Math 2280
Quiz 7
1. Find the general solution to the system
~x′ =
6
1
−7
−2
~x.
First we must find the eigenvalues of the coefficient matrix. To do this, we solve the characteristic equation
6−λ
−7 0 = = (6 − λ)(−2 − λ) + 7 = −12 − 6λ + 2λ + λ2 + 7
1
−2 − λ = λ2 − 4λ − 5 ⇒ λ2 − 4λ + 4 = 5 + 4 ⇒ (λ − 2)2 = 9 ⇒ λ = 2 ± 3 = 5 or − 1.
Now we need the eigenvectors corresponding to these eigenvalues. For λ = 5 we have
1 −7
1 −7
A − 5I =
∼
⇒ a = 7b.
1 −7
0 0
If we choose b = 1, then an eigenvector corresponding to λ = 5 is
7
~v1 =
.
1
For λ = −1, we have
A+I =
7 −7
1 −1
∼
1 −1
0 0
⇒ a = b.
1
1
If we choose b = 1, then
~v2 =
1
1
.
Our general solution is
~x(t) = e
5t
7
1
+e
−t
.
2. Find the general solution to the system


1
0 0
~x′ =  1
3 0  ~x.
−2 −4 1
Since A is lower triangular, it is clear that its eigenvalues are λ = 1, 1, and 3. Since 3 is not a repeated eigenvalue,
let’s deal with it first. For λ = 3 we have
 

 

1 0
0
1 0 0
−2 0
0
1
0  ∼  0 1 21  ⇒ a = 0, b = − c.
0
0 ∼ 0 0
A − λI =  1
2
0 −4 −2
0 0 0
−2 −4 −2
If we choose c = 2, then an eigenvecto for λ = 3 is


0
~v1 =  −1  .
2
For λ = 1 we have

 
0
0 0
1



1
2
0
0
A − λI =
−2 −4 0
0
1

2 0
0 0  ⇒ a = 2b,
0 0
so if we choose b = 0 and c = 1, then we get


0
~v2 =  0  .
1
If instead we choose b = 1 and c = 0, we get


−2
~v3 =  1  ,
0
so even though λ = 1 is repeated with algebraic multiplicity 2, we were able to find two linearly independent
eigenvectors. Thus the geometric multiplicity of λ = 1 is 2 and its defect is zero. Our general solution is
~x(t) = c1 e3t~v1 + c2 et~v2 + c3 et~v3 .
2