Answer
See below
Work Step by Step
Given: $\begin{bmatrix}
0 & -3 & 1\\
-2 & -1 & 1\\
0 & 0 & -2
\end{bmatrix}$
A straightforward calculation yields
$\det (A-\lambda I)=\begin{vmatrix}
-\lambda & -3 & 1\\
-2 & -1-\lambda & 1\\
0 & 0 & -2-\lambda
\end{vmatrix}=(2-\lambda)(\lambda+3)(\lambda-2)$
so that A has eigenvalues $\lambda_1=2\\
\lambda_2=2\\
\lambda_3=-3$
Eigenvalue $λ_1 =2$: In this case, the system $(A − \lambda I)v = 0$ is $v_3=2v_1+3v_2\\
v_1=0\\
v_2=1\\
\rightarrow v_3=3$
The solution is $v = r(0,1,3)$. Therefore,
$v_1=e^{2t} \begin{bmatrix}
0\\
1 \\
3
\end{bmatrix}$
Eigenvalue $\lambda_2 =-2$: In this case, the system $(A − \lambda I)v = 0$ is $v_3=2v_1+3v_2\\
v_1=1\\
v_2=0\\
\rightarrow v_3=2$
The solution is $v = r(1, 0,2)$. Therefore,
$v_2=e^{-2t} \begin{bmatrix}
1\\
0\\
2
\end{bmatrix}$
Eigenvalue $\lambda_3 =-3$: In this case, the system $(A − \lambda I)v = 0$ is $v_1=v_2=1\\
v_3=0$
The solution is $v = r(1,1,0)$. Therefore,
$v_3=e^{-3t} \begin{bmatrix}
1\\
1\\
0
\end{bmatrix}$
Hence, the general solution to the given equation is
$x(t)=c_1v_1+c_2v_2+c_3v_3=c_1e^{2t}\begin{bmatrix}
0\\
1\\
3
\end{bmatrix}+c_2e^{2t}\begin{bmatrix}
1\\
0\\
2
\end{bmatrix}+c_3e^{-3t}\begin{bmatrix}
1\\
1\\
0
\end{bmatrix}=\begin{bmatrix}
c_2e^{2t}+c_3e^{-3t}\\
c_1e^{2t}+c_3e^{-3t}\\
3c_1e^{2t}+2c_2e^{2t}
\end{bmatrix}$
