Answer
See below
Work Step by Step
Given: $\begin{bmatrix}
-1 & 2\\
2 & 2
\end{bmatrix}$
A straightforward calculation yields
$\det (A-\lambda I)=\begin{vmatrix}
-1-\lambda & 2\\
2 & 2-\lambda
\end{vmatrix}=\lambda^2-\lambda -6=(\lambda-3)(\lambda +2)$
so that A has eigenvalues $\lambda_1=3\\
\lambda_2=-2$
Eigenvalue $λ_1 =3$: In this case, the system $(A − \lambda I)v = 0$ is
$-4v_1+2v_2=0\\
2v_1-v_2=0$
with solution $v = r(1, 2)$. Therefore,
$x_1(t)=e^{3t} \begin{bmatrix}
1\\
2
\end{bmatrix}$
Eigenvalue $λ_1 =-2$: In this case, the system $(A − \lambda I)v = 0$ is
$v_1+2v_2=0\\
2v_1+4v_2=0$
with solution $v = r(-2, 1)$. Therefore,
$x_2(t)=e^{-2t} \begin{bmatrix}
-2\\
1
\end{bmatrix}$
Hence, the general solution to the given equation is
$x(t)=c_1x_1+c_2x_2=c_1e^{3t}\begin{bmatrix}
1\\
2
\end{bmatrix}+c_2e^{-2t}\begin{bmatrix}
-2\\
1
\end{bmatrix}=\begin{bmatrix}
c_1e^{3t}-2c_2e^{-2t}\\
2c_1e^{3t}+c_2e^{-2t}
\end{bmatrix}$
