Math 2250 (Jason Underdown)
Exam 2
November 13, 2014
Name
1. [10 pts.] If possible, compute A−1 for


1 0 2
A = 2 3 0  .
0 1 −1
Solution:


1 0 2 1 0 0
2 3 0 0 1 0
0 1 −1 0 0 1


1 0 2
1 0 0
0 1 −1 0 0 1
0 3 −4 −2 1 0


1 0 2 1 0 0
0 1 −1 0 0 1
0 0 1 2 −1 3


1 0 0 −3 2 −6
0 1 0 2 −1 4 
0 0 1 2 −1 3
Thus A−1


3 2 −6
= 2 −1 4 .
2 −1 3
−2R1 +R2
−→


1 0 0
1 0 2
0 3 −4 −2 1 0
0 1 −1 0 0 1
−→

−3R2 +R3
−→
R3 +R2
−→

1 0 2
1 0 0
0 1 −1 0 0 1 
0 0 −1 −2 1 −3
R2 ↔R3


1 0 2 1 0 0
0 1 0 2 −1 4
0 0 1 2 −1 3
(−1)R3
−→
−2R3 +R1
−→
Math 2250
Exam 2, Page 2 of 10
November 13, 2014
2. Solve the following homogeneous, linear DEs with constant coefficients:
(a) [5 pts.] y 00 − 5y 0 + 6y = 0
Solution: Computing the characteristic equation yields:
r2 − 5r + 6 = 0
(r − 2)(r − 3) = 0
Thus y = c1 e2x + c2 e3x .
(b) [5 pts.] y 000 − 5y 00 + 7y 0 − 3y = 0
Solution: By inspection we see that 1 is a root of the characteristic equation.
After polynomial long division (omitted) we obtain the full factorization:
r3 − 5r2 + 7r − 3 = 0
(r − 1)(r2 − 4r + 3) = 0
(r − 1)(r − 1)(r − 3) = 0
Thus y = (c1 + c2 x)ex + c3 e3x .
(c) [5 pts.] y 00 + 4y 0 + 40y = 0
Solution: Completing the square on the characteristic equation yields:
r2 + 4r + 40 = 0
r2 + 4r + 4 = −40 + 4
(r + 2)2 = −36
√
r + 2 = ± −36
r = −2 ± 6i
Thus y = c1 e−2x cos 6x + c2 e−2x sin 6x.
Math 2250
Exam 2, Page 3 of 10
November 13, 2014
3. [8 pts.] Is the following set of vectors from R4 linearly dependent or linearly independent?
Circle one answer and justify your answer.
       
2
1
1
5 


       

