Answer
See below
Work Step by Step
Given: $\begin{bmatrix}
-1 & -6\\
3 & -5
\end{bmatrix}$
A straightforward calculation yields
$\det (A-\lambda I)=\begin{vmatrix}
-1-\lambda & -6\\
3 & -5-\lambda
\end{vmatrix}=\lambda^2-4\lambda+13$
so that A has eigenvalues $\lambda_1=2+3i\\
\lambda_2=2-3i$
Eigenvalue $λ_1 =2+3i$: In this case, the system $(A − \lambda I)v = 0$ is $v_1=-1+i\\
v_2=1$
The solution is $v = r(-1+i,1)$. Therefore,
$v_1= e^{(2+3i)t}\begin{bmatrix}
-1+i\\
1
\end{bmatrix}=e^{2t}(\cos 3t+i\sin 3t)\begin{bmatrix}
-1+i\\
1
\end{bmatrix}=e^{2t}\begin{bmatrix}
-\cos 3t-\sin 3t+i(\cos 3t-\sin 3t)\\
\cos 3t+i\sin 3t
\end{bmatrix}$
Hence, the general solution to the given equation is
$x(t)=c_1v_1+c_2v_2=c_1e^{2t}\begin{bmatrix}
-\cos 3t-\sin 3t\\
\cos 3t
\end{bmatrix}+c_2e^{2t}\begin{bmatrix}
-\cos 3t-\sin 3t\\
\sin 3t
\end{bmatrix}$
Since \begin{bmatrix}
2\\
2
\end{bmatrix}
we obtain: $-c_1+c_2=2\\
c_1=2$
then $c_1=2\\
c_2=4$
Hence,
$x(t)=2e^{2t}\begin{bmatrix}
-\cos 3t-\sin 3t\\
\cos 3t
\end{bmatrix}+4e^{2t}\begin{bmatrix}
-\cos 3t-\sin 3t\\
\sin 3t
\end{bmatrix}=e^{2t}\begin{bmatrix}
-2\cos 3t-6\sin 3t\\
2\cos 3t+4\sin 3t
\end{bmatrix}$
