Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix} -3-\lambda & -2 \\ 2 & 1-\lambda \end{bmatrix}\begin{bmatrix} v_1\\ v_2 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \end{bmatrix}$
$\begin{bmatrix} -3-\lambda & -2 \\ 2 & 1-\lambda \end{bmatrix}=0$
$(\lambda+1)^2=0$
$\lambda_1=\lambda_2=-1$
2. Find eigenvectors:
For $\lambda=-1$
let $B=A-\lambda_1I$
$B=\begin{bmatrix} -3-\lambda & -2 \\ 2 & 1-\lambda \end{bmatrix}=\begin{bmatrix} -2& -2 \\ 2 & 0 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \end{bmatrix} $
Let $r,s$ be free variables.
$\vec{V}=r(-2,2)+s(1,0)\\
E_1=\{(-2,2);(1,0)\} \\
\rightarrow dim(E_2)=2$
Hence, $S=\begin{bmatrix} -2 & 2\\ 1 & 0 \end{bmatrix} $
We have $x=Sy \rightarrow x'=Ax \rightarrow y'=\begin{bmatrix} -1 & 1 \\ 0 & -1 \end{bmatrix}y $
then $y_1=c_2e^{-t}+c_1te^{-t}\\
y_2=c_1e^{-t}$
Hence,
$x(t)=S.y(t)=\begin{bmatrix} -2 & 1\\ 2 & 0 \end{bmatrix} \begin{bmatrix} c_1te^{-t}+c_2e^{-t}\\ c_1e^{-t}\end{bmatrix} \\
=\begin{bmatrix} e^{-t}(-2c_2-2c_1t+c_1)\\ e^{-t}(2c_2+2c_1t)\end{bmatrix} $
