(S)
det(S)
det(S)
det(S)
det(S)
Substituting this result into Equation (33.1), we obtain
MC
MC
AT
=
diag(λ1 , λ2 , . . . , λn )
det(S)
det(S)
T
=⇒ A MC = MC diag(λ1 , λ2 , . . . , λn )
=⇒ AT [v1 , v2 , v3 , . . . , vn ] = [v1 , v2 , v3 , . . . , vn ] diag(λ1 , λ2 , . . . , λn )
=⇒ [AT v1 , AT v2 , AT v3 , . . . , AT vn ] = [λ1 v1 , λ2 v2 , λ3 v3 , . . . , λn vn ]
=⇒ AT vi = λvi for i ∈ {1, 2, 3, . . . , n}.
Hence, the column vectors of MC are linearly independent eigenvectors of AT .
−2 − λ
4 = 0 ⇐⇒ (λ + 3)(λ − 2) = 0 ⇐⇒ λ1 = −3 or λ2 = 2.
34. det(A − λI) = 0 ⇐⇒ 1
1−λ If λ1 = −3 the corresponding eigenvectors are of the form v1 = r(−4, 1) where r ∈ R.
If λ2 = 2 the corresponding eigenvectors are of the form v2 = s(1, 1) where s ∈R.
−4 1
Thus, a complete set of eigenvectors is {(−4, 1), (1, 1)} so that S =
. If MC denotes the matrix
1 1
1 −1
. Consequently, from Problem 33, (1, −1) is an eigenvector
of cofactors of S, then MC =
−1 −4
corresponding to λ = −3 and (−1, −4) is an eigenvector corresponding to λ = 2 for the matrix AT .
λ 1
−1
35. S AS = Jλ ⇐⇒ AS = SJλ ⇐⇒ A[v1 , v2 ] = [v1 , v2 ]
0 λ
=⇒ [Av1 , Av2 ] = [λv1 , v1 + λv2 ] =⇒ Av1 = λv1 and Av2 = v1 + λv2
=⇒ (A − λI)v1 = 0 and (A − λI)v2 = v1 .
2−λ
1 36. det(A − λI) = 0 ⇐⇒ = 0 ⇐⇒ (λ − 3)2 = 0 ⇐⇒ λ1 = 3 of multiplicity two.
−1
4−λ If λ1 = 3 the corresponding eigenvectors are of the v1 = r(1, 1) where r ∈ R. Consequently, A does not have
a complete set of eigenvectors, so itis a defective
matrix.
3 1
By the preceding problem, J3 =
is similar to A. Hence, there exists S = [v1 , v2 ] such that
0 3
−1
S AS = J3 . From the first part of the problem, we can let v1 = (1, 1). Now consider (A − λI)v2 = v1
where v1 = (a, b) for a, b ∈ R. Upon substituting, we obtain
−1 1
a
1
=
=⇒ −a + b = 1.
−1 1
b
1
1 b−1
1 −1
Thus, S takes the form S =
where b ∈ R, and if b = 0, then S =
.
1
b
1
0
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⎡
37.
λ
S −1 AS = ⎣ 0
0
1
λ
0
⎤
⎡
0
λ
1 ⎦ ⇐⇒ A[v1 , v2 , v3 ] = [v1 , v2 , v3 ] ⎣ 0
λ
0
1
λ
0
⎤
0
1 ⎦
λ
⇐⇒ [Av1 , Av2 , Av3 ] = [λv1 , v1 + λv2 , v2 + λv3 ]
⇐⇒ Av1 = λv1 , Av2 = v1 + λv2 , and Av3 = v2 + λv3
⇐⇒ (A − λI)v1 = 0, (A − λI)v2 = v1 , and (A − λI)v3 = v2 .
⎡
⎤
n
n
,
,
⎣ci
38. (a). From (7.3.15), c1 f1 + c2 f2 + · · · + cn fn = 0 ⇐⇒
sji ej ⎦ = 0
⇐⇒
n
n
,
,
j=1
i=1
sji ci ej = 0 ⇐⇒
i=1
n
,
j=1
sji ci = 0, j = 1, 2, . . . , n, since {ei } is a linearly independent set. The
i=1
latter equation is just the component form of the linear system Sc = 0. Since {fi } is a linearly independent
set, the only solution to this system is the trivial solution c = 0, so det(S) = 0. Consequently, S is invertible.
(b). From (7.3.14) and (7.3.15), we have
T (fk ) =
n
,
bik
i=1
n
,
sji ej =
j=1
n
n
,
,
j=1
sji bik ej , k = 1, 2, . . . , n.
i=1
Replacing i with j and j with i yields
⎡
⎤
n
n
,
,
⎣
T (fk ) =
sij bjk ⎦ ei , k = 1, 2, . . . , n.
i=1
(∗)
j=1
(c). From (7.3.15) and (7.3.13), we have
T (fk ) =
n
,
sjk T (ej ) =
j=1
that is,
n
,
j=1
sjk
n
,
aij ei ,
i=1
⎡
⎤
n
n
,
,
⎣
T (fk ) =
aij sjk ⎦ ei , k = 1, 2, . . . , n.
i=1
(∗∗)
j=1
(d). Subtracting (∗) from (∗∗) yields
⎡
⎤
n
n
,
,
⎣ (sij bjk − aij sjk )⎦ ei = 0.
i=1
j=1
Thus, since {e1 } is a linearly independent set,
n
,
j=1
sij bjk =
n
,
aij sjk , i = 1, 2, . . . , n.
j=1
But this is just the index form of the matrix equation SB = AS. Multiplying both sides of the preceding
equation on the left by S −1 yields B = S −1 AS.
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Solutions to Section 7.4
True-False Review:
(a): TRUE. In the definition of the matrix exponential function eAt , we see that powers of the matrix A
must be computed:
(At)2
(At)3
eAt = In + (At) +
+
+ ....
2!
3!
In order to do this, A must be a square matrix.
(b): TRUE. We see this by plugging in t = 1 into the definition of the matrix exponential function. All
terms containing A3 , A4 , A5 , . . . must be zero, leaving us with the result given in this statement.
(c): FALSE. The inverse of the matrix exponential function eAt is the matrix exponential function e−At ,
and this will exist for all square matrices A, not just invertible ones.
(d): TRUE. The matrix exponential function eAt converges to a matrix the same size as A, for all t ∈ R.
This is asserted, but not proven, directly beneath Definition 7.4.1.
(e): FALSE. The correct statement is
(SDS −1 )k = SDk S −1 .
The matrices S and S −1 on the right-hand side of this equation do not get raised to the power k.
(f ): FALSE. According to Property 1 of the Matrix Exponential Function, we have
(eAt )2 = (eAt )(eAt ) = e2At .
Problems:
1. We have
det(A − λI) = 0 ⇐⇒
1−λ
0
2
= 0 ⇐⇒ λ2 − 4λ + 3 = 0 ⇐⇒ λ = 1 or λ = 3.
3−λ
1
.
0
−2 2
1
Eigenvalue λ = 3: We have A − 3I =
, so we can choose the eigenvector v2 =
.
0 0
1
1 1
1 0
We form the matrices S =
and D =
. From Theorem 7.4.3, we have
0 1
0 3
t
t
0
1 −1
e e3t − et
1 1
e
=
.
eAt = SeDt S −1 =
0 e3t
0
1
0
e3t
0 1
Eigenvalue λ = 1: We have A − I =
0
0
2
2
, so we can choose the eigenvector v1 =
2. We have
det(A − λI) = 0 ⇐⇒
3−λ
1
Eigenvalue λ = 4: We have A − 4I =
1
= 0 ⇐⇒ λ2 − 6λ + 8 = 0 ⇐⇒ λ = 2 or λ = 4.
3−λ
−1
1
1 −1
, so we can choose the eigenvector v1 =
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.
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1 1
1
, so we can choose the eigenvector v2 =
.
1 1
−1
1
4 0
and D =
. From Theorem 7.4.3, we have
−1
0 2
Eigenvalue λ = 2: We have A − 2I =
We form the matrices S =
1
1
e4t
0
0
e2t
det(A − λI) = 0 ⇐⇒
−λ
−2
eAt = SeDt S −1 =
1
1
1 −1
3. We have
−2i
−2
Eigenvalue λ = 2i: We have A − 2iI =
1/2
1/2
1/2 −1/2
=
1 4t
2t
2 (e + e
1 4t
2t
2 (e − e )
1 4t
2t
2 (e − e )
1 4t
2t
2 (e + e )
.
2
= 0 ⇐⇒ λ2 + 4 = 0 ⇐⇒ λ = ±2i.
−λ
2
−2i
1
i
, so we can choose the eigenvector v1 =
.
Eigenvalue λ = −2i: By taking the conjugate of the eigenvector obtained above, we can choose v2 =
We form the matrices S =
e
At
= Se
Dt
S
−1
=
1
i
1
−i
1
i
1
−i
2i
0
and D =
e2it
0
0
e−2it
0
−2i
1
−i
.
. From Theorem 7.4.3, we have
1/2
1/2
−i/2
i/2
=
cos 2t sin 2t
− sin 2t cos 2t
.
The simplification in the last step uses the well-known identities cos x = 12 (eix + e−ix ) and sin x = − 2i (eix −
e−ix ).
4. We have
det(A − λI) = 0 ⇐⇒
−1 − λ
−3
3
= 0 ⇐⇒ λ2 + 2λ + 10 = 0 ⇐⇒ λ = −1 ± 3i.
−1 − λ
Eigenvalue λ = −1 + 3i: We have A − (−1 + 3i)I =
1
.
i
−3i
3
−3 −3i
, so we can choose the eigenvector v1 =
Eigenvalue λ = −1 − 3i: By taking the conjugate of the eigenvector obtained above, we can choose v2 =
1
.
−i
1 1
−1 + 3i
0
We form the matrices S =
and D =
. From Theorem 7.4.3, we have
i −i
0
−1 − 3i
e
At
= Se
Dt
S
−1
=
1
i
1
−i
e(−1+3i)t
0
0
e(−1−3i)t
1/2 −i/2
1/2
i/2
=e
−t
cos 3t sin 3t
− sin 3t cos 3t
.
The simplification in the last step uses the well-known identities cos x = 12 (eix + e−ix ) and sin x = − 2i (eix −
e−ix ).
Alternative Solution: Apply Problem 5 below.
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5. We have
det(A − λI) = 0 ⇐⇒
a−λ
−b
b
= 0 ⇐⇒ λ2 − 2aλ + a2 + b2 = 0 ⇐⇒ λ = a ± bi.
a−λ
Eigenvalue λ = a + bi: We have A − (a + bi)I =
−bi
−b
b
−bi
, so we can choose the eigenvector v1 =
1
i
.
Eigenvalue λ = a − bi: By taking the conjugate of the eigenvector obtained above, we can choose v2 =
1
.
−i
1 1
a + bi
0
We form the matrices S =
and D =
. From Theorem 7.4.3, we have
i −i
0
a − bi
(a+bi)t
0
1/2 −i/2
cos bt sin bt
1
1
e
At
Dt −1
at
=e
.
e = Se S =
1/2
i/2
− sin bt cos bt
i −i
0
e(a−bi)t
The simplification in the last step uses the well-known identities cos x = 21 (eix + e−ix ) and sin x = − 2i (eix −
e−ix ).
6. We have
det(A − λI) = 0 ⇐⇒
3−λ
1
0
−2
−λ
0
−2
−2 = 0 ⇐⇒ (3 − λ)(λ2 − 3λ + 2) = 0 ⇐⇒ λ = 1 or λ = 2 or λ = 3.
3−λ
⎡
⎤
⎡
⎤
2 −2 −2
1
Eigenvalue λ = 1: We have A − I = ⎣ 1 −1 −2 ⎦, and we may choose the vector v1 = ⎣ 1 ⎦ in
0
0
2
0
nullspace(A − I) as an eigenvector corresponding to λ = 1.
⎤
⎡
⎤
⎡
1 −2 −2
2
Eigenvalue λ = 2: We have A − 2I = ⎣ 1 −2 −2 ⎦, and we may choose the vector v2 = ⎣ 1 ⎦ in
0
0
1
0
nullspace(A − 2I) as an eigenvector corresponding to λ = 2.
⎡
⎤
⎡
⎤
0 −2 −2
1
Eigenvalue λ = 3: We have A − 3I = ⎣ 1 −3 −2 ⎦, and we may choose the vector v3 = ⎣ 1 ⎦ in
0
0
0
−1
nullspace(A − 3I) as an eigenvector corresponding to λ = 3.
⎡
⎤
⎡
⎤
1 2
1
1 0 0
1 ⎦ and D = ⎣ 0 2 0 ⎦. From Theorem 7.4.3, we have
We form the matrices S = ⎣ 1 1
0 0 −1
0 0 3
eAt = SeDt S −1
⎤⎡
⎤
⎤⎡
⎡
−1
2
1
1 0 0
1 2
1
0 ⎦
1 ⎦ ⎣ 0 2 0 ⎦ ⎣ 1 −1
=⎣ 1 1
0 0 3
0
0 −1
0 0 −1
⎡ t t
⎤
t t
t 2t
e (2e − 1) −2e (e − 1) −e (e − 1)
−et (et − 2) −et (e2t − 1) ⎦ .
= ⎣ et (et − 1)
0
0
e3t
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7. We have
det(A − λI) = 0 ⇐⇒
6−λ
8
4
−2
−2 − λ
−2
−1
−2 = 0 ⇐⇒ (λ − 2)2 (λ − 1) = 0 ⇐⇒ λ = 2 or λ = 1.
1−λ
⎡
⎤
⎡
⎤
4 −2 −1
1
Eigenvalue λ = 2: We have A − 2I = ⎣ 8 −4 −2 ⎦, and we may choose two vectors, v1 = ⎣ 2 ⎦ and
4 −2 −1
0
⎡
⎤
1
v2 = ⎣ 0 ⎦ in nullspace(A − 2I) as an eigenvector corresponding to λ = 2.
4
⎡
⎤
5 −2 −1
Eigenvalue λ = 1: We have A − I = ⎣ 8 −3 −2 ⎦. The reduced row-echelon form of this matrix is
4 −2
0
⎡
⎤
⎡
⎤
1
1 0 −1
⎣ 0 1 −2 ⎦, so we make choose v3 = ⎣ 2 ⎦ in nullspace(A − I) as an eigenvector corresponding to
1
0 0
0
λ = 1.
⎡
1
We form the matrices S = ⎣ 2
0
1
0
4
⎤
⎡
1
2
2 ⎦ and D = ⎣ 0
1
0
0
2
0
⎤
0
0 ⎦. From Theorem 7.4.3, we have
1
eAt = SeDt S −1
⎤⎡
⎤
⎡
⎤ ⎡ 2t
0
0
4 −3/2 −1
1 1 1
e
0 ⎦
= ⎣ 2 0 2 ⎦ ⎣ 0 e2t 0 ⎦ ⎣ 1 −1/2
t
−4
2
1
0
0 e
0 4 1
⎤
⎡ t t
t
t
t
t
e (5e − 4) 2e (1 − e ) e (1 − e )
= ⎣ 8et (et − 1) et (4 − 3et ) 2et (1 − et ) ⎦ .
4et (et − 1) 2et (1 − et )
et
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8. Note that
1
1
eAt = I + At + (At)2 + · · · + (At)k + . . .
2!
k!
⎤ ⎡
⎡
1 0 ... 0
d1 t 0 . . .
⎢ 0 1 . . . 0 ⎥ ⎢ 0 d2 t . . .
⎢
⎥ ⎢
=⎢ . . .
+⎢ .
..
..
. . ... ⎥
⎣ .. ..
⎦ ⎣ ..
.
.
0 0 ... 1
⎡ ∞ 1
k
k=0 k! (d1 t)
⎢
0
⎢
=⎢
..
⎣
.
0
...
0
∞
..
.
0
⎤
ed1 t
⎢ 0
⎢
=⎢ .
⎣ ..
0
ed2 t
..
.
...
...
..
.
0
0
..
.
0
0
...
edn t
⎡
(d1 t)k
⎥
0
1 ⎢
⎥
⎢
⎥ + ··· + ⎢
..
k! ⎣
⎦
.
dn t
...
...
..
.
...
1
k
k=0 k! (d2 t)
0
⎡
0
⎤
0
0
..
.
∞
...
...
..
.
0
0
..
.
0
...
(dn t)k
0
⎤
0
0
..
.
0
(d2 t)k
..
.
⎥
⎥
⎥
⎦
1
k
k=1 k! (dn t)
⎥
⎥
⎥
⎦
= diag(ed1 t , ed2 t , . . . , edn t ).
9. Using Problem 8, we have
e−3t
0
0
e5t
eλt
⎢ 0
⎢
eλIn t = ⎢ .
⎣ ..
0
eλt
..
.
...
...
..
.
0
0
..
.
0
0
...
eλt
0
0
b
0
eAt =
10. We have
⎡
and
e−At =
⎤
⎡
e3t
0
0
.
e−5t
1
⎥
⎢ 0
⎥
⎢
⎥ = eλt ⎢ ..
⎦
⎣ .
0 ...
1 ...
.. . .
.
.
0 0 ...
⎤
0
0 ⎥
⎥
λt
.. ⎥ = e In .
. ⎦
1
11.
(a). We have
BC =
2
(b). We have C =
0
0
b
0
a
0
0
a
0
0
b
0
=
e
Ct
= I + Ct =
=
0
0
0
0
1
0
0
1
0
0
ab
0
=
0
0
b
0
a
0
0
a
= CB.
= 02 . Therefore,
+
0
0
b
0
t=
1
0
bt
1
(c). Using the results in the first two parts of this problem, it follows that
at
at
0
1 bt
e
e
eAt = e(B+C)t = eBt eCt =
=
0 eat
0 1
0
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btebt
eat
.
⎤
⎥
⎥
⎥
⎦
482
a 0
0 b
0 ab
and C =
, so that A = B + C. Now BC =
= CB, so
0 a
−b 0
−ab 0
Property (1) of the matrix exponential function implies that eAt = e(B+C)t = eBt eCt . Now
at
0
1 0
e
Bt
at
=e
,
e =
0 eat
0 1
12. Define B =
by Problem 8. To determine eCt , observe first that
4
0
−b2
0 b3
b
2
3
4
, C =
, C =
C =
0 −b2
0
−b3 0
In general, we have
C
2n
=b
2n
(−1)n
0
0
(−1)n
and
C
2n+1
=b
2n+1
for each integer n ≥ 0. Therefore, from Definition 7.4.1, we have
∞ (−1)n 2n+1
∞ (−1)n 2n
n=0 (2n)! b
n=0 (2n+1)! b
Ct
e =
∞ (−1)n 2n+1
∞ (−1)n 2n
− n=0 (2n+1)! b
n=0 (2n)! b
Thus,
eAt = eBt eCt = eat
cos bt
− sin bt
0
b4
0
(−1)n+1
(−1)n
0
−b5
0
, C =
cos bt sin bt
− sin bt cos bt
=
sin bt
cos bt
0
b5
5
,...
.
.
13. Direct computation shows that A2 = 02 . Therefore,
1 0
1
1
1+t
At
+
t=
e = I + At =
0 1
−1 −1
−t
t
1−t
.
14. Direct computation shows that A2 = 02 . Therefore,
1 0
−3 9
−3t + 1
9t
+
t=
.
eAt = I + At =
0 1
−1 3
−t
3t + 1
⎡
⎤
−4
8 −4
2 −1 ⎦ and A3 = 03 . Therefore,
15. Direct computation shows that A2 = ⎣ −1
2 −4
2
⎡
⎤ ⎡
⎤
⎡
⎤
1 0 0
−1 −6 −5
−4
8 −4
t2
2 −1 ⎦
eAt = ⎣ 0 1 0 ⎦ + ⎣ 0 −2 −1 ⎦ t + ⎣ −1
2
0 0 1
1
2
3
2 −4
2
⎤
⎡
1 − t − 2t2 −6t + 4t2 −5t − 2t2
−t2 /2
1 − 2t + t2 −t − t2 /2 ⎦ .
=⎣
2
t+t
2t − 2t2
1 + 3t + t2
⎡
⎤
0 0 0
16. Direct computation shows that A2 = ⎣ 0 0 0 ⎦ and A3 = 03 . Therefore,
1 0 0
⎤ ⎡
⎤
⎤
⎡
⎡
⎤
⎡
1 0 0
0 0 0
0 0 0
1
0 0
2
t
1 0 ⎦.
eAt = ⎣ 0 1 0 ⎦ + ⎣ 1 0 0 ⎦ t + ⎣ 0 0 0 ⎦ = ⎣ t
2
2
0 0 1
0 1 0
1 0 0
t /2 t 1
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⎡
0
⎢ 0
2
⎢
17. Direct computation shows that A = ⎣
0
0
0
0
0
0
1
0
0
0
⎤
⎡
0
0
⎢
1 ⎥
⎥ , A3 = ⎢ 0
⎣ 0
0 ⎦
0
0
0
0
0
0
0
0
0
0
⎤
1
0 ⎥
⎥, and A4 = 04 . Therefore,
0 ⎦
0
(At)2
(At)3
eAt = I4 + At +
+
2 ⎤ ⎡6
⎤ ⎡
⎤
⎤ ⎡
⎡
0 0 t2 /2
0 t 0 0
0 0 0 t3 /6
0
1 0 0 0
⎢
⎢ 0 1 0 0 ⎥ ⎢ 0 0 t 0 ⎥ ⎢ 0 0
0
t2 /2 ⎥
0 ⎥
⎥+⎢
⎥
⎥+⎢
⎥+⎢ 0 0 0
=⎢
⎣ 0 0 1 0 ⎦ ⎣ 0 0 0 t ⎦ ⎣ 0 0
0 ⎦
0
0 ⎦ ⎣ 0 0 0
0 0 0 0
0 0 0
0
0 0 0 0
0 0
0
0
⎡
⎤
2
3
1 t t /2 t /6
⎢ 0 1
t
t2 /2 ⎥
⎥.
=⎢
⎣ 0 0
1
t ⎦
0 0
0
1
18. It is easy to see (and can be formally verified by induction on k) that the matrix Ak contains all zero
entries except for the (k + i, i) entry for 1 ≤ i ≤ n − k. In particular, An = An+1 = An+2 = · · · = 0n .
Therefore, by Definition 7.4.1, we have
eAt = In + At +
⎡
1
⎢
t
⎢
⎢ t2
⎢ 2!3
=⎢
⎢ t3!
⎢
..
⎢
.
⎣
tn−1
(n−1)!
(At)2
(At)3
(At)n−1
+
+ ··· +
2!
3!
(n − 1)!
⎤
0
0
0
... 0
1
0
0
... 0 ⎥
⎥
t
1
0
... 0 ⎥
⎥
⎥.
t2
t
1
.
.
.
0
⎥
2!
⎥
..
..
..
.
..
.
. . ⎥
.
.
.
⎦
tn−2
tn−3
tn−4
.
.
.
1
(n−2)!
(n−3)!
(n−4)!
19. Assume that A0 is m × m and B0 is n × n. Note that
two matrices must be square in order for
these
Ak0 0
A0 t
B0 t
k
. Therefore,
and e
to make sense. The key point is that A =
e
0 B0k
(At)3
(At)2
eAt = I + At +
+
+ ...
2! 3!
2
2 3
3
t
t
A0 0
A0 0
Im 0
A0 0
+
t+
=
+
+ ...
3
0 In
0 B0
0 B02
0
B
2!
3!
0
2
3
0 t)
0 t)
+ (A3!
+ ...
0
Im + A0 t + (A2!
=
2
3
0 t)
0 t)
0
In + B0 t + (B2!
+ (B3!
+ ...
At
0
0
e
.
=
0
e B0 t
Solutions to Section 7.5
True-False Review:
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(a): TRUE. This is immediate from Definition 7.5.1.
(b): TRUE. The roots of p(λ) = λ3 + λ are λ = 0 and λ = ±i. Therefore, the matrix has complex
eigenvalues. But Theorem 7.5.4 indicates that a real matrix that is symmetric must have only real eigenvalues.
(c): FALSE. The zero matrix is a real, symmetric matrix for which both v1 and v2 are eigenvectors.
(d): TRUE. This is the statement of the Principal Axes Theorem (Theorem 7.5.6).
(e): TRUE. This is a direct application of Lemma 7.5.9.
(f ): TRUE. Assuming that A and B are orthogonal matrices, then A and B are invertible, and hence AB
is invertible. Moreover,
(AB)−1 = B −1 A−1 = B T AT = (AB)T ,
so that AB meets the requirements of orthogonality spelled out in Definition 7.5.1.
(g): TRUE. This is essentially the content of Theorem 7.5.3, since a set of orthogonal unit vectors are
precisely a set of orthonormal vectors.
(h): TRUE. Let S be the matrix consisting of a complete set of orthonormal eigenvectors of A. Then
S T AS = diag(λ1 , λ2 , . . . , λn ) = D, where the λi are the corresponding eigenvalues. Then S is an orthogonal
matrix: S T = S −1 . Hence, A = SDS T and AT = (SDS T )T = SDT S T = SDS T = A, so A is symmetric.
Problems:
4
= 0 ⇐⇒ λ2 + 4λ − 21 = 0 ⇐⇒ (λ + 7)(λ − 3) = 0 ⇐⇒ λ =
−5 − λ 0
8 4
v1
=
=⇒ 8v1 + 4v2 = 0. Therefore,
If λ = −7 then (A − λI)v = 0 assumes the form
v2
0
4 2
2
v1
1
is a unit
= −√ , √
v1 = (−1, 2) is an eigenvector of A corresponding to λ = −7, and w1 =
||v1 ||
5
5
eigenvector corresponding to λ = −7.
0
−2
4
v1
=
=⇒ −2v1 + 4v2 = 0.
If λ = 3 then (A − λI)v = 0 assumes the form
v2
0
4 −8
1
v2
2
is a
= √ ,√
Therefore, v2 = (2, 1) is an eigenvector of A corresponding to λ = 3, and w2 =
||v2 ||
5
5
⎡
⎤
1
2
√
−√
⎢
5
5 ⎥
T
unit eigenvector corresponding to λ = 3. Thus, S = ⎣
2
1 ⎦ and S AS = diag(−7, 3).
√
√
5
5
2−λ
2
= 0 ⇐⇒ λ2 − λ − 6 = 0 ⇐⇒ (λ + 2)(λ − 3) = 0 ⇐⇒ λ = −2 or
2. det(A − λI) = 0 ⇐⇒ 2
−1 − λ λ = 3.
0
4 2
v1
=
=⇒ 2v1 + v2 = 0. v1 = (1, −2), is
If λ = −2 then (A − λI)v = 0 assumes the form
v
0
2 1
2
2
v1
1
is a unit eigenvector corresponding
an eigenvector corresponding to λ = −2 and w1 =
= √ , −√
||v1 ||
5
5
to λ = −2.
0
−1
2
v1
If λ = 3 then (A − λI)v = 0 assumes the form
=
=⇒ v1 + 2v2 = 0. v2 = (2, 1),
2 −4 v2
0
v2
2
1
is a unit eigenvector corresponding
= √ ,√
is an eigenvector corresponding to λ = 3 and w2 =
||v2 ||
5
5
1−λ
1. det(A − λI) = 0 ⇐⇒ 4
−7 or λ = 3.
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to λ = 3. ⎡
2 ⎤
√
5 ⎥
T
1 ⎦ and S AS = diag(−2, 3).
√
5
4−λ
6 = 0 ⇐⇒ λ2 − 13λ = 0 ⇐⇒ λ(λ − 13) = 0 ⇐⇒ λ = 0 or λ = 13.
3. det(A − λI) = 0 ⇐⇒ 6
9−λ 4 6
v1
0
=
=⇒ 2v1 + 3v2 = 0. v1 = (−3, 2), is
If λ = 0 then (A − λI)v = 0 assumes the form
0 6 9 v2
v1
3
2
is a unit eigenvector corresponding
= −√ , √
an eigenvector corresponding to λ = 0 and w1 =
||v1 ||
13
13
to λ = 0.
0
−9
6
v1
If λ = 13 then (A − λI)v = 0 assumes the form
=
=⇒ −3v1 + 2v2 = 0.
v2
0
6 −4
v2
2
3
is a unit eigenvector
= √ ,√
v2 = (2, 3), is an eigenvector corresponding to λ = 13 and w2 =
||v2 ||
13
13
corresponding
⎡ to 3λ = 13.2 ⎤
√
−√
⎢
13
13 ⎥
T
Thus, S = ⎣
2
3 ⎦ and S AS = diag(0, 13).
√
√
13
13
1−λ
2 = 0 ⇐⇒ (λ − 1)2 − 4 = 0 ⇐⇒ (λ + 1)(λ − 3) = 0 ⇐⇒ λ = −1 or
4. det(A − λI) = 0 ⇐⇒ 2
1−λ λ = 3.
0
2 2
v1
=
=⇒ v1 + v2 = 0. v1 = (−1, 1), is
If λ = −1 then (A − λI)v = 0 assumes the form
0
v
2 2
2
v1
1
1
is a unit eigenvector corresponding
= −√ , √
an eigenvector corresponding to λ = −1 and w1 =
||v1 ||
2
2
to λ = −1.
0
−2
2
v1
If λ = 3 then (A − λI)v = 0 assumes the form
=
=⇒ v1 − v2 = 0. v2 = (1, 1), is
v2 0
2 −2
1
v2
1
an eigenvector corresponding to λ = 3 and w2 =
is a unit eigenvector corresponding to
= √ ,√
||v2 ||
2
2
λ = 3.
⎡
1 ⎤
1
√
−√
⎢
2
2 ⎥
T
Thus, S = ⎣
1 ⎦ and S AS = diag(−1, 3).
1
√
√
2
2
−λ
0
3 0 = 0 ⇐⇒ (λ + 2)(λ − 3)(λ + 3) = 0 ⇐⇒ λ = −3, λ = −2, or
5. det(A − λI) = 0 ⇐⇒ 0 −2 − λ
3
0
−λ λ = 3.
⎤ ⎡
⎡
⎤⎡
⎤
3 0 3
v1
0
If λ = −3 then (A − λI)v = 0 assumes the form ⎣ 0 1 0 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ v1 + v3 = 0 and v2 = 0.
3 0 3
v3
0
v1
1
1
is a unit
v1 = (−1, 0, 1), is an eigenvector corresponding to λ = −3 and w1 =
= − √ , 0, √
||v1 ||
2
2
eigenvector corresponding to λ = −3.
1
√
⎢
5
Thus, S = ⎣
2
−√
5
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⎤ ⎡
⎤⎡
⎤
3
v1
0
0 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ v1 = v3 = 0 and v2 ∈ R.
2
v3
0
v2
v2 = (0, 1, 0), is an eigenvector corresponding to λ = −2 and w2 =
= (0, 1, 0) is a unit eigenvector
||v2 ||
corresponding to λ = −2.
⎤
⎤
⎡
⎤⎡
⎡
−3
0
3
v1
0
0 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ v1 − v3 = 0 and
If λ = 3 then (A − λI)v = 0 assumes the form ⎣ 0 −5
0
3
0 −3
v3
v3
1
1
is a unit
= √ , 0, √
v2 = 0. v3 = (1, 0, 1), is an eigenvector corresponding to λ = 3 and w3 =
||v3 ||
2
2
eigenvector⎡corresponding to λ
= 3.
1
1 ⎤
−√
0 √
⎢
2
2 ⎥
⎢
⎥
0 1
0 ⎥ and S T AS = diag(−3, −2, 3).
Thus, S = ⎢
⎣
1
1 ⎦
√
0 √
2
2
1−λ
2
1 4−λ
2 = 0 ⇐⇒ λ3 − 6λ2 = 0 ⇐⇒ λ2 (λ − 6) = 0 ⇐⇒ λ = 6 or
6. det(A − λI) = 0 ⇐⇒ 2
1
2
1−λ λ = 0 of multiplicity two.
⎤
⎡
⎤⎡
⎡
⎤
1 2 1
v1
0
If λ = 0 then (A − λI)v = 0 assumes the form ⎣ 2 4 2 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ v1 + 2v2 + v3 = 0.
1 2 1
v3
0
v1 = (−1, 0, 1) and v2 = (−2, 1, 0) are linearly independent eigenvectors corresponding to λ = 0. v1 and v2
are not orthogonal since v1 , v2 = 2 = 0, so we will use the Gram-Schmidt procedure.
Let u1 = v1 = (−1, 0, 1), so
⎡
2
If λ = −2 then (A − λI)v = 0 assumes the form ⎣ 0
3
0
0
0
v2 , u1 2
u1 = (−2, 1, 0) − (−1, 0, 1) = (−1, 1, −1).
||u1 ||2
2
1
1
1
u1
u2
1
1
and w2 =
are orthonormal eigenvectors
Now w1 =
= − √ , 0, √
= −√ , √ , √
||u1 ||
||u2 ||
2
2
3
3
3
corresponding to λ = 0.
