Answer
See below
Work Step by Step
Given: $\begin{bmatrix}
0 & 1 & 0\\
-1 & 0 & 0\\
0 & 0 & 5
\end{bmatrix}$
A straightforward calculation yields
$\det (A-\lambda I)=\begin{vmatrix}
-\lambda & 1 & 0\\
-1 & -\lambda & 0\\
0 & 0 & 5-\lambda
\end{vmatrix}=(\lambda^2+1)(5-\lambda)$
so that A has eigenvalues $\lambda_1=5\\
\lambda_2=i\\
\lambda_3=-i$
Eigenvalue $λ_1 =5$: In this case, the system $(A − \lambda I)v = 0$ is $v_1=v_2=0\\
v_3=1$
The solution is $v = r(0, 0, 1)$. Therefore,
$v_1=e^{5t} \begin{bmatrix}
0\\
0 \\
1
\end{bmatrix}$
Eigenvalue $\lambda_2 =i$: In this case, the system $(A − \lambda I)v = 0$ is $v_1=ib\\
v_2=1\\
v_3=0\\
\rightarrow v_1=i$
The solution is $v = r(i, 1, 0)$. Therefore,
$v_2=e^{-t} \begin{bmatrix}
i\\
1\\
0
\end{bmatrix}$
Obtain $w(t)=e^{it}\begin{bmatrix}
i \\
1\\
0
\end{bmatrix}=(\cos t +i \sin t)\begin{bmatrix}
i \\
1\\
0
\end{bmatrix}=\begin{bmatrix}
-\sin t\\
\cos t\\
0
\end{bmatrix}+i\begin{bmatrix}
\cos t \\
\sin t\\
0
\end{bmatrix}$
Hence, the general solution to the given equation is
$x(t)=c_1v_1+c_2v_2+c_3v_3=c_1e^{5t}\begin{bmatrix}
0\\
0\\
1
\end{bmatrix}+c_2\begin{bmatrix}
-\sin t\\
\cos t\\
0
\end{bmatrix}+c_3\begin{bmatrix}
\cos t\\
\sin t\\
1
\end{bmatrix}=\begin{bmatrix}
-c_2\sin t+c_3\cos t\\
c_2\cos t+c_3\sin t\\
c_1e^{5t}
\end{bmatrix}$
