Answer
See below
Work Step by Step
Given: $\begin{bmatrix}
2 &-1 & 3\\
3 & 1 & 0 \\
2 & -1 & 3
\end{bmatrix}$
A straightforward calculation yields
$\det (A-\lambda I)=\begin{vmatrix}
-2-\lambda & 3\\
3 & 1-\lambda & 0 \\
2 & -1 & 3-\lambda
\end{vmatrix}=\lambda(\lambda-4)(\lambda-2)$
so that A has eigenvalues $\lambda_1=0\\
\lambda_2=4\\
\lambda_3=2$
Eigenvalue $λ_1 =0$: In this case, the system $(A − \lambda I)v = 0$ is $5v_2-9v_3=0\\
2v_1-v_2+3v_3=0$
With $v_3=5\\
\rightarrow v_1=-3\\
v_2=9$
The solution is $v = r(-3,9,5)$. Therefore,
$v_1= \begin{bmatrix}
-3\\
9\\
5
\end{bmatrix}$
Eigenvalue $\lambda_2 =4$: In this case, the system $(A − \lambda I)v = 0$ is $v_2-v_3=0\\
v_1-v_3=0$
With $v_3=1\\
\rightarrow v_1=1\\
v_2=1$
The solution is $v = r(1,1,1)$. Therefore,
$v_2=e^{4t} \begin{bmatrix}
1\\
1\\
1
\end{bmatrix}$
Eigenvalue $\lambda_3 =2$: In this case, the system $(A − \lambda I)v = 0$ is $v_2-3v_3=0\\
v_1-v_3=0$
With $v_3=1\\
\rightarrow v_1=1\\
v_2=3$
The solution is $v = r(1,3,1)$. Therefore,
$v_3=e^{2t} \begin{bmatrix}
1\\
3\\
1
\end{bmatrix}$
Hence, the general solution to the given equation is
$x(t)=c_1v_1+c_2v_2=c_1\begin{bmatrix}
-3\\
9\\
5
\end{bmatrix}+c_2e^{4t}\begin{bmatrix}
1\\
1\\
1
\end{bmatrix}+c_3e^{2t}\begin{bmatrix}
1\\
3\\
1
\end{bmatrix}=\begin{bmatrix}
-3c_1+c_2e^{4t}+c_3e^{2t}\\
9c_1+c_2e^{4t}+3c_3e^{2t}\\
5c_1+c_2e^{4t}+c_3e^{2t}
\end{bmatrix}$
Since \begin{bmatrix}
-4\\
4\\
4
\end{bmatrix}
we obtain: $c_3=-2\\
2c_2+3c_3=-4\\
-3c_1+c_2+c_3=-4$
then $c_1=1\\
c_2=1\\
c_3=-2$
Hence,
$x(t)=\begin{bmatrix}
-3+e^{4t}-2e^{2t}\\
9+e^{4t}-6e^{2t}\\
5+e^{4t}-2e^{2t}
\end{bmatrix}$
