Answer
See below
Work Step by Step
Given:
$\begin{bmatrix}
-7& -6 & -7\\
-3 & -3 & -3\\
7& 6 & 7
\end{bmatrix}$
A straightforward calculation yields
$\det (A-\lambda I)=\begin{vmatrix}
-7-\lambda & -6 & -7\\
-3 & -3-\lambda & -3\\
7 & 6 & 7-\lambda
\end{vmatrix}=\lambda^2(\lambda+3)$
so that A has eigenvalues $\lambda_1=0\\
\lambda_2=0\\
\lambda_3=-3$
Eigenvalue $\lambda_1 =0$: In this case, the system $(A − \lambda I)v = 0$ is $v_1=t\\
v_2=-1\\
v_3=1+t$
The solution is $v = r(t,-1,1+t)$. Therefore,
$v_1=\begin{bmatrix}
t\\
-1\\
1+t\end{bmatrix}$
Do the same for $v_2$ and $v_3$,
$v_2=\begin{bmatrix}
-1\\
0 \\
1
\end{bmatrix}$
$v_3= e^{-3t}\begin{bmatrix}
-2\\
-1\\
2
\end{bmatrix}$
Hence, the general solution is:
$x(t)=c_1\begin{bmatrix}
t\\
-1 \\
t+1\end{bmatrix}+c_2\begin{bmatrix}
-1\\
0\\
1
\end{bmatrix}+c_3e^{-3t}\begin{bmatrix}
-2\\
-1\\
2 \end{bmatrix}=\begin{bmatrix}
-c_1+c_2t -2c_3e^{-3t}\\
-c_2-c_3e^{-3t}\\
c_1+(1+t)c_2+2c_3e^{-3t}
\end{bmatrix}$
