Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix} -\lambda & 1 & 0 \\ 0 & -\lambda & 1\\ 1 & 1 & -1-\lambda\end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \\0 \end{bmatrix}$
$\begin{bmatrix} -\lambda & 1 & 0 \\ 0 & -\lambda & 1\\ 1 & 1 & -1-\lambda\end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3 \end{bmatrix}=0$
$(\lambda-1)(\lambda+1)^2=0$
$\lambda_1=\lambda_2=-1, \lambda_3=1$
2. Find eigenvectors:
For $\lambda=-1$
let $B=A-\lambda_1I$
$B=\begin{bmatrix} -\lambda & 1 & 0 \\ 0 & -\lambda & 1\\ 1 & 1 & -1-\lambda\end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3 \end{bmatrix}=\begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1\\ 1 & 1 & 0 \end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \\0 \end{bmatrix} $
Let $r,s$ be free variables.
$\vec{V}=r(1,0,1)+s(1,0,0)\\
E_1=\{(1,0,1);(1,0,0)\} \\
\rightarrow dim(E_2)=2$
For $\lambda=2$
let $B=A-\lambda_1I$
$B=\begin{bmatrix} -\lambda & 1 & 0 \\ 0 & -\lambda & 1\\ 1 & 1 & -1-\lambda\end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3 \end{bmatrix}=\begin{bmatrix} -2 & 1 & 0 \\ 0 & -2 & 1\\ 1 & 1 & -3 \end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \\0 \end{bmatrix} $
Let $r$ be a free variable.
$\vec{V}=r(1,1,1)\\
E_1=\{(1,1,1)\} \\
\rightarrow dim(E_2)=2$
Hence, $S=\begin{bmatrix} 1 & 1 & 1\\ 0 & 0 & 1 \\ 1 & 0 & 1 \end{bmatrix} $
We have $x=Sy \rightarrow x'=Ax \rightarrow y'=\begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 &0 \\ 0 & 0 & 1 \end{bmatrix}y $
then $y_1=c_3e^{-t}+c_2te^{-t}\\
y_2=c_2e^{-t}\\
y_3=c_1e^t$
Hence,
$x(t)=S.y(t)=\begin{bmatrix} 1 & 1 & 1\\ 0 & 0 & 1 \\ 1 & 0 & 1 \end{bmatrix} \begin{bmatrix} c_3e^{-t}+c_2te^{-t}\\ c_2e^{-t} \\ c_1e^t \end{bmatrix} \\
=\begin{bmatrix} c_3e^{-t}+c_2te^{-t} +c_2e^{-t}+c_1e^t\\ c_1e^t \\ c_3e^{-t}+c_2te^{-t}+c_1e^t\end{bmatrix} $