⎤
⎡
⎤⎡
⎡
⎤
−5
2
1
v1
0
2 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ v1 − v3 = 0 and
If λ = 6 then (A − λI)v = 0 assumes the form ⎣ 2 −2
1
2 −5
v3
0
2
1
v3
1
is
v2 − 2v3 = 0. v3 = (1, 2, 1), is an eigenvector corresponding to λ = 6 and w3 =
= √ ,√ ,√
||v3 ||
6
6
6
a unit eigenvector
corresponding to
⎡
⎤ λ = 6.
1
1
1
√
−√
−√
⎢
2
3
6 ⎥
⎢
1
2 ⎥
⎢
√
√ ⎥
0
Thus, S = ⎢
⎥ and S T AS = diag(0, 0, 6).
⎢
3
6 ⎥
⎣
1
1 ⎦
1
√
√
−√
2
3
6
2−λ
0
0 3−λ
1 = 0 ⇐⇒ (λ − 2)2 (λ − 4) = 0 ⇐⇒ λ = 4 or λ = 2 of
7. det(A − λI) = 0 ⇐⇒ 0
0
1
3−λ u2 = v2 −
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multiplicity two.
⎤
⎤⎡
⎡
⎤
0 0 0
v1
0
If λ = 2 then (A − λI)v = 0 assumes the form ⎣ 0 1 1 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ v2 + v3 = 0 and
0 1 1
v3
0
linearly
independent
eigenvectors
corresponding
to λ = 2, and
v1 ∈ R. v1 = (1, 0, 0) and v2 = (0, −1, 1) are
v1
v2
1
1
are unit eigenvectors corresponding to λ = 2. w1
w1 =
= (1, 0, 0) and w2 =
= 0, − √ , √
||v1 ||
||v2 ||
2
2
and w2 are also orthogonal because w1 , w2 =⎡0.
⎤ ⎡
⎤⎡
⎤
−2
0
0
v1
0
1 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ v1 = 0 and
If λ = 4 then (A − λI)v = 0 assumes the form ⎣ 0 −1
0
1 −1
v3
0
v3
1
1
is a
v2 − v3 = 0. v3 = (0, 1, 1), is an eigenvector corresponding to λ = 4 and w3 =
= 0, √ , √
||v3 ||
2
2
unit eigenvector
corresponding⎤ to λ = 4.
⎡
1
0
0
1 ⎥
1
⎢
√ ⎥
⎢ 0 −√
Thus, S = ⎢
and S T AS = diag(2, 2, 4).
2
2 ⎥
⎣
⎦
1
1
√
√
0
2
2
−λ
1
0 0 = 0 ⇐⇒ (λ − 1)2 (λ + 1) = 0 ⇐⇒ λ = −1 or λ = 1 of
8. det(A − λI) = 0 ⇐⇒ 1 −λ
0
0 1−λ multiplicity two.
⎤
⎡
⎤⎡
⎡
⎤
−1
1 0
v1
0
If λ = 1 then (A − λI)v = 0 assumes the form ⎣ 1 −1 0 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ v1 − v2 = 0 and
0
0 0
v3
0
1,
0)
and
v
=
(0,
0,
1)
are
linearly
independent
eigenvectors
corresponding
to λ = 1, and
v3 ∈ R. v1 = (1,
2
v1
v2
1
1
w1 =
= √ , √ , 0 and w2 =
= (0, 0, 1) are unit eigenvectors corresponding to λ = 1. w1
||v1 ||
||v2 ||
2
2
and w2 are also orthogonal because w1 , w2 = 0.
⎤ ⎡
⎡
⎤⎡
⎤
1 1 0
v1
0
If λ = −1 then (A − λI)v = 0 assumes the form ⎣ 1 1 0 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ v1 + v2 = 0 and v3 = 0.
0 0 2
v3
0
v3
1
1
v3 = (−1, 1, 0), is an eigenvector corresponding to λ = −1 and w3 =
= − √ , √ , 0 is a unit
||v3 ||
2
2
eigenvector⎡corresponding to λ
=
−1.
1 ⎤
1
0 √
−√
⎢
2
2 ⎥
⎢
1 ⎥
and S T AS = diag(−1, 1, 1).
Thus, S = ⎢ √1
0 √ ⎥
⎦
⎣
2
2
0 1
0
1−λ
1
−1 1−λ
1 = 0 ⇐⇒ (λ − 2)2 (λ + 1) = 0 ⇐⇒ λ = 2 of multiplicity two
9. det(A − λI) = 0 ⇐⇒ 1
−1
1
1−λ λ = −1.
⎤ ⎡
⎤
⎡
⎤⎡
−1
1 −1
v1
0
1 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ v1 − v2 + v3 = 0.
If λ = 2 then (A − λI)v = 0 assumes the form ⎣ 1 −1
0
−1
1 −1
v3
⎡
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v1 = (1, 1, 0) and v2 = (−1, 0, 1) are linearly independent eigenvectors corresponding to λ = 2. v1 and v2
are not orthogonal since v1 , v2 = −1 = 0, so we will use the Gram-Schmidt procedure.
Let u1 = v1 = (1, 1, 0), so
v2 , u1 1
1 1
u2 = v2 −
u1 = (−1, 0, 1) + (1, 1, 0) = − , , 1 .
||u1 ||2
2
2 2
u1
u2
1
1
1
1
2
are orthonormal eigenvectors
Now w1 =
= √ , √ , 0 and w2 =
= −√ , √ , √
||u1 ||
||u
||
2
2
6
6
6
2
corresponding to λ = 2.
⎤
⎤
⎡
⎤⎡
⎡
2 1 −1
v1
0
1 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ v1 − v3 = 0 and
If λ = −1 then (A − λI)v = 0 assumes the form ⎣ 1 2
−1 1
2
0
v3
v3
1
1
1
v2 + v3 = 0. v3 = (1, −1, 1), is an eigenvector corresponding to λ = −1 and w3 =
= √ , −√ , √
||v3 ||
3
3
3
is a unit eigenvector
corresponding⎤ to λ = −1.
⎡
1
1
1
√
√
−√
⎢ 2
6
3 ⎥
⎢ 1
1
1 ⎥
⎢ √
√
−√ ⎥
Thus, S = ⎢
⎥ and S T AS = diag(2, 2, −1).
⎢ 2
6
3 ⎥
⎣
1 ⎦
2
√
√
0
6
3
1−λ
0
−1 1−λ
1 = 0 ⇐⇒ (1 − λ)(λ + 1)(λ − 2) = 0 ⇐⇒ λ = −1, λ = 1, or
10. det(A − λI) = 0 ⇐⇒ 0
−1
1
−λ λ = 2.
⎤ ⎡
⎡
⎤⎡
⎤
2 0 −1
v1
0
1 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ 2v1 − v3 = 0 and
If λ = −1 then (A − λI)v = 0 assumes the form ⎣ 0 2
−1 1
1
v3
0
v1
1
1
2
2v2 +v3 = 0. v1 = (1, −1, 2), is an eigenvector corresponding to λ = −1 and w1 =
= √ , −√ , √
||v1 ||
6
6
6
is a unit eigenvector corresponding to λ = −1. ⎡
⎤
⎤⎡
⎡
⎤
0 0 −1
v1
0
1 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ v1 − v2 = 0 and
If λ = 1 then (A − λI)v = 0 assumes the form ⎣ 0 0
−1 1 −1
v3
0 v2
1
1
v3 = 0. v2 = (1, 1, 0), is an eigenvector corresponding to λ = 1 and w2 =
= √ , √ , 0 is a unit
||v2 ||
2
2
eigenvector corresponding to λ = 1.
⎤
⎤
⎡
⎤⎡
⎡
−1
0 −1
v1
0
1 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ v1 + v3 = 0 and
If λ = 2 then (A − λI)v = 0 assumes the form ⎣ 0 −1
0
−1
1 −2
v3
1
1
v3
1
= −√ , √ , √
v2 − v3 = 0. v3 = (−1, 1, 1), is an eigenvector corresponding to λ = 2 and w3 =
||v3 ||
3
3
3
is a unit eigenvector
corresponding⎤ to λ = 2.
⎡
1
1
1
√
√
−√
⎢
6
2
3 ⎥
⎢
1
1 ⎥
⎢ − √1
√
√ ⎥
Thus, S = ⎢
⎥ and S T AS = diag(−1, 1, 2).
⎢
6
2
3 ⎥
⎣
2
1 ⎦
√
√
0
6
3
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3−λ
11. det(A − λI) = 0 ⇐⇒ 3
4
or λ = 8.
4 0 = 0 ⇐⇒ (λ − 3)(λ + 2)(λ − 8) = 0 ⇐⇒ λ = −2, λ = 3,
3−λ ⎤ ⎡
⎡
⎤⎡
⎤
5 3 4
v1
0
If λ = −2 then (A − λI)v = 0 assumes the form ⎣ 3 5 0 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ 4v1 + 5v3 = 0 and 4v2 −
4 0 5
v3
0
v1
1
3
4
3v3 = 0. v1 = (−5, 3, 4), is an eigenvector corresponding to λ = −2 and w1 =
= −√ , √ , √
||v1 ||
2 5 2 5 2
is a unit eigenvector corresponding to λ = −2. ⎡
⎤ ⎡
⎤⎡
⎤
0 3 4
v1
0
If λ = 3 then (A − λI)v = 0 assumes the form ⎣ 3 0 0 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ v1 = 0 and 3v2 + 4v3 = 0.
4 0 0
v3
0
v2
4 3
is a unit eigenvector
v2 = (0, −4, 3), is an eigenvector corresponding to λ = 3 and w2 =
= 0, − ,
||v2 ||
5 5
corresponding to λ = 3.
⎤ ⎡
⎡
⎤⎡
⎤
−5
3
4
v1
0
0 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ 4v1 − 5v3 = 0 and
If λ = 8 then (A − λI)v = 0 assumes the form ⎣ 3 −5
4
0 −5
v3
0
v3
1
3
4
4v2 − 3v3 = 0. v3 = (5, 3, 4), is an eigenvector corresponding to λ = 8 and w3 =
= √ , √ , √
||v3 ||
2 5 2 5 2
is a unit eigenvector
corresponding
to
λ
=
8.
⎡
⎤
1
1
−√
0 √
⎢
2
2 ⎥
⎢
3
3 ⎥
4
⎢ √
√ ⎥
−
Thus, S = ⎢
⎥ and S T AS = diag(−2, 3, 8).
5 5 2 ⎥
⎢ 5 2
⎣
3
4
4 ⎦
√
√
5 5 2
5 2
3
3−λ
0
2
= 0 ⇐⇒ (λ + 5)2 (λ − 1) = 0 ⇐⇒ λ = −5 of
2
−3 − λ ⎤
⎡
⎤⎡
⎡
⎤
−4
2
2
v1
0
2 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ v1 − v3 = 0 and
If λ = 1 then (A − λI)v = 0 assumes the form ⎣ 2 −4
2
2 −4
v3
0
v1
1
1
1
is a
v2 − v3 = 0. v1 = (1, 1, 1) is an eigenvector corresponding to λ = 1 and w1 =
= √ ,√ ,√
||v1 ||
3
3
3
unit eigenvector corresponding to λ = 1.
⎤
⎡
⎤⎡
⎡
⎤
2 2 2
v1
0
If λ = −5 then (A − λI)v = 0 assumes the form ⎣ 2 2 2 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ v1 + v2 + v3 = 0.
2 2 2
v3
0
v2 = (−1, 1, 0) and v3 = (−1, 0, 1) are linearly independent eigenvectors corresponding to λ = −5. v2 and
v3 are not orthogonal since v2 , v3 = −1 = 0, so we will use the Gram-Schmidt procedure.
Let u2 = v2 = (−1, 1, 0), so
−3 − λ
2
12. det(A − λI) = 0 ⇐⇒ 2
multiplicity two, or λ = 1.
u3 = v3 −
2
−3 − λ
2
v3 , u2 1
u2 = (−1, 0, 1) − (−1, 1, 0) =
||u2 ||2
2
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− ,− ,1 .
2 2
490
1
1
2
u2
u3
1
1
are orthonormal eigenvectors
= − √ , √ , 0 and w3 =
= −√ , −√ , √
||u2 ||
||u3 ||
2
2
6
6
6
corresponding
⎤
⎡ to λ = −5.
1
1
1
√
−√
−√
⎢ 3
2
6 ⎥
⎢ 1
1
1 ⎥
⎢ √
√
−√ ⎥
Thus, S = ⎢
⎥ and S T AS = diag(1, −5, −5).
⎢ 3
2
6 ⎥
⎣ 1
2 ⎦
√
√
0
3
6
−λ
1
1 1 = 0 ⇐⇒ (λ + 1)2 (λ − 2) = 0 ⇐⇒ λ = −1 of multiplicity two, or
13. det(A − λI) = 0 ⇐⇒ 1 −λ
1
1 −λ λ = 2.
⎤⎡
⎤
⎤
⎡
⎡
v1
1 1 1
0
If λ = −1 then (A − λI)v = 0 assumes the form ⎣ 1 1 1 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ v1 + v2 + v3 = 0.
1 1 1
0
v3
v1 = (−1, 0, 1) and v2 = (−1, 1, 0) are linearly independent eigenvectors corresponding to λ = −1. v1 and
v2 are not orthogonal since v1 , v2 = 1 = 0, so we will use the Gram-Schmidt procedure.
Let u1 = v1 = (−1, 0, 1), so
1
v2 , u1 1
1
.
u2 = v2 −
u1 = (−1, 1, 0) − (−1, 0, 1) = − , 1, −
||u1 ||2
2
2
2
u1
u2
1
1
1
2
1
√
√
√
√
√
Now w1 =
and w2 =
are orthonormal eigenvectors
= −
= −
, 0,
,
,−
||u1 ||
||u2 ||
2
2
6
6
6
corresponding to λ = −1.
⎤
⎤
⎡
⎤⎡
⎡
−2
1
1
v1
0
1 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ v1 − v3 = 0 and
If λ = 2 then (A − λI)v = 0 assumes the form ⎣ 1 −2
0
1
1 −2
v3
v3
1
1
1
is a
= √ ,√ ,√
v2 − v3 = 0. v3 = (1, 1, 1), is an eigenvector corresponding to λ = 2 and w3 =
||v3 ||
3
3
3
unit eigenvector
corresponding to ⎤λ = 2.
⎡
1
1
1
√
−√
−√
⎢
2
6
3 ⎥
⎢
2
1 ⎥
⎢
√
√ ⎥
0
Thus, S = ⎢
⎥ and S T AS = diag(−1, −1, 2).
⎢
6
3 ⎥
⎣
1
1 ⎦
1
√
√
−√
2
6
3
eigenvectors v1 = (1, 1) and v2 = (−1, 1).
14. A has eigenvalues λ1 = 4 and λ2 = −2 with corresponding
1
1
Therefore, a set of principal axes is √ (1, 1), √ (−1, 1) . Relative to these principal axes, the quadratic
2
2
form reduces to 4y12 − 2y22 .
Now w2 =
15. A has eigenvalues λ1 = 7 and λ2 = 3 with corresponding
eigenvectors v1 = (1, 1) and v2 = (1, −1).
1
1
Therefore, a set of principal axes is √ (1, 1), √ (1, −1) . Relative to these principal axes, the quadratic
2
2
form reduces to 7y12 + 3y22 .
16. A has eigenvalue λ = 2 of multiplicity two with corresponding linearly independent eigenvectors
v1 = (1, 0, −1) and v2 = (0, 1, 1). Using the Gram-Schmidt procedure, an orthogonal basis in this eigenspace
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is {u1 , u2 } where
1
1
u1 = (1, 0, −1), u2 = (0, 1, 1) + (1, 0, −1) = (1, 2, 1).
2
2
1
1
An orthonormal basis for the eigenspace is √ (1, 0, −1), √ (1, 2, 1) . The remaining eigenvalue of A is
2
6
a
set of principal axes for the given quadratic
λ = −1,with eigenvector v3 = (−1, 1, −1). Consequently,
1
1
1
form is √ (1, 0, −1), √ (1, 2, 1), √ (−1, 1, −1) . Relative to these principal axes, the quadratic form
2
6
3
reduces to 2y12 + 2y22 − y32 .
17. A has eigenvalue λ = 0 of multiplicity two with corresponding linearly independent eigenvectors
v1 = (0, 1, 0, −1) and v2 = (1, 0, −1, 0). Notice that these are orthogonal, hence we do not need to apply
the Gram-Schmidt procedure. The remaining eigenvalues of A are λ = 4 and λ = 8, with corresponding
eigenvectors v3 = (−1, 1, −1, 1) and v4 = (1, 1, 1, 1), respectively. Consequently, a set of principal axes for
the given quadratic form is
1
1
1
1
√ (0, 1, 0, −1), √ (1, 0, −1, 0), (−1, 1, −1, 1), (1, 1, 1, 1) .
2
2
2
2
Relative to these principal axes, the quadratic form reduces to 4y32 + 8y42 .
a b
18. A =
where a, b, c ∈ R.
b c
a−λ
c Consider det(A − λI) = 0 ⇐⇒ = 0 ⇐⇒ λ2 − (a + c)λ + (ac − b2 ) = 0
c
b−λ (a + c) ± (a − c)2 + 4b2
⇐⇒ λ =
. Now A has repeated eigenvalues
2
a 0
⇐⇒ (a − c)2 + 4b2 = 0 ⇐⇒ a = c and b = 0 (since a, b, c ∈ R) ⇐⇒ A =
0 a
⇐⇒ A = aI2 ⇐⇒ A is a scalar matrix.
19. (a). A is a real n × n symmetric matrix =⇒ A possesses a complete set of eigenvectors
=⇒ A is similar to a diagonal matrix
=⇒ there exists an invertible matrix S such that S −1 AS = diag(λ, λ, λ, . . . , λ) where λ occurs n times.
But diag(λ, λ, λ, . . . , λ) = λIn , so S −1 AS = λIn =⇒ AS = S(λIn ) =⇒ AS = λS =⇒ A = λSS −1
=⇒ A = λIn =⇒ A is a scalar matrix
(b). Theorem: Let A be a nondefective n × n matrix. If λ is an eigenvalue of multiplicity n, then A is a
scalar matrix.
Proof: The proof is the same as that in part (a) since A has a complete set of eigenvectors.
20. Since real eigenvectors of A that correspond to distinct eigenvalues are orthogonal, it must be the case
that if y = (y1 , y2 ) corresponds to λ2 where Ay = λ2 y, then
x, y = 0 =⇒ (1, 2), (y1 , y2 ) = 0 =⇒ y1 + 2y2 = 0 =⇒ y1 = −2y2 =⇒ y = (−2y2 , y2 ) = y2 (−2, 1).
Consequently, (−2, 1) is an eigenvector corresponding to λ2 .
21. (a). Let A be a real symmetric 2 × 2 matrix with two distinct eigenvalues, λ1 and λ2 , where v1 = (a, b)
is an eigenvector corresponding to λ1 . Since real eigenvectors of A that correspond to distinct eigenvalues
are orthogonal, it follows that if v2 = (c, d) corresponds to λ2 where av2 = λ2 v2 , then
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v1 , v2 = 0 =⇒ (a, b), (c, d) = 0, that is, ac + bd = 0. By inspection, we see that v2 = (−b, a). An
orthonormal set of eigenvectors for A is
√
⎡
1
a
⎢ a 2 + b2
Thus, if S = ⎣
1
√
b
a 2 + b2
√
1
1
√
(a, b),
(−b, a) .
a 2 + b2
a 2 + b2
⎤
1
b
a 2 + b2 ⎥
⎦, then S T AS = diag(λ1 , λ2 ).
1
√
a
a 2 + b2
−√
T
−1
(b). S T AS = diag(λ1 , λ2 ) =⇒ AS = S diag(λ
1 , λ2 ), since
S = S . Thus,
1
a −b
λ1 0
a b
A = S diag(λ1 , λ2 )S T = 2
a
0 λ2
−b a
(a + b2 ) b
1
a −b
λ1 a λ1 b
= 2
a
−λ2 b λ2 a
(a + b2 ) b
2
2
1
λ1 a + λ2 b
ab(λ1 − λ2 )
.
= 2
(a + b2 ) ab(λ1 − λ2 ) λ1 b2 + λ2 a2
22. A is a real symmetric 3 × 3 matrix with eigenvalues λ1 and λ2 of multiplicity two.
(a). Let v1 = (1, −1, 1) be an eigenvector of A that corresponds to λ1 . Since real eigenvectors of A that
correspond to distinct eigenvalues are orthogonal, it must be the case that if v = (a, b, c) corresponds to λ2
where Av = λ2 v, then v1 , v = 0 =⇒ (1, −1, 1), (a, b, c) = 0 =⇒ a−b+c = 0 =⇒ v = r(1, 1, 0)+s(−1, 0, 1)
where r and s are free variables. Consequently, v2 = (1, 1, 0) and v3 = (−1, 0, 1) are linearly independent
eigenvectors corresponding to λ2 . Thus, {(1, 1, 0), (−1, 0, 1)} is a basis for E2 . v2 and v3 are not orthogonal
since v2 , v3 = −1 = 0, so we will apply the Gram-Schmidt procedure to v2 and v3 . Let u2 = v2 = (1, 1, 0)
and
v3 , u2 1
u3 = v3 −
u2 = (−1, 0, 1) + (1, 1, 0) =
||u2 ||2
2
1 1
− , ,1 .
2 2
v1
1
1
1
is a unit eigenvector corresponding to λ1 , and
= √ , −√ , √
||v
3 3
3
1 ||
u2
u3
1
1
1
1
2
are orthonormal eigenvectors corresponding
= √ , √ , 0 , w3 =
= −√ , √ , √
w2 =
||u2 ||
||u3 ||
2
2
6
6
6
to λ2 .
⎡
⎤
1
1
1
√
√
−√
⎢
3
2
6 ⎥
⎢
1
1 ⎥
⎢ − √1
√
√ ⎥
Consequently, S = ⎢
⎥ and S T AS = diag(λ1 , λ2 , λ2 ).
⎢
3
2
6 ⎥
⎣
2 ⎦
1
√
√
0
3
6
Now, w1 =
(b). Since S is an orthogonal matrix, S T AS = diag(λ1 , λ2 , λ2 ) =⇒ AS = S diag(λ1 , λ2 , λ3 )
=⇒ A = S diag(λ1 , λ2 , λ3 )S T =⇒
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493
⎡
1
√
⎢
3
⎢
⎢ − √1
A=⎢
⎢
3
⎣
1
√
3
⎡
1
√
⎢
3
⎢
⎢ − √1
=⎢
⎢
3
⎣
1
√
3
1
√
2
1
√
2
0
1
√
2
1
√
2
0
⎤
⎤
⎡
1
1
1
1
√
√
−√
−√
⎡
⎤
⎢
6 ⎥ λ1 0
3
3
3 ⎥
0
⎥
⎢
1
1 ⎥
1
⎥
⎢
√ ⎥ ⎣ 0 λ2 0 ⎦ ⎢ √
√
0 ⎥
⎥
⎥
⎢
6 ⎥
2
2
0
0 λ2 ⎣
1
1
2 ⎦
2 ⎦
√
√
√
−√
6
6
6
6
⎤⎡
⎤
λ1
λ1
λ1
1
√
√
−√
−√
⎡
⎢
3
3
3 ⎥
6 ⎥
λ1 + 2λ2
⎥
⎢
⎥
1
λ2
λ2
⎢ √
⎥ 1⎣
√ ⎥
√
0
−λ
⎥⎢
⎥=
1 + λ2
⎥ 3
6 ⎥⎢
2
2
λ
1 − λ2
2 ⎦⎣
λ2
λ
2λ2 ⎦
√
√2
√
−√
6
6
6
6
−λ1 + λ2
λ1 + 2λ2
−λ1 + λ2
⎤
λ 1 − λ2
−λ1 + λ2 ⎦ .
λ1 + 2λ2
23. (a). Let v1 , v2 ∈ Cn and recall that v1T v2 = [v1 , v2 ].
[Av1 , v2 ] = (Av1 )T v2 = (v1T AT )v2 = v1T (−A)v2 = −v1T (A)v2 = −v1T (Av2 ) = −v1T Av2 = [−v1 , Av2 ].
Thus,
Av1 , v2 = −v1 , Av2 .
(23.1)
(b). Let v1 be an eigenvector corresponding to the eigenvalue λ1 , so
Av1 = λ1 v1 .
(23.2)
Taking the inner product of (23.2) with v1 yields Av1 , v1 = λ1 v1 , v1 , that is
Av1 , v1 = λ1 ||v1 ||2 .
(23.3)
Taking the complex conjugate of (23.3) gives Av1 , v1 = λ1 ||v1 ||2 , that is
v1 , Av1 = λ1 ||v1 ||2 .
(23.4)
Adding (23.3) and (23.4), and using (23.1) with v2 = v1 yields (λ1 + λ1 )||v1 ||2 = 0. But ||v1 ||2 = 0, so
λ1 + λ1 = 0, or equivalently, λ1 = −λ1 , which means that all nonzero eigenvalues of A are pure imaginary.
24. Let A be an n × n real skew-symmetric matrix where n is odd. Since A is real, the characteristic
equation, det(A − λI) = 0, has real coefficients, so its roots come in conjugate pairs. By Problem 23, all
nonzero solutions of det(A − λI) = 0 are pure imaginary, hence when n is odd, zero will be one of the
eigenvalues of A.
−λ
4 −4 25. det(A − λI) = 0 ⇐⇒ −4 −λ −2 = 0 ⇐⇒ λ3 + 36λ = 0 ⇐⇒ λ = 0 or λ = ±6i.
4
2 −λ ⎤
⎤
⎡
⎤⎡
⎡
0 4 −4
v1
0
If λ = 0 then (A − λI)v = 0 assumes the form ⎣ −4 0 −2 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ 2v1 + v3 = 0 and
0
4 2
0
v3
v2 − v3 = 0. If we let v3 = 2r ∈ C, then the solution set of this system is {(−r, 2r, 2r) : r ∈ C} so the
eigenvectors corresponding to λ = 0 are v1 = r(−1,
⎡ 2, 2) where r⎤∈⎡ C. ⎤ ⎡
⎤
6i 4 −4
v1
0
If λ = −6i then (A−λI)v = 0 assumes the form ⎣ −4 6i −2 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ 5v1 +(−2+6i)v3 = 0
4 2
6i
v3
0
and 5v2 + (4 + 3i)v3 = 0. If we let v3 = 5s ∈ C, then the solution set of this system is
{(2 − 6i)s, (−4 − 3i)s, 5s : s ∈ C}, so the eigenvectors corresponding to λ = −6i are v2 = s(2 − 6i, −4 − 3i, 5)
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494
where s ∈ C. By Theorem 7.1.8, since A is a matrix with real entries, the eigenvectors corresponding to
λ = 6i are of the form v3 = t(2 + 6i, −4 + 3i, 5) where t ∈ C.
−λ −1 −6 √
5 = 0 ⇐⇒ −λ3 − 62λ = 0 ⇐⇒ λ = 0 or λ = ± 62i.
26. det(A − λI) = 0 ⇐⇒ 1 −λ
6 −5 −λ ⎤ ⎡
⎡
⎤⎡
⎤
0 −1 −6
v1
0
0
5 ⎦ ⎣ v2 ⎦ = ⎣ 0 ⎦ =⇒ v3 = t, v2 = −6t, v1 =
If λ = 0 then (A − λI)v = 0 assumes the form ⎣ 1
6 −5
0
v3
0
−5t, where t ∈ C. Thus, the solution set of the system is {(−5t, −6t, t) : t ∈ C}, so the eigenvectors
corresponding to λ = 0 are v = t(−5, −6, 1), where t ∈ C.
For the other eigenvalues, it is best to use technology to generate the corresponding eigenvectors.
27. A is a real n×n orthogonal matrix ⇐⇒ A−1 = AT ⇐⇒ AT A = In . Suppose that A = [v1 , v2 , v3 , . . . , vn ].
Since the ith row of AT is equal to viT , the matrix multiplication assures us that the ij th entry of AT A is
equal to viT , vi . Thus from the equality AT A = In = [δij ], an n × n matrix A = [v1 , v2 , v3 , . . . , vn ] is
orthogonal if an only if viT , vi = δij , that is, if and only if the columns (rows) of A, {v1 , v2 , v3 , . . . , vn },
form an orthonormal set of vectors.
Solutions to Section 7.6
True-False Review:
(a): TRUE. See Remark 1 following Definition 7.6.1.
(b): TRUE. Each Jordan block corresponds to a cycle of generalized eigenvectors, and each such cycle
contains exactly one eigenvector. By construction, the eigenvectors (and generalized eigenvectors) are chosen
to form a linearly independent set of vectors.
(c): FALSE. For example, in a diagonalizable n × n matrix, the n linearly independent eigenvectors can
be arbitrarily placed in the columns of the matrix S. Thus, an ample supply of invertible matrices S can be
constructed.
0 1
(d): FALSE. For instance, if J1 = J2 =
, then J1 and J2 are in Jordan canonical form, but
0 0
0 2
is not in Jordan canonical form.
J1 + J 2 =
0 0
(e): TRUE. This is simply a restatement of the definition of a generalized eigenvector.
(f ): FALSE. The number of Jordan blocks corresponding to λ in the Jordan canonical form of A is the
number of linearly independent eigenvectors of A corresponding to λ, which is dim[Eλ ], the dimension of the
eigenspace corresponding to λ, not dim[Kλ ].
(g): TRUE. This is the content of Theorem 7.6.8.
1 1
, then J1 and J2 are in Jordan canonical form, but
(h): FALSE. For instance, if J1 = J2 =
0 1
1 2
J1 J2 =
is not in Jordan canonical form.
0 1
(i): TRUE. If we place the vectors in a cycle of generalized eigenvectors of A (see Equation (7.6.3)) in
the columns of the matrix S formulated in this section in the order they appear in the cycle, then the
corresponding columns of the matrix S −1 AS will form a Jordan block.
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(j): TRUE. The assumption here is that all Jordan blocks have size 1 × 1, which precisely says that the
Jordan canonical form of A is a diagonal matrix. This means that A is diagonalizable.
(k): TRUE. Suppose that S −1 AS = B and that J is a Jordan canonical form of A. So there exists an
invertible matrix T such that T −1 AT = J. Then B = S −1 AS = S −1 (T JT −1 )S = (T −1 S)−1 J(T −1 S), and
hence,
(T −1 S)B(T −1 S)−1 = J,
which shows that J is also a Jordan canonical form of B.
0 1
0
(l): FALSE. For instance, if J =
and r = 2, then J is in Jordan canonical form, but rJ =
0 0
0
is not in Jordan canonical form.
Problems:
1. There is just one Jordan canonical form (disregarding rearrangements of the Jordan blocks):
⎡
⎤
−4 0 0
⎣ 0 0 0 ⎦.
0 0 9
2. There are 3 possible Jordan canonical forms:
⎡
⎤
⎡
1 0 0
1
⎣ 0 1 0 ⎦, ⎣ 0
0 0 1
0
1
1
0
3. There are 4 possible Jordan canonical forms:
⎤
⎤
⎡
⎡
1 1 0 0
1 0 0 0
⎢ 0 1 0 0 ⎥
⎢ 0 1 0 0 ⎥
⎥
⎥
⎢
⎢
⎣ 0 0 3 0 ⎦, ⎣ 0 0 3 0 ⎦,
0 0 0 3
0 0 0 3
⎤
0
0 ⎦,
1
⎡
⎤
0
1 ⎦.
1
1
⎣ 0
0
1
1
0
0
1
0
0
0
0
3
0
⎤
0
0 ⎥
⎥,
1 ⎦
3
2
⎢ 0
⎢
⎢ 0
⎢
⎣ 0
0
1
2
0
0
0
0
0
2
0
0
0
0
1
2
0
0
0
1
2
0
⎤
0
0 ⎥
⎥
0 ⎥
⎥,
0 ⎦
2
⎡
1
⎢ 0
⎢
⎣ 0
0
1
1
0
0
0
0
3
0
⎤
0
0 ⎥
⎥.
