Answer
See below
Work Step by Step
Given: $\begin{bmatrix}
1 & -2\\
5 & -5
\end{bmatrix}$
A straightforward calculation yields
$\det (A-\lambda I)=\begin{vmatrix}
1-\lambda & -2\\
5 & -5-\lambda
\end{vmatrix}=(\lambda+2-i)(\lambda-2-i)$
so that A has eigenvalues $\lambda_1=-2+i\\
\lambda_2=-2-i$
Eigenvalue $λ_1 =-2+i$: In this case, the system $(A − \lambda I)v = 0$ is $(3-i)v_1-2v_2=0$
Assume that $v_1=2 \rightarrow v_2=3-i$
The solution is $v = r(2, 3-i)$. Therefore,
$x_1(t)=e^{(-2+i)t} \begin{bmatrix}
2\\
3-i
\end{bmatrix}=e^{-2t} \begin{bmatrix}
3\cos t\\
3\cos t+\sin t
\end{bmatrix}$
Eigenvalue $λ_1 =-2-i$: In this case, the system $(A − \lambda I)v = 0$ is $(3+i)v_1-2v_2=0$
Assume that $v_1=2 \rightarrow v_2=3+i$
The solution is $v = r(2, 3+i)$. Therefore,
$x_2(t)=e^{(-2-i)t} \begin{bmatrix}
2\\
3+i
\end{bmatrix}= \begin{bmatrix}
3\sin t\\
-\cos t+3\sin t
\end{bmatrix}$
Hence, the general solution to the given equation is
$x(t)=c_1x_1+c_2x_2=c_1e^{-2t}\begin{bmatrix}
3\cos t\\
3\cos t+\sin t
\end{bmatrix}+c_2e^{-2t}\begin{bmatrix}
3\sin t\\
-\cos t+3\sin t
\end{bmatrix}=\begin{bmatrix}
e^{-2t}(3c_1\cos t+3c_2\sin t)\\
e^{-2t}((3\cos t+\sin t)c_1+(-\cos t+3\sin t)c_2)
\end{bmatrix}$
