Answer
See below
Work Step by Step
Given: $\begin{bmatrix}
-1 & 0 & 0\\
1 & 5 & -1\\
1 & 6 & -2
\end{bmatrix}$
A straightforward calculation yields
$\det (A-\lambda I)=\begin{vmatrix}
-1-\lambda & 0 & 0\\
1 & 5-\lambda & -1\\
1 & 6 & -2-\lambda
\end{vmatrix}=(\lambda+1)^2(\lambda-4)$
so that A has eigenvalues $\lambda_1=-1\\
\lambda_2=-1\\
\lambda_3=4$
Eigenvalue $λ_1 =-1$: In this case, the system $(A − \lambda I)v = 0$ is $v_1=v_3-6v_2\\
v_2=0\\
v_3=1 \\
\rightarrow v_1=1$
The solution is $v = r(1, 0, 1)$. Therefore,
$v_1=e^{-t} \begin{bmatrix}
1\\
0 \\
1
\end{bmatrix}$
Eigenvalue $\lambda_2 =-1$: In this case, the system $(A − \lambda I)v = 0$ is $v_1=v_3-6v_2\\
v_2=1\\
v_3=0\\
\rightarrow v_1=-6$
The solution is $v = r(-6, 0, 1)$. Therefore,
$v_2=e^{-t} \begin{bmatrix}
-6\\
1\\
0
\end{bmatrix}$
Eigenvalue $\lambda_3 =4$: In this case, the system $(A − \lambda I)v = 0$ is $v_1=v_3-v_2\\
v_2=v_3=1\\
\rightarrow v_1=0$
The solution is $v = r(0,1,1)$. Therefore,
$v_3=e^{4t} \begin{bmatrix}
0 \\
1\\
1
\end{bmatrix}$
Hence, the general solution to the given equation is
$x(t)=c_1v_1+c_2v_2+c_3v_3=c_1e^{-t}\begin{bmatrix}
1\\
0\\
1
\end{bmatrix}+c_2e^{-t}\begin{bmatrix}
-6\\
1\\
0
\end{bmatrix}+c_3e^{4t}\begin{bmatrix}
0\\
1\\
1
\end{bmatrix}=\begin{bmatrix}
c_1e^{-t}-6c_2e^{-t}\\
c_2e^{-t}+c_3e^{4t}\\
c_2e^{-t}+c_3e^{4t}
\end{bmatrix}$