1 ⎦
3
2
⎢ 0
⎢
⎢ 0
⎢
⎣ 0
0
1
2
0
0
0
0
1
2
0
0
0
0
1
2
0
⎤
0
0 ⎥
⎥
0 ⎥
⎥
1 ⎦
2
⎡
1
⎢ 0
⎢
⎣ 0
0
4. There are 7 possible Jordan canonical forms:
⎡
2
⎢ 0
⎢
⎢ 0
⎢
⎣ 0
0
0
2
0
0
0
0
0
2
0
0
0
0
0
2
0
⎡
2
⎢ 0
⎢
⎢ 0
⎢
⎣ 0
0
⎡
⎤
0
0 ⎥
⎥
0 ⎥
⎥,
0 ⎦
2
1
2
0
0
0
0
1
2
0
0
2
⎢ 0
⎢
⎢ 0
⎢
⎣ 0
0
1
2
0
0
0
0
0
2
0
0
0
0
0
2
0
⎤
0
0 ⎥
⎥
0 ⎥
⎥,
1 ⎦
2
0
0
0
2
0
⎡
⎤
0
0 ⎥
⎥
0 ⎥
⎥,
0 ⎦
2
⎡
2
⎢ 0
⎢
⎢ 0
⎢
⎣ 0
0
1
2
0
0
0
0
1
2
0
0
5. There are 10 possible Jordan canonical forms:
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⎡
2
⎢ 0
⎢
⎢ 0
⎢
⎣ 0
0
⎡
⎤
0
0 ⎥
⎥
0 ⎥
⎥,
0 ⎦
2
1
2
0
0
0
0
1
2
0
0
0
0
0
2
0
⎤
0
0 ⎥
⎥
0 ⎥
⎥
0 ⎦
2
2
0
496
⎡
3
⎢ 0
⎢
⎢ 0
⎢
⎢ 0
⎢
⎣ 0
0
⎡
3
⎢ 0
⎢
⎢ 0
⎢
⎢ 0
⎢
⎣ 0
0
0
3
0
0
0
0
0
0
3
0
0
0
0
0
0
3
0
0
0
0
0
0
9
0
1
3
0
0
0
0
0
1
3
0
0
0
0
0
1
3
0
0
0
0
0
1
9
0
⎤
0
0 ⎥
⎥
0 ⎥
⎥,
0 ⎥
⎥
0 ⎦
9
⎤
0
0 ⎥
⎥
0 ⎥
⎥,
0 ⎥
⎥
0 ⎦
9
⎡
3
⎢ 0
⎢
⎢ 0
⎢
⎢ 0
⎢
⎣ 0
0
⎡
3
⎢ 0
⎢
⎢ 0
⎢
⎢ 0
⎢
⎣ 0
0
⎡
3
⎢ 0
⎢
⎢ 0
⎢
⎢ 0
⎢
⎣ 0
0
1
3
0
0
0
0
0
0
3
0
0
0
0
0
0
3
0
0
0
0
0
0
9
0
0
3
0
0
0
0
0
0
3
0
0
0
0
0
0
3
0
0
0
0
0
0
9
0
1
3
0
0
0
0
0
1
3
0
0
0
0
0
0
3
0
0
0
0
0
0
9
0
⎤
0
0 ⎥
⎥
0 ⎥
⎥,
0 ⎥
⎥
0 ⎦
9
⎤
0
0 ⎥
⎥
0 ⎥
⎥,
0 ⎥
⎥
1 ⎦
9
⎤
0
0 ⎥
⎥
0 ⎥
⎥,
0 ⎥
⎥
1 ⎦
9
⎡
3
⎢ 0
⎢
⎢ 0
⎢
⎢ 0
⎢
⎣ 0
0
⎡
3
⎢ 0
⎢
⎢ 0
⎢
⎢ 0
⎢
⎣ 0
0
⎡
3
⎢ 0
⎢
⎢ 0
⎢
⎢ 0
⎢
⎣ 0
0
1
3
0
0
0
0
0
0
3
0
0
0
0
0
1
3
0
0
0
0
0
0
9
0
1
3
0
0
0
0
0
0
3
0
0
0
0
0
0
3
0
0
0
0
0
0
9
0
1
3
0
0
0
0
0
1
3
0
0
0
0
0
1
3
0
0
0
0
0
1
9
0
⎤
0
0 ⎥
⎥
0 ⎥
⎥,
0 ⎥
⎥
0 ⎦
9
⎤
0
0 ⎥
⎥
0 ⎥
⎥,
0 ⎥
⎥
1 ⎦
9
⎤
0
0 ⎥
⎥
0 ⎥
⎥.
0 ⎥
⎥
1 ⎦
9
⎡
3
⎢ 0
⎢
⎢ 0
⎢
⎢ 0
⎢
⎣ 0
0
⎡
3
⎢ 0
⎢
⎢ 0
⎢
⎢ 0
⎢
⎣ 0
0
1
3
0
0
0
0
0
1
3
0
0
0
0
0
0
3
0
0
0
0
0
0
9
0
1
3
0
0
0
0
0
0
3
0
0
0
0
0
1
3
0
0
0
0
0
0
9
0
⎤
0
0 ⎥
⎥
0 ⎥
⎥,
0 ⎥
⎥
0 ⎦
9
⎤
0
0 ⎥
⎥
0 ⎥
⎥,
0 ⎥
⎥
1 ⎦
9
6. Since λ = 2 occurs with multiplicity 4, it can give rise to the following possible Jordan block sizes:
(a) 4
(b) 3,1
(c) 2,2
(d) 2,1,1
(e) 1,1,1,1
Likewise, λ = 6 occurs with multiplicity 4, so it can give rise to the same five possible Jordan block sizes.
Finally, λ = 8 occurs with multiplicity 3, so it can give rise to three possible Jordan block sizes:
(a) 3
(b) 2,1
(c) 1,1,1
Since the block sizes for each eigenvalue can be independently determined, we have 5 · 5 · 3 = 75 possible
Jordan canonical forms.
7. Since λ = 2 occurs with multiplicity 4, it can give rise to the following possible Jordan block sizes:
(a) 4
(b) 3,1
(c) 2,2
(d) 2,1,1
(e) 1,1,1,1
Next, λ = 5 occurs with multiplicity 6, so it can give rise to the following possible Jordan block sizes:
(a) 6
(b) 5,1
(c) 4,2
(d) 4,1,1
(e) 3,3
(f) 3,2,1
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(g) 3,1,1,1
(h) 2,2,2
(i) 2,2,1,1
(j) 2,1,1,1,1
(k) 1,1,1,1,1,1
There are 5 possible Jordan block sizes corresponding to λ = 2 and 11 possible Jordan block sizes
corresponding to λ = 5. Multiplying these results, we have 5 · 11 = 55 possible Jordan canonical forms.
8. Since (A − 5I)2 = 0, no cycles of generalized eigenvectors corresponding to λ = 5 can have length greater
than 2, and hence, only Jordan block sizes 2 or less are possible. Thus, the possible block sizes under this
restriction (corresponding to λ = 5) are:
2,2,2
2,2,1,1
2,1,1,1,1
1,1,1,1,1,1
There are four such. There are still five possible block size lists corresponding to λ = 2. Multiplying
these results, we have 5 · 4 = 20 possible Jordan canonical forms under this restriction.
9.
(a).
⎡
⎤ ⎡
⎤ ⎡
⎤
0
0
0
0
0
0
0
0
0
λ1 0
λ1 1
λ1 1
⎢ 0 λ1 0
⎢
⎢
0
0 ⎥
0
0 ⎥
0
0 ⎥
⎢
⎥ ⎢ 0 λ1 0
⎥ ⎢ 0 λ1 1
⎥
⎢ 0
⎥
⎢
⎥
⎢
0 λ1 0
0 λ1 0
0 λ1 0
0 ⎥,⎢ 0
0 ⎥,⎢ 0
0 ⎥
⎢
⎥,
⎣ 0
0
0 λ2 0 ⎦ ⎣ 0
0
0 λ2 0 ⎦ ⎣ 0
0
0 λ2 0 ⎦
0
0
0
0 λ2
0
0
0
0 λ2
0
0
0
0 λ2
⎤ ⎡
⎤
⎡
⎤ ⎡
0
0
0
0
0
0
0
0
0
λ1 1
λ1 1
λ1 0
⎢
⎢
⎢ 0 λ1 0
0
0 ⎥
0
0 ⎥
0
0 ⎥
⎥
⎥ ⎢ 0 λ1 0
⎥ ⎢ 0 λ1 1
⎢
⎥
⎥,⎢ 0
⎥,⎢ 0
⎢ 0
0
λ
0
0
0
λ
0
λ
0
0
0
0
1
1
1
⎥
⎥ ⎢
⎥ ⎢
⎢
⎣ 0
0
0 λ2 1 ⎦
0
0 λ2 1 ⎦ ⎣ 0
0
0 λ2 1 ⎦ ⎣ 0
0
0
0
0 λ2
0
0
0
0 λ2
0
0
0
0 λ2
(b). The assumption that (A − λ1 I)2 = 0 implies that there can be no Jordan blocks corresponding to λ1
of size 3 × 3 (or greater). Thus, the only possible Jordan canonical forms for this matrix now are
⎤ ⎡
⎤
⎡
λ1 1
λ1 0
0
0
0
0
0
0
⎢
⎢ 0 λ1 0
0
0 ⎥
0
0 ⎥
⎥ ⎢ 0 λ1 0
⎥
⎢
⎢
⎥
⎢ 0
0 λ1 0
0 λ1 0
0 ⎥,⎢ 0
0 ⎥
⎥,
⎢
⎣ 0
0
0 λ2 0 ⎦ ⎣ 0
0
0 λ2 0 ⎦
0
0
0
0 λ2
0
0
0
0 λ2
⎡
⎤ ⎡
⎤
λ1 0
0
0
0
0
0
0
λ1 1
⎢ 0 λ1 0
⎢
0
0 ⎥
0
0 ⎥
⎢
⎥ ⎢ 0 λ1 0
⎥
⎢ 0
⎥
⎥
⎢
0
0
λ
0
λ
0
0
0
0
,
1
1
⎢
⎥
⎥ ⎢
⎣ 0
0
0 λ2 1 ⎦ ⎣ 0
0
0 λ2 1 ⎦
0
0
0
0 λ2
0
0
0
0 λ2
10. The assumption that (A − λI)3 = 0 implies no Jordan blocks of size greater than 3 × 3 are possible.
The fact that (A − λI)2 = 0 implies that there is at least one Jordan block of size 3 × 3. Thus, the possible
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498
block size combinations for a 6 × 6 matrix with eigenvalue λ of multiplicity 6 and no blocks of size greater
than 3 × 3 with at least one 3 × 3 block are:
3,3
3,2,1
3,1,1,1
Thus, there are 3 possible Jordan canonical forms. (We omit the list itself; it can be produced simply
from the list of block sizes above.)
11. The eigenvalues of the matrix with this characteristic polynomial are λ = 4, 4, −6. The possible Jordan
canonical forms in this case are therefore:
⎤ ⎡
⎤
⎡
4 0
0
4 1
0
⎣ 0 4
0 ⎦,⎣ 0 4
0 ⎦.
0 0 −6
0 0 −6
12. The eigenvalues of the matrix with this characteristic polynomial are λ = 4, 4, 4, −1, −1. The possible
Jordan canonical forms in this case are therefore:
⎡
4
⎢ 0
⎢
⎢ 0
⎢
⎣ 0
0
⎡
4
⎢ 0
⎢
⎢ 0
⎢
⎣ 0
0
0
4
0
0
0
0
4
0
0
0
⎤ ⎡
4
0
0
0
⎢ 0
0
0
0 ⎥
⎥ ⎢
⎢
4
0
0 ⎥
⎥,⎢ 0
⎣ 0
⎦
0 −1
0
0
0
0 −1
⎤ ⎡
4
0
0
0
⎢ 0
0
0
0 ⎥
⎥ ⎢
⎢
4
0
0 ⎥
⎥,⎢ 0
0 −1
1 ⎦ ⎣ 0
0
0
0 −1
1
4
0
0
0
1
4
0
0
0
⎤ ⎡
4
0
0
0
⎢ 0
0
0
0 ⎥
⎥ ⎢
⎢
4
0
0 ⎥
⎥,⎢ 0
⎣ 0
⎦
0 −1
0
0
0 −1
0
⎤ ⎡
4
0
0
0
⎢ 0
0
0
0 ⎥
⎥ ⎢
⎢
4
0
0 ⎥
⎥,⎢ 0
0 −1
1 ⎦ ⎣ 0
0
0
0 −1
⎤
1 0
0
0
4 1
0
0 ⎥
⎥
0 4
0
0 ⎥
⎥,
0 0 −1
0 ⎦
0 0
0 −1
⎤
1 0
0
0
4 1
0
0 ⎥
⎥
0 4
0
0 ⎥
⎥.
0 0 −1
1 ⎦
0 0
0 −1
13. The eigenvalues of the matrix with this characteristic polynomial are λ = −2, −2, −2, 0, 0, 3, 3. The
possible Jordan canonical forms in this case are therefore:
⎤ ⎡
−2
0
0 0 0 0 0
−2
0
0
⎢ 0 −2
⎢
0 0 0 0 0 ⎥
0
⎥ ⎢ 0 −2
⎢
⎢ 0
⎢
0 −2 0 0 0 0 ⎥
0 −2
⎥ ⎢ 0
⎢
⎥
⎢ 0
0
0 0 0 0 0 ⎥,⎢
0
0
⎢
⎢ 0
⎢ 0
⎢
0
0 0 0 0 0 ⎥
0
0
⎥ ⎢ 0
⎢
⎣ 0
0
0 0 0 3 0 ⎦ ⎣ 0
0
0
0
0
0 0 0 0 3
0
0
0
⎡
0
0
0
0
0
0
0
⎤
⎤ ⎡
−2
0
0 0 0 0 0
0 0 0
⎢
0 0 0 0 0 ⎥
0 0 0 ⎥
⎥
⎥ ⎢ 0 −2
⎥
⎢
0 −2 0 0 0 0 ⎥
0 0 0 ⎥ ⎢ 0
⎥
⎢
0
0 0 0 0 0 ⎥
1 0 0 ⎥
⎥,
⎥,⎢ 0
⎢
0
0 0 0 0 0 ⎥
0 0 ⎥
⎥
⎥ ⎢ 0
0
0 0 0 3 1 ⎦
0 3 0 ⎦ ⎣ 0
0
0
0 0 0 0 3
0 0 3
⎤ ⎡
⎤ ⎡
⎤
−2
1
0 0 0 0 0
−2
1
0 0 0 0 0
−2
0
0 0 0 0 0
⎢
⎢
⎢ 0 −2
0 0 0 0 0 ⎥
0 0 0 0 0 ⎥
0 0 0 0 0 ⎥
⎥ ⎢ 0 −2
⎥ ⎢ 0 −2
⎢
⎥
⎢
⎥
⎢
⎥
⎢ 0
0 −2 0 0 0 0 ⎥
0 −2 0 0 0 0 ⎥ ⎢ 0
0 −2 0 0 0 0 ⎥ ⎢ 0
⎢
⎥
⎢
⎢
⎢ 0
0
0 0 1 0 0 ⎥
0
0 0 0 0 0 ⎥
0
0 0 1 0 0 ⎥
⎥,⎢ 0
⎥,⎢ 0
⎢
⎥,
⎥ ⎢ 0
⎥ ⎢ 0
⎢ 0
⎥
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
⎥ ⎢
⎥ ⎢
⎢
⎥
⎣
⎦
⎣
⎦
⎣ 0
0
0
0 0 0 3 0 ⎦
0
0
0 0 0 3 0
0
0 0 0 3 1
0
0
0 0 0 0 3
0
0
0 0 0 0 3
0
0
0 0 0 0 3
⎡
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⎡
−2
1
0 0
⎢ 0 −2
0 0
⎢
⎢ 0
0 −2 0
⎢
⎢ 0
0
0 0
⎢
⎢ 0
0
0 0
⎢
⎣ 0
0
0 0
0
0
0 0
⎡
−2
1
0 0
⎢ 0 −2
1
0
⎢
⎢ 0
0
−2
0
⎢
⎢ 0
0
0 0
⎢
⎢ 0
0
0 0
⎢
⎣ 0
0
0 0
0
0
0 0
0
0
0
0
0
0
0
0
0
0
0
0
3
0
0 0
0 0
0 0
1 0
0
0 3
0 0
⎤ ⎡
−2
1
0
0
⎢
0
0 ⎥
⎥ ⎢ 0 −2
⎢
0 −2
0 ⎥
⎥ ⎢ 0
⎥
0
0
0 ⎥,⎢
⎢ 0
⎢
0
0
0 ⎥
⎥ ⎢ 0
0
0
1 ⎦ ⎣ 0
0
0
0
3
⎤ ⎡
0
−2
1
0
⎢ 0 −2
0 ⎥
1
⎥ ⎢
⎢
0 −2
0 ⎥
⎥ ⎢ 0
⎥
0
0
0
0 ⎥,⎢
⎢
⎢ 0
0
0
0 ⎥
⎥ ⎢
0
0
0 ⎦ ⎣ 0
0
0
0
3
⎤ ⎡
−2
1
0
0 0 0 0
⎢
1
0 0 0 0 ⎥
⎥ ⎢ 0 −2
⎢
0 −2
0 0 0 0 ⎥
⎥ ⎢ 0
⎥
0
0
0 1 0 0 ⎥,⎢
⎢ 0
⎢
0
0
0 0 0 0 ⎥
⎥ ⎢ 0
0
0
0 0 3 1 ⎦ ⎣ 0
0
0
0
0 0 0 3
⎤ ⎡
−2
1
0
0 0 0 0
⎢ 0 −2
1
0 0 0 0 ⎥
⎥ ⎢
⎢
0 −2
0 0 0 0 ⎥
⎥ ⎢ 0
⎥
0
0
0
0 0 0 0 ⎥,⎢
⎢
⎢ 0
0
0
0 0 0 0 ⎥
⎥ ⎢
0
0
0 0 3 1 ⎦ ⎣ 0
0
0
0
0 0 0 3
⎤
0 0 0 0
0 0 0 0 ⎥
⎥
0 0 0 0 ⎥
⎥
0 0 0 0 ⎥
⎥,
0 0 0 0 ⎥
⎥
0 0 3 0 ⎦
0 0 0 3
⎤
0 0 0 0
0 0 0 0 ⎥
⎥
0 0 0 0 ⎥
⎥
0 1 0 0 ⎥
⎥.
0 0 0 0 ⎥
⎥
0 0 3 1 ⎦
0 0 0 3
14. The eigenvalues of the matrix with this characteristic polynomial are λ = −2, −2, 6, 6, 6, 6, 6. The
possible Jordan canonical forms in this case are therefore:
⎡
−2
0
⎢ 0 −2
⎢
⎢ 0
0
⎢
⎢ 0
0
⎢
⎢ 0
0
⎢
⎣ 0
0
0
0
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
0
0
6
0
0
0
0
−2
1 0
0 −2 0
0
0 6
0
0 0
0
0 0
0
0 0
0
0 0
⎤ ⎡
⎤ ⎡
0
−2
−2
0 0 0 0 0 0
⎢ 0 −2 0 0 0 0 0 ⎥ ⎢ 0
0 ⎥
⎥ ⎢
⎥ ⎢
⎢
⎢
0 ⎥
0 6 0 0 0 0 ⎥
⎥ ⎢ 0
⎥ ⎢ 0
⎥
⎥
⎢
0 ⎥,⎢ 0
0 0 6 1 0 0 ⎥,⎢
⎢ 0
⎥
⎢
⎢
0 ⎥ ⎢ 0
0 0 0 6 0 0 ⎥
⎥ ⎢ 0
⎦
⎣ 0
⎦
⎣
0
0
0 0 0 0 6 0
6
0
0
0 0 0 0 0 6
⎤ ⎡
−2
0 0 0 0
−2
0 0 0 0 0 0
⎢ 0 −2 0 0 0 0 0 ⎥ ⎢ 0 −2 0 0 0
⎥ ⎢
⎢
⎢
⎢ 0
0 6 1 0
0 6 1 0 0 0 ⎥
⎥ ⎢ 0
⎢
⎥
⎢ 0
⎢ 0
0 0 6 1
0
0
6
1
0
0
,
⎥ ⎢
⎢
⎥ ⎢ 0
⎢ 0
0 0 0 6
0
0
0
6
0
0
⎥ ⎢
⎢
⎣ 0
0 0 0 0
0 0 0 0 6 0 ⎦ ⎣ 0
0
0 0 0 0
0
0 0 0 0 0 6
⎤ ⎡
⎡
−2
0 0 0 0
−2
0 0 0 0 0 0
⎢ 0 −2 0 0 0 0 0 ⎥ ⎢ 0 −2 0 0 0
⎥ ⎢
⎢
⎢
⎢ 0
0 6 1 0
0 6 1 0 0 0 ⎥
⎥ ⎢ 0
⎢
⎥,⎢ 0
⎢ 0
0 0 6 1
0
0
6
1
0
0
⎥ ⎢
⎢
⎥ ⎢ 0
⎢ 0
0 0 0 6
0
0
0
6
1
0
⎥ ⎢
⎢
⎣ 0
0 0 0 0
0 0 0 0 6 0 ⎦ ⎣ 0
0
0 0 0 0 0 6
0
0 0 0 0
⎤ ⎡
⎤ ⎡
−2
−2
1 0 0 0 0 0
0 0 0 0
⎢ 0 −2 0 0 0 0 0 ⎥ ⎢ 0
0 0 0 0 ⎥
⎥ ⎢
⎥ ⎢
⎢
⎢
0 6 1 0 0 0 ⎥
0 0 0 0 ⎥
⎥ ⎢ 0
⎥ ⎢ 0
⎥
⎥
⎢
0 0 6 0 0 0 ⎥,⎢
6 0 0 0 ⎥,⎢ 0
⎢ 0
⎢
⎥
⎢
0 0 0 6 0 0 ⎥
0 6 0 0 ⎥ ⎢ 0
⎥ ⎢ 0
⎦
⎣ 0
⎦
⎣
0
0 0 0 0 6 0
0 0 6 0
0
0
0 0 0 0 0 6
0 0 0 6
0
0
0
6
0
0
0
⎡
0
0
0
0
6
0
0
0
0
0
0
0
6
0
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⎤
0 0 0 0 0 0
−2 0 0 0 0 0 ⎥
⎥
0 6 1 0 0 0 ⎥
⎥
0 0 6 0 0 0 ⎥
⎥,
0 0 0 6 1 0 ⎥
⎥
0 0 0 0 6 0 ⎦
0 0 0 0 0 6
⎤
0 0
0 0 ⎥
⎥
0 0 ⎥
⎥
0 0 ⎥
⎥
0 0 ⎥
⎥
6 1 ⎦
0 6
⎤
0 0
0 0 ⎥
⎥
0 0 ⎥
⎥
0 0 ⎥
⎥
1 0 ⎥
⎥
6 1 ⎦
0 6
⎤
1 0 0 0 0 0
−2 0 0 0 0 0 ⎥
⎥
0 6 1 0 0 0 ⎥
⎥
0 0 6 0 0 0 ⎥
⎥,
0 0 0 6 1 0 ⎥
⎥
0 0 0 0 6 0 ⎦
0 0 0 0 0 6
500
⎡
−2
1
⎢ 0 −2
⎢
⎢ 0
0
⎢
⎢ 0
0
⎢
⎢ 0
0
⎢
⎣ 0
0
0
0
⎡
−2
1
⎢ 0 −2
⎢
⎢ 0
0
⎢
⎢ 0
0
⎢
⎢ 0
0
⎢
⎣ 0
0
0
0
0 0
0 0
6 1
0 6
0 0
0 0
0 0
0 0
0 0
6 1
0 6
0 0
0 0
0 0
⎤
⎤ ⎡
−2
1 0 0 0 0 0
0 0 0
⎥
⎢
0 0 0 ⎥
⎥ ⎢ 0 −2 0 0 0 0 0 ⎥
⎢
⎥
0 6 1 0 0 0 ⎥
0 0 0 ⎥ ⎢ 0
⎥
⎢
⎥
0 0 6 1 0 0 ⎥
1 0 0 ⎥,⎢ 0
⎥
⎢
0 0 0 6 0 0 ⎥
6 0 0 ⎥
⎥
⎥ ⎢ 0
0 0 0 0 6 1 ⎦
0 6 0 ⎦ ⎣ 0
0
0 0 0 0 0 6
0 0 6
⎤
⎤ ⎡
−2
1 0 0 0 0 0
0 0 0
⎥
⎢
0 0 0 ⎥
⎥ ⎢ 0 −2 0 0 0 0 0 ⎥
⎥
⎢ 0
0
6
1
0
0
0
0 0 0 ⎥
⎥
⎥ ⎢
⎥
⎢ 0
0
0
6
1
0
0
1 0 0 ⎥
,
⎥
⎥ ⎢
⎥
⎢ 0
0
0
0
6
1
0
6 1 0 ⎥
⎥
⎥ ⎢
0 0 0 0 6 1 ⎦
0 6 0 ⎦ ⎣ 0
0
0 0 0 0 0 6
0 0 6
15. Of the Jordan canonical forms in Problem 14, we are asked here to find the ones that contain exactly
five Jordan blocks, since there is a correspondence between the Jordan blocks and the linearly independent
eigenvectors. There are three:
⎡
⎤
⎤ ⎡
⎤ ⎡
−2
0 0 0 0 0 0
−2
0 0 0 0 0 0
−2
1 0 0 0 0 0
⎢ 0 −2 0 0 0 0 0 ⎥ ⎢ 0 −2 0 0 0 0 0 ⎥ ⎢ 0 −2 0 0 0 0 0 ⎥
⎥
⎥ ⎢
⎥ ⎢
⎢
⎢
⎢
⎢ 0
0 6 1 0 0 0 ⎥
0 6 1 0 0 0 ⎥
0 6 1 0 0 0 ⎥
⎥
⎥ ⎢ 0
⎥ ⎢ 0
⎢
⎢
⎢
⎢ 0
0 0 6 1 0 0 ⎥
0 0 6 0 0 0 ⎥
0 0 6 0 0 0 ⎥
⎥
⎥,⎢ 0
⎥,⎢ 0
⎢
⎢
⎢
⎢ 0
0 0 0 6 0 0 ⎥
0 0 0 6 1 0 ⎥
0 0 0 6 0 0 ⎥
⎥
⎥ ⎢ 0
⎥ ⎢ 0
⎢
⎣ 0
0 0 0 0 6 0 ⎦
0 0 0 0 6 0 ⎦ ⎣ 0
0 0 0 0 6 0 ⎦ ⎣ 0
0
0 0 0 0 0 6
0
0 0 0 0 0 6
0
0 0 0 0 0 6
0
0
1
0
16. Many examples are possible here. Let A =
. The only eigenvalue of A is 0. The vector
0
1
v=
is not an eigenvector since Av =
= λv. However A2 = 02 , so every vector is a generalized
1
0
eigenvector of A corresponding to λ = 0.
⎡
⎤
0 1 0
17. Many examples are possible here. Let A = ⎣ 0 0 0 ⎦. The only eigenvalue of A is 0. The vector
0 0 0
⎡
⎤
⎡
⎤
0
1
v = ⎣ 1 ⎦ is not an eigenvector since Av = ⎣ 0 ⎦ = λv. However, A2 = 03 , so every vector is a generalized
0
0
eigenvector of A corresponding to λ = 0.
18. Since the given matrix is already a diagonal matrix, it is already in Jordan canonical form. Thus, we
can simply take J = A and S = I2 .
19. The characteristic polynomial is
det(A − λI) = det
4−λ
−4
4
12 − λ
= λ2 − 16λ + 64 = (λ − 8)2 ,
with roots λ = 8, 8. Let us determine the eigenvectors corresponding to this eigenvalue:
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Eigenvalue λ = 8: Consider
nullspace(A − 8I) = nullspace
−4
−4
4
4
,
which is equivalent to the equation
−4x + 4y = 0. Therefore, if we set y = t, then x = t. Thus, setting t = 1
1
. Hence, we see that A is not diagonalizable, and therefore,
we obtain the eigenvector
1
8 1
JCF(A) =
.
0 8
To determine the matrix S, we must find a cycle of generalized eigenvectors of length 2. Therefore, it suffices
0
2
to find a vector v in R such that (A − 8I)v = 0. Many choices are possible here. We take v =
.
1
4
Then, (A − 8I)v =
. Thus, we have
4
4 0
S=
.
4 1
20. The characteristic polynomial is
1−λ
det(A − λI) = det
−1
1
3−λ
with roots λ = 2, 2. We have
A − 2I =
= (1 − λ)(3 − λ) + 1 = λ2 − 4λ + 4 = (λ − 2)2 ,
−1
−1
1
1
∼
1 −1
0
0
.
Because there is only one unpivoted column in this latter matrix, we only have one eigenvector for A. Hence,
A is not diagonalizable, and therefore
2 1
JCF(A) =
.
0 2
To determine the matrix S, we must find a cycle of generalized eigenvectors of length 2. Therefore,
itsuffices
0
2
. Then
to find a vector v in R such that (A − 2I)v = 0. Many choices are possible here. We take v =
1
1
(A − 2I)v =
. Thus, we have
1
1 0
S=
.
1 1
21. The characteristic polynomial is
⎡
⎤
1−λ
1
1
1−λ
1 ⎦ = (1 − λ) (1 − λ)2 − 1 = (1 − λ)(λ2 − 2λ) = λ(1 − λ)(λ − 2),
det(A − λI) = det ⎣ 0
0
1
1−λ
with roots λ = 0, 1, 2. Since A is a 3×3 matrix with three distinct eigenvalues, it is diagonalizable. Therefore,
it’s Jordan canonical form is simply a diagonal matrix with the eigenvalues as its diagonal entries:
⎤
⎡
0 0 0
JCF(A) = ⎣ 0 1 0 ⎦ .
0 0 2
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To determine the invertible matrix S, we must find eigenvectors associated with each eigenvalue.
Eigenvalue λ = 0: Consider
⎡
⎤
1 1 1
nullspace(A) = nullspace ⎣ 0 1 1 ⎦ ,
0 1 1
⎡
⎤
1 1 1
and this latter matrix can be row reduced to ⎣ 0 1 1 ⎦. The equations corresponding to the rows of this
0 0 0
matrix are x + y + z = 0 and y + z = 0. Setting z = t, then y = −t and x = 0. With t = 1 this gives us the
eigenvector (0, −1, 1).
Eigenvalue λ = 1: Consider
⎡
0
nullspace(A − I) = nullspace ⎣ 0
0
1
0
1
⎤
1
1 ⎦.
0
By inspection, we see that z = 0 and y = 0 are required, but x = t is free. Thus, an eigenvector associated
with λ = 1 may be chosen as (1, 0, 0).
Eigenvalue λ = 2: Consider
⎡
⎤
−1
1
1
1 ⎦,
nullspace(A − 2I) = nullspace ⎣ 0 −1
0
1 −1
⎡
⎤
1 −1 −1
1 −1 ⎦. Setting z = t, we have y = t and x = 2t. Thus, with t = 1
which can be row-reduced to ⎣ 0
0
0
0
we obtain the eigenvector (2, 1, 1) associated with λ = 2.
Placing the eigenvectors obtained as the columns of S (with columns corresponding to the eigenvalues of
JCF(A) above), we have
⎤
⎡
0 1 2
S = ⎣ −1 0 1 ⎦ .
1 0 1
22. We can get the characteristic polynomial by using cofactor expansion along the second column as follows:
⎡
5−λ
nullspace(A−λI) = det ⎣ 1
1
0
4−λ
0
⎤
−1
−1 ⎦ = (4−λ) [(5 − λ)(3 − λ) + 1] = (4−λ)(λ2 −8λ+16) = (4−λ)(λ−4)2 ,
3−λ
with roots λ = 4, 4, 4. ⎡
⎤
1 0 −1
We have A − 4I = ⎣ 1 0 −1 ⎦, and so vectors (x, y, z) in the nullspace of this matrix must satisfy
1 0 −1
x − z = 0. Setting z = t and y = s, we have x = t. Hence, we obtain two linearly independent eigenvectors
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⎡
⎤
⎡
⎤
1
0
of A corresponding to λ = 4: ⎣ 0 ⎦ and ⎣ 1 ⎦. Therefore, JCF(A) contains exactly two Jordan blocks.
1
0
This uniquely determines JCF(A), up to a rearrangement of the Jordan blocks:
⎡
⎤
4 1 0
JCF(A) = ⎣ 0 4 0 ⎦ .
0 0 4
To determine the matrix S, we must seek a generalized eigenvector. It is easy to verify that (A − 4I)2 = 03 ,
so every nonzero vector v is a generalized eigenvector. We must choose one such that (A −
⎡ 4I)v
⎤ = 0 in
1
order to form a cycle of length 2. There are many choices here, but let us choose v = ⎣ 0 ⎦. Then
0
⎡
⎤
1
(A − 4I)v = ⎣ 1 ⎦. Notice that this is an eigenvector of A corresponding to λ = 4. To complete the matrix
1
S, we will need a second
independent eigenvector. Again, there are a multitude of choices. Let us
⎡ linearly
⎤
0
choose the eigenvector ⎣ 1 ⎦ found above. Thus,
0
⎡
⎤
1 1 0
S = ⎣ 1 0 1 ⎦.