0 1  0 1

S =  , , , 
0
0
−1
0 





0
0
0
1
Linearly Dependent
Linearly Independent
Solution: Theorem 2 in §4.3 says that these four vectors are linearly independent if
and only if the determinant of the 4 × 4 matrix formed by these vectors is nonzero.
Since the matrix formed by these vectors is an upper triangular matrix, the determinant is computed by just multiplying the entries on the diagonal:
2 1 1 5
0 1 0 1
0 0 −1 0 = 2 · 1 · (−1) · 1 = −2 6= 0
0 0 0 1
Thus these four vectors are linearly independent.
4. [5 points (bonus)] Is the set S a basis for the vector space R4 ? Circle one answer and
justify your answer. (You won’t receive any credit without justification.)
YES
NO
Solution: By Example 2 in §4.4, any n linearly independent vectors in Rn span Rn ,
thus since we already showed linear independence of S, we see that S also spans R4
and is thus a basis for R4 .
Math 2250
Exam 2, Page 4 of 10
November 13, 2014
5. [15 pts.] Use the method of undetermined coefficients to find a general solution to
y 00 + 4y = 8 cos 2x.
(1)
Solution: The characteristic equation r2 + 4 = 0 has roots r = ±2i, thus the
homogeneous solution is:
yh = c1 cos 2x + c2 sin 2x.
(2)
The annhilating operator for the RHS (right hand side) of the equation is (D2 + 4)
which is the same as the operator on the LHS, thus a trial particular solution yp
should be a solution to:
(D2 + 4)2 yp = 0,
(D − 2i)2 (D + 2i)2 yp = 0.
Thus,
yp = (A + Bx) cos 2x + (C + Dx) sin 2x
(
(
(2x
2x
yp = (
A(
cos
+ Bx cos 2x + C
sin
+ Dx sin 2x
yp = Bx cos 2x + Dx sin 2x,
where the cancellation is due to overlap with terms in the homogeneous solution (2).
Next we compute yp00 .
yp0 = B cos 2x − 2Bx sin 2x + D sin 2x + 2Dx cos 2x
yp00 = −2B sin 2x − 2B sin 2x − 4Bx cos 2x + 2D cos 2x + 2D cos 2x − 4Dx sin 2x
yp00 = 4D cos 2x − 4B sin 2x − 4Bx cos 2x − 4Dx sin 2x
Plugging yp and yp00 into the governing equation, equation (1) yields:
yp00 + 4yp = 4D cos 2x − 4B sin 2x = 8 cos 2x.
Equating the coefficients of like terms on both sides yields the two equations:
−4B = 0
4D = 8
This system has solution B = 0 and D = 2, thus a particular solution of equation (1)
is
yp = 2x sin 2x.
Finally the general solution is
y = yh + yp
y = c1 cos 2x + c2 sin 2x + 2x sin 2x
Math 2250
Exam 2, Page 5 of 10
November 13, 2014
6. [9 pts.] Recall that M2×2 , the set of 2 × 2 matrices, where
a b M2×2 =
a, b, c, d ∈ R
c d forms a vector space. Show that U , the subset of upper triangular 2 × 2 matrices, where
a b U=
a, b, c ∈ R ⊂ M2×2 ,
0 c is a subspace of M2×2 .
Solution: According to §4.2 Theorem 1 (Conditions for a Subspace) we must show
three things:
1. U is nonempty.
2. U is closed under addition.
3. U is closed under scalar multiplication.
0 0
1. Since
∈ U , U 6= ∅.
0 0
a
2. Let A, B ∈ U , say A =
0
a
A+B =
0
b
and B
c
b
p
+
c
0
p q
=
. Then
0 r
q
a+p b+q
=
∈ U.
r
0
c+r
3. Let α ∈ R, and let A be defined as before, then
a b
α·a α·b
α·a α·b
αA = α
=
=
∈ U.
0 c
α·0 α·c
0
α·c
7. [1 pts.] What is the dimension of U ?
Solution: Since one basis for U is:
1 0
0 1
0 0
,
,
,
0 0
0 0
0 1
and this set contains three elements, dim U = 3.
Math 2250
Exam 2, Page 6 of 10
November 13, 2014
8. Circle or box the letter corresponding to the correct trial particular solution, yp , for each
given equation.
(a) [3 pts.] y 00 − 6y 0 − 7y = x
A.
B.
C.
D.
yp
yp
yp
yp
= Ax
= A + Bx + Cx2
= (A + Bx)ex
= A + Bx
(b) [3 pts.] y 00 − 4y 0 + 4y = e2x
A.
B.
C.
D.
yp
yp
yp
yp
= Ae2x
= Axe2x
= Ax2 e2x
= Ax3 e2x
Math 2250
Exam 2, Page 7 of 10
(c) [3 pts.] y 000 − y 0 = x
A.
B.
C.
D.
yp
yp
yp
yp
= Bx + Cx2
= A + Bx
= Bx
=A
(d) [3 pts.] (D2 + 9)2 y = 5 cos 3x
A.
B.
C.
D.
yp
yp
yp
yp
= A cos 3x + B sin 3x
= (A + Bx) cos 3x + (C + Dx) sin 3x
= Bx cos 3x + Dx sin 3x
= Cx2 cos 3x + F x2 sin 3x
November 13, 2014
Math 2250
Exam 2, Page 8 of 10
November 13, 2014
9. [20 pts.] True or False. Circle one.
(a) T
F
The following algebraic

1
0
0
system has infinitely many solutions.
   
0 0 x
0




1 0
y = 0
0 1
z
0
(b) T
F
The following algebraic

1
0
0
system has infinitely many solutions.
   
0 0 x
0
1 0 y  = 0
0 0
z
1
(c) T
F
If the determinant of a matrix is 0 then that matrix is invertible.
F A fourth order, linear, homogeneous DE with constant coefficients has
(d) T
exactly four linearly independent solutions.
(e) T
F
The solution space of a linear differential equation is a vector subspace
of F, the vector space of all real valued functions.
(f) T
F The driven mass—spring—dashpot system corresponding to the following
DE is underdamped.
x00 + 5x0 + 25x = cos 2t
(g) T
F
If A and B are n × n matrices, then AB = BA.
(h) T
F
The following two linear operators are equivalent, that is L1 = L2 .
L1 = (D − 3)(D + 2)
L2 = (D + 2)(D − 3)
(i) T
F Suppose A is an n × n matrix. If A is invertible, then for every n–vector
~b, the system A~x = ~b has a unique solution.
(j) T
F
The following system will exhibit resonance.
x00 + 9x = sin 3t
Math 2250
Exam 2, Page 9 of 10
November 13, 2014
10. The are two ways you can fail to earn the points for the last two questions:
1. not providing an answer,
2. not writing an integer between 1 and 10 inclusive.
(a) [5 pts.] Rate the difficulty of this exam on a scale of 1 to 10, where 1 is very easy
and 10 is very difficult.
(a)
7.4
(b) [5 pts.] Rate the fairness of this exam on a scale of 1 to 10, where 1 is completely
unfair and 10 is completely fair.
(b)
6.7
Math 2250
Exam 2, Page 10 of 10
November 13, 2014
Scratch Paper
Question:
1
2
3
4
5
6
7
8
9
10
Total
Points:
10
15
8
0
15
9
1
12
20
10
100
Bonus Points:
0
0
0
5
0
0
0
0
0
0
5
Score: