Math 20D
Final Exam
8 December 2009
1. (10 points) Show that the matrix
4

2
2
1
1
2

1
2
1
has eigenvalues 3
,
3
,
0 and find the eigenvectors associated with 3.
Solution:
det
4

λ

2
2
1
1

λ
2

1
2
1

λ
= (4

λ
) det
1

λ
2
2
1

λ

(

2) det
1
2

1
1

λ
+ 2 det
1
1

λ

1
2
= (4

λ
)((1

λ
)
2

4) + 2(1

λ
+ 2) + 2(2 + 1

λ
)
= (4

λ
)(

3

2
λ
+
λ
2
) + 12

4
λ
=

12

8
λ
+ 4
λ
2
+ 3
λ
+ 2
λ
2

λ
3
+ 12

4
λ
=

λ
3
+ 6
λ
2
+ 9
λ
=

λ
(
λ

3)
2
.
Thus the eigenvalues are 3, 3, and 0. Perform Guassian elimination on
A

3
I
1

2
2
1

2
2

1
2

2
→
1

2
2
0
0
0
0
0
0
.
Thus the eigenvalues are
2
1
0

2
0
1
.
2. Find the general form of the solution for each of the following differential equations.
(a) (5 points) 3
y
00

y
0

2
y
= 0
Solution:
The characteristic polynomial is 3
r
2

r

2 = 0 which has roots
1
±
√
1+24
6
=
1
±
5
6
. Thus the general form of the solution is
c
1
e
t
+
c
2
e

2
3
t
.
(b) (5 points)
y
00
+ 4
y
0
+ 13
y
= 0
Solution:
The characteristic polynomial is
r
2
+4
r
+13 = 0 which has roots

4
±
√
16

52
2
=

4
±
6
i
2
=

2
±
3
i
. Thus the general form of the solution is
e
3
t
(
c
1
cos(2
t
) +
c
2
sin(2
t
)).
(c) (5 points) 4
y
00
+ 4
y
0
+
y
= 0.
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Math 20D
Final Exam
8 December 2009
Solution:
The characteristic polynomial is 4
r
2
+4
r
+1 = 0 which has roots

4
±
√
16

16
8
=

1
2
.
Thus the general form of the solution is
c
1
e

t
2
+
c
2
te

t
2
.
3. Find the general form of the solution for each of the following systems of differential equations
of the form
x
0
=
Ax
.
(a) (15 points)
A
=

1
1

2

3
Solution:
det

1

λ
1

2

3

λ
= (1 +
λ
)(3 +
λ
) + 2 =
λ
2
+ 4
λ
+ 5
.
Thus the eigenvalues are

2
±
i
and the eigenvectors are
1
2
∓
i
. Thus the general
form of the solution is
c
1
e

2
t
cos(
t
)
2 cos(
t
)

sin(
t
)
+
c
2
e

2
t
sin(
t
)
2 sin(
t
) + cos(
t
)
.
(b) (15 points)
A
=
1
3
4
2
Solution:
det
1

λ
3
4
2

λ
= (1

λ
)(2

λ
)

12 =
λ
2

3
λ

10 = (
λ

5)(
λ
+ 2)
.
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 Fall '09
 Eggers,John
 Equations, Elementary algebra, representative, Eigenvalue, eigenvector and eigenspace

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