Answer
See below
Work Step by Step
Given: $\begin{bmatrix}
-1 & 2\\
-2 & -1
\end{bmatrix}$
A straightforward calculation yields
$\det (A-\lambda I)=\begin{vmatrix}
-1-\lambda & 2\\
-2 & -1-\lambda
\end{vmatrix}=(-1-\lambda+2i)(-1-\lambda-2i)$
so that A has eigenvalues $\lambda_1=-1+2i\\
\lambda_2=-1-2i$
Eigenvalue $λ_1 =-1+2i$: In this case, the system $(A − \lambda I)v = 0$ is $-2iv_1+2v_2=0\\
\rightarrow v_2=iv_1$
Assume that $v_1=1 \rightarrow v_2=i$
The solution is $v = r(1, i)$. Therefore,
$v=e^{(-2+i)t} \begin{bmatrix}
1\\
i
\end{bmatrix}$
Obtain $w(t)=e^{(-1+2i)t} \begin{bmatrix}
1\\
i
\end{bmatrix}=e^{-t}[\cos 2t +i\sin 2t] \begin{bmatrix}
1\\
i
\end{bmatrix}=e^{-t} \begin{bmatrix}
\cos 2t\\
\sin 2t
\end{bmatrix}+ie^{-t} \begin{bmatrix}
\sin 2t\\
\cos 2t
\end{bmatrix}$
Hence, the general solution to the given equation is
$x(t)=c_1x_1+c_2x_2=e^{-t}\begin{bmatrix}
c_1\cos 2t+c_2\sin 2t\\
c_1\sin 2t+c_2\cos 2t
\end{bmatrix}=\begin{bmatrix}
c_1e^{-t}\cos 2t+c_2e^{-t}\sin 2t\\
c_1e^{-t}\sin 2t+c_2e^{-t}\cos 2t
\end{bmatrix}$
