Answer
See below
Work Step by Step
Given: $\begin{bmatrix}
0 & -4\\
4 & 0
\end{bmatrix}$
A straightforward calculation yields
$\det (A-\lambda I)=\begin{vmatrix}
-\lambda & -4\\
4 & -\lambda
\end{vmatrix}=\lambda^2+16$
so that A has eigenvalues $\lambda_1=4i\\
\lambda_2=-4i$
Eigenvalue $λ_1 =4i$: In this case, the system $(A − \lambda I)v = 0$ is $iv_1-v_2=0$
Assume that $v_2=1 \rightarrow v_1=-i$
The solution is $v = r(-i, 1)$. Therefore,
$x_1(t)=e^{4it} \begin{bmatrix}
-i\\
1
\end{bmatrix}= \begin{bmatrix}
\cos 4t\\
-\sin 4t
\end{bmatrix}$
Eigenvalue $λ_1 =-4i$: In this case, the system $(A − \lambda I)v = 0$ is $iv_1-v_2=0$
Assume that $v_1=1 \rightarrow v_2=i$
The solution is $v = r(1, i)$. Therefore,
$x_2(t)=e^{-4it} \begin{bmatrix}
1\\
i
\end{bmatrix}= \begin{bmatrix}
\sin 4t\\
\cos 4t
\end{bmatrix}$
Hence, the general solution to the given equation is
$x(t)=c_1x_1+c_2x_2=c_1\begin{bmatrix}
\cos 4t\\
-\sin 4t
\end{bmatrix}+c_2\begin{bmatrix}
\sin 4t\\
\cos 4t
\end{bmatrix}=\begin{bmatrix}
c_1\cos 4t+c_2\sin 4t\\
- c_1\sin 4t+c_2\cos 4t
\end{bmatrix}$