1 0 0
23. We will do cofactor expansion along the first column of the matrix to obtain the characteristic polynomial:
⎡
⎤
4−λ
−4
5
4−λ
2 ⎦
det(A − λI) = det ⎣ −1
−1
2
4−λ
= (4 − λ)(λ2 − 8λ + 12) + (−4)(4 − λ) − 10 − (−8 − 5(4 − λ))
= (4 − λ)(λ2 − 8λ + 12) + (2 − λ)
= (4 − λ)(λ − 2)(λ − 6) + (2 − λ)
= (λ − 2) [(4 − λ)(λ − 6) − 1]
= (λ − 2)(−λ2 + 10λ − 25)
= −(λ − 2)(λ − 5)2 ,
with eigenvalues λ = 2, 5, 5.
Since λ = 5 is a repeated
eigenvalue, ⎤we consider this eigenvalue first.⎡ We must consider
the nullspace
⎡
⎤
−1 −4
5
1 1 −2
2 ⎦, which can be row-reduced to ⎣ 0 3 −3 ⎦, a matrix that has
of the matrix A − 5I = ⎣ −1 −1
−1
2 −1
0 0
0
only one unpivoted column, and hence λ = 5 only yields one linearly independent eigenvector. Thus,
⎤
⎡
2 0 0
JCF(A) = ⎣ 0 5 1 ⎦ .
0 0 5
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To determine an invertible matrix S, we first proceed to find a cycle of eigenvectors of length 2 corresponding
to λ = 5. Therefore, we must find a vector v in R3 such that (A − 5I)v = 0, but (A − 5I)2 v =⎡0 (in
⎤ order
1
0 18 −18
, so if we set v = ⎣ 0 ⎦, then
that v is a generalized eigenvector). Note that (A − 5I)2 =
0
9 −9
0
⎡
⎤
−1
(A − 5I)v = ⎣ −1 ⎦ and (A − 5I)2 v = 0. Thus, we obtain the cycle
−1
⎧⎡
⎤ ⎡
⎤⎫
1 ⎬
⎨ −1
⎣ −1 ⎦ , ⎣ 0 ⎦ .
⎩
⎭
−1
0
Next, corresponding
to ⎤
λ = 2, we must find an eigenvector. We need to ⎡find a nonzero ⎤vector (x, y, z)
⎡
2 −4 5
1 −2 −2
2 2 ⎦, and this latter matrix can be row-reduced to ⎣ 0
0
1 ⎦. The middle
in nullspace⎣ −1
−1
2 2
0
0
0
row requires that z = 0, and if we set y = t, then x = 2t. Thus, by using t = 1, we obtain the eigenvector
(2, 1, 0).
Thus, we can form the matrix
⎤
⎡
2 −1 1
S = ⎣ 1 −1 0 ⎦ .
0 −1 0
24. We are given that λ = −5 occurs with multiplicity 2 as a root of the characteristic polynomial of A. To
search for corresponding eigenvectors, we consider
⎤
⎡
−1 1
0
1 ⎦
nullspace(A + 5I) = nullspace ⎣ − 12 21
,
2
1
1
1
−2 2 −2
⎡
⎤
1 −1 0
0 1 ⎦. Since there is only one unpivoted column in this row-echelon
and this matrix row-reduces to ⎣ 0
0
0 0
form of A, the eigenspace corresponding to λ = −5 is only one-dimensional. Thus, based on the eigenvalues
λ = −5, −5, −6, we already know that
⎡
⎤
−5
1
0
0 ⎦.
JCF(A) = ⎣ 0 −5
0
0 −6
Next, we seek a cycle of generalized eigenvectors of length 2 corresponding to λ = −5. The cycle
2
takes the form
+ 5I)v, v},
⎤ where v is a vector such that (A + 5I) v = 0. We readily compute that
⎡ {(A
1
1
1
−2 2
2
0 0 ⎦. An obvious vector that is killed by (A + 5I)2 (although other choices are
(A + 5I)2 = ⎣ 0
1
1
−⎡2 12⎤
2
⎤
⎡
0
1
also possible) is v = ⎣ 1 ⎦. Then (A + 5I)v = ⎣ 1 ⎦. Hence, we have a cycle of generalized eigenvectors
1
0
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corresponding to λ = −5:
⎧⎡
⎤ ⎡
⎤⎫
0 ⎬
⎨ 1
⎣ 1 ⎦,⎣ 1 ⎦ .
⎩
⎭
0
1
Now consider the eigenspace corresponding to λ = −6. We need only find one eigenvector (x, y, z) in this
eigenspace. To do so, we must compute
⎤
⎡
0 1 0
nullspace(A + 6I) = nullspace ⎣ − 12 32 12 ⎦ ,
− 12 12 12
⎡
⎤
1 −3 −1
1
0 ⎦. We see that y = 0 and x − 3y − z = 0, which is equivalent
and this matrix row-reduces to ⎣ 0
0
0
0
to x − z = 0. Setting z = t, we have x = t. With t = 1, we obtain the eigenvector (1, 0, 1). Hence, we can
form the matrix
⎡
⎤
1 0 1
S = ⎣ 1 1 0 ⎦.
0 1 1
25. Because the matrix is upper triangular, the eigenvalues of A appear along the main diagonal: λ = 2, 2, 3.
Let us first consider the eigenspace corresponding to λ = 2: We consider
⎡
⎤
0 −2 14
1 −7 ⎦ ,
nullspace(A − 2I) = nullspace ⎣ 0
0
0
0
⎡
⎤
0 1 −7
0 ⎦. There are two unpivoted columns, so this eigenspace is two-dimensional.
which row reduces to ⎣ 0 0
0 0
0
Therefore, the matrix A is diagonalizable:
⎤
⎡
2 0 0
JCF(A) = ⎣ 0 2 0 ⎦ .
0 0 3
Next, we must determine an invertible matrix S such that S −1 AS is the Jordan canonical form just obtained.
Using the row-echelon form of A − 2I obtained above, vectors (x, y, z) in nullspace(A − 2I) must satisfy
y − 7z = 0. Setting z = t, y = 7t, and x = s, we obtain the eigenvectors (1, 0, 0) and (0, 7, 1).
Next, consider the eigenspace corresponding to λ = 3. We consider
⎡
⎤
−1 −2 14
0 −7 ⎦ ,
nullspace(A − 3I) = nullspace ⎣ 0
0
0 −1
⎡
⎤
1 2 −14
1 ⎦. Vectors (x, y, z) in the nullspace of this matrix must satisfy z = 0
which row reduces to ⎣ 0 0
0 0
0
and x + 2y − 14z = 0 or x + 2y = 0. Setting y = t and x = −2t. Hence, setting t = 1 gives the eigenvector
(−2, 1, 0).
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Thus, using the eigenvectors obtained above, we obtain the matrix
⎤
⎡
1 0 −2
1 ⎦.
S=⎣ 0 7
0 1
0
26. We use the characteristic polynomial to determine the eigenvalues of A:
⎡
⎤
7−λ
−2
2
4−λ
−1 ⎦
det(A − λI) = det ⎣ 0
−1
1
4−λ
= (4 − λ) [(7 − λ)(4 − λ) + 2] + (7 − λ − 2)
= (4 − λ)(λ2 − 11λ + 30) + (5 − λ)
= (4 − λ)(λ − 5)(λ − 6) + (5 − λ)
= (5 − λ) [1 − (4 − λ)(6 − λ)]
= (5 − λ)(λ − 5)2
= −(λ − 5)3 .
Hence, the eigenvalues are λ = 5, 5, 5. Let us consider the eigenspace corresponding to λ = 5. We consider
⎡
⎤
2 −2
2
nullspace(A − 5I) = nullspace ⎣ 0 −1 −1 ⎦ ,
−1
1 −1
⎡
⎤
1 −1 1
1 1 ⎦, which contains one unpivoted column. Therefore, the
and this latter matrix row-reduces to ⎣ 0
0
0 0
eigenspace corresponding to λ = 5 is one-dimensional. Therefore, the Jordan canonical form of A consists
of one Jordan block:
⎡
⎤
5 1 0
JCF(A) = ⎣ 0 5 1 ⎦ .
0 0 5
A corresponding invertible matrix S in this case must have columns that consist of one cycle of generalized
eigenvectors, which will take the form {(A − 5I)2 v, (A − 5I)v, v}, where v is a generalized eigenvector. Now,
we can verify quickly that
⎡
⎤
⎡
⎤
2 −2
2
2 0
4
2 ⎦ , (A − 5I)3 = 03 .
A − 5I = ⎣ 0 −1 −1 ⎦ , (A − 5I)2 = ⎣ 1 0
−1
1 −1
−1 0 −2
we
The fact that (A − 5I)3 = 03 means that every nonzero vector v is a generalized eigenvector.
⎡ Hence,
⎤
1
simply choose v such that (A − 5I)2 v = 0. There are many choices. Let us take v = ⎣ 0 ⎦. Then
0
⎡
⎤
⎡
⎤
2
2
(A − 5I)v = ⎣ 0 ⎦ and (A − 5I)2 v = ⎣ 1 ⎦. Thus, we have the cycle of generalized eigenvectors
−1
−1
⎧⎡
⎤ ⎡
⎤ ⎡
⎤⎫
2
2
1 ⎬
⎨
⎣ 1 ⎦,⎣ 0 ⎦,⎣ 0 ⎦ .
⎩
⎭
−1
−1
0
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Hence, we have
⎡
⎤
2
2 1
0 0 ⎦.
S=⎣ 1
−1 −1 0
27. Because the matrix is upper triangular, the eigenvalues of A appear along the main diagonal: λ =
−1, −1, −1. Let us consider the eigenspace corresponding to λ = −1. We consider
⎡
⎤
0 −1
0
0 −2 ⎦ ,
nullspace(A + I) = nullspace ⎣ 0
0
0
0
and it is straightforward to see that the nullspace here consists precisely of vectors that are multiples of
(1, 0, 0). Because only one linearly independent eigenvector was obtained, the Jordan canonical form of this
matrix consists of only one Jordan block:
⎡
⎤
−1
1
0
1 ⎦.
JCF(A) = ⎣ 0 −1
0
0 −1
A corresponding invertible matrix S in this case must have columns that consist of one cycle of generalized
eigenvectors, which will take the form {(A + I)2 v, (A + I)v, v}. Here, we have
⎡
⎤
⎡
⎤
0 0 2
0 −1
0
0 −2 ⎦
A+I =⎣ 0
and (A + I)3 = 03 .
and (A + I)2 = ⎣ 0 0 0 ⎦
0 0 0
0
0
0
Therefore, every nonzero vector is a generalized eigenvector. We⎡ wish
⎤ to choose a vector ⎡v such⎤ that
0
0
(A + I)2 v = 0. There are many choices, but we will choose v = ⎣ 0 ⎦. Then (A + I)v = ⎣ −2 ⎦ and
1
0
⎡
⎤
2
(A + I)2 v = ⎣ 0 ⎦. Hence, we form the matrix S as follows:
0
⎤
2
0 0
S = ⎣ 0 −2 0 ⎦ .
0
0 1
⎡
28. We use the characteristic polynomial to determine the eigenvalues of A:
⎤
⎡
2−λ
−1
0
1
⎢ 0
3−λ
−1
0 ⎥
⎥
det(A − λI) = det ⎢
⎣ 0
1
1−λ
0 ⎦
0
−1
0
3−λ
= (2 − λ)(3 − λ) [(3 − λ)(1 − λ) + 1]
= (2 − λ)(3 − λ)(λ2 − 4λ + 4) = (2 − λ)(3 − λ)(λ − 2)2 ,
and so the eigenvalues are λ = 2, 2, 2, 3.
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First, consider the eigenspace corresponding to λ = 2. We consider
⎤
⎡
0 −1
0 1
⎢ 0
1 −1 0 ⎥
⎥,
nullspace(A − 2I) = nullspace ⎢
⎣ 0
1 −1 0 ⎦
0 −1
0 1
⎤
⎡
0 1 −1
0
⎥
⎢ 0 0
1
−1
⎥. There are two unpivoted columns, and
and this latter matrix can be row-reduced to ⎢
⎣ 0 0
0
0 ⎦
0 0
0
0
therefore two linearly independent eigenvectors corresponding to λ = 2. Thus, we will obtain two Jordan
blocks corresponding to λ = 2, and they necessarily will have size 2 × 2 and 1 × 1. Thus, we are already in
a position to write down the Jordan canonical form of A:
⎤
⎡
2 1 0 0
⎢ 0 2 0 0 ⎥
⎥
JCF(A) = ⎢
⎣ 0 0 2 0 ⎦.
0 0 0 3
We continue in order to obtain an invertible matrix S such that S −1 AS is in Jordan canonical form. To
this end, we see a generalized eigenvector v such that (A − 2I)v = 0 and (A − 2I)2 v = 0. Note that
⎤
⎤
⎡
⎡
0 −1
0 1
0 −2 1 1
⎢ 0
⎢ 0
1 −1 0 ⎥
0 0 0 ⎥
⎥
⎥.
and (A − 2I)2 = ⎢
A − 2I = ⎢
⎦
⎣ 0
⎣
1 −1 0
0
0 0 0 ⎦
0 −1
0 1
0 −2 1 1
⎤
⎡
0
⎢ 0 ⎥
⎥
By inspection, we see that by taking v = ⎢
⎣ −1 ⎦ (there are many other valid choices, of course), then
1
⎤
⎡
1
⎢ 1 ⎥
2
⎥
(A − 2I)v = ⎢
⎣ 1 ⎦ and (A − 2I) v = 0. We also need a second eigenvector corresponding to λ = 2 that is
1
linearly independent from (A − 2I)v just obtained.⎡ From
⎤ the row-echelon form of A −⎡ 2I, ⎤we see that all
1
s
⎢ 0 ⎥
⎢ t ⎥
⎥
⎥
⎢
eigenvectors corresponding to λ = 2 take the form ⎢
⎣ t ⎦, so for example, we can take ⎣ 0 ⎦.
0
t
Next, we consider the eigenspace corresponding to λ = 3. We consider
⎤
⎡
−1 −1
0 1
⎢ 0
0 −1 0 ⎥
⎥.
nullspace(A − 3I) = nullspace ⎢
⎣ 0
1 −2 0 ⎦
0 −1
0 0
Now, if (x, y, z, w) is an eigenvector corresponding to λ = 3, the last three rows of the matrix imply that
y = z = 0. Thus, the first row becomes −x + w = 0. Setting w = t, then x = t, so we obtain eigenvectors in
the form (t, 0, 0, t). Setting t = 1 gives the eigenvector (1, 0, 0, 1).
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Thus, we can now form the matrix S such that S −1 AS is the Jordan canonical form we obtained above:
⎤
⎡
1
0 1 1
⎢ 1
0 0 0 ⎥
⎥
S=⎢
⎣ 1 −1 0 0 ⎦ .
1
1 0 1
29. From the characteristic polynomial, we have eigenvalues λ = 2, 2, 4, 4. Let us consider the associated
eigenspaces.
Corresponding to λ = 2, we seek eigenvectors (x, y, z, w) by computing
⎤
⎡
0 −4 2 2
⎢ −2 −2 1 3 ⎥
⎥
nullspace(A − 2I) = nullspace ⎢
⎣ −2 −2 1 3 ⎦ ,
−2 −6 3 5
⎤
⎡
2 2 −1 −3
⎢ 0 2 −1 −1 ⎥
⎥. Setting w = 2t and z = 2s, we obtain y = s + t
and this matrix can be row-reduced to ⎢
⎣ 0 0
0
0 ⎦
0 0
0
0
and x = 2t, so we obtain the eigenvectors (2, 1, 0, 2) and (0, 1, 2, 0) corresponding to λ = 2.
Next, corresponding to λ = 4, we seek eigenvectors (x, y, z, w) by computing
⎤
⎡
−2 −4
2 2
⎢ −2 −4
1 3 ⎥
⎥
nullspace(A − 4I) = nullspace ⎢
⎣ −2 −2 −1 3 ⎦ .
−2 −6
3 3
⎤
⎡
1 2 −1 −1
⎢ 0 2 −1 −1 ⎥
⎥. Since there is only one unpivoted column, this eigenspace
This matrix can be reduced to ⎢
⎣ 0 0
1 −1 ⎦
0 0
0
0
is only one-dimensional, despite λ = 4 occurring with multiplicity 2 as a root of the characteristic equation.
Therefore, we must seek a generalized eigenvector v such that (A − 4I)v is an eigenvector. This in turn
requires that (A − 4I)2 v = 0. We find that
⎤
⎤
⎡
⎡
4 8 −4 −4
−2 −4
2 2
⎢ 4 4
⎢ −2 −4
0 −4 ⎥
1 3 ⎥
⎥.
⎥
and (A − 4I)2 = ⎢
A − 4I = ⎢
⎣ 4 0
⎣ −2 −2 −1 3 ⎦
4 −4 ⎦
4 8 −4 −4
−2 −6
3 3
⎤
⎡
⎤
⎡
1
0
⎢ 0 ⎥
⎢ 1 ⎥
2
⎥
⎥
⎢
Note that the vector v = ⎢
⎣ 0 ⎦ satisfies (A − 4I) v = 0 and (A − 4I)v = ⎣ 1 ⎦. Thus, we have the cycle
1
1
of generalized eigenvectors corresponding to λ = 4 given by
⎧⎡
⎤⎫
⎤ ⎡
1 ⎪
⎪
⎪
⎪ 0
⎨
⎢ 1 ⎥ ⎢ 0 ⎥⎬
⎥ .
⎥,⎢
⎢
⎣ 1 ⎦ ⎣ 0 ⎦⎪
⎪
⎪
⎪
⎭
⎩
1
1
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Hence, we can form the matrix
⎡
2
⎢ 1
S=⎢
⎣ 0
2
0
1
2
0
0
1
1
1
⎤
1
0 ⎥
⎥
0 ⎦
1
⎡
and
2
⎢ 0
JCF(A) = ⎢
⎣ 0
0
0
2
0
0
0
0
4
0
⎤
0
0 ⎥
⎥.
1 ⎦
4
30. Since A is upper triangular, the eigenvalues appear along the main diagonal: λ = 2, 2, 2, 2, 2. Looking
at
⎤
⎡
0 1 1 1 1
⎢ 0 0 0 0 1 ⎥
⎥
⎢
⎥
nullspace(A − 2I) = nullspace ⎢
⎢ 0 0 0 0 1 ⎥,
⎣ 0 0 0 0 1 ⎦
0 0 0 0 0
we see that the row-echelon form of this matrix will contain two pivots, and therefore, three unpivoted
columns. That means that the eigenspace corresponding to λ = 2 is three-dimensional. Therefore, JCF(A)
consists of three Jordan blocks. The only list of block sizes for a 5 × 5 matrix with three blocks are (a) 3,1,1
and (b) 2,2,1. In this case, note that
⎤
⎡
0 0 0 0 4
⎢ 0 0 0 0 1 ⎥
⎥
⎢
⎥
(A − 2I)2 = ⎢
⎢ 0 0 0 0 1 ⎥ = 05 ,
⎣ 0 0 0 0 1 ⎦
0 0 0 0 1
so that it is possible to find a vector v that generates a cycle of generalized eigenvectors of length 3:
{(A − 2I)2 v, (A − 2I)v, v}. Thus, JCF(A) contains a Jordan block of size 3 × 3. We conclude that the
correct list of block sizes for this matrix is 3,1,1:
⎤
⎡
2 1 0 0 0
⎢ 0 2 1 0 0 ⎥
⎥
⎢
⎥
JCF(A) = ⎢
⎢ 0 0 2 0 0 ⎥.
⎣ 0 0 0 2 0 ⎦
0 0 0 0 2
31. Since A is upper triangular, the eigenvalues appear along the main diagonal: λ = 0, 0, 0, 0, 0. Looking at
nullspace(A − 0I) = nullspace(A), we see that eigenvectors (x, y, z, u, v) corresponding to λ = 0 must satisfy
z = u = v = 0 (since the third row gives 6u = 0, the first row gives u + 4v = 0, and the second row gives
z + u + v = 0). Thus, we have only two free variables, and thus JCF(A) will consist of two Jordan blocks.
The only list of block sizes for a 5 × 5 matrix with two blocks are (a)4,1 and (b) 3,2. In this case, it is easy
to verify that A3 = 0, so that the longest possible cycle of generalized eigenvectors {A2 v, Av, v} has length
3. Therefore, case (b) holds: JCF(A) consists of one Jordan block of size 3 × 3 and one Jordan block of size
2 × 2:
⎤
⎡
0 1 0 0 0
⎢ 0 0 1 0 0 ⎥
⎥
⎢
⎥
JCF(A) = ⎢
⎢ 0 0 0 0 0 ⎥.
⎣ 0 0 0 0 1 ⎦
0 0 0 0 0
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32. Since A is upper triangular, the eigenvalues appear along the main diagonal: λ = 1, 1, 1, 1, 1, 1, 1, 1.
Looking at
⎤
⎡
0 1 0 1 0 1 0 1
⎢ 0 0 0 0 0 0 0 0 ⎥
⎥
⎢
⎢ 0 0 0 1 0 1 0 1 ⎥
⎥
⎢
⎢ 0 0 0 0 0 0 0 0 ⎥
⎥
nullspace(A − I) = nullspace ⎢
⎢ 0 0 0 0 0 1 0 1 ⎥,
⎥
⎢
⎢ 0 0 0 0 0 0 0 0 ⎥
⎥
⎢
⎣ 0 0 0 0 0 0 0 1 ⎦
0 0 0 0 0 0 0 0
we see that if (a, b, c, d, e, f, g, h) is an eigenvector of A, then b = d = f = h = 0, and a, c, e, and g are
free variables. Thus, we have four linearly independent eigenvectors of A, and hence we expect four Jordan
blocks. Now, an easy calculation shows that (A − I)2 = 0, and thus, no Jordan blocks of size greater than
2 × 2 are permissible. Thus, it must be the case that JCF(A) consists of four Jordan blocks, each of which
is a 2 × 2 matrix:
⎤
⎡
1 1 0 0 0 0 0 0
⎢ 0 1 0 0 0 0 0 0 ⎥
⎥
⎢
⎢ 0 0 1 1 0 0 0 0 ⎥
⎥
⎢
⎢ 0 0 0 1 0 0 0 0 ⎥
⎥
JCF(A) = ⎢
⎢ 0 0 0 0 1 1 0 0 ⎥.
⎥
⎢
⎢ 0 0 0 0 0 1 0 0 ⎥
⎥
⎢
⎣ 0 0 0 0 0 0 1 1 ⎦
0 0 0 0 0 0 0 1
33. NOT SIMILAR. We will compute JCF(A) and JCF(B). If they are the same (up to a rearrangement
of the Jordan blocks), then A and B are similar; otherwise they are not. Both matrices have eigenvalues
λ = 6, 6, 6. Consider
⎡
⎤
1
1 0
nullspace(A − 6I) = nullspace ⎣ −1 −1 0 ⎦ .
1
0 0
We see that eigenvectors (x, y, z) corresponding to λ = 6 must satisfy x + y = 0 and x = 0. Therefore,
x = y = 0, while z is a free variable. Since we obtain only one free variable, JCF(A) consists of just one
Jordan block corresponding to λ = 6:
⎡
⎤
6 1 0
JCF(A) = ⎣ 0 6 1 ⎦ .
0 0 6
Next, consider
⎡
⎤
0 −1 1
0 0 ⎦.
nullspace(B − 6I) = nullspace ⎣ 0
0
0 0
In this case, eigenvectors (x, y, z) corresponding to λ = 6 must satisfy −y + z = 0. Therefore, both x and z
are free variables, and hence, JCF(B) consists of two Jordan blocks corresponding to λ = 6:
⎤
⎡
6 1 0
JCF(B) = ⎣ 0 6 0 ⎦ .
0 0 6
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Since JCF(A) = JCF(B), we conclude that A and B are not similar.
34. SIMILAR. We will compute JCF(A) and JCF(B). If they are the same (up to a rearrangement of the
Jordan blocks), then A and B are similar; otherwise they are not. Both matrices have eigenvalues λ = 5, 5, 5.
(For A, this is easiest to compute by expanding det(A − λI) along the middle row, and for B, this is easiest
to compute by expanding det(B − λI) along the second column.) In Problem 26, we computed
⎡
⎤
5 1 0
JCF(A) = ⎣ 0 5 1 ⎦ .
0 0 5
Next, consider
⎡
⎤
−2 −1 −2
1
1 ⎦,
nullspace(B − 5I) = nullspace ⎣ 1
1
0
1
⎡
⎤
1 1 1
and this latter matrix can be row-reduced to ⎣ 0 1 0 ⎦, which has only one unpivoted column. Thus,
0 0 0
the eigenspace of B corresponding to λ = 5 is only one-dimensional, and so
⎡
⎤
5 1 0
JCF(B) = ⎣ 0 5 1 ⎦ .
0 0 5
Since A and B each had the same Jordan canonical form, they are similar matrices.
35. SIMILAR. We will compute JCF(A) and JCF(B). If they are the same (up to a rearrangement
of the Jordan blocks), then A and B are similar; otherwise they are not. Both matrices have eigenvalues
λ = −1, −1, 2. For λ = −1, consider
⎡
⎤
4 0
4
0 ⎦.
nullspace(A + I) = nullspace ⎣ 0 3
−4 0 −4
We see that the eigenvectors (x, y, z) corresponding to λ = −1 must satisfy x + z = 0 and y = 0. Therefore,
we can choose z = t as a free variable, from which x = −t. Hence, we only obtain one linearly independent
eigenvector (−1, 0, 1) corresponding to λ − 1, from which we conclude that
⎡
⎤
−1
1 0
JCF(A) = ⎣ 0 −1 0 ⎦ .
0
0 2
Next, consider
⎡
⎤
0 −1 3
0 1 ⎦.
nullspace(B + I) = nullspace ⎣ 0
0
0 3
In this case, eigenvectors (x, y, z) corresponding to λ = −1 must satisfy y = z = 0. Therefore, only x = t
is a free variable, resulting in just one linearly independent eigenvector (1, 0, 0) corresponding to λ = −1.
Therefore, we conclude that
⎡
⎤
−1
1 0
JCF(B) = ⎣ 0 −1 0 ⎦ .
0
0 2
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Since JCF(A) = JCF(B), we conclude that A and B are similar.
36. The eigenvalues of A are λ = −5, −5, and the eigenspace corresponding to λ = −5 is only onedimensional. Thus, we seek a generalized eigenvector v of A corresponding
to λ = −5 such that {(A+5I)v, v}
1
1
is a cycle of generalized eigenvectors. Note that A + 5I =
and (A + 5I)2 = 02 . Thus, every
−1 −1
1
nonzero vector v is a generalized eigenvector of A corresponding to λ = −5. Let us choose v =
. Then
0
1
(A + 5I)v =
. Form the matrices
−1
1 1
−5
1
S=
and J =
.
−1 0
0 −5
Via the substitution x = Sy, the system x = Ax is transformed into y = Jy. The corresponding equations
are
y1 = −5y1 + y2 and y2 = −5y2 .
The solution to the second equation is
y2 (t) = c1 e−5t .
Substituting this solution into y1 = −5y1 + y2 gives
y1 + 5y1 = c1 e−5t .
This is a first order linear equation with integrating factor I(t) = e5t . When we multiply the differential
equation for y1 (t) by I(t), it becomes (y1 · e5t ) = c1 . Integrating both sides yields y1 · e5t = c1 t + c2 . Thus,
y1 (t) = c1 te−5t + c2 e−5t .
Thus, we have
y(t) =
y1 (t)
y2 (t)
=
c1 te−5t + c2 e−5t
c1 e−5t
.
Finally, we solve for x(t):
c1 te−5t + c2 e−5t
x(t) = Sy(t) =
c1 e−5t
c1 te−5t + c2 e−5t + c1 e−5t
=
−(c1 te−5t + c2 e−5t )
t+1
1
−5t
−5t
+ c2 e
.
= c1 e
−t
−1
1
−1
1
0
This is an acceptable answer, or we can write the individual equations comprising the general solution:
x1 (t) = c1 te−5t + c2 e−5t + c1 e−5t
and
x2 (t) = −(c1 te−5t + c2 e−5t ).
37. The eigenvalues of A are λ = −1, −1, and the eigenspace corresponding to λ = −1 is only onedimensional. Thus, we seek a generalized eigenvector v of A corresponding
to λ = −1 such that {(A+I)v, v}
−2 −2
is a cycle of generalized eigenvectors. Note that A + I =
and (A + I)2 = 02 . Thus, every
2
2
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nonzero vector v is a generalized eigenvector of A corresponding to λ = −1. Let us choose v =
−2
(A + I)v =
. Form the matrices
2
−2 1
−1
1
S=
and J =
.
2 0
0 −1
1
0
. Then
Via the substitution x = Sy, the system x = Ax is transformed into y = Jy. The corresponding equations
are
y1 = −y1 + y2 and y2 = −y2 .
The solution to the second equation is
y2 (t) = c1 e−t .
Substituting this solution into y1 = −y1 + y2 gives
y1 + y1 = c1 e−t .
This is a first order linear equation with integrating factor I(t) = et . When we multiply the differential
equation for y1 (t) by I(t), it becomes (y1 · et ) = c1 . Integrating both sides yields y1 · et = c1 t + c2 . Thus,
y1 (t) = c1 te−t + c2 e−t .
Thus, we have
y(t) =
y1 (t)
y2 (t)
=
c1 te−t + c2 e−t
c1 e−t
.
Finally, we solve for x(t):
x(t) = Sy(t) =
−2
2
1
0
c1 te−t + c2 e−t
c1 e−t
−2(c1 te−t + c2 e−t ) + c1 e−t
=
2(c1 te−t + c2 e−t )
−2t + 1
−2
−t
−t
+ c2 e
.
= c1 e
2t
2
This is an acceptable answer, or we can write the individual equations comprising the general solution:
x1 (t) = −2(c1 te−t + c2 e−t ) + c1 e−t
and
x2 (t) = 2(c1 te−t + c2 e−t ).
38. The eigenvalues of A are λ = −1, −1, 1. The eigenspace corresponding to λ = −1 is
⎡
⎤
1 1 0
nullspace(A + I) = nullspace ⎣ 0 1 1 ⎦ ,
1 1 0
which is only one-dimensional, spanned by the vector (1, −1, 1). Therefore, we seek a generalized eigenvector
v of A corresponding to λ = −1 such that {(A + I)v, v} is a cycle of generalized eigenvectors. Note that
⎡
⎡
⎤
⎤
1 1 0
1 2 1
A+I =⎣ 0 1 1 ⎦
and (A + I)2 = ⎣ 1 2 1 ⎦ .
1 1 0
1 2 1
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In order that v be a generalized eigenvector of A corresponding to λ = −1, we should choose
that
⎡ v such
⎤
1
(A + I)2 v = 0 and (A + I)v = 0. There are many valid choices; let us choose v = ⎣ 0 ⎦. Then
−1
⎡
⎤
1
(A + I)v = ⎣ −1 ⎦. Hence, we obtain the cycle of generalized eigenvectors corresponding to λ = −1:
1
⎧⎡
⎤ ⎡
⎤⎫
1
1 ⎬
⎨
⎣ −1 ⎦ , ⎣ 0 ⎦ .
⎩
⎭
1
−1
Next, consider the eigenspace corresponding to λ = 1. For this, we compute
⎤
−1
1
0
1 ⎦.
nullspace(A − I) = nullspace ⎣ 0 −1
1
1 −2
⎡
⎡
⎤
⎤
1
1 −1
0
1 −1 ⎦. We find the eigenvector ⎣ 1 ⎦ as a basis for this eigenspace.
This can be row-reduced to ⎣ 0
1
0
0
0
Hence, we are ready to form the matrices S and J:
⎡
⎡
⎤
1
1 1
0 1 ⎦
S = ⎣ −1
1 −1 1
⎡
and
⎤
−1
1 0
J = ⎣ 0 −1 0 ⎦ .
0
0 1
Via the substitution x = Sy, the system x = Ax is transformed into y = Jy. The corresponding equations
are
y1 = −y1 + y2 ,
y2 = −y2 ,
y3 = y3 .
The third equation has solution y3 (t) = c3 et , the second equation has solution y2 (t) = c2 e−t , and so the first
equation becomes
y1 + y1 = c2 e−t .
This is a first-order linear equation with integrating factor I(t) = et . When we multiply the differential
equation for y1 (t) by I(t), it becomes (y1 · et ) = c2 . Integrating both sides yields y1 · et = c2 t + c1 . Thus,
y1 (t = c2 te−t + c1 e−t .
