Answer
See below
Work Step by Step
Given: $\begin{bmatrix}
2 & -1 & 3\\
2& -1 & 3\\
2 & -1 & 3
\end{bmatrix}$
A straightforward calculation yields
$\det (A-\lambda I)=\begin{vmatrix}
2-\lambda & -1 & 3\\
2 & -1-\lambda & 3\\
2 & -1 & 3-\lambda
\end{vmatrix}=-\lambda^2(\lambda-4)$
so that A has eigenvalues $\lambda_1=0\\
\lambda_2=0\\
\lambda_3=4$
Eigenvalue $λ_1 =0$: In this case, the system $(A − \lambda I)v = 0$ is $v_1=0\\
v_3=1\\
v_2=2v_1+3v_3\\
\rightarrow v_2=3$
The solution is $v = r(0,3,1)$. Therefore,
$v_1= \begin{bmatrix}
0\\
3 \\
1
\end{bmatrix}$
Eigenvalue $\lambda_2 =0$: In this case, the system $(A − \lambda I)v = 0$ is $v_1=1\\
v_3=0\\
v_2=2v_1+3v_3\\
\rightarrow v_2=2$
The solution is $v = r(1, 2,0)$. Therefore,
$v_2=e^{2t} \begin{bmatrix}
1\\
2\\
0
\end{bmatrix}$
Eigenvalue $\lambda_3 =4$: In this case, the system $(A − \lambda I)v = 0$ is $v_1=1\\
v_2=v_3=1$
The solution is $v = r(1,1,1)$. Therefore,
$v_3=e^{4t} \begin{bmatrix}
1\\
1\\
1
\end{bmatrix}$
Hence, the general solution to the given equation is
$x(t)=c_1v_1+c_2v_2+c_3v_3=c_1\begin{bmatrix}
0\\
3\\
1
\end{bmatrix}+c_2\begin{bmatrix}
1\\
2\\
0
\end{bmatrix}+c_3e^{4t}\begin{bmatrix}
1\\
1\\
1
\end{bmatrix}=\begin{bmatrix}
c_2e^{2t}+c_3e^{4t}\\
3c_1+2c_2+c_3e^{4t}\\
c_1+c_3e^{4t}
\end{bmatrix}$
