Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix} -4-\lambda & 1 \\ -1 & -6-\lambda \end{bmatrix}\begin{bmatrix} v_1\\ v_2 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \end{bmatrix}$
$\begin{bmatrix} -4-\lambda & 1 \\ -1 & -6-\lambda \end{bmatrix}=0$
$(\lambda+5)^2=0$
$\lambda_1=\lambda_2=-5$
2. Find eigenvectors:
For $\lambda=-5$
let $B=A-\lambda_1I$
$B=\begin{bmatrix} -4-\lambda & 1 \\ -1 & -6-\lambda \end{bmatrix}=\begin{bmatrix} 1 & 1 \\ -1 & -1 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \end{bmatrix} $
Let $r,s$ be free variables.
$\vec{V}=r(1,-1)+s(\frac{1}{2},\frac{1}{2})\\
E_1=\{(1,-1);(\frac{1}{2},\frac{1}{2})\} \\
\rightarrow dim(E_2)=2$
Hence, $S=\begin{bmatrix} 1 & -1\\ \frac{1}{2} & \frac{1}{2}\\
0 & 1 & 1 \end{bmatrix} $
We have $x=Sy \rightarrow x'=Ax \rightarrow y'=\begin{bmatrix} -5 & 1 \\ 0 & -5 \end{bmatrix}y $
then $y_1=c_2e^{-5t}+c_1te^{-5t}\\
y_2=c_1e^{-5t}$
Hence,
$x(t)=S.y(t)=\begin{bmatrix} 1 & -1\\ \frac{1}{2} & \frac{1}{2}\\
0 & 1 & 1 \end{bmatrix} \begin{bmatrix} c_1e^{-5t}+c_2e^{-5t}\\ c_1e^{-5t}\end{bmatrix} \\
=\begin{bmatrix} e^{-5t}(c_1t+c_2+\frac{1}{2}c_1)\\ e^{-5t}(c_1-c_1t-c_2)\\
0 & 1 & 1 \end{bmatrix} $