Thus, we have
⎡
⎤
⎤ ⎡
y1 (t)
c2 te−t + c1 e−t
⎦.
c2 e−t
y(t) = ⎣ y2 (t) ⎦ = ⎣
t
y3 (t)
c3 e
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Finally, we solve for x(t):
⎤
⎤⎡
1
1 1
c2 te−t + c1 e−t
⎦
c2 e−t
0 1 ⎦⎣
x(t) = Sy(t) = ⎣ −1
c3 et
1 −1 1
⎤
⎡
c2 te−t + c1 e−t + c2 e−t + c3 et
⎦
−c2 te−t − c1 e−t + c3 et
=⎣
c2 te−t + c1 e−t − c2 e−t + c3 et
⎡
⎤
⎡
⎤
⎡
⎤
1
t+1
1
= c1 e−t ⎣ −1 ⎦ + c2 e−t ⎣ −t ⎦ + c3 et ⎣ 1 ⎦ .
1
t−1
1
⎡
39. The eigenvalues of A are λ = −2, −2, −2. The eigenspace corresponding to λ = −2 is
⎡
⎤
0
0
0
nullspace(A + 2I) = nullspace ⎣ 1 −1 −1 ⎦ ,
−1
1
1
and there are two linearly independent vectors in this nullspace, corresponding to the unpivoted columns of
the row-echelon form of this matrix. Therefore, the Jordan canonical form of A is
⎡
⎤
−2
1
0
0 ⎦.
J = ⎣ 0 −2
0
0 −2
To form an invertible matrix S such that S −1 AS = J, we must find a cycle of generalized eigenvectors
corresponding to λ = −2 of length 2: {(A + 2I)v, v}. Now
⎡
⎤
0
0
0
A + 2I = ⎣ 1 −1 −1 ⎦
and (A + 2I)2 = 03 .
−1
1
1
Since (A + 2I)2 = 03 , every nonzero vector in R3 is a generalized eigenvector corresponding to λ = −2. We
need ⎡only ⎤find a nonzero vector v⎡ such ⎤that (A + 2I)v = 0. There are many valid choices; let us choose
1
0
v = ⎣ 0 ⎦. Then (A + 2I)v = ⎣ 1 ⎦, an eigenvector of A corresponding to λ = −2. We also need a
0
−1
second
⎡
⎤ linearly independent eigenvector corresponding to λ = −2. There are many choices; let us choose
1
⎣ 1 ⎦. Therefore, we can form the matrix
0
⎡
0
S=⎣ 1
−1
1
0
0
⎤
1
1 ⎦.
0
Via the substitution x = Sy, the system x = Ax is transformed into y = Jy. The corresponding equations
are
y1 = −2y1 + y2 , y2 = −2y2 , y3 = −2y3 .
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The third equation has solution y3 (t) = c3 e−2t , the second equation has solution y2 (t) = c2 e−2t , and so the
first equation becomes
y1 + 2y1 = c2 e−2t .
This is a first-order linear equation with integrating factor I(t) = e2t . When we multiply the differential
equation for y1 (t) by I(t), it becomes (y1 · e2t ) = c2 . Integrating both sides yields y1 · e2t = c2 t + c1 . Thus,
y1 (t) = c2 te−2t + c1 e−2t .
Thus, we have
⎤
⎤ ⎡
y1 (t)
c2 te−2t + c1 e−2t
⎦.
c2 e−2t
y(t) = ⎣ y2 (t) ⎦ = ⎣
−2t
y3 (t)
c3 e
⎡
Finally, we solve for x(t):
⎤
⎤⎡
0 1 1
c2 te−2t + c1 e−2t
⎦
c2 e−2t
x(t) = Sy(t) = ⎣ 1 0 1 ⎦ ⎣
−1 0 0
c3 e−2t
⎤
⎡
c2 e−2t + c3 e−2t
= ⎣ c2 te−2t + c1 e−2t + c3 e−2t ⎦
−(c2 te−2t + c1 e−2t )
⎡
⎤
⎡
⎤
⎡
⎤
0
1
1
= c1 e−2t ⎣ 1 ⎦ + c2 e−2t ⎣ t ⎦ + c3 e−2t ⎣ 1 ⎦ .
−1
−t
0
⎡
40. The eigenvalues of A are λ = 4, 4, 4. The eigenspace corresponding to λ = 4 is
⎡
⎤
0 0 0
nullspace(A − 4I) = nullspace ⎣ 1 0 0 ⎦ ,
0 1 0
and there is only one eigenvector. Therefore, the Jordan canonical form of A is
⎡
⎤
4 1 0
J = ⎣ 0 4 1 ⎦.
0 0 4
Next, we need to find an invertible matrix S such that S −1 AS = J. To do this, we must find a cycle of
generalized eigenvectors {(A − 4I)2 v, (A − 4I)v, v} of length 3 corresponding to λ = 4. We have
⎡
⎤
⎡
⎤
0 0 0
0 0 0
A − 4I = ⎣ 1 0 0 ⎦
and (A − 4I)2 = ⎣ 0 0 0 ⎦
and (A − 4I)3 = 03 .
0 1 0
1 0 0
From (A − 4I)3 = 03 , we
⎡ know
⎤ that every nonzero vector is a generalized eigenvector corresponding to
1
λ = 4. We choose v = ⎣ 0 ⎦ (any multiple of the chosen vector v would be acceptable as well). Thus,
0
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⎡
⎤
⎤
0
0
(A − 4I)v = ⎣ 1 ⎦ and (A − 4I)2 v = ⎣ 0 ⎦. Thus, we have the cycle of generalized eigenvectors
1
0
⎡
⎧⎡
⎤ ⎡
⎤ ⎡
⎤⎫
0
1 ⎬
⎨ 0
⎣ 0 ⎦,⎣ 1 ⎦,⎣ 0 ⎦ .
⎩
⎭
1
0
0
Thus, we can form the matrix
⎡
0
S=⎣ 0
1
0
1
0
⎤
1
0 ⎦.
0
Via the substitution x = Sy, the system x = Ax is transformed into y = Jy. The corresponding equations
are
y1 = 4y1 + y2 ,
y2 = 4y2 + y3 ,
y3 = 4y3 .
The third equation has solution y3 (t) = c3 e4t , and the second equation becomes
y2 − 4y2 = c3 e4t .
This is a first-order linear equation with integrating factor I(t) = e−4t . When we multiply the differential
equation for y2 (t) by I(t), it becomes (y2 · e−4t ) = c3 . Integrating both sides yields y2 · e−4t = c3 t + c2 .
Thus,
y2 (t) = c3 te4t + c2 e4t = e4t (c3 t + c2 ).
Therefore, the differential equation for y1 (t) becomes
y1 − 4y1 = e4t (c3 t + c2 ).
This equation is first-order linear with integrating factor I(t) = e−4t . When we multiply the differential
equation for y1 (t) by I(t), it becomes
(y1 · e−4t ) = c3 t + c2 .
Integrating both sides, we obtain
y1 · e−4t = c3
Hence,
Thus, we have
t2
+ c2 t + c1 .
2
2
t
y1 (t) = e4t c3 + c2 t + c1 .
2
⎤
⎤ ⎡ 4t t2
y1 (t)
c3 2 + c2 t + c1
e
⎥
⎢
y(t) = ⎣ y2 (t) ⎦ = ⎣
⎦.
e4t (c3 t + c2 )
4t
y3 (t)
c3 e
⎡
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Finally, we solve for x(t):
⎡
0
x(t) = Sy(t) = ⎣ 0
1
⎡
⎢
=⎣
⎤
⎤ ⎡ 4t t2
1
c3 2 + c2 t + c1
e
⎥
⎢
0 ⎦⎣
⎦
e4t (c3 t + c2 )
4t
0
c3 e
⎤
4t
c3 e
4t
⎥
e (c3 t + c2 ) ⎦
0
1
0
2
e4t c3 t2 + c2 t + c1
⎡
⎤
⎡
⎤
⎡
⎤
0
0
1
= c1 e4t ⎣ 0 ⎦ + c2 e4t ⎣ 1 ⎦ + c3 e4t ⎣ t ⎦ .
1
t
t2 /2
41. The eigenvalues of A are λ = −1, −1, 2. The eigenspace corresponding to λ = −1 is
⎡
⎤
4 0
4
0 ⎦.
nullspace(A + I) = nullspace ⎣ 0 3
−4 0 −4
We see that the eigenvectors (x, y, z) corresponding to λ = −1 must satisfy x + z = 0 and y = 0. Therefore,
we can choose z = t as a free variable, from which x = −t. Hence, we only obtain one linearly independent
eigenvector (−1, 0, 1) corresponding to λ = −1, from which we conclude that
⎡
⎤
−1
1 0
JCF(A) = J = ⎣ 0 −1 0 ⎦ .
0
0 2
To form an invertible matrix S such that S −1 AS = J, we must find a cycle of generalized eigenvectors
corresponding to λ = −1 of length 2: {(A + I)v, v}. Now
⎡
⎤
⎡
⎤
4 0
4
0 0 0
0 ⎦
A+I =⎣ 0 3
and (A + I)2 = ⎣ 0 9 0 ⎦ .
−4 0 −4
0 0 0
We need to find
v such that (A ⎡
+ I)v ⎤
= 0 but (A + I)2 v = 0. There are many valid choices; let
⎡ a vector
⎤
1
4
us choose v = ⎣ 0 ⎦. Then (A + I)v = ⎣ 0 ⎦, an eigenvector of A corresponding to λ = −1. Thus, we
−4
⎧0⎡
⎤ ⎡
⎤⎫
4
1
⎨
⎬
obtain the cycle ⎣ 0 ⎦ , ⎣ 0 ⎦ .
⎩
⎭
−4
0
Next, we seek an eigenvector corresponding to λ = 2. The eigenspace corresponding to λ = 2 is
⎡
⎤
1 0
4
0 ⎦.
nullspace(A − 2I) = nullspace ⎣ 0 0
−4 0 −7
We see that the eigenvectors (x, y, z) corresponding to λ = 2 must satisfy x + 4z = 0 and −4x − 7z = 0. We
quickly find that x = z = 0. However, y = t is a free variable. Therefore, we obtain the eigenvector (0, 1, 0)
corresponding to λ = 2.
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Therefore, we can form the matrix
⎡
4
S=⎣ 0
−4
1
0
0
⎤
0
1 ⎦.
0
Via the substitution x = Sy, the system x = Ax is transformed into y = Jy. The corresponding equations
are
y1 = −y1 + y2 , y2 = −y2 , y3 = 2y3 .
The third equation has solution y3 (t) = c3 e2t , the second equation has solution y2 (t) = c2 e−t , and so the
first equation becomes
y1 + y1 = c2 e−t .
This is a first-order linear equation with integrating factor I(t) = et . We we multiply the differential equation
for y1 (t) through by I(t), it becomes (y · et ) = c2 . Integrating both sides yields y1 · et = c2 t + c1 . Thus,
y1 (t) = c2 te−t + c1 e−t .
Thus, we have
⎤
⎤ ⎡
y1 (t)
c2 te−t + c1 e−t
⎦.
c2 e−t
y(t) = ⎣ y2 (t) ⎦ = ⎣
2t
y3 (t)
c3 e
⎡
Finally, we solve for x(t):
⎤
⎤⎡
4 1 0
c2 te−t + c1 e−t
⎦
c2 e−t
x(t) = Sy(t) = ⎣ 0 0 1 ⎦ ⎣
c3 e2t
−4 0 0
⎤
⎡
4(c2 te−t + c1 e−t ) + c2 e−t
⎦
c3 e2t
=⎣
−4(c2 te−t + c1 e−t )
⎡
⎤
⎡
⎤
⎡
⎤
4
4t + 1
0
⎦ + c3 e2t ⎣ 1 ⎦ .
0
= c1 e−t ⎣ 0 ⎦ + c2 e−t ⎣
−4
−4t
0
⎡
42. The eigenvalues of A are λ = −3, −3. The eigenspace corresponding to λ = −3 is only one-dimensional,
and therefore the Jordan canonical form of A contains one 2 × 2 Jordan block:
−3
1
J=
.
0 −3
Next, we look for a cycle of generalized eigenvectors of the form {(A + 3I)v, v}, where v is a generalized
2
1 −1
eigenvector of A. Since (A+3I)2 =
= 02 , every nonzero vector in R2 is a generalized eigenvector.
1
−1
1
1
Let us choose v =
. Then (A + 3I)v =
. Thus, we form the matrix
0
1
S=
1
1
1
0
.
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Via the substitution x = Sy, the system x = Ax is transformed into y = Jy. The corresponding equations
are
y1 = −3y1 + y2 and y2 = −3y2 .
The second equation has solution y2 (t) = c2 e−3t . Substituting this expression for y2 (t) into the differential
equation for y1 (t) yields
y1 + 3y1 = c2 e−3t .
An integrating factor for this first-order linear differential equation is I(t) = e3t . Multiplying the differential
equation for y1 (t) by I(t) gives us (y1 · e3t ) = c2 . Integrating both sides, we obtain y1 · e3t = c2 t + c1 . Thus,
y1 (t) = c2 te−3t + c1 e−3t .
Thus,
y(t) =
y1 (t)
y2 (t)
=
c2 te−3t + c1 e−3t
c2 e−3t
.
Finally, we solve for x(t):
x(t) = Sy(t) =
1
1
1
0
c2 te−3t + c1 e−3t
c2 e−3t
c2 te−3t + c1 e−3t + c2 e−3t
=
c2 te−3t + c1 e−3t
1
t+1
−3t
−3t
+ c2 e
.
= c1 e
1
t
Now, we must apply the initial condition:
1
1
0
+ c2
.
= x(0) = c1
1
0
−1
Therefore, c1 = −1 and c2 = 1. Hence, the unique solution to the given initial-value problem is
1
t+1
−3t
−3t
+e
.
x(t) = −e
1
t
43. Let J = JCF(A) = JCF(B). Thus, there exist invertible matrices S and T such that
S −1 AS = J
Thus,
and so
and
T −1 BT = J.
S −1 AS = T −1 BT,
B = T S −1 AST −1 = (ST −1 )−1 A(ST −1 ),
which implies by definition that A and B are similar matrices.
44. Since the characteristic polynomial has degree 3, we know that A is a 3 × 3 matrix. Moreover, the roots
of the characteristic equation has roots λ = 0, 0, 0. Hence, the Jordan canonical form J of A must be one of
the three below:
⎡
⎤ ⎡
⎤ ⎡
⎤
0 0 0
0 1 0
0 1 0
⎣ 0 0 0 ⎦,⎣ 0 0 0 ⎦,⎣ 0 0 1 ⎦.
0 0 0
0 0 0
0 0 0
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In all three cases, note that J 3 = 03 . Moreover, there exists an invertible matrix S such that S −1 AS = J.
Thus, A = SJS −1 , and so
A3 = (SJS −1 )3 = SJ 3 S −1 = S03 S −1 = 03 ,
which implies that A is nilpotent.
45.
(a). Let J be an n × n Jordan block with eigenvalue λ. Then the eigenvalues of J T are λ (with multiplicity
n). The matrix J T − λI consists of 1’s on the subdiagonal (the diagonal parallel and directly beneath the
main diagonal) and zeros elsewhere. Hence, the null space of J T − λI is one-dimensional (with a free variable
corresponding to the right-most column of J T −λI). Therefore, the Jordan canonical form of J T consists of a
single Jordan block, since there is only one linearly independent eigenvector corresponding to the eigenvalue
λ. However, a single Jordan block with eigenvalue λ is precisely the matrix J. Therefore,
JCF(J T ) = J.
(b). Let JCF(A) = J. Then there exists an invertible matrix S such that S −1 AS = J. Transposing both
sides, we obtain (S −1 AS)T = J T , or S T AT (S −1 )T = J T , or S T AT (S T )−1 = J T . Hence, the matrix AT is
similar to J T . However, by applying part (a) to each block in J T , we find that JCF(J T ) = J. Hence, J T
is similar to J. By Problem 29 in Section 7.3, we conclude that AT is similar to J. Hence, AT and J have
the same Jordan canonical form. However, since JCF(J) = J, we deduce that JCF(AT ) = J = JCF(A), as
required.
Solutions to Section 7.7
Additional Problems:
1. Since A is lower triangular, its eigenvalues lie along the main diagonal, λ1 = 3 and λ2 = −1. Since A is
2 × 2 with two distinct eigenvalues, A is diagonalizable. To get an invertible matrix S such that S −1 AS = D,
we need to find an eigenvector associated with each eigenvector:
Eigenvalue λ1 = 3: To get an eigenvector, we consider
nullspace(A − 3I) = nullspace
and we see that one possible eigenvector is
1
4
0
0
16 −4
,
.
Eigenvalue λ2 = −1: To get an eigenvector, we consider
nullspace(A + I) = nullspace
and we see that one possible eigenvector is
0
1
Putting the above results together, we form
1 0
S=
4 1
4
16
0
0
3
0
0
−1
,
.
and
D=
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2. To compute the eigenvalues, we find the characteristic equation
13 − λ
−9
det(A − λI) = det
= (13 − λ)(−17 − λ) + 225 = λ2 + 4λ + 4,
25
−17 − λ
and the roots of this equation are λ = −2, −2.
Eigenvalue λ = 2: We compute
nullspace(A − 2I) = nullspace
15
−9
25 −15
,
but since there is only one linearly independent solution to the corresponding system (one free variable), the
eigenvalue λ = 2 does not have two linearly independent solutions. Hence, A is not diagonalizable.
3. To compute the eigenvalues, we find the characteristic equation
⎡
⎤
−4 − λ
3
0
⎦ = (−1 − λ) [(−4 − λ)(5 − λ) + 18]
5−λ
0
det(A − λI) = det ⎣ −6
3
−3
−1 − λ
= (−1 − λ)(λ2 − λ − 2)
= −(λ + 1)2 (λ − 2),
so the eigenvalues are λ1 = −1 and λ2 = 2.
Eigenvalue λ1 = −1: To get eigenvectors, we consider
⎡
−3
nullspace(A + I) = nullspace ⎣ −6
3
⎤ ⎡
⎤
3 0
1 −1 0
6 0 ⎦∼⎣ 0
0 0 ⎦.
−3 0
0
0 0
There are two free variables, z = t and y = s. From the first equation x = s. Thus, two linearly independent
eigenvectors can be obtained corresponding to λ1 = −1:
⎡
⎤
⎡
⎤
1
0
⎣ 1 ⎦
⎣ 0 ⎦.
and
0
1
Eigenvalue λ2 = 2: To get an eigenvector, we consider
⎡
−6
nullspace(A − 2I) = nullspace ⎣ −6
3
⎤ ⎡
⎤
3
0
1 −1 −1
3
0 ⎦∼⎣ 0
1
2 ⎦.
−3 −3
0
0
0
We let z = t. Then y = −2t from the middle line, and x = −t from the top line. Thus, an eigenvector
corresponding to λ2 = 2 may be chosen as
⎡
⎤
−1
⎣ −2 ⎦ .
1
Putting the above results together, we form
⎡
⎤
−1 1 0
S = ⎣ −2 1 0 ⎦
1 0 1
⎡
and
⎤
2
0
0
0 ⎦.
D = ⎣ 0 −1
0
0 −1
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4. To compute the eigenvalues, we find the characteristic equation
⎡
⎤
1−λ
1
0
⎦ = (−2 − λ) [(1 − λ)(5 − λ) + 4]
5−λ
0
det(A − λI) = det ⎣ −4
17
−11 −2 − λ
= (−2 − λ)(λ2 − 6λ + 9)
= (−2 − λ)(λ − 3)2 ,
so the eigenvalues are λ1 = 3 and λ2 = −2.
Eigenvalue λ1 = 3: To get eigenvectors, we consider
⎡
⎤ ⎡
⎤ ⎡
⎤ ⎡
−2
1
0
−2
1
0
1 −3 −5
1
2
0 ⎦ ∼ ⎣ 17 −11 −5 ⎦ ∼ ⎣ −2
1
0 ⎦∼⎣ 0
nullspace(A−3I) = nullspace ⎣ −4
17 −11 −5
0
0
0
0
0
0
0
⎤
−3 −5
1
2 ⎦.
0
0
The latter matrix contains only one unpivoted column, so that only one linearly independent eigenvector
can be obtained. However, λ1 = 3 occurs with multiplicity 2 as a root of the characteristic equation for the
matrix. Therefore, the matrix is not diagonalizable.
5. We are given that the only eigenvalue of A is λ = 2.
Eigenvalue λ = 2: To get eigenvectors, we consider
⎡
⎤ ⎡
⎤ ⎡
−3 −1
3
1
0 −1
1
2 −4 ⎦ ∼ ⎣ −3 −1
3 ⎦∼⎣ 0
nullspace(A − 2I) = nullspace ⎣ 4
−1
0
1
4
2 −4
0
⎤
0 −1
1
0 ⎦.
2
0
We see that only one unpivoted column will occur in a row-echelon form of A − 2I, and thus, only one
linearly independent eigenvector can be obtained. Since the eigenvalue λ = 2 occurs with multiplicity 3 as
a root of the characteristic equation for the matrix, the matrix is not diagonalizable.
6. We are given that the eigenvalues of A are λ1 = 4 and λ2 = −1.
Eigenvalue λ1 = 4: We consider
⎡
5
nullspace(A − 4I) = nullspace ⎣ 0
10
⎤
5 −5
−5
0 ⎦.
5 −10
The middle row tells us that nullspace vectors (x, y, z) must have y = 0. From this information, the first and
last rows of the matrix tell us the same thing: x = z. Thus, an eigenvector corresponding to λ1 = 4 may be
chosen as
⎡
⎤
1
⎣ 0 ⎦.
1
Eigenvalue λ2 = −1: We consider
⎤ ⎡
10 5 −5
10
0 ⎦∼⎣ 0
nullspace(A + I) = nullspace ⎣ 0 0
10 5 −5
0
⎡
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5 −5
0
0 ⎦.
0
0
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From the first row, a vector (x, y, z) in the nullspace must satisfy 10x + 5y − 5z = 0. Setting z = t and y = s,
we get x = 12 t − 12 s. Hence, the eigenvectors corresponding to λ2 = −1 take the form ( 12 t − 12 s, s, t), and so
a basis for this eigenspace is
⎧⎡
⎤ ⎡
⎤⎫
−1/2 ⎬
⎨ 1/2
⎣ 0 ⎦,⎣ 1 ⎦ .
⎩
⎭
1
0
Putting the above results together, we form
⎡
⎤
1 1/2 −1/2
1 ⎦
S=⎣ 0 0
1 1
0
⎡
and
⎤
4
0
0
0 ⎦.
D = ⎣ 0 −1
0
0 −1
11. We will compute the dimension of each of the two eigenspaces associated with the matrix A.
For λ1 = 1, we compute as follows:
⎡
⎤ ⎡
⎤
4
8
16
1 2 4
0
8 ⎦ ∼ ⎣ 0 1 1 ⎦,
nullspace(A − I) = nullspace ⎣ 4
−4 −4 −12
0 0 0
which has only one unpivoted column. Thus, this eigenspace is 1-dimensional.
For λ2 = −3, we compute as follows:
⎡
⎤ ⎡
8
8 16
1 1
4
8 ⎦∼⎣ 0 0
nullspace(A + 3I) = nullspace ⎣ 4
−4 −4 −8
0 0
⎤
2
0 ⎦,
0
which has two unpivoted columns. Thus, this eigenspace is 2-dimensional. Between the two eigenspaces, we
have a complete set of linearly independent eigenvectors. Hence, the matrix A in this case is diagonalizable.
Therefore, A is diagonalizable, and we may take
⎡
⎤
1
0
0
0 ⎦.
J = ⎣ 0 −3
0
0 −3
(Of course, the eigenvalues of A may be listed in any order along the main diagonal of the Jordan canonical
form, thus yielding other valid Jordan canonical forms for A.)
12. We will compute the dimension of each of the two eigenspaces associated with the matrix A.
For λ1 = −1, we compute as follows:
⎡
⎤ ⎡
⎤
3 1
1
1 0
1
nullspace(A + I) = nullspace ⎣ 2 2 −2 ⎦ ∼ ⎣ 0 1 −2 ⎦ ,
−1 0 −1
0 0
0
which has one unpivoted column. Thus, this eigenspace is 1-dimensional.
For λ2 = 3, we compute as follows:
⎡
⎤ ⎡
⎤
−1
1
1
1 −1 −1
1
6 ⎦,
nullspace(A − 3I) = nullspace ⎣ 2 −2 −2 ⎦ ∼ ⎣ 0
−1
0 −5
0
0
0
which has one unpivoted column. Thus, this eigenspace is also one-dimensional.
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Since we have only generated two linearly independent eigenvectors from the eigenvalues of A, we know
that A is not diagonalizable, and hence, the Jordan canonical form of A is not a diagonal matrix. We must
have one 1 × 1 Jordan block and one 2 × 2 Jordan block. To determine which eigenvalue corresponds to the
1 × 1 block and which corresponds to the 2 × 2 block, we must determine the multiplicity of the eigenvalues
as roots of the characteristic equation of A.
A short calculation shows that λ1 = −1 occurs with multiplicity 2, while λ2 = 3 occurs with multiplicity
1. Thus, the Jordan canonical form of A is
⎡
⎤
−1
1 0
J = ⎣ 0 −1 0 ⎦ .
0
0 3
13. There are 3 different possible Jordan canonical forms, up to a rearrangement of the Jordan blocks:
Case 1:
⎤
−1
0
0 0
⎢ 0 −1
0 0 ⎥
⎥.
J =⎢
⎣ 0
0 −1 0 ⎦
0
0
0 2
⎡
In this case, the matrix has four linearly independent eigenvectors, and because all Jordan blocks have size
1 × 1, the maximum length of a cycle of generalized eigenvectors for this matrix is 1.
Case 2:
⎤
−1
1
0 0
⎢ 0 −1
0 0 ⎥
⎥.
J =⎢
⎣ 0
0 −1 0 ⎦
0
0
0 2
⎡
In this case, the matrix has three linearly independent eigenvectors (two corresponding to λ = −1 and one
corresponding to λ = 2). There is a Jordan block of size 2 × 2, and so a cycle of generalized eigenvectors can
have a maximum length of 2 in this case.
Case 3:
⎤
−1
1
0 0
⎢ 0 −1
1 0 ⎥
⎥.
J =⎢
⎣ 0
0 −1 0 ⎦
0
0
0 2
⎡
In this case, the matrix has two linearly independent eigenvectors (one corresponding to λ = −1 and one
corresponding to λ = 2). There is a Jordan block of size 3 × 3, and so a cycle of generalized eigenvectors can
have a maximum length of 3 in this case.
14. There are 7 different possible Jordan canonical forms, up to a rearrangement of the Jordan blocks:
Case 1:
⎡
4
⎢ 0
⎢
J =⎢
⎢ 0
⎣ 0
0
0
4
0
0
0
0
0
4
0
0
0
0
0
4
0
⎤
0
0 ⎥
⎥
0 ⎥
⎥.
0 ⎦
4
In this case, the matrix has five linearly independent eigenvectors, and because all Jordan blocks have size
1 × 1, the maximum length of a cycle of generalized eigenvectors for this matrix is 1.
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Case 2:
⎡
4
⎢ 0
⎢
J =⎢
⎢ 0
⎣ 0
0
1
4
0
0
0
0
0
4
0
0
0
0
0
4
0
⎤
0
0 ⎥
⎥
0 ⎥
⎥.
0 ⎦
4
In this case, the matrix has four linearly independent eigenvectors, and because the largest Jordan block is
of size 2 × 2, the maximum length of a cycle of generalized eigenvectors for this matrix is 2.
Case 3:
⎡
4
⎢ 0
⎢
J =⎢
⎢ 0
⎣ 0
0
1
4
0
0
0
0
0
4
0
0
0
0
1
4
0
⎤
0
0 ⎥
⎥
0 ⎥
⎥.
0 ⎦
4
In this case, the matrix has three linearly independent eigenvectors, and because the largest Jordan block is
of size 2 × 2, the maximum length of a cycle of generalized eigenvectors for this matrix is 2.
Case 4:
⎡
4
⎢ 0
⎢
J =⎢
⎢ 0
⎣ 0
0
1
4
0
0
0
0
1
4
0
0
0
0
0
4
0
⎤
0
0 ⎥
⎥
0 ⎥
⎥.
0 ⎦
4
In this case, the matrix has three linearly independent eigenvectors, and because the largest Jordan block is
of size 3 × 3, the maximum length of a cycle of generalized eigenvectors for this matrix is 3.
Case 5:
⎡
4
⎢ 0
⎢
J =⎢
⎢ 0
⎣ 0
0
1
4
0
0
0
0
1
4
0
0
0
0
0
4
0
⎤
0
0 ⎥
⎥
0 ⎥
⎥.
1 ⎦
4
In this case, the matrix has two linearly independent eigenvectors, and because the largest Jordan block is
of size 3 × 3, the maximum length of a cycle of generalized eigenvectors for this matrix is 3.
Case 6:
⎡
4
⎢ 0
⎢
J =⎢
⎢ 0
⎣ 0
0
1
4
0
0
0
0
1
4
0
0
0
0
1
4
0
⎤
0
0 ⎥
⎥
0 ⎥
⎥.
0 ⎦
4
In this case, the matrix has two linearly independent eigenvectors, and because the largest Jordan block is
of size 4 × 4, the maximum length of a cycle of generalized eigenvectors for this matrix is 4.
Case 7:
⎡
4
⎢ 0
⎢
J =⎢
⎢ 0
⎣ 0
0
1
4
0
0
0
0
1
4
0
0
0
0
1
4
0
⎤
0
0 ⎥
⎥
0 ⎥
⎥.
1 ⎦
4
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In this case, the matrix has only one linearly independent eigenvector, and because the largest Jordan block
is of size 5 × 5, the maximum length of a cycle of generalized eigenvectors for this matrix is 5.
15. There are 10 different possible Jordan canonical forms, up to a rearrangement of the Jordan blocks:
Case 1:
⎡
6
⎢ 0
⎢
⎢ 0
J =⎢
⎢ 0
⎢
⎣ 0
0
0
6
0
0
0
0
0
0
6
0
0
0
⎤
0
0
0
0
0
0 ⎥
⎥
0
0
0 ⎥
⎥.
6
0
0 ⎥
⎥
0 −3
0 ⎦
0
0 −3
In this case, the matrix has six linearly independent eigenvectors, and because all Jordan blocks have size
1 × 1, the maximum length of a cycle of generalized eigenvectors for this matrix is 1.
Case 2:
⎡
6
⎢ 0
⎢
⎢ 0
J =⎢
⎢ 0
⎢
⎣ 0
0
1
6
0
0
0
0
0
0
6
0
0
0
⎤
0
0
0
0
0
0 ⎥
⎥
0
0
0 ⎥
⎥.
6
0
0 ⎥
⎥
0 −3
0 ⎦
0
0 −3
In this case, the matrix has five linearly independent eigenvectors (three corresponding to λ = 6 and two
corresponding to λ = −3), and because the largest Jordan block is of size 2 × 2, the maximum length of a
cycle of generalized eigenvectors for this matrix is 2.
Case 3:
⎡
6
⎢ 0
⎢
⎢ 0
J =⎢
⎢ 0
⎢
⎣ 0
0
1
6
0
0
0
0
0
0
6
0
0
0
⎤
0
0
0
0
0
0 ⎥
⎥
1
0
0 ⎥
⎥.
6
0
0 ⎥
⎥
0 −3
0 ⎦
0
0 −3
In this case, the matrix has four linearly independent eigenvectors (two corresponding to λ = 6 and two
corresponding to λ = −3), and because the largest Jordan block is of size 2 × 2, the maximum length of a
cycle of generalized eigenvectors for this matrix is 2.
Case 4:
⎡
6
⎢ 0
⎢
⎢ 0
J =⎢
⎢ 0
⎢
⎣ 0
0
1
6
0
0
0
0
0
1
6
0
0
0
⎤
0
0
0
0
0
0 ⎥
⎥
0
0
0 ⎥
⎥.
6
0
0 ⎥
⎥
0 −3
0 ⎦
0
0 −3
In this case, the matrix has four linearly independent eigenvectors (two corresponding to λ = 6 and two
corresponding to λ = −3), and because the largest Jordan block is of size 3 × 3, the maximum length of a
cycle of generalized eigenvectors for this matrix is 3.
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Case 5:
⎡
6
⎢ 0
⎢
⎢ 0
J =⎢
⎢ 0
⎢
⎣ 0
0
1
6
0
0
0
0
0
1
6
0
0
0
⎤
0
0
0
0
0
0 ⎥
⎥
1
0
0 ⎥
⎥.
6
0
0 ⎥
⎥
0 −3
0 ⎦
0
0 −3
In this case, the matrix has three linearly independent eigenvectors (one corresponding to λ = 6 and two
corresponding to λ = −3), and because the largest Jordan block is of size 4 × 4, the maximum length of a
cycle of generalized eigenvectors for this matrix is 4.
