Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix} -2-\lambda & -1 \\ 1 & -4-\lambda \end{bmatrix}\begin{bmatrix} v_1\\ v_2 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \end{bmatrix}$
$\begin{bmatrix} -2-\lambda & -1 \\ 1 & -4-\lambda \end{bmatrix}=0$
$(\lambda+3)^2=0$
$\lambda_1=\lambda_2=-3$
2. Find eigenvectors:
For $\lambda=-3$
let $B=A-\lambda_1I$
$B=\begin{bmatrix} 1 & -1 \\ 1 & -1 \end{bmatrix}=\begin{bmatrix} 0 \\0 \end{bmatrix} $
Let $r,s$ be a free variable.
$\vec{V}=r(1,1)+s(1,0)\\
E_1=\{(1,1);(1,0)\} \\
\rightarrow dim(E_2)=2$
Hence, $S=\begin{bmatrix} 1 & 1\\ 1 & 0 \end{bmatrix} $
We have $x=Sy \rightarrow x'=Ax \rightarrow y'=\begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 2 \end{bmatrix}y $
then $y_1=c_1e^{-3t}\\
y_2=c_2e^{-3t}$
Hence,
$x(t)=S.y(t)=\begin{bmatrix} 1 & 1\\ 1 & 0 \end{bmatrix} \begin{bmatrix} c_1e^{-3t}+c_2te^{-3t}\\ c_2e^{-3t} \end{bmatrix} \\
=\begin{bmatrix} c_1e^{-3}+c_2(1+t)e^{-3t}\\ c_1e^{-3t}+c_2te^{-3t} \end{bmatrix} $
We are given: $\begin{bmatrix} 0 \\ -1 \end{bmatrix}\rightarrow $$c_1=-1\\
c_2=1$
Hence, $x_1(t)=te^{-3t}\\
x_2(t)=(t-1)e^{-3t}$
