Math 2270 Spring 2004
Homework 9 Solutions
Section 3.4) 2, 6, 8, 12, 14, 16, 18, 22, 36, 43
(2) We must solve the system




#
−5
−1 −2 "


 c1

1 
= 1 
 0
c2
3
1 0




1 0 |3
−1 −2 | −5



1 | 1 → 0 1 |1 

 0
0 0 |0
1 0 | 3
The solution is c1 = 3 and c2 = 1, so
[~x]B =
"
3
1
#
(6)


(8) We must solve




11
5
8






~x = 2  4  − 1  2  =  6 
−1
−1
−1


1


2~v1 − 1~v2 = ~x =  −1 
1
such that the vectors ~v1 and ~v2 form a basis for the plane defined by x1 + 2x2 + x3 =0. Any

−2


vector on the plane must satisfy x1 = −2x2 − x3 , so two possible basis vectors are  1 
0




−2
−1


v2 such that 2v~1 − 1~v2 = ~x. Since the
v1 = 
and 
 1 . It remains to find ~
 0 . Let ~
0
1
vector ~x is itself on the plane we know that a linear combination of ~x and ~v1 will yield a
vector that is still on the plane. Therefore, we can solve the last equation for ~v2 and define






−5
1
−2




 
~v2 = 2~v1 − ~x = 2  1  −  −1  =  3 
−1
1
0
1
Math 2270 Spring 2004
Homework 9 Solutions
Notice that the third entry of ~v1 is zero and the third entry of ~v2 is non-zero, guaranteeing
their independence. Thus, one possible basis is







(12) We must solve



 
−5 
−2



1 
, 3 

−1 
0






3
0
0
1








c1  1  + c2  1  + c2  0  =  7 
13
1
1
1
By inspection, c1 must equal 3 as the first vector is the only one with a non-zero value for
the first entry. Then, only the first and second vectors have non-zero values for the second
entry so c1 + c2 = 3 + c2 = 7, so c2 = 4. Then, we solve for the remaining variable using
c1 + c2 + c3 = 3 + 4 + c3 = 13, so c3 = 6. Thus


3


[~x]B =  4 
6
(14) Define the matrix S to have columns equal to the basis vectors and find its inverse using
the determinant formula.
S=
"
1 −1
2 3
#
S −1 =
1
5
"
3 1
−2 1
#
Then, using Fact 3.4.4, we can calculate the desired matrix B = S −1 AS.
1
B=
5
"
3 1
−2 1
#"
7 −1
−6 8
#"
1 −1
2 3
#
1
=
5
"
3 1
−2 1
#"
5 −10
10 30
#
1
=
5
"
25 0
0 50
#
=
"
5 0
0 10
(16) To find the matrix B of the linear transformationwith respect to the
given
basis, we


−1
1




first notice that the first two vectors of the basis, ~v1 =  1  and ~v2 =  2  are on the
−1
−1
plane x1 + 2x2 + 3x3 = 0. Therefore, they will not change when reflected in the plane. Hence,
the first two columns
of matrix B will be the first two columns of the identity matrix. The


1


third vector ~v3 =  2  is perpendicular to the plane. Therefore, it will map to the negative
3
2
#
Math 2270 Spring 2004
Homework 9 Solutions
of itself (when written with respect to the given basis), so [T (~v3 )]B = −~e3 .


1 0 0


B= 0 1 0 
0 0 −1
Now, we define the matrix S to have columns equal to the given basis vectors and find its
inverse.




8 2 −4
1 −1 1



1 
S −1 = 14
S= 1 2 2 
 −5 4 −1 
1 2 3
−1 −1 3
Then, by Fact 3.4.4 the standard matrix of the transformation is A = SBS −1 .







8 2 −4
1 0 0
8 2 −4
1 −1 1
1 −1 1
1 
1 




2 2   0 1 0   −5 4 −1  =
2 2   −5 4 −1 
A=

 1
 1
14
14
1 2 3
0 0 −1
−1 −2 −3
−1 −1 3
−1 −1 3




12 −4 −6
6 −2 −3
1 
1

=
6 −12  =  −2 3 −6 

 −4
14
7
−6 −12 −4
−3 −6 −2
(18) We use Fact 3.4.4 with
B=
A = SBS
−1
=
"
"
=
1 9
9 7
"
#
3 5
5 8
#"
S=
"
1 9
9 7
#"
#
S
−8 5
5 −3
#
3 5
5 8
3(37) − 5(37) 3(−22) + 5(24)
5(37) − 8(37) 5(−22) + 8(24)
−1
=
#
=
"
"
3 5
5 8
=
"
#
−8 5
5 −3
#"
37 −22
−37 24
−74 54
−111 82
#
(36) We must solve AS = SB for S where
A=
"
1 2
4 3
#
B=
"
5 0
0 −1
#
S=
"
and S is invertible, or det(S) = ad − bc 6= 0.
"
1 2
4 3
#"
a b
c d
3
#
=
"
a b
c d
#"
5 0
0 −1
#
a b
c d
#
#
Math 2270 Spring 2004
Homework 9 Solutions
"
a + 2c b + 2d
4a + 3c 4b + 3d
#
=
"
5a −b
5c −d
#
This gives us a linear system with four equations that reduce to two:
a + 2c = 5a ⇒ 2a = c
4a + 3c = 5c ⇒ 2a = c
b + 2d = −b ⇒ d = −b
4b + 3d = −d ⇒ b = −d
Therefore, the general form for the matrix S is
"
a b
2a −b
#
with det(S) = −ab − 2ab = −3ab. As long as a and b are both non-zero, the matrix S will
be invertible. Let a = b = 1. Then, one possible basis is
("
1
2
# "
,
1
−1
#)
(43) We are given AS = SD where, on the right, the matrices S and D have already been
multiplied together.
AS# = SD
"
"
#
1 2
3 −2
A
=
2 1
6 −1
We must find a diagonal matrix D that satisfies the above equation. such that AS = SD.
If S is invertible, we can multiply both sides of the equation by S −1 on the left leaving us
with the desired matrix D on the right: S −1 AS = S −1 SD = D. We see that S is invertible
as det(S) = 1 − 4 = −3. Therefore,
D=S
−1
1
SD =
−3
"
1 −2
−2 1
4
#"
3 −2
6 −1
#
1
=
−3
"
−9 0
0 3
#
=
"
3 0
0 −1
#