Case 6:
⎡
6
⎢ 0
⎢
⎢ 0
J =⎢
⎢ 0
⎢
⎣ 0
0
0
6
0
0
0
0
0
0
6
0
0
0
⎤
0
0
0
0
0
0 ⎥
⎥
0
0
0 ⎥
⎥.
6
0
0 ⎥
⎥
0 −3
1 ⎦
0
0 −3
In this case, the matrix has five linearly independent eigenvectors (four corresponding to λ = 6 and one
corresponding to λ = −3), and because the largest Jordan block is of size 2 × 2, the maximum length of a
cycle of generalized eigenvectors for this matrix is 2.
Case 7:
⎡
6
⎢ 0
⎢
⎢ 0
J =⎢
⎢ 0
⎢
⎣ 0
0
1
6
0
0
0
0
0
0
6
0
0
0
⎤
0
0
0
0
0
0 ⎥
⎥
0
0
0 ⎥
⎥.
6
0
0 ⎥
⎥
0 −3
1 ⎦
0
0 −3
In this case, the matrix has four linearly independent eigenvectors (three corresponding to λ = 6 and one
corresponding to λ = −3), and because the largest Jordan block is of size 2 × 2, the maximum length of a
cycle of generalized eigenvectors for this matrix is 2.
Case 8:
⎡
6
⎢ 0
⎢
⎢ 0
J =⎢
⎢ 0
⎢
⎣ 0
0
1
6
0
0
0
0
0
0
6
0
0
0
⎤
0
0
0
0
0
0 ⎥
⎥
1
0
0 ⎥
⎥.
6
0
0 ⎥
⎥
0 −3
1 ⎦
0
0 −3
In this case, the matrix has three linearly independent eigenvectors (two corresponding to λ = 6 and one
corresponding to λ = −3), and because the largest Jordan block is of size 2 × 2, the maximum length of a
cycle of generalized eigenvectors for this matrix is 2.
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Case 9:
⎡
6
⎢ 0
⎢
⎢ 0
J =⎢
⎢ 0
⎢
⎣ 0
0
1
6
0
0
0
0
0
1
6
0
0
0
⎤
0
0
0
0
0
0 ⎥
⎥
0
0
0 ⎥
⎥.
6
0
0 ⎥
⎥
0 −3
1 ⎦
0
0 −3
In this case, the matrix has three linearly independent eigenvectors (two corresponding to λ = 6 and one
corresponding to λ = −3), and because the largest Jordan block is of size 3 × 3, the maximum length of a
cycle of generalized eigenvectors for this matrix is 3.
Case 10:
⎡
6
⎢ 0
⎢
⎢ 0
J =⎢
⎢ 0
⎢
⎣ 0
0
1
6
0
0
0
0
0
1
6
0
0
0
⎤
0
0
0
0
0
0 ⎥
⎥
1
0
0 ⎥
⎥.
6
0
0 ⎥
⎥
0 −3
1 ⎦
0
0 −3
In this case, the matrix has two linearly independent eigenvectors (one corresponding to λ = 6 and one
corresponding to λ = −3), and because the largest Jordan block is of size 4 × 4, the maximum length of a
cycle of generalized eigenvectors for this matrix is 4.
16. There are 15 different possible Jordan canonical forms, up to a rearrangement of Jordan blocks:
Case 1:
⎡
2
⎢ 0
⎢
⎢ 0
⎢
J =⎢
⎢ 0
⎢ 0
⎢
⎣ 0
0
0
2
0
0
0
0
0
0
0
2
0
0
0
0
⎤
0
0
0
0
0
0
0
0 ⎥
⎥
0
0
0
0 ⎥
⎥
2
0
0
0 ⎥
⎥.
0 −4
0
0 ⎥
⎥
0
0 −4
0 ⎦
0
0
0 −4
In this case, the matrix has seven linearly independent eigenvectors (four corresponding to λ = 2 and three
corresponding to λ = −4), and because all Jordan blocks are size 1 × 1, the maximum length of a cycle of
generalized eigenvectors for this matrix is 1.
Case 2:
⎡
2
⎢ 0
⎢
⎢ 0
⎢
J =⎢
⎢ 0
⎢ 0
⎢
⎣ 0
0
1
2
0
0
0
0
0
0
0
2
0
0
0
0
⎤
0
0
0
0
0
0
0
0 ⎥
⎥
0
0
0
0 ⎥
⎥
2
0
0
0 ⎥
⎥.
0 −4
0
0 ⎥
⎥
0
0 −4
0 ⎦
0
0
0 −4
In this case, the matrix has six linearly independent eigenvectors (three corresponding to λ = 2 and three
corresponding to λ = −4), and because the largest Jordan block is of size 2 × 2, the maximum length of a
cycle of generalized eigenvectors for this matrix is 2.
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Case 3:
⎡
2
⎢ 0
⎢
⎢ 0
⎢
J =⎢
⎢ 0
⎢ 0
⎢
⎣ 0
0
1
2
0
0
0
0
0
0
0
2
0
0
0
0
⎤
0
0
0
0
0
0
0
0 ⎥
⎥
1
0
0
0 ⎥
⎥
2
0
0
0 ⎥
⎥.
0 −4
0
0 ⎥
⎥
0
0 −4
0 ⎦
0
0
0 −4
In this case, the matrix has five linearly independent eigenvectors (two corresponding to λ = 2 and three
corresponding to λ = −4), and because the largest Jordan block is of size 2 × 2, the maximum length of a
cycle of generalized eigenvectors for this matrix is 2.
Case 4:
⎤
2 1 0 0
0
0
0
⎢ 0 2 1 0
0
0
0 ⎥
⎥
⎢
⎢ 0 0 2 0
0
0
0 ⎥
⎥
⎢
0
0
0 ⎥
J =⎢
⎥.
⎢ 0 0 0 2
⎥
⎢ 0 0 0 0 −4
0
0
⎥
⎢
⎣ 0 0 0 0
0 −4
0 ⎦
0 0 0 0
0
0 −4
⎡
In this case, the matrix has five linearly independent eigenvectors (two corresponding to λ = 2 and three
corresponding to λ = −4), and because the largest Jordan block is of size 3 × 3, the maximum length of a
cycle of generalized eigenvectors for this matrix is 3.
Case 5:
⎤
2 1 0 0
0
0
0
⎢ 0 2 1 0
0
0
0 ⎥
⎥
⎢
⎢ 0 0 2 1
0
0
0 ⎥
⎢
⎥
0
0
0 ⎥
J =⎢
⎢ 0 0 0 2
⎥.
⎢ 0 0 0 0 −4
⎥
0
0
⎢
⎥
⎣ 0 0 0 0
0 −4
0 ⎦
0 0 0 0
0
0 −4
⎡
In this case, the matrix has four linearly independent eigenvectors (one corresponding to λ = 2 and three
corresponding to λ = −4), and because the largest Jordan block is of size 4 × 4, the maximum length of a
cycle of generalized eigenvectors for this matrix is 4.
Case 6:
⎡
2
⎢ 0
⎢
⎢ 0
⎢
J =⎢
⎢ 0
⎢ 0
⎢
⎣ 0
0
0
2
0
0
0
0
0
0
0
2
0
0
0
0
⎤
0
0
0
0
0
0
0
0 ⎥
⎥
0
0
0
0 ⎥
⎥
2
0
0
0 ⎥
⎥.
0 −4
1
0 ⎥
⎥
0
0 −4
0 ⎦
0
0
0 −4
In this case, the matrix has six linearly independent eigenvectors (four corresponding to λ = 2 and two
corresponding to λ = −4), and because the largest Jordan block is of size 2 × 2, the maximum length of a
cycle of generalized eigenvectors for this matrix is 2.
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Case 7:
⎡
2
⎢ 0
⎢
⎢ 0
⎢
J =⎢
⎢ 0
⎢ 0
⎢
⎣ 0
0
1
2
0
0
0
0
0
0
0
2
0
0
0
0
⎤
0
0
0
0
0
0
0
0 ⎥
⎥
0
0
0
0 ⎥
⎥
2
0
0
0 ⎥
⎥.
0 −4
1
0 ⎥
⎥
0
0 −4
0 ⎦
0
0
0 −4
In this case, the matrix has five linearly independent eigenvectors (three corresponding to λ = 2 and two
corresponding to λ = −4), and because the largest Jordan block is of size 2 × 2, the maximum length of a
cycle of generalized eigenvectors for this matrix is 2.
Case 8:
⎡
2
⎢ 0
⎢
⎢ 0
⎢
J =⎢
⎢ 0
⎢ 0
⎢
⎣ 0
0
1
2
0
0
0
0
0
0
0
2
0
0
0
0
⎤
0
0
0
0
0
0
0
0 ⎥
⎥
1
0
0
0 ⎥
⎥
2
0
0
0 ⎥
⎥.
0 −4
1
0 ⎥
⎥
0
0 −4
0 ⎦
0
0
0 −4
In this case, the matrix has four linearly independent eigenvectors (two corresponding to λ = 2 and two
corresponding to λ = −4), and because the largest Jordan block is of size 2 × 2, the maximum length of a
cycle of generalized eigenvectors for this matrix is 2.
Case 9:
⎡
2
⎢ 0
⎢
⎢ 0
⎢
J =⎢
⎢ 0
⎢ 0
⎢
⎣ 0
0
1
2
0
0
0
0
0
0
1
2
0
0
0
0
⎤
0
0
0
0
0
0
0
0 ⎥
⎥
0
0
0
0 ⎥
⎥
2
0
0
0 ⎥
⎥.
0 −4
1
0 ⎥
⎥
0
0 −4
0 ⎦
0
0
0 −4
In this case, the matrix has four linearly independent eigenvectors (two corresponding to λ = 2 and two
corresponding to λ = −4), and because the largest Jordan block is of size 3 × 3, the maximum length of a
cycle of generalized eigenvectors for this matrix is 3.
Case 10:
⎡
2
⎢ 0
⎢
⎢ 0
⎢
J =⎢
⎢ 0
⎢ 0
⎢
⎣ 0
0
1
2
0
0
0
0
0
0
1
2
0
0
0
0
⎤
0
0
0
0
0
0
0
0 ⎥
⎥
1
0
0
0 ⎥
⎥
2
0
0
0 ⎥
⎥.
0 −4
1
0 ⎥
⎥
0
0 −4
0 ⎦
0
0
0 −4
In this case, the matrix has three linearly independent eigenvectors (one corresponding to λ = 2 and two
corresponding to λ = −4), and because the largest Jordan block is of size 4 × 4, the maximum length of a
cycle of generalized eigenvectors for this matrix is 4.
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Case 11:
⎡
2
⎢ 0
⎢
⎢ 0
⎢
J =⎢
⎢ 0
⎢ 0
⎢
⎣ 0
0
0
2
0
0
0
0
0
0
0
2
0
0
0
0
⎤
0
0
0
0
0
0
0
0 ⎥
⎥
0
0
0
0 ⎥
⎥
2
0
0
0 ⎥
⎥.
0 −4
1
0 ⎥
⎥
0
0 −4
1 ⎦
0
0
0 −4
In this case, the matrix has five linearly independent eigenvectors (four corresponding to λ = 2 and one
corresponding to λ = −4), and because the largest Jordan block is of size 3 × 3, the maximum length of a
cycle of generalized eigenvectors for this matrix is 3.
Case 12:
⎤
2 1 0 0
0
0
0
⎢ 0 2 0 0
0
0
0 ⎥
⎥
⎢
⎢ 0 0 2 0
0
0
0 ⎥
⎥
⎢
0
0
0 ⎥
J =⎢
⎥.
⎢ 0 0 0 2
⎥
⎢ 0 0 0 0 −4
1
0
⎥
⎢
⎣ 0 0 0 0
0 −4
1 ⎦
0 0 0 0
0
0 −4
⎡
In this case, the matrix has four linearly independent eigenvectors (three corresponding to λ = 2 and one
corresponding to λ = −4), and because the largest Jordan block is of size 3 × 3, the maximum length of a
cycle of generalized eigenvectors for this matrix is 3.
Case 13:
⎤
2 1 0 0
0
0
0
⎢ 0 2 0 0
0
0
0 ⎥
⎥
⎢
⎢ 0 0 2 1
0
0
0 ⎥
⎥
⎢
0
0
0 ⎥
J =⎢
⎢ 0 0 0 2
⎥.
⎢ 0 0 0 0 −4
1
0 ⎥
⎢
⎥
⎣ 0 0 0 0
0 −4
1 ⎦
0 0 0 0
0
0 −4
⎡
In this case, the matrix has three linearly independent eigenvectors (two corresponding to λ = 2 and one
corresponding to λ = −4), and because the largest Jordan block is of size 3 × 3, the maximum length of a
cycle of generalized eigenvectors for this matrix is 3.
Case 14:
⎤
2 1 0 0
0
0
0
⎢ 0 2 1 0
0
0
0 ⎥
⎥
⎢
⎢ 0 0 2 0
0
0
0 ⎥
⎥
⎢
0
0
0 ⎥
J =⎢
⎥.
⎢ 0 0 0 2
⎥
⎢ 0 0 0 0 −4
1
0
⎥
⎢
⎣ 0 0 0 0
0 −4
1 ⎦
0 0 0 0
0
0 −4
⎡
In this case, the matrix has three linearly independent eigenvectors (two corresponding to λ = 2 and one
corresponding to λ = −4), and because the largest Jordan block is of size 3 × 3, the maximum length of a
cycle of generalized eigenvectors for this matrix is 3.
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Case 15:
⎡
2
⎢ 0
⎢
⎢ 0
⎢
J =⎢
⎢ 0
⎢ 0
⎢
⎣ 0
0
1
2
0
0
0
0
0
0
1
2
0
0
0
0
⎤
0
0
0
0
0
0
0
0 ⎥
⎥
1
0
0
0 ⎥
⎥
2
0
0
0 ⎥
⎥.
0 −4
1
0 ⎥
⎥
0
0 −4
1 ⎦
0
0
0 −4
In this case, the matrix has two linearly independent eigenvectors (one corresponding to λ = 2 and one
corresponding to λ = −4), and because the largest Jordan block is of size 4 × 4, the maximum length of a
cycle of generalized eigenvectors for this matrix is 4.
1 1
1 0
17. FALSE. For instance, if A =
and B =
, then we have eigenvalues λA = λB = 1, but
0 1
1 1
0 1
the matrix A − B =
is invertible, and hence, zero is not an eigenvalue of A − B. This can also
−1 0
be verified directly.
1 1
, then A2 = B 2 = I2 , but the matrices A and B are
18. FALSE. For instance, if A = I2 and B =
0 1
not similar. (Otherwise, there would exist an invertible matrix S such that S −1 AS = B. But since A = I2
this reduces to I2 = B, which is clearly not the case. Thus, no such invertible matrix S exists.)
19. We have
Ai v = λi v
for each i = 1, 2, . . . , k. Thus,
(A1 A2 . . . Ak )v = (A1 A2 . . . Ak−1 )(Ak v) = (A1 A2 . . . Ak−1 )(λk v) = λk (A1 A2 . . . Ak−1 )v
= λk (A1 A2 . . . Ak−2 )(Ak−1 v) = λk (A1 A2 . . . Ak−2 )(λk−1 v) = λk−1 λk (A1 A2 . . . Ak−2 )v
..
.
= λ2 λ3 . . . λk (A1 v) = λ2 λ3 . . . λk (λ1 v)
= (λ1 λ2 . . . λk )v,
which shows that v is an eigenvector of A1 A2 . . . Ak with corresponding eigenvalue λ1 λ2 . . . λk .
20. We are given that Av = λ1 v and Bv = λ2 v. Note that v = 0 (since v is an eigenvector), and
(AB − BA)v = A(Bv) − B(Av) = A(λ2 v) − B(λ1 v) = λ2 (Av) − λ1 (Bv) = λ2 (λ1 v) − λ1 (λ2 v) = 0,
which shows that v belongs to the null space of AB − BA. Thus, nullspace(AB − BA) = 0. Therefore, by
the Invertible Matrix Theorem, AB − BA cannot be invertible.
Chapter 8 Solutions
Solutions to Section 8.1
True-False Review:
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(a): TRUE. This is essentially the statement of Theorem 8.1.4.
(b): FALSE. As stated in Theorem 8.1.6, if there is any point x0 in I such that W [y1 , y2 , . . . , yn ](x0 ) = 0,
then {y1 , y2 , . . . , yn } is linearly dependent on I.
(c): FALSE. Many counterexamples are possible. Note that
(xD − Dx)(x) = xD(x) − D(x2 ) = x − 2x = (−1)(x).
Therefore, xD − Dx = −1. Setting L1 = x and L2 = D, we therefore see that L1 L2 = L2 L1 in this example.
(d): TRUE. By assumption, L1 (y1 + y2 ) = L1 (y1 ) + L1 (y2 ) and L1 (cy) = cL1 (y) for all functions y, y1 , y2 .
Likewise, L2 (y1 + y2 ) = L2 (y1 ) + L2 (y2 ) and L2 (cy) = cL2 (y) for all functions y, y1 , y2 . Therefore,
(L1 + L2 )(y1 + y2 ) = L1 (y1 + y2 ) + L2 (y1 + y2 )
= (L1 (y1 ) + L1 (y2 )) + (L2 (y1 ) + L2 (y2 ))
= (L1 (y1 ) + L2 (y1 )) + (L1 (y2 ) + L2 (y2 ))
= (L1 + L2 )(y1 ) + (L1 + L2 )(y2 )
and
(L1 + L2 )(cy) = L1 (cy) + L2 (cy)
= cL1 (y) + cL2 (y)
= c(L1 (y) + L2 (y))
= c(L1 + L2 )(y).
Therefore, L1 + L2 is a linear differential operator.
(e): TRUE. By assumption L(y1 + y2 ) = L(y1 ) + L(y2 ) and L(ky) = kL(y) for all scalars k. Therefore, for
all constants c, we have
(cL)(y1 + y2 ) = cL(y1 + y2 ) = c(L(y1 ) + L(y2 )) = cL(y1 ) + cL(y2 ) = (cL)(y1 ) + (cL)(y2 )
and
(cL)(ky) = c(L(ky)) = c(k(L(y))) = k(cL)(y).
Therefore, cL is a linear differential operator.
(f ): TRUE. We have
L(yp + u) = L(yp ) + L(u) = F + 0 = F.
(g): TRUE. We have
L(y1 + y2 ) = L(y1 ) + L(y2 ) = F1 + F2 .
(h):TRUE. It follows from the results of this section that Ker(L) is a vector space of dimension 2. Therefore
any two linearly independent functions in Ker(L) will form a basis for Ker(L). Since L(ex ) = (D2 − 1)(ex ) =
D2 (ex ) − ex = ex − ex = 0, and L(e−x ) = (D2 − 1)(e−x ) = D2 (e−x ) − e−x = e−x − e−x = 0 it follows
that both ex and e−x are in Ker(L). Further, ex and e−x are not proportional and therefore are linearly
independent on any interval. Consequently, {ex , e−x } is a basis for Ker(L).
(i): TRUE. It follows from the results of this section that the solution space to the given differential
equation is a vector space of dimension 2. Therefore a basis for this solution space will consist of precisely
two linearly independent solutions to the differential equation. If we set y1 (x) = ex , andy2 (x) = e2x , then
y1 − 3y1 + 2y1 = ex − 3ex + 2ex = 0,
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and
y2 − 3y2 + 2y2 = 4e−2x − 6e−2x + 2e−2x = 0.
Hence, y1 (x) and y2 (x) are both solutions to the differential equation. Further, they are not proportional
and so are linearly independent on any interval. Consequently, {ex , e−2x } is a basis for the solution space of
the differential equation.
(j): FALSE. It follows from the results of this section that the solution space to the given differential
equation is a vector space of dimension 3. Since the given set of vectors contains only two functions, it
cannot form a basis for the solution space.
Problems:
1. (a). L(y(x)) = (D − x)(2e3x ) = D(2e3x ) − 2xe3x = 6e3x − 2xe3x .
(b). L(y(x)) = (D − x)(3 ln x) = D(3 ln x) − 3x ln x =
3
− 3x ln x.
x
(c). Since L is a linear operator, we have L(2e3x + 3 ln x) = L(2e3x ) + L(3 ln x) = 6e3x − 2xe3x +
where we have used the results from (a) and (b).
3
− 3x ln x,
x
2. (a). L(y(x)) = (D2 − x2 D + x)(2e3x ) = D2 (2e3x ) − x2 D(2e3x ) + 2xe3x = 18e3x − 6x2 e3x + 2xe3x .
3
(b). L(y(x)) = (D2 − x2 D + x)(3 ln x) = D2 (3 ln x) − x2 D(3 ln x) + x(3 ln x) = − 2 − 3x + 3x ln x.
x
(c). Since L is a linear operator, we have
L(2e3x + 3 ln x) = L(2e3x ) + L(3 ln x) = 18e3x − 6x2 e3x + 2xe3x −
3
− 3x + 3x ln x,
x2
where we have used the results from (a) and (b).
3. (a). L(y(x)) = (D3 − 2xD2 )(2e3x ) = D3 (2e3x ) − 2xD2 (2e3x ) = 54e3x − 36xe3x .
(b). L(y(x)) = (D3 − 2xD2 )(3 ln x) = D3 (3 ln x) − 2xD2 (3 ln x) =
6
6
+ .
3
x
x
(c). Since L is a linear operator, we have
L(2e3x + 3 ln x) = L(2e3x ) + L(3 ln x) = 54e3x − 36xe3x +
6
6
+ ,
3
x
x
where we have used the results from (a) and (b).
4. (a). L(y(x)) = (D3 − D + 4)(2e3x ) = D3 (2e3x ) − D(2e3x ) + 8e3x = 54e3x − 6e3x + 8e3x = 56e3x .
(b). L(y(x)) = (D3 − D + 4)(3 ln x) = D3 (3 ln x) − 2D(3 ln x) + 12 ln x =
6
3
− + 12 ln x.
x3
x
(c). Since L is a linear operator, we have
L(2e3x + 3 ln x) = L(2e3x ) + L(3 ln x) = 56e3x +
where we have used the results from (a) and (b).
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3
− + 12 ln x,
3
x
x
537
5.
L(y(x)) = (D2 − 4D + 4)(xe2x ) = D2 (xe2x ) − 4D(xe2x ) + 4xe2x
= D(e2x + 2xe2x ) − 4(e2x + 2xe2x ) + 4xe2x
= 4e2x + 4xe2x − 4(e2x + 2xe2x ) + 4xe2x
= 0.
Thus, xe2x ∈ Ker(L).
6.
L(y(x)) = (x2 D2 + 2xD − 2)(x−2 ) = x2 D2 (x−2 ) + 2xD(x−2 ) − 2x−2
= x2 D(−2x−3 ) − 4x−2 − 2x−2
= 6x−2 − 6x−2 = 0.
Thus, f ∈ Ker(L).
7.
L(y(x)) = (D2 − x−1 D + 4x2 )(sin(x2 )) = D2 (sin (x2 )) − x−1 D(sin (x2 )) + 4x2 (sin (x2 ))
= D(2x cos (x2 )) − x−1 (2x cos (x2 )) + 4x2 (sin (x2 ))
= 2[cos (x2 ) − 2x2 sin (x2 )] − 2 cos (x2 ) + 4x2 (sin (x2 ))
= 0.
Thus, f ∈ Ker(L).
8.
L(y(x)) = (D3 + D2 + D + 1)(sin x + cos x)
= D3 (sin x + cos x) + D2 (sin x + cos x) + D(sin x + cos x) + (sin x + cos x)
= D2 (cos x − sin x) + D(cos x − sin x) + cos x − sin x + sin x + cos x
= D(− sin x − cos x) + (− sin x − cos x) + 2 cos x
= − cos x + sin x − sin x − cos x + 2 cos x
= 0.
Thus, f ∈ Ker(L).
9.
L(y(x)) = (−D2 + 2D − 1)(xex )
= −D2 (xex ) + 2D(xex ) − xex
= −D(ex + xex ) + 2(ex + xex ) − xex
= −(ex + ex + xex ) + 2ex + 2xex − xex
= 0.
Thus, f ∈ Ker(L).
10. L(y(x)) = 0 ⇐⇒ (D − 3x2 )y = 0 ⇐⇒ y − 3x2 y = 0. This linear equation has integrating factor
d −x3
2
3
y) = 0, which has a general
I = e− 3x dx = e−x , so that the differential equation can be written as
(e
dx
3
3
solution e−x y = c, that is, y(x) = cex . Consequently,
3
Ker(L) = {y ∈ C 1 (R) : y(x) = cex , c ∈ R}.
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11. L(y(x)) = 0 ⇐⇒ (D2 +1)y = 0 ⇐⇒ y +y = 0. This is a second-order homogeneous linear differential
equation, so by Theorem 8.1.4, the solution set to this equation forms a 2-dimensional vector space. A little
thought shows that both y1 = cos x and y2 = sin x are solutions to the equation. Since they are linearly
independent, these two functions form a basis for the solution set. Therefore,
Ker(L) = {a cos x + b sin x : a, b ∈ R}.
12. L(y(x)) = 0 ⇐⇒ (D2 + 2D − 15)y = 0 ⇐⇒ y + 2y − 15y = 0. This is a second-order homogeneous
linear differential equation, so by Theorem 8.1.4, the solution set to this equation forms a 2-dimensional
vector space. Following the hint given in the text, we try for solutions of the form y(x) = erx . Substituting
this into the differential equation, we get r2 erx + 2rerx − 15erx = 0, or erx (r2 + 2r − 15) = 0. It follows that
r2 + 2r − 15 = 0. That is (r + 5)(r − 3) = 0, and hence, r = −5 and r = 3 are the solutions. Therefore,
we obtain the solutions y1 = e−5x and y2 = e3x . Since they are linearly independent (by computing the
Wronskian, for instance), these two functions form a basis for the solution set. Therefore,
Ker(L) = {ae−5x + be3x : a, b ∈ R}.
13. Ly = 0 ⇐⇒ (x2 D + x)y = 0 ⇐⇒ x2 y + xy = 0. Dividing by x2 , we can express this as y + x1 y = 0.
This differential equation is separable: y1 dy = − x1 dx. Integrating both sides, we obtain ln |y| = − ln |x| + c1 .
c3
Therefore |y| = e− ln |x|+c1 = c2 e− ln |x| . Hence, y(x) = c3 e− ln |x| = |x|
. Thus,
c
Ker(L) =
:c∈R .
|x|
14. We have
(L1 L2 )(f ) = L1 (f − 2x2 f )
= (D + 1)(f − 2x2 f )
= f + f − 4xf − 2x2 f − 2x2 f
= f + (1 − 2x2 )f − (4x + 2x2 )f,
so that
Furthermore,
L1 L2 = D2 + (1 − 2x2 )D − 2x(2 + x).
(L2 L1 )(f ) = L2 (f + f )
= (D − 2x2 )(f + f )
= f + f − 2x2 f − 2x2 f,
so that
L2 L1 = D2 + (1 − 2x2 )D − 2x2 .
Therefore, L1 L2 = L2 L1 .
15. We have
(L1 L2 )(f ) = L1 (f + (2x − 1)f )
= (D + x)(f + (2x − 1)f )
= f + xf + 2f + (2x − 1)f + x(2x − 1)f
= (D2 + (3x − 1)D + (2x2 − x + 2))(f ),
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so that
L1 L2 = D2 + (3x − 1)D + (2x2 − x + 2).
Furthermore,
(L2 L1 )(f ) = L2 (f + xf )
= (D + (2x − 1))(f + xf )
= f + (2x − 1)f + f + xf + (2x − 1)xf
= (D2 + (3x − 1)D + (2x2 − x + 1))(f ),
so that
L2 L1 = D2 + (3x − 1)D + (2x2 − x + 1).
Therefore, L1 L2 = L2 L1 .
16. We have
(L1 L2 )(f ) = L1 (D + b1 )f
= (D + a1 )(f + b1 f )
= f + [b1 + a1 ]f + (b1 + a1 b1 )f.
Thus
L1 L2 = D2 + (b1 + a1 )D + (b1 + a1 b1 ).
Similarly,
L2 L1 = D2 + (a1 + b1 )D + (a1 + b1 a1 ).
Thus L1 L2 −L2 L1 = b1 −a1 , which is the zero operator if and only if b1 = a1 which can be integrated directly
to obtain b1 = a1 + c2 , where c2 is an arbitrary constant. Consequently we must have L2 = D + [a1 (x) + c2 ].
17. (D3 + x2 D2 − (sin x)D + ex )y = x3
and
18. (D2 + 4xD − 6x2 )y = x2 sin x
y + 4xy − 6x2 y = 0.
and
y + x2 y − (sin x)y + ex y = 0.
19. x2 and ex are both continuous functions for all x ∈ R and in particular on any interval I containing
x0 = 0. Thus, y(x) = 0 is clearly a solution to the given differential equation and also satisfies the initial
conditions. Thus, by the existence-uniqueness theorem, y(x) = 0 is the only solution to the initial-value
problem.
20. Let a1 , ..., an be functions that are continuous on the interval I. Then, for any x0 in I, the initial-value
problem y (n) + a1 (x)y (n−1) + ... + an−1 (x)y + an (x)y = 0, y(x0 ) = 0, y (x0 ) = 0, ..., y (n−1) (x0 ) = 0, has
only the trivial solution y(x) = 0.
Proof:
All of the conditions of the existence-uniqueness theorem are satisfied and y(x) = 0 is a solution; consequently,
it is the only solution.
21. Since the differential equation is second order, linear, and homogenous, any basis for the solution space
will consist of precisely two linearly independent solutions. We can therefore immediately rule out S1 (too few
functions), S2 (too many functions), and S5 (the set is linearly dependent). We now consider the remaining
sets of functions.
S3 = {e4x , e2x } : Since the two functions in S3 are not proportional, they are linearly independent on any
interval. Therefore, S3 will be a basis for the solution space of the differential equation provide y1 (x) = e4x
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and y2 (x) = e2x , are both solutions of the differential equation. Since
y1 − 16y1 = 16e4x − 16e4x = 0,
it follows that y1 (x) = e4x is a solution to the differential equation. However,
y2 − 16y2 = 4e2x − 16e2x = −12e2x = 0,
so that y2 (x) = e2x is not a solution to the differential equation. Consequently, S3 is not a basis for the
solution space of the differential equation.
S4 = {e4x , e−4x }: We have already shown above that y1 (x) = e4x is a solution to the differential equation.
Further, if y2 (x) = e−4x , then
y2 − 16y2 = 16e−4x − 16e−4x = 0,
so that y2 (x) = e−2x is also a solution to the differential equation. Since y1 (x) and y2 (x) are not proportional,
they are linearly independent on any interval. Consequently, S4 is a basis for the solution space of the
differential equation.
1 4x
1 4x
S5 = {cosh 4x, sinh 4x}: Since cosh 4x =
e + e−4x , and sinh 4x =
e − e−4x , are both linear
2
2
combinations of the solutions e4x , e−4x , it follows that they are themselves solutions. Further since cosh x
and sinh 4x are not proportional, they are linearly independent on any interval. Consequently, S5 is a basis
for the solution space of the differential equation.
22. Since the differential equation is second order, linear, and homogenous, any basis for the solution space
will consist of precisely two linearly independent solutions. We can therefore immediately rule out S1 (too
few functions). We now consider the remaining sets of functions.
S2 = {x2 , x2 ln x}: Let y1 (x) = x2 . Then,
x2 y1 − 3xy1 + 4y1 = x2 (2) − 3x(2x) + 4x2 = 0,
so that y1 (x) = x2 is a solution to the differential equation. Further, if y2 (x) = x2 ln x, then
y2 (x) = 2x ln x + x, and y2 (x) = 2 ln x + 3,
so that
x2 y2 − 3xy2 + 4y2 = x2 (2 ln x + 3) − 3x(2x ln x + x) + 4x2 ln x = 0.
Hence, y2 (x) = x2 ln x is also a solution to the differential equation. Since y1 (x) and y2 (x) are not proportional, they are linearly independent on any interval. Consequently, S2 is a basis for the solution space of
the differential equation.
S3 = {2x2 , 3x2 ln x} : The elements of S3 are scalar multiples of the elements of S2 , and therefore are
solutions to the differential equation. Further, since they are not proportional, they are linearly independent
on (0, ∞). Hence S3 is a basis for the solution space to the differential equation.
S4 = {x2 (2 + 3 ln x), x2 (2 − 3 ln x)} : The elements of S4 are linear combinations of the elements of S2 , and
therefore are solutions to the differential equation. Further, since they are not proportional, they are linearly
independent on (0, ∞). Hence S4 is a basis for the solution space to the differential equation.
