Answer
See below
Work Step by Step
Given: $\begin{bmatrix}
-2 & -7\\
-1 & 4
\end{bmatrix}$
A straightforward calculation yields
$\det (A-\lambda I)=\begin{vmatrix}
-2-\lambda & -7\\
-1 & 4-\lambda
\end{vmatrix}=\lambda^2-2\lambda -15=(\lambda+3)(\lambda -5)$
so that A has eigenvalues $\lambda_1=-3\\
\lambda_2=5$
Eigenvalue $λ_1 =-3$: In this case, the system $(A − \lambda I)v = 0$ is
$v_1-7v_2=0\\
-v_1+7v_2=0$
with solution $v = r(7, 1)$. Therefore,
$x_1(t)=e^{-3t} \begin{bmatrix}
7\\
1
\end{bmatrix}$
Eigenvalue $λ_1 =5$: In this case, the system $(A − \lambda I)v = 0$ is
$-7v_1-7v_2=0\\
-v_1-v_2=0$
with solution $v = r(-1, 1)$. Therefore,
$x_2(t)=e^{5t} \begin{bmatrix}
-1\\
1
\end{bmatrix}$
Hence, the general solution to the given equation is
$x(t)=c_1x_1+c_2x_2=c_1e^{-3t}\begin{bmatrix}
7\\
1
\end{bmatrix}+c_2e^{5t}\begin{bmatrix}
-1\\
1
\end{bmatrix}=\begin{bmatrix}
7c_1e^{-3t}-c_2e^{5t}\\
c_1e^{-3t}+c_2e^{5t}
\end{bmatrix}$