23. Given y − 2y − 3y = 0 then r2 − 2r − 3 = 0 =⇒ r ∈ {−1, 3} =⇒ y(x) = c1 e−x + c2 e3x .
24. Given y + 7y + 10y = 0 then r2 + 7r + 10 = 0 =⇒ r ∈ {−5, −2} =⇒ y(x) = c1 e−5x + c2 e−2x .
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25. Given y − 36y = 0 then r2 − 36 = 0 =⇒ r ∈ {−6, 6} =⇒ y(x) = c1 e−6x + c2 e6x .
26. Given y + 4y = 0 then r2 + 4r = 0 =⇒ r ∈ {−4, 0} =⇒ y(x) = c1 e−4x + c2 .
27. Substituting y(x) = erx into the given differential equation yields erx (r3 − 3r2 − r + 3) = 0, so that
we will have a solution provided that r satisfies r3 − 3r2 − r + 3 = 0, that is, (r − 1)(r + 1)(r − 3) = 0.
Consequently, three solutions to the given differential equation are y1 (x) = ex , y2 (x) = e−x , y3 (x) = e3x .
Further, the Wronskian of these solutions is
x
e
e−x e3x x
−e−x 3e3x = −16e3x .
W [y1 , y2 , y3 ] = e
x
e
e−x 9e3x Since the Wronskian is never zero, the solutions are linearly independent on any interval. Hence the general
solution to the differential equation is y(x) = c1 ex + c2 e−x + c3 e3x .
28. Substituting y(x) = erx into the given differential equation yields erx (r3 + 3r2 − 4r − 12) = 0, so that
we will have a solution provided that r satisfies r3 + 3r2 − 4r − 12 = 0, that is, (r − 2)(r + 2)(r + 3) = 0.
Consequently, three solutions to the given differential equation are y1 (x) = e2x , y2 (x) = e−2x , y3 (x) = e−3x .
Further, the Wronskian of these solutions is
2x
e
e−2x
e−3x 2x
−2e−2x −3e−3x = −20e−3x .
W [y1 , y2 , y3 ] = 2e
2x
4e
4e−2x
9e−3x Since the Wronskian is never zero, the solutions are linearly independent on any interval. Hence the general
solution to the differential equation is y(x) = c1 e2x + c2 e−2x + c3 e−3x .
29. Substituting y(x) = erx into the given differential equation yields erx (r3 + 3r2 − 18r − 40) = 0, so
that we will have a solution provided r satisfies r3 + 3r2 − 18r − 40 = 0, that is, (r + 5)(r + 2)(r − 4) = 0.
Consequently, three solutions to the differential equation are y1 (x) = e−5x , y2 (x) = e−2x , y3 (x) = e4x .
Further, the Wronskian of the solution is
e−5x
e−2x
e4x 4e4x = 162e−3x .
W [y1 , y2 , y3 ] = −5e−5x −2e−2x
−5x
−2x
25e
4e
16e4x Since the Wronskian is never zero, the solutions are linearly independent on any interval. Hence the general
solution to the differential equation is y(x) = c1 e−5x + c2 e−2x + c3 e4x .
30. Given y − y − 2y = 0 then r3 − r2 − 2r = 0 =⇒ r(r2 − r − 2) = 0 =⇒ r ∈ {−1, 0, 2} =⇒ y(x) =
c1 e−x + c2 + c3 e2x .
31. Given y + y − 10y + 8y = 0 then r3 + r2 − 10r + 8 = 0 =⇒ (r − 2)(r − 1)(r + 4) = 0 =⇒ r ∈
{−4, 1, 2} =⇒ y(x) = c1 e−4x + c2 ex + c3 e2x .
32. Given y (iv) − 2y − y + 2y = 0 then r4 − 2r3 − r2 − 2r = 0 =⇒ r(r3 − 2r2 − r − 2) = 0 =⇒
r(r − 2)(r − 1)(r + 1) = 0 =⇒ r ∈ {−1, 0, 1, 2} =⇒ y(x) = c1 e−x + c2 + c3 ex + c4 e2x .
33. Given y (iv) − 13y + 36y = 0 then r4 − 13r2 + 36 = 0 =⇒ (r2 − 9)(r2 − 4) = 0 =⇒ r2 ∈ {4, 9} =⇒ r ∈
{−3, −2, 2, 3} =⇒ y(x) = c1 e−3x + c2 e−2x + c3 e2x + c4 e3x .
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34. Given x2 y + 3xy − 8y = 0, the trial solution gives x2 r(r − 1)xr−2 + 3xrxr−1 − 8xr = 0, or
xr [r(r − 1) + 3r − 8] = 0. Therefore, r2 + 2r − 8 = 0, which factors as (r + 4)(r − 2) = 0. Therefore, r = −4
or r = 2. Hence, we obtain the solutions y1 (x) = x−4 and y2 (x) = x2 . Furthermore,
W [x−4 , x2 ] = (x−4 )(2x) − (−4x−5 )(x2 ) = 6x−3 = 0,
so that {x−4 , x2 } is a linearly independent set of solutions to the given differential equation on (0, ∞).
Consequently, from Theorem 8.1.4, the general solution is given by y(x) = c1 x−4 + c2 x2 .
35. Given 2x2 y + 5xy + y = 0, the trial solution gives 2x2 r(r − 1)xr−2 + 5xrxr−1 + xr = 0, or
xr [2r(r − 1) + 5r + 1] = 0. Therefore, 2r2 + 3r + 1 = 0, which factors as (2r + 1)(r + 1) = 0. Therefore,
1
r = − 12 and r = −1. Hence, we obtain the solutions y1 (x) = x− 2 and y2 (x) = x−1 . Furthermore,
1
1
1
1 3
1 5
W [x− 2 , x−1 ] = (x− 2 )(− 2 ) − (− x− 2 )(x−1 ) = − x− 2 = 0,
x
2
2
1
so that {x− 2 , x−1 } is a linearly independent set of solutions to the given differential equation on (0, ∞).
1
Consequently, from Theorem 8.1.4, the general solution is given by y(x) = c1 x− 2 + c2 x−1 .
36. Substituting y(x) = xr into the given differential equation yields xr [r(r −1)(r −2)+r(r −1)−2r +2] = 0,
so that r must satisfy (r − 1)(r − 2)(r + 1) = 0. If follows that three solutions to the differential equation
are y1 (x) = x, y2 (x) = x2 , y3 (x) = x−1 . Further, the Wronskian of these solutions is
x x2
x−1 W [y1 , y2 , y3 ] = 1 2x −x−2 = 6x−1 .
0 2
2x−3 Since the Wronskian is nonzero on (0, ∞), the solutions are linearly independent on this interval. Consequently, the general solution to the differential equation is y(x) = c1 x + c2 x2 + c3 x−1 .
37. Given x3 y + 3x2 y − 6xy = 0, the trial solution gives
x3 r(r − 1)(r − 2)xr−3 + 3x2 r(r − 1)xr−2 − 6xrxr−1 = 0,
or
xr [r(r − 1)(r − 2) + 3r(r − 1) − 6r] = 0.
Therefore, r(r − 1)(r − 2) + 3r(r − 1) − 6r = 0, or r[(r − 1)(r − 2) + 3(r − 1) − 6] = 0. Therefore,
r[r2 − 7] =√0,
√
√
so that r = 0 or r = ± 7. Hence, we obtain the solutions y1 (x) = 1, y2 (x) = x 7 , and y3 (x) = x− 7 .
Furthermore,
√
√
7
1
x √7
x− √
√
√
√
√
7
− 7
W [1, x , x
]= 0
7x 7−1√
− 7x− 7−1 √
√ √
√
√
7−2
0
7( 7 − 1)x
− 7(− 7 − 1)x− 7−2
√
√
= −7(− 7 − 1)x−3 + 7( 7 − 1)x−3
√
= 14 7x−3 = 0,
√
√
so that {1, x 7 , x− 7 } is a linearly independent set of solutions to the given differential
equation
on (0, ∞).
√
√
Consequently, from Theorem 8.1.4, the general solution is given by y(x) = c1 + c2 x 7 + c3 x− 7 .
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38. To determine a particular solution of the form yp (x) = A0 e5x , we substitute this solution into the
differential equation:
(A0 e5x ) + (A0 e5x ) − 6(A0 e5x ) = 18e5x .
Therefore
25A0 e5x + 5A0 e5x − 6A0 e5x = 18e5x ,
3 5x
e .
4
To obtain the general solution, we need to find the complementary function yc (x), the solution to the
associated homogeneous differential equation: y + y − 6y = 0. Seeking solutions of the form y(x) = erx ,
we obtain r2 erx + rerx − 6erx = 0, or erx (r2 + r − 6) = 0. Therefore, r2 + r − 6 = 0, which factors as
(r + 3)(r − 2) = 0. Hence, r = −3 or r = 2. Therefore, we obtain the solutions y1 (x) = e−3x and y2 (x) = e2x .
Since
W [e−3x , e2x ] = (e−3x )(2e2x ) − (−3e−3x )(e2x ) = 2e−x + 3e−x = 5e−x = 0,
which forces A0 = 3/4. Therefore, yp (x) =
{e−3x , e2x } is linearly independent. Therefore, the complementary function is yc (x) = c1 e−3x + c2 e2x . By
Theorem 8.1.8, the general solution to the differential equation is
y(x) = c1 e−3x + c2 e2x + 3e3x .
39. Substituting y(x) = A0 + A1 x + A2 x2 into the differential equation yields
(A0 + A1 x + A2 x2 ) + (A0 + A1 x + A2 x2 ) − 2(A0 + A1 x + A2 x2 ) = 4x2 + 5
or
(2A2 + A1 − 2A0 ) + (2A2 − 2A1 )x − 2A2 x2 = 4x2 + 5.
Equating the powers of x on each side of this equation yields
2A2 + A1 − 2A0 = 5,
2A2 − 2A1 = 0,
−2A2 = 4.
11
−2x−2x2 .
2
To obtain the general solution, we need to find the complementary function yc (x), the solution to the
associated homogeneous differential equation: y + y − 2y = 0. Seeking solutions of the form y(x) = erx ,
we obtain r2 erx + rerx − 2erx = 0, or erx (r2 + r − 2) = 0. Therefore, r2 + r − 2 = 0, which factors as
(r + 2)(r − 1) = 0. Hence, r = −2 or r = 1. Therefore, we obtain the solutions y1 (x) = e−2x and y2 (x) = ex .
Since
W [e−2x , ex ] = (e−2x )(ex ) − (−2e−2x )(ex ) = 3e−x = 0,
Solving for A0 , A1 , and A2 , we find that A0 = −11/2, A1 = −2, and A2 = −2. Thus, yp (x) = −
{e−2x , ex } is linearly independent. Therefore, the complementary function is yc (x) = c1 e−2x + c2 ex . By
Theorem 8.1.8, the general solution to the differential equation is
y(x) = c1 e−2x + c2 ex −
11
− 2x − 2x2 .
2
40. Substituting y(x) = A0 e2x into the given differential equation yields A0 e2x (8 + 8 − 2 − 2) = 4e2x , so
1
1
that A0 = e2x . Hence, a particular solution to the differential equation is yp (x) = e2x . To determine the
3
3
general solution we need to solve the associated homogeneous differential equation y + 2y − y − 2y = 0.
We try for solutions of the form y(x) = erx . Substituting into the homogeneous differential equation gives
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erx (r3 +2r2 −r−2) = 0, so that we choose r to satisfy r3 +2r2 −r−2 = 0, or equivalently, (r+2)(r−1)(r+1) =
0. It follows that three solutions to the differential equation are y1 (x) = e−2x , y2 (x) = ex , y3 (x) = e−x .
Further, the Wronskian of these solutions is
e−2x ex
e−x W [y1 , y2 , y3 ] = −2e−2x ex −e−x = −6e−2x .
4e−2x ex
e−x Since the Wronskian is nonzero, the solutions are linearly independent on any interval. Thus, the complementary function is yc (x) = c1 e−2x + c2 ex + c3 e−x . By Theorem 8.1.8, the general solution to the differential
equation is
1
y(x) = yc (x) + yp (x) = c1 e−2x + c2 ex + c3 e−x + e2x .
3
41. Substituting y(x) = A0 e−3x into the differential equation yields
A0 e−3x (−27 + 9 + 30 + 8) = 24e−3x .
6
6
. Thus, yp (x) = e−3x .
5
5
To obtain the general solution, we need to find the complementary function yc (x), the solution to the
associated homogeneous differential equation: y + y − 10y + 8y = 0. Seeking solutions of the form
y(x) = erx , we obtain r3 erx + r2 erx − 10rerx + 8erx = 0, or erx (r3 + r2 − 10r + 8) = 0. Therefore,
r3 + r2 − 10r + 8 = 0, which has roots r = 1, r = 2, and r = −4. Therefore, we obtain the solutions
y1 (x) = ex , y2 (x) = e2x , and y3 (x) = e−4x . Since
Therefore, 20A0 = 24. Hence, A0 =
x
2x
W [e , e , e
−4x
ex
] = ex
ex
e2x
2e2x
4e2x
e−4x
−4e−4x = 30e−x = 0,
16e−4x
{ex , e2x , e−4x } is linearly independent. Thus, the complementary function is yc (x) = c1 ex + c2 e2x + c3 e−4x .
By Theorem 8.1.8, the general solution to the differential equation is
6
y(x) = c1 ex + c2 e2x + c3 e−4x + e−3x .
5
42. Substituting y(x) = A0 e−x into the differential equation yields
A0 e−x (−1 + 5 − 6) = 6e−x .
Therefore, −2A0 = 6. Hence, A0 = −3. Thus, yp (x) = −3e−x .
To obtain the general solution, we need to find the complementary function yc (x), the solution to the
associated homogeneous differential equation: y + 5y + 6y = 0. Seeking solutions of the form y(x) = erx ,
we obtain r3 erx + 5r2 erx + 6rerx = 0, or erx (r3 + 5r2 + 6r) = 0. Therefore, r3 + 5r2 + 6r = 0, which has roots
r = 0, r = −2, and r = −3. Therefore, we obtain the solutions y1 (x) = 1, y2 (x) = e−2x , and y3 (x) = e−3x .
Since
1
e−2x
e−3x
−2x −3x
−2x
−3e−3x = −6e−5x = 0,
W [1, e
,e
] = 0 −2e
−2x
0
4e
9e−3x
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{1, e−2x , e−3x } is linearly independent. Therefore, the complementary function is yc (x) = c1 +c2 e−2x +c3 e−3x .
By Theorem 8.1.8, the general solution to the differential equation is
y(x) = c1 + c2 e−2x + c3 e−3x − 3e−x .
43. Let y1 and y2 belong to C n (I), and let c be a scalar. We must show that
L(y1 + y2 ) = L(y1 ) + L(y2 )
and
L(cy1 ) = cL(y1 ).
We have
L(y1 + y2 ) = (Dn + a1 Dn−1 + · · · + an−1 D + an )(y1 + y2 )
= Dn (y1 + y2 ) + a1 Dn−1 (y1 + y2 ) + · · · + an−1 D(y1 + y2 ) + an (y1 + y2 )
= (Dn y1 + a1 Dn−1 y1 + · · · + an−1 Dy1 + an y1 ) + (Dn y2 + a1 Dn−1 y2 + · · · + an−1 Dy2 + an y2 )
= (Dn + a1 Dn−1 + · · · + an−1 D + an )(y1 ) + (Dn + a1 Dn−1 + · · · + an−1 D + an )(y2 )
= L(y1 ) + L(y2 ),
and
L(cy1 ) = (Dn + a1 Dn−1 + · · · + an−1 D + an )(cy1 )
= Dn (cy1 ) + a1 Dn−1 (cy1 ) + · · · + an−1 D(cy1 ) + an (cy1 )
= cDn (y1 ) + ca1 Dn−1 (y1 ) + · · · + can−1 D(y1 ) + can (y1 )
= c(Dn (y1 ) + a1 Dn−1 (y1 ) + · · · + an−1 D(y1 ) + an (y1 ))
= c(Dn + a1 Dn−1 + · · · + an−1 D + an )(y1 )
= cL(y1 ).
44. Prior to the statement of Theorem 8.1.4 it was shown that the set of all solutions forms a vector space.
We now show that th dimension of this solution space is n, by constructing a basis. Let y1 , y2 , ..., yn be the
unique solutions of the n initial-value problems:
(n)
a0 (x)yi
(n−1)
+ a1 (x)yi
(k−1)
yi
+ ... + an−1 (x)yi + an (x)yi = 0,
(x0 ) = δik , k = 1, 2, ..., n,
respectively. The Wronskian of these functions at x0 is W [y1 , y2 , ..., yn ](x0 ) = det[In ] = 1 = 0 so that the
solutions are linearly independent on I. We now show that they span the solution space.
Let u(x) be any solution of the differential equation on I, and suppose that u(x0 ) = u1 , u (x0 ) =
u2 , ..., u(n−1) (x0 ) = un , where u1 , u2 , ..., un are constants. It follows that y = u(x) is the unique solution to
the initial-value problem:
Ly = 0,
y(x0 ) = u1 , y (x0 ) = u2 , ..., y (n−1) (x0 ) = un .
However, if we define w(x) = u1 y1 (x) + u2 y2 (x) + ... + un yn (x), then w(x) also satisfies this initial value
problem . Thus, by uniqueness, we must have u(x) = w(x), that is, u(x) = u1 y1 (x) + u2 y2 (x) + ... + un yn (x).
Thus we have shown that {y1 , y2 , ..., yn } forms a basis for the solution space and hence the dimension of this
solution space is n.
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45. Consider the linear system
c1 y1 (x0 ) + c2 y2 (x0 ) + ... + cn yn (x0 )
c1 y1 (x0 ) + c2 y2 (x0 ) + ... + cn yn (x0 )
..
.
(n−1)
c 1 y1
(n−1)
(x0 ) + c2 y2
(n−1)
(x0 ) + ... + cn yn
=0
=0
(x0 )
= 0,
where we are solving for c1 , c2 , ..., cn . The determinant of the matrix of coefficients of this system is
W [y1 , y2 , ..., yn ](x0 ) = 0, so that the system has non-trivial solutions. Let (α1 , α2 , ..., αn ) be one such
non-trivial solution for (c1 , c2 , . . . , cn ). Therefore, not all of the αi are zero. Define the function u(x) by
u(x) = α1 y1 (x) + α2 y2 (x) + ... + αn yn (x). It follows that y = u(x) satisfies the initial-value problem:
a0 y (n) + a1 y (n−1) + ... + an−1 y + an y = 0
y(x0 ) = 0, y (x0 ) = 0, ..., y (n−1) (x0 ) = 0.
However, y(x) = 0 also satisfies the above initial value problem and hence, by uniqueness, we must have
u(x) = 0, that is, α1 y1 (x) + α2 y2 (x) + ... + αn yn (x) = 0, where not all of the αi are zero. Thus, the functions
y1 , y2 , ..., yn are linearly dependent on I.
46. Let vp be a particular solution to Equation (8.1.15), and let v be any solution to this equation. Then
T (vp ) = w
and
T (v) = w.
Subtracting these two equations gives
T (v) − T (vp ) = 0,
or equivalently, since T is a linear transformation,
T (v − vp ) = 0.
But this latter equation implies that the vector v − vp is an element of Ker(T ) and therefore can be written
as
v − vp = c1 v1 + c2 v2 + ... + cn vn ,
for some scalars c1 , c2 , . . . , cn . Hence,
v = c1 v1 + c2 v2 + ... + cn vn + vp .
Solutions to Section 8.2
True-False Review:
(a): FALSE. Even if the auxiliary polynomial fails to have n distinct roots, the differential equation still has
n linearly independent solutions. For example, if L = D2 + 2D + 1, then the differential equation Ly = 0 has
auxiliary polynomial with (repeated) roots r = −1, −1. Yet we do have two linearly independent solutions
y1 (x) = e−x and y2 (x) = xe−x to the differential equation.
(b): FALSE. Theorem 8.2.2 only applies to polynomial differential operators. However, in general, many
counterexamples can be given. For example, note that
(xD − Dx)(x) = xD(x) − D(x2 ) = x − 2x = (−1)(x).
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Therefore, xD − Dx = −1. Setting L1 = x and L2 = D, we therefore see that L1 L2 = L2 L1 in this example.
(c): TRUE. This is really just the statement that a polynomial of degree n always has n roots, with
multiplicities counted.
(d): TRUE. Since 0 is a root of multiplicity four, each term of the polynomial differential operator must
contain a factor of D4 , so that any polynomial of degree three or less becomes zero after taking four (or
more) derivatives. Therefore, for a homogeneous differential equation of this type, a polynomial of degree
three or less must be a solution.
(e): FALSE. Note that r = 0 is a root of the auxiliary polynomial, but only of multiplicity 1. The expression
c1 + c2 x in the solution reflects r = 0 as a root of multiplicity 2.
(f ): TRUE. The roots of the auxiliary polynomial are r = −3, −3, 5i, −5i. The portion of the solution
corresponding to the repeated root r = −3 is c1 e−3x +c2 xe−3x , and the portion of the solution corresponding
to the complex conjugate pair r = ±5i is c3 cos 5x + c4 sin 5x.
(g): TRUE. The roots of the auxiliary polynomial are r = 2 ± i, 2 ± i. The terms corresponding to the
first pair 2 ± i are c1 e2x cos x and c2 e2x sin x, and the repeated root gives two more terms: c3 xe2x cos x and
c4 xe2x sin x.
(h): FALSE. Many counterexamples can be given. For instance, if P (D) = D − 1, then the general solution
is y(x) = cex . However, the differential equation (P (D))2 y = (D − 1)2 y has auxiliary equation with roots
r = 1, 1 and general solution z(x) = c1 ex + c2 xex = xy(x).
Problems:
1. The auxiliary polynomial is P (r) = r2 + 2r − 3 = (r + 3)(r − 1). Therefore, the auxiliary equation has
roots r = −3 and r = 1. Therefore, two linearly independent solutions to the given differential equation are
y1 (x) = e−3x
and
y2 (x) = ex .
By Theorem 8.1.4, the solution space to this differential equation is 2-dimensional, and hence {e−3x , ex }
forms a basis for the solution space.
2. The auxiliary polynomial is P (r) = r2 + 6r + 9 = (r + 3)2 . Therefore, the auxiliary equation has roots
r = −3 (with multiplicity 2). Therefore, two linearly independent solutions to the given differential equation
are
y1 (x) = e−3x and y2 (x) = xe−3x .
By Theorem 8.1.4, the solution space to this differential equation is 2-dimensional, and hence {e−3x , xe−3x }
forms a basis for the solution space.
3. The auxiliary polynomial is P (r) = r2 − 6r + 25. According to the quadratic equation, the auxiliary
equation has roots r = 3 ± 4i. Therefore, two linearly independent solutions to the given differential equation
are
y1 (x) = e3x cos 4x and y2 (x) = e3x sin 4x.
By Theorem 8.1.4, the solution space to this differential equation is 2-dimensional, and hence {e3x cos 4x, e3x sin 4x}
forms a basis for the solution space.
4. The general vector in S takes the form y(x) = c(sin 4x + 5 cos 4x). The full solution space for this
differential equation is 2-dimensional by Theorem 8.1.4. Note that sin 4x belongs to the solution space and
is linearly independent from sin 4x + 5 cos 4x since
W [sin 4x + 5 cos 4x, sin 4x] = 20 = 0.
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Therefore, we can extend the basis for S to a basis for the entire solution space with {sin 4x+5 cos 4x, sin 4x}.
Many other extensions are also possible, of course.
5. P (r) = r2 + r − 2 = (r + 2)(r − 1) =⇒ two linearly independent solutions to the given differential equation
are y1 (x) = ex , y2 (x) = e−2x . According to the Gram-Schmidt process an orthogonal basis for the solution
space to the given differential equation is {u1 (x), u2 (x)} where u1 (x) = y1 (x) = ex and
1 −2x x
e
· e dx x
< y2 , u1 >
2(e−1 − 1) x
2
−2x
−2x
0
u
(x)
=
e
−
=
e
+
u2 (x) = y2 (x) −
·
e
e = e−2x −
ex .
1
1
2
2
2x
||u1 ||
e − 1)
e(e − 1)
e dx
0
6. P (r) = r2 + 4 = 0, when r = ±2i =⇒ two linearly independent solutions to the given differential equation
are y1 (x) = cos 2x, y2 (x) = sin 2x. According to the Gram-Schmidt process an orthogonal basis for the
solution space to the given differential equation is {u1 (x), u2 (x)} where u1 (x) = y1 (x) = cos 2x and
π/4
sin 2x · cos 2x dx
< y2 , u1 >
2
0
u
(x)
=
sin
2x
−
· cos 2x = sin 2x − cos 2x.
u2 (x) = y2 (x) −
1
π/4
2
||u1 ||2
π
cos 2x dx
0
7. P (r) = r2 + 1 =⇒ two linearly independent solutions to the given differential equation are y1 ((x) = cos x,
π
y2 (x) = sin x. Since < y1 , y2 >= 0 cos x sin x dx = 0, {y1 (x), y2 (x)} is an orthogonal basis for the solution
√
π
π
2 x dx =
cos
π,
and
||y
||
=
sin2 x dx =
space to the given differential equation. Further, ||y1 || =
2
0
0
√
1
1
π, so that an orthonormal basis is √ cos x, √ sin x .
π
π
8. We have r2 − r − 2 = 0 =⇒ r ∈ {−1, 2} =⇒ y(x) = c1 e−x + c2 e2x .
9. We have r2 − 6r + 9 = 0 =⇒ r ∈ {3, 3} =⇒ y(x) = c1 e3x + c2 xe3x .
10. We have r2 + 6r + 25 = 0 =⇒ r ∈ {−3 − 4i, −3 + 4i} =⇒ y(x) = c1 e−3x cos 4x + c2 e−3x sin 4x.
11. We have (r + 1)(r − 5) = 0 =⇒ r ∈ {−1, 5} =⇒ y(x) = c1 e−x + c2 e5x .
12. We have (r + 2)2 = 0 =⇒ r ∈ {−2, −2} =⇒ y(x) = c1 e−2x + c2 xe−2x .
13. We have r2 − 6r + 34 = 0 =⇒ r ∈ {3 − 5i, 3 + 5i} =⇒ y(x) = c1 e3x cos 5x + c2 e3x sin 5x.
14. We have r2 + 10r + 25 = 0 =⇒ r ∈ {−5, −5} =⇒ y(x) = c1 e−5x + c2 xe−5x .
√
√
√ √
15. We have r2 − 2 = 0 =⇒ r ∈ {− 2, 2} =⇒ y(x) = c1 e− 2x + c2 e 2x .
16. We have r2 + 8r + 20 = 0 =⇒ r ∈ {−4 − 2i, −4 + 2i} =⇒ y(x) = c1 e−4x cos 2x + c2 e−4x sin 2x.
17. We have r2 + 2r + 2 = 0 =⇒ r ∈ {−1 − i, −1 + i} =⇒ y(x) = c1 e−x cos x + c2 e−x sin x.
18. We have (r − 4)(r + 2) = 0 =⇒ r ∈ {−2, 4} =⇒ y(x) = c1 e−2x + c2 e4x .
19. We have r2 − 14r + 58 = 0 =⇒ r ∈ {7 − 3i, 7 + 3i} =⇒ y(x) = c1 e7x cos 3x + c2 e7x sin 3x.
20. We have r3 − r2 + r − 1 = 0 =⇒ r ∈ {1, i, −i} =⇒ y(x) = c1 ex + c2 cos x + c3 sin x.
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21. We have r3 − 2r2 − 4r + 8 = 0 =⇒ r ∈ {−2, 2, 2} =⇒ y(x) = c1 e−2x + (c2 + c3 x)e2x .
22. We have (r − 2)(r2 − 16) = 0 =⇒ r ∈ {2, 4, −4} =⇒ y(x) = c1 e2x + c2 e4x + c3 e−4x .
23. We have (r2 + 2r + 10)2 = 0 =⇒ r ∈ {−1 + 3i, −1 + 3i, −1 − 3i, −1 − 3i} =⇒
y(x) = e−x [c1 cos 3x + c2 sin 3x + x(c4 cos 3x + c4 sin 3x)].
24. We have (r2 + 4)2 (r + 1) = 0 =⇒ r ∈ {2i, 2i, −2i, −2i, −1} =⇒
y(x) = c1 e−x + c2 cos 2x + c3 sin 2x + x(c4 cos 2x + c5 sin 2x).
√ √
−i 3, i 3} =⇒
25. We have (r2 + 3)(r + 1) = 0√=⇒ r ∈ {−1, −1,
√
y(x) = c1 e−x + c2 xe−x + c3 cos ( 3x) + c4 sin ( 3x).
26. We have r2 (r − 1) = 0 =⇒ r ∈ {0, 0, 1} =⇒ y(x) = c1 + c2 x + c3 ex .
27. We have r4 − 8r2 + 16 = (r − 2)2 (r + 2)2 = 0 =⇒ r ∈ {2, 2, −2, −2} =⇒
y(x) = e2x (c1 + c2 x) + e−2x (c3 + c4 x).
28. We have r4 − 16 = 0 =⇒ r ∈ {2, −2, 2i, −2i} =⇒ y(x) = c1 e2x + c2 e−2x + c3 cos 2x + c4 sin 2x.
29. We have r3 +8r2 +22r +20 = 0 =⇒ r ∈ {−2, −3+i, −3−i} =⇒ y(x) = c1 e−2x +e−3x (c2 cos x+c3 sin x).
30. We have r4 − 16r2 + 40r − 25 = 0 =⇒ r ∈ {1, −5, 2 + i, 2 − i} =⇒
y(x) = c1 ex + c2 e−5x + e2x (c3 cos x + c4 sin x).
31. We have (r −1)3 (r2 +9) = 0 =⇒ r ∈ {1, 1, 1, −3i, 3i} =⇒ y(x) = ex (c1 +c2 x+c3 x2 )+c4 cos 3x+c5 sin 3x.
32. We have (r2 − 2r + 2)2 (r2 − 1) = 0 =⇒ r ∈ {1 + i, 1 + i, 1 − i, 1 − i, 1, −1} =⇒
y(x) = ex (c1 cos x + c2 sin x) + xex (c3 cos x + c4 sin x) + c5 e−x + c6 ex .
33. We have (r + 3)(r − 1)(r + 5)3 = 0 =⇒ r ∈ {−3, 1, −5, −5, −5} =⇒
y(x) = c1 e−3x + c2 ex + e−5x (c3 + c4 x + c5 x2 ).
34. We have (r2 + 9)3 = 0 =⇒ r ∈ {3i, 3i, 3i, −3i, −3i, −3i} =⇒
y(x) = c1 cos 3x + c2 sin 3x + x(c3 cos 3x + c4 sin 3x) + x2 (c5 cos 3x + c6 sin 3x).
35. We have r2 − 8r + 16 = 0 =⇒ r ∈ {4, 4} =⇒ y(x) = c1 e4x + c2 xe4x . Now 2 = y(0) = c1 and
7 = y (0) = 4c1 + c2 . Therefore, c1 = 2 and c2 = −1 Hence, the solution to this initial-value problem is
y(x) = 2e4x − xe4x .
36. We have r2 − 4r + 5 = 0 =⇒ r ∈ {2 + i, 2 − i} =⇒ y(x) = c1 e2x cos x + c2 e2x sin x. Now 3 = y(0) = c1
and 5 = y (0) = 2c1 + c2 . Therefore, c1 = 3 and c2 = −1. Hence, the solution to this initial-value problem
is y(x) = 3e2x cos x − e2x sin x.
37. We have r3 − r2 + r − 1 = 0 =⇒ r ∈ {1, i, −i} =⇒ y(x) = c1 ex + c2 cos x + c3 sin x. Then since y(0) = 0,
we have c1 + c2 = 0. Moreover, since y (0) = 1, we have c1 + c3 = 1. Finally, y (0) = 2 implies that
c1 − c2 = 2. Solving these equations yields c1 = 1, c2 = −1, c3 = 0. Hence, the solution to this initial-value
problem is y(x) = ex − cos x.
38. We have r3 + 2r2 − 4r − 8 = (r − 2)(r + 2)2 = 0 =⇒ r ∈ {2, −2, −2} =⇒ y(x) = c1 e2x + c2 e−2x + c3 xe−2x .
Then since y(0) = 0, we have c1 + c2 = 0. Moreover, since y (0) = 6, we have 2c1 − 2c2 + c3 = 6. Further,
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y (0) = 8 implies that 4c1 + 4c2 − 4c3 = 8. Solving these equations yields c1 = 2, c2 = c3 = −2. Hence, the
solution to this initial-value problem is y(x) = 2(e2x − e−2x − xe−2x ).
39. Given m > 0 and k > 0, we have auxiliary polynomial P (r) = r2 − 2mr + (m2 + k 2 ) = 0 =⇒ r = m ± ki.
Therefore, y(x) = emx (c1 cos kx + c2 sin kx). Differentiating, we obtain y (x) = memx (c1 cos kx + c2 sin kx) +
emx (−c1 k sin kx + c2 k cos kx). Now 0 = y(0) = c1 and k = y (0) = c2 k. Therefore, c1 = 0 and c2 = 1. Hence,
the solution to this initial-value problem is y(x) = emx sin kx.
40. Given m > 0 and k > 0, we have auxiliary polynomial P (r) = r2 − 2mr + (m2 − k 2 ) = 0 =⇒ r = m ± k.
Therefore, the general solution to this differential equation is y(x) = ae(m+k)x +be(m−k)x = emx (aekx +be−kx ).
2
2
Letting a = c1 +c
and b = c1 −c
in the last equality gives
2
2
y(x) = emx
c1 + c2 kx c1 − c2 −kx
e +
e
2
2
kx
e + e−kx
ekx − e−kx
= emx c1
+ c2
2
2
= emx (c1 cosh kx + c2 sinh kx).
41. The auxiliary
polynomial is P (r) = r2 + 2cr + k 2 . The quadratic formula supplies the roots of P (r) = 0:
√
2
2
r = −c ± c − k .
(a). √If c2 < k 2 , then c2 − k 2 < 0 and the roots above are complex. We can write r = −c ± ωi, where
ω = k 2 − c2 . Thus,
y(t) = e−ct (c1 cos ωt + c2 sin ωt)
and
y (t) = e−ct [(ωc2 − cc1 ) cos ωt − (cc2 + ωc1 ) sin ωt] .
Using the initial conditions y(0) = y0 and y (0) = 0, we find that c1 = y0 and ωc2 − y0 c = 0, or c2 = yω0 c .
Therefore,
y0
y(t) = e−ct (ω cos ωt + c sin ωt).
ω
√
√
(b). Since ω = k 2 − c2 , we find k = ω 2 + c2 (note k is assumed positive, so we take the positive square
root). Therefore, continuing from part (a), we have
y0 −ct
e (ω cos ωt + c sin ωt)
ω
ω
y0
c
= k e−ct
cos ωt + sin ωt
ω
k
k
ω
c
y0 −ct
√
cos ωt + √
sin ωt .
=k e
ω
ω 2 + c2
ω 2 + c2
y(t) =
Now since
ω
√
2
ω + c2
there exists φ such that
cos φ = √
2
c
ω 2 + c2
+
c
√
2
ω + c2
and
2
= 1,
sin φ = √
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;
ω 2 + c2
551
that is, φ = tan−1
ω
c
. Hence, y(t) can be written as
ky0 −ct
ky0
e−ct sin(ωt + φ).
y(t) =
e (sin φ cos ωt + cos φ sin ωt) =
ω
ω
ky0
e−ct . Since this tends to zero as t → ∞, the vibrations
The amplitude is not constant, but is given by
ω
tend to die out as time goes on. The damping is exponential and the motion is reasonable.
42. Let u(x, y) = ex/α f (ξ), where ξ = βx − αy, and assume that α > 0 and β > 0.
(a). Let us compute the partial derivatives of u:
∂u
df
1
= ex/α
+ ex/α f
∂x
dx α
1
x/α df ∂ξ
=e
+ ex/α f
dξ ∂x α
df
1
= ex/α β + ex/α f
dξ
α
and
2
df
df ∂ξ
β
1
1
∂2u
x/α d f ∂ξ
=
e
β + ex/α β + ex/α
+ 2 ex/α f
2
2
∂x
dξ ∂x
α
dξ
α
dξ ∂x α
2
d
f
df
f
β
= β 2 ex/α 2 + 2 ex/α
+ ex/α 2 .
dξ
α
dξ
α
Moreover,
∂f
∂u
= ex/α
∂y
∂y
df
dξ
= ex/α
dξ dy
df
= ex/α (−α)
dξ
df
= −αex/α
dξ
and
From the formulas for
df
∂2u
x/α ∂
=
−αe
2
∂y
∂y dξ
d2 f ∂ξ
= −αex/α 2
dξ ∂y
2
d
f
= −αex/α 2 (−α)
dξ
2
d
f
= α2 ex/α 2 .
dξ
∂2u
∂2u
and
, we have
2
∂x
∂y 2
2
2
∂2u ∂2u
β df
f
x/α
2d f
2d f
β
.
+
=
e
+
2
+
α
+
∂x2
∂y 2
dξ 2
α dξ
α2
dξ 2
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Now if
∂2u ∂2u
+ 2 = 0, then the last equation becomes
∂x2
∂y
β2
2
d2 f
β df
f
2d f
+
2
+
α
=0
+
dξ 2
α dξ
α2
dξ 2
or
(α2 + β 2 )
Therefore,
Letting p =
d2 f
β df
f
+2
+ 2 = 0.
dξ 2
α dξ
α
2β
d2 f
df
f
+
+ 2 2
= 0.
dξ 2
α(α2 + β 2 ) dξ
α (α + β 2 )
d2 f
df
β
1
q
,
the
last
equation
reduces
to
+ 2p + 2 f = 0.
and
q
=
2
2
2
2
2
α(α + β )
α +β
dξ
dξ
α
q
(b). The auxiliary equation associated with Equation (8.2.11) is r2 + 2pr + 2 = 0. The quadratic formula
α
yields the roots
q
r = −p ± p2 − 2 .
α
Therefore,
1/2
β
α2 + β 2
β2
r=−
− 2 2
.
±
α(α2 + β 2 )
α2 (α2 + β 2 )2
α (α + β 2 )2
Hence,
r=
Therefore,
−β ± iα
= −p ± iq.
α(α2 + β 2 )
f (ξ) = e−pξ [A sin qξ + B cos qξ] .
Since u(x, y) = ex/α f (ξ), we conclude that
u(x, y) = ex/α e−pξ (A sin qξ + B cos qξ) = ex/α−pξ (A sin qξ + B cos qξ).
43. The auxiliary equation is r2 + a1 r + a2 = 0.
(a). Depending on whether r1 = r2 , we know that the solution to the differential equation takes one of these
two forms:
y(x) = c1 er1 x + c2 er2 x or y(x) = er1 x (c1 + c2 x).
In order for lim y(x) = 0, we must have that the roots are negative: r1 < 0 and r2 < 0.
x→+∞
(b). Complex roots guarantee a solution of the form y(x) = eax (c1 cos bx + c2 sin bx). In order for
lim y(x) = 0, we must have a < 0. Therefore, the complex conjugate roots must have negative real
x→+∞
part.
(c). By the quadratic formula, the roots of the auxiliary equation are
−a1 ± a21 − 4a2
r=
.
2
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Case 1: If a21 − 4a2 = 0, then r = − a21 < 0 is a double root and y(x) = erx (c1 + c2 x) goes to zero as x → ∞
(see part (a)).
√
−a ± a2 −4a2
Case 2: If a21 − 4a2 > 0, then both roots 1 2 1
are negative; thus, the solution y(x) = c1 er1 x + c2 er2 x
goes to zero as x → ∞ (see part (a)).
Case 3: If a21 − 4a2 < 0, then the roots of the auxiliary polynomial are complex conjugates with negative real
part. Therefore, the conclusion follows from part (b).
(d). In this case, the auxiliary equation is r2 + a1 r = 0, with roots r = 0 and r = −a1 . The general solution
to the differential equation in this case is y(x) = c1 + c2 e−a1 x . Thus, lim y(x) = c1 , a constant.
x→∞
√
√
= ± a2 i. The general
(e). In this case, the auxiliary equation is r2 + a2 = 0, with roots
√ r =
± −a2√
solution to the differential equation in this case is y(x) = c1 cos a2 x + c2 sin a2 x , which is clearly
bounded for all x ∈ R, since the sine and cosine functions are both bounded between ±1 in value.
44. All the roots must have a negative real part.
45. Multiplication of polynomial differential operators is identical to the multiplication of polynomials in
general, an operation that is widely known to be commutative: P (D)Q(D) = Q(D)P (D).
46. This is the special case of Problem 47 for m = 1. Therefore, the solution to Problem 47 can be applied
to solve this exercise as well.
47. The given equations are linearly independent if and only if
m−1
,
ck xk eax cos bx +
k=0
m
,
dk xk eax sin bx = 0 =⇒ ck = 0 and dk = 0
k=0
for all k ∈ {0, 1, 2, ..., m}. If we let
P (x) =
m
,
ck xk and Q(x) =
k=0
m
,
dk xk
k=0
then the first equation can be written as
eax [P (x) cos bx + Q(x) sin bx] = 0,
which is equivalent to
P (x) cos bx + Q(x) sin bx = 0.
nπ nπ
Now x =
where n is an integer implies cos bx = ±1 and sin bx = 0 so P
= 0 for all integers
b
b
(2n + 1)π
, where n is an integer, then cos bx = 0 and sin bx = ±1 which implies that
n. Also if x =
b
(2n + 1)π
= 0 for all integers n. This in turn implies that P (x) and Q(x) are identically zero so ck = 0
Q
b
and dk = 0 for all k ∈ {0, 1, 2, ..., m}. Thus the given functions are linearly independent.
48. y(x) = c1 e−5x + c2 e−7x + c3 e19x .
49. y(x) = c1 e4x + c2 e−4x + e−2x (c3 cos 3x + c4 sin 3x).
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√
√
√
√
50. y(x) = e−5x/2 [c1 cos ( 11x/2) + c2 sin ( 11x/2)] + e−3x/2 [c3 cos ( 7x/2) + c4 sin ( 7x/2)].
51. y(x) = c1 e−4x + c2 cos 5x + c3 sin 5x + x(c4 cos 5x + c5 sin 5x).
52. y(x) = c1 e−3x + c2 cos x + c3 sin x + x(c4 cos x + c5 sin x) + x2 (c6 cos x + c7 sin x).
Solutions to Section 8.3
True-False Review:
(a): FALSE. Under the given assumptions, we have A1 (D)F1 (x) = 0 and A2 (D)F2 (x) = 0. However, this
means that
(A1 (D) + A2 (D))(F1 (x) + F2 (x)) = A1 (D)F1 (x) + A1 (D)F2 (x) + A2 (D)F1 (x) + A2 (D)F2 (x)
= A1 (D)F2 (x) + A2 (D)F1 (x),
which is not necessarily zero. As a specific example, if A1 (D) = D − 1 and A2 (D) = D − 2, then A1 (D)
annihilates F1 (x) = ex and A2 (D) annihilates F2 (x) = e2x . However, A1 (D) + A2 (D) = 2D − 3 does not
annihilate ex + e2x .
(b): FALSE. The annihilator of F (x) in this case is A(D) = Dk+1 , since it takes k + 1 derivatives in order
to annihilate xk .
(c): TRUE. We apply rule 1 in this section with k = 1, or we can compute directly that (D −a)2 annihilates
xeax .
(d): FALSE. Some functions cannot be annihilated by a polynomial differential operator. Only those of
the forms listed in 1-4 can be annihilated. For example, F (x) = ln x does not have an annihilator.
(e): FALSE. For instance, if F (x) = x, A1 (D) = A2 (D) = D, then although A1 (D)A2 (D)F (x) = 0, neither
A1 (D) nor A2 (D) annihilates F (x) = x.
(f ): FALSE. The annihilator of F (x) = 3 − 5x is D2 , but since r = 0 already occurs twice as a root of the
auxiliary equation, the appropriate trial solution here is yp (x) = A0 x2 + A0 x3 .
(g): FALSE. The annihilator of F (x) = x4 is D5 , but since r = 0 already occurs three times as a root of
the auxiliary equation, the appropriate trial solution here is yp (x) = A0 x3 + A1 x4 + A2 x5 + A3 x6 + A4 x7 .
(h): TRUE. The annihilator of F (x) = cos x is D2 + 1, but since r = ±i already occurs once as a
complex conjugate pair of roots of the auxiliary equation, the appropriate trial solution is not yp (x) =
A0 cos x + B0 sin x; we must multiply by a factor of x to occur for the fact that r = ±i is a pair of roots of
the auxiliary equation.
Problems:
Note: In Problems 1-16, we use the four boxed formulas on pages 518-519 of the text.
1. A(D) = D + 3; (D + 3)(5e−3x ) = −15e−3x + 15e−3x = 0.
2. A(D) = D2 (D − 1); (D − 1)(2ex ) = 0 and D2 (3x) = 0 =⇒ D2 (D − 1)(2ex − 3x) = 0.
3. A(D) = (D2 + 1)(D − 2)2 ; (D2 + 1)(sin x) = 0 and (D − 2)2 (3xe2x ) = 0 =⇒
(D2 + 1)(D − 2)2 (sin x + 3xe2x ) = 0.
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4. A(D) = (D − 7)4 (D2 + 16); (D − 7)4 (x3 e7x ) = 0 and (D2 + 16)(5 sin 4x) = 0 =⇒
(D − 7)4 (D2 + 16)(x3 e7x + 5 sin 4x) = 0.
5. A(D) = D2 + 4D + 5; In the expression 4e−2x sin x, a = −2 and b = 1 so that D2 − 2aD + (a2 + b2 ) =
D2 + 4D + 5 =⇒ (D2 + 4D + 5)(4e−2x sin x) = 0.
6. A(D) = (D2 − 2D + 5)(D2 + 4); In the expression ex sin 2x, a = 1 and b = 2, so that D2 − 2aD +
(a2 + b2 ) = D2 − 2D + 5 =⇒ (D2 − 2D + 5)(ex sin 2x) = 0. Moreover, in the expression 3 cos 2x, a = 0
and b = 2, so that D2 − 2aD + (a2 + b2 ) = D2 + 4 =⇒ (D2 + 4)(3 cos 2x) = 0. Thus we conclude that
(D2 − 2D + 5)(D2 + 4)(ex sin 2x + 3 cos 2x) = 0.
7. A(D) = D3 (D − 4)2 ; Note that D3 (2x2 ) = 0 and (D − 4)2 [(1 − 3x)e4x ] = 0 =⇒
D3 (D − 4)2 ((1 − 3x)e4x + 2x2 ) = 0.
8. A(D) = (D2 − 10D + 26)3 ; In the expression e5x cos x, a = 5 and b = 1 so that D2 − 2aD + (a2 + b2 ) =
D2 − 10D + 26 =⇒ (D2 − 10D + 26)3 (e5x (2 − x) cos x) = 0.
9. A(D) = D2 + 6D + 10; In the expression 2e−3x sin x + 7e−3x cos x, a = −3 and b = 1 so that D2 − 2aD +
(a2 + b2 ) = D2 + 6D + 10 =⇒ (D2 + 6D + 10)(e−3x (2 sin x + 7 cos x)) = 0.
10. A(D) = (D − 4)2 (D2 − 8D + 41)D2 (D2 + 4D + 5)3 ; Note that
(D−4)2 (xe4x ) = 0,
(D2 −8D+41)(−2e4x sin 5x) = 0,
D2 (3x) = 0,
and (D2 +4D+5)3 (x2 e−2x cos x) = 0.
Therefore,
(D − 4)2 (D2 − 8D + 41)D2 (D2 + 4D + 5)3 (e4x (x − 2 sin 5x) + 3x − x2 e−2x cos x) = 0.
11. A(D) = (D2 + 1)3 ; Here we have applied Rule 3 (page 474) with a = 0, b = 1, and k = 2.
12. A(D) = (D2 + 9)2 ; Here we have applied Rule 3 (page 474) with a = 0, b = 3, and k = 1.
1 + cos 2x
1
13. Following the hint, we write cos2 x =
. We use A1 (D) = D to annihilate
and we use
2
2
1
A2 (D) = D2 + 4 to annihilate (cos 2x). Hence, A(D) = A1 (D)A2 (D) = D3 + 4D.
2
14. First we simplify
4
2
2
sin x = (sin x) =
1 − cos 2x
2
2
1
1 + cos 4x
3 − 4 cos 2x + cos 4x
1 − 2 cos 2x +
=
=
.
4
2
8
3
1
We use A1 (D) = D to annihilate , A2 (D) = D2 + 4 to annihilate cos 2x, and A3 (D) = D2 + 16 to
8
2
1
2
annihilate cos 4x. Hence, A(D) = A1 (D)A2 (D)A3 (D) = D(D + 4)(D2 + 16) = D5 + 20D3 + 64D.
8
15. To find the annihilator of F (x), let us use the identities
cos2 x =
1 + cos 2x
2
and
sin2 x =
1 − cos 2x
2
and
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556
to rewrite the formula for F (x):
1
sin x cos x(1 + cos 2x)
2
1
= (sin 2x)(1 + cos 2x)
4
1
1
= sin 2x + (sin 2x)(cos 2x)
4
4
1
1
= sin 2x + sin 4x.
4
8
F (x) =
1
1
sin 2x is A1 (D) = D2 +4, and the annihilator of sin 4x is A2 (D) = D2 +16. Therefore,
4
8
the annihilator for F (x) is A(D) = A1 (D)A2 (D) = (D2 + 4)(D2 + 16) = D4 + 20D2 + 64.
The annihilator of
16. To find the annihilator of F (x), let us use the identities
cos2 x =
1 + cos 2x
2
and
sin2 x =
1 − cos 2x
2
to rewrite the formula for F (x):
F (x) = sin2 x cos2 x cos2 2x
1 − cos 2x 1 + cos 2x 1 + cos 4x
=
·
·
2
2
2
1
2
=
(1 − cos 2x)(1 + cos 4x)
8
1
= sin2 2x(1 + cos 4x)
8
1
=
(1 − cos 4x)(1 + cos 4x)
16
1
=
(1 − cos2 4x)
16
1
1
1
1
=
sin2 4x =
(1 − cos 8x) =
−
cos 8x.
16
32
32 32
1
1
is A1 (D) = D, and the annihilator for
cos 8x is A2 (D) = D2 + 64. Therefore, the
32
32
annihilator for F (x) is A(D) = A1 (D)A2 (D) = D(D2 + 64) = D3 + 64D.
The annihilator for
17. We have P (r) = (r − 1)(r + 2) =⇒ yc (x) = c1 ex + c2 e−2x . Now, A(D) = (D − 3). Operating on the given
differential equation with A(D) yields the homogeneous differential equation (D − 3)(D − 1)(D − 2)y = 0,
with solution y(x) = c1 ex + c2 e−2x + A0 e3x . Therefore, yp (x) = A0 e3x . Differentiating yp yields yp =
3A0 e3x , yp = 9A0 e3x . Substituting into the given differential equation and simplifying yields 10A0 = 5, so
1
1
1
that A0 = . Hence, yp (x) = e3x , and therefore y(x) = c1 ex + c2 e−2x + e3x .
2
2
2
18. We have P (r) = (r + 5)(r − 2) =⇒ yc (x) = c1 e2x + c2 e−5x . Now, A(D) = (D − 2). Operating
on the given differential equation with A(D) yields the homogeneous differential equation (D − 2)2 (D +
5)(D − 2)y = 0, with solution y(x) = c1 e2x + c2 e−5x + A0 xe2x . Therefore, yp (x) = A0 xe2x . Differentiating
yp yields yp = A0 e2x (2x + 1), yp = A0 e2x (4x + 4). Substituting into the given differential equation
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yields A0 e2x [4x + 4 + 3(2x + 1) − 10x] = 14e2x , so that A0 = 2. Hence, yp (x) = 2xe2x , and therefore
y(x) = c1 e2x + c2 e−5x + 2xe2x .
19. We have P (r) = r2 + 16 =⇒ yc (x) = c1 cos 4x + c2 sin 4x. Now, A(D) = D2 + 1. Operating on the
given differential equation with A(D) yields (D2 + 1)(D2 + 16)y = 0, so that y(x) = c1 cos 4x + c2 sin 4x +
A0 cos x + B0 sin x. We therefore choose yp (x) = A0 cos x + B0 sin x. Substitution into the given differential
4
equation and simplification yields 15A0 cos x + 15B0 sin x = 4 cos x, so that A0 =
, B0 = 0. Hence,
15
4
4
yp (x) =
cos x, so that y(x) = c1 cos 4x + c2 sin 4x +
cos x.
15
15
20. We have P (r) = (r − 1)2 =⇒ yc (x) = c1 ex + c2 xex . Now, A(D) = (D − 1). Operating on the given
differential equation with A(D) yields the homogeneous differential equation (D − 1)3 y = 0, with solution
y(x) = c1 ex +c2 xex +A0 x2 ex . Therefore, yp (x) = A0 x2 ex . Differentiating yp yields yp = A0 ex (x2 +2x), yp =
A0 ex (x2 +4x+2). Substituting into the given differential equation yields A0 ex [(x2 +4x+2)−2(x2 +2x)+x2 ] =
6ex , so that A0 = 3. Hence, yp (x) = 3x2 ex , and therefore y(x) = c1 ex + c2 xex + 3x2 ex .
21. We have P (r) = (r − 2)(r + 1) =⇒ yc (x) = c1 e2x + c2 e−x . Now, A(D) = D3 . Operating on the given
differential equation with A(D) yields the homogeneous differential equation D3 (D − 2)(D + 1)y = 0, with
solution y(x) = c1 e2x + c2 e−x + A0 + A1 x + A2 x2 . Therefore, yp (x) = A0 + A1 x + A2 x2 . Differentiating yp
yields yp = A1 +2A2 x, yp = 2A2 . Substituting into the given differential equation yields 2A2 −(A1 +2A2 x)−
2(A0 + A1 x + A2 x2 ) = 4x2 − 8x, so that 2A2 − A1 − 2A0 = 0, −2A2 − 2A1 = −8, −2A2 = 4 =⇒A0 = −5,
A1 = 6, A2 = −2. Hence, yp (x) = −5 + 6x − 2x2 , and therefore y(x) = c1 e2x + c2 e−x − 5 + 6x − 2x2 .
22. We have P (r) = r2 −1 = (r−1)(r+1) =⇒ yc (x) = c1 ex +c2 e−x . Now, A(D) = (D−2)(D−3). Operating
on the given differential equation with A(D) yields the homogeneous differential equation (D −2)(D −3)(D −
1)(D + 1)y = 0, with solution y(x) = c1 ex + c2 e−x + A0 e2x + B0 e3x . Therefore, yp (x) = A0 e2x + B0 e3x .
Differentiating yp yields yp = 2A0 e2x + 3B0 e3x , yp = 4A0 e2x + 9B0 e3x . Substituting into the given
differential equation yields 4A0 e2x + 9B0 e3x − 2A0 e2x − 3B0 e3x = 3e2x − 8e3x , so that A0 = 1, B0 = −1.
Hence, yp (x) = e2x − e3x , and therefore y(x) = c1 ex + c2 e−x + e2x − e3x .
23. We have P (r) = (r + 1)(r − 3) =⇒ yc (x) = c1 e−x + c2 e3x . Now, A(D) = (D2 + 1)(D + 1). Operating
on the given differential equation with A(D) yields the homogeneous differential equation (D2 + 1)(D +
1)2 (D − 3)y = 0, with solution y(x) = c1 e−x + c2 e3x + A0 xe−x + A1 cos x + B1 sin x. Therefore, yp (x) =
A0 xe−x + A1 cos x + B1 sin x. Differentiating yp yields yp = A0 e−x (1 − x) − A1 sin x + B1 cos x, yp =
A0 e−x (x − 2) + A1 cos x + B1 sin x. Substituting into the given differential equation yields A0 e−x (x −
2) + A1 cos x + B1 sin x − 2 [A0 e−x (1 − x) − A1 sin x + B1 cos x] − 3 (A0 xe−x + A1 cos x + B1 sin x) = 4(e−x −
2 cos x) =⇒ −4A0 e−x − 2(2A1 + B1 ) cos x + 2(A1 − 2B1 ) sin x = 4(e−x − 2 cos x) so that A0 = −1, and
8
4
8
4
A1 − 2B1 = 0, 2A1 + B1 = 4 =⇒ A1 = , B1 = . Hence, yp (x) = −xe−x + cos x + sin x, and therefore
5
5
5
5
8
4
y(x) = c1 e−x + c2 e3x − xe−x + cos x + sin x.
5
5
24. We have P (r) = r(r + 3) =⇒ yc (x) = c1 + c2 e−3x . Now, A(D) = D2 (D − 1)2 . Operating on the given
differential equation with A(D) yields the homogeneous differential equation D3 (D − 1)2 (D + 3)y = 0, with
solution y(x) = c1 + c2 e−3x + A1 x + A2 x2 + B1 ex + B2 xex . Therefore, yp (x) = A1 x + A2 x2 + B1 ex + B2 xex .
Differentiating yp yields yp = A1 + 2A2 x + B1 ex + B2 ex (x + 1), yp = 2A2 + B1 ex + B2 ex (x + 2). Substituting
into the given differential equation yields 2A2 + B1 ex + B2 ex (x + 2) + 3[A1 + 2A2 x + B1 ex + B2 ex (x + 1)] =
5x + xex , so that 2A2 + 3A1 = 0, 6A2 = 5, 4B1 + 5B2 = 0, 4B2 = 1 =⇒ A1 = −5/9, A2 = 5/6, B1 = −5/16,
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5
5
1
5
5
5
5
B2 = 1/4. Hence, yp (x) = − x + x2 − ex + xex , and therefore y(x) = c1 + c2 e−3x − x + x2 − ex +
9
6
16
4
9
6
16
1 x
xe .
4
25. We have P (r) = r2 + 1 =⇒ yc (x) = c1 cos x + c2 sin x. Now, A(D) = D − 1. Operating on the given
differential equation with A(D) yields (D − 1)(D2 + 1)y = 0, with solution y(x) = c1 cos x + c2 sin x + A0 ex .
Hence, yp (x) = A0 ex . Substitution into the given differential equation yields A0 = 3, so that yp (x) = 3ex .
Consequently, y(x) = c1 cos x + c2 sin x + 3ex .
26. We have P (r) = r2 +4r+4 =⇒ yc (x) = c1 e−2x +c2 xe−2x . Now, A(D) = (D+2)2 . Operating on the given
differential equation with A(D) yields the homogeneous differential equation (D + 2)4 y = 0, with solution
y(x) = c1 e−2x +c2 xe−2x +A0 x2 e−2x +A1 x3 e−2x . Therefore, yp (x) = A0 x2 e−2x +A1 x3 e−2x . Differentiating yp
yields yp = A0 e−2x (−2x2 +2x)+A1 e−2x (−2x3 +3x2 ), yp = A0 e−2x (4x2 −8x+2)+A1 e−2x (4x3 −12x2 +6x).
Substituting into the given differential equation and simplifying yields 2A0 + 6A1 x = 5x, so that A0 =
5
5
5
0, A1 = . Hence, yp (x) = x3 e−2x , and therefore y(x) = c1 e−2x + c2 xe−2x + x3 e−2x .
6
6
6
27. We have P (r) = r2 + 4 =⇒ yc (x) = c1 cos 2x + c2 sin 2x. Now, A(D) = D2 + 4. Operating on
the given differential equation with A(D) yields the homogeneous differential equation (D2 + 4)2 y = 0,
with solution y(x) = c1 cos 2x + c2 sin 2x + x(A0 cos 2x + B0 sin 2x). Therefore, yp (x) = x(A0 cos 2x +
B0 sin 2x). Differentiating yp yields yp = (A0 cos 2x + B0 sin 2x) + x(−2A0 sin 2x + 2B0 cos 2x), yp =
−4A0 sin 2x + 4B0 cos 2x − 4x(A0 cos 2x + B0 sin 2x). Substituting into the given differential equation yields
−4A0 sin 2x + 4B0 cos 2x − 4x(A0 cos 2x + B0 sin 2x) + 4x(A0 cos 2x + B0 sin 2x) = 8 sin 2x, so that A0 = −2,
B0 = 0. Hence, yp (x) = −2x cos 2x, and therefore y(x) = c1 cos 2x + c2 sin 2x − 2x cos 2x.
28. We have P (r) = (r − 2)(r + 1) =⇒ yc (x) = c1 e2x + c2 e−x . Now, A(D) = D − 2. Operating on the given
differential equation with A(D) yields (D−2)2 (D+1) = 0 with general solution y(x) = c1 e2x +c2 e−x +A0 xe2x .
We therefore choose yp (x) = A0 xe2x . Differentiating yp yields yp (x) = A0 e2x (2x+1), yp (x) = A0 e2x (4x+4).
5
5
Substituting into the given differential equation and simplifying we find A0 = . Hence, yp (x) = xe2x , and
3
3
5 2x
2x
−x
so y(x) = c1 e + c2 e + xe .
3
29. We have P (r) = r2 + 2r + 5 =⇒ yc (x) = e−x [c1 cos 2x + c2 sin 2x]. Now, A(D) = D2 + 4. Operating on
the given differential equation with A(D) yields (D2 + 4)(D2 + 2D + 5)y = 0 with general solution y(x) =
e−x (c1 cos 2x + c2 sin 2x) + A0 cos 2x + B0 sin 2x. We therefore choose yp (x) = A0 cos 2x + B0 sin 2x. Substitution into the given differential equation and simplification yields (A0 + 4B0 ) cos 2x + (−4A0 + B0 ) sin 2x =
12
3
3 sin 2x. Consequently A0 + 4B0 = 0, −4A0 + B0 = 3. This system has a solution A0 = − , B0 =
.
17
17
12
12
3
3
Hence, yp (x) = − cos 2x +
sin 2x, and so y(x) = e−x (c1 cos 2x + c2 sin 2x) −
cos 2x +
sin 2x.
17
17
17
17
30. We have P (r) = r3 + 2r2 − 5r − 6 = (r − 2)(r + 3)(r + 1) =⇒ yc (x) = c1 e2x + c2 e−3x + c3 e−x . Now,
A(D) = D3 . Operating on the given differential equation with A(D) yields D3 (D − 2)(D + 3)(D + 1)y = 0,
with general solution y(x) = c1 e2x + c2 e−3x + c3 e−x + A0 + A1 x + A2 x2 . We therefore choose yp (x) =
A0 + A1 x + A2 x2 . Substituting into the given differential equation and simplifying yields
4A2 − 5(A1 + 2A2 x) − 6(A0 + A1 x + A2 x2 ) = 4x2 ,
so that 4A2 − 5A1 − 6A0 = 0, −10A2 − 6A1 = 0, −6A2 = 4. This system has a solution A0 = −
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, A1 =
27
559
2
37 10
37 10
10
2
2
, A2 = − . Hence, yp (x) = − + x − x2 , and so y(x) = c1 e2x + c2 e−3x + c3 e−x −
+ x − x2 .
9
3
27
9
3
27
9
3
31. We have P (r) = (r − 1)(r2 + 1) =⇒ yc (x) = c1 ex + c2 cos x + c3 sin x. Now, A(D) = D + 1. Operating
on the given differential equation with A(D) yields (D + 1)(D − 1)(D2 + 1)y = 0, with general solution
y(x) = c1 ex + c2 cos x + c3 sin x + A0 e−x . We therefore choose yp (x) = A0 e−x . Substitution into the given
9
9
differential equation and simplification yields A0 = − . Hence, yp (x) = − e−x , and so the general solution
4
4
9 −x
x
is y(x) = c1 e + c2 cos x + c3 sin x − e .
4
32. We have P (r) = (r + 1)3 =⇒ yc (x) = c1 e−x + c2 xe−x + c3 x2 e−x . Now, A(D) = (D − 2)(D + 1).
Operating on the given differential equation with A(D) yields (D − 2)(D + 1)4 y = 0, with general solution
y(x) = c1 e−x + c2 xe−x + c3 x2 e−x + A0 x3 e−x + A1 e2x . We therefore choose yp (x) = A0 x3 e−x + A1 e2x .
Differentiating yp yields
yp = A0 e−x (−x3 + 3x2 ) + 2A1 e2x , yp = A0 e−x (x3 − 6x2 + 6x) + 4A1 e2x ,
yp = A0 e−x (−x3 + 9x2 − 18x + 6) + 8A1 e2x .
Substituting into the given differential equation and simplifying we find 6A0 e−x + 27A1 e2x = 2e−x + 3e2x , so
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that A0 = , A1 = . Hence, yp (x) = x3 e−x + e2x , and therefore, the general solution to the differential
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equation is y(x) = c1 e−x + c2 xe−x + c3 x2 e−x + x3 e−x + e2x .
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33. We have P (r) = r2 + 9 =⇒ yc (x) = c1 cos 3x + c2 sin 3x. Now, A(D) = D2 + 4. Operating on the
given differential equation with A(D) yields the homogeneous differential equation (D2 + 4)(D2 + 9)y = 0,
with solution y(x) = c1 cos 3x + c2 sin 3x + A0 cos 2x + B0 sin 2x. Therefore, yp (x) = A0 cos 2x + B0 sin 2x.
Differentiating yp yields yp = −2A0 sin 2x + 2B0 cos 2x, yp = −4(A0 cos 2x + B0 sin 2x). Substituting into
the given differential equation yields −4(A0 cos 2x + B0 sin 2x) + 9(A0 cos 2x + B0 sin 2x) = 5 cos 2x, so that
A0 = 1, B0 = 0. Hence, yp