Answer
See below
Work Step by Step
Given: $\begin{bmatrix}
3 & 2 & 6\\
-2 & 1 & -2\\
-1 & -2 & -4
\end{bmatrix}$
A straightforward calculation yields
$\det (A-\lambda I)=\begin{vmatrix}
3-\lambda & 2 & 6\\
-2 & 1-\lambda & -2\\
-1 & -2 & -4-\lambda
\end{vmatrix}=(1-\lambda)(\lambda+3)(\lambda-2)$
so that A has eigenvalues $\lambda_1=1\\
\lambda_2=-3\\
\lambda_3=2$
Eigenvalue $λ_1 =1$: In this case, the system $(A − \lambda I)v = 0$ is $v_3=1\\
v_1=-v_3=-1\\
v_2=-2v_3=-2$
The solution is $v = r(-1,-2,1)$. Therefore,
$v_1=e^{t} \begin{bmatrix}
-1\\
-2 \\
1
\end{bmatrix}$
Eigenvalue $\lambda_2 =-2$: In this case, the system $(A − \lambda I)v = 0$ is $v_3=3\\
v_1=\frac{2}{3}v_3=2\\
v_2=-\frac{10}{3}v_3=-10$
The solution is $v = r(2, -10,3)$. Therefore,
$v_2=e^{2t} \begin{bmatrix}
2\\
-10\\
3
\end{bmatrix}$
Eigenvalue $\lambda_3 =-3$: In this case, the system $(A − \lambda I)v = 0$ is $v_1=1\\
v_3=-1\\
v_2=0$
The solution is $v = r(1, 0,-1)$. Therefore,
$v_3=e^{-3t} \begin{bmatrix}
1\\
0\\
-1
\end{bmatrix}$
Hence, the general solution to the given equation is
$x(t)=c_1v_1+c_2v_2+c_3v_3=c_1e^{t}\begin{bmatrix}
-1\\
-2\\
1
\end{bmatrix}+c_2e^{2t}\begin{bmatrix}
2\\
-10\\
3
\end{bmatrix}+c_3e^{-3t}\begin{bmatrix}
1\\
0\\
1
\end{bmatrix}=\begin{bmatrix}
-c_1e^t+2c_2e^{2t}+c_3e^{-3t}\\
-2c_1e^t-10c_2e^{2t}\\
c_1e^{t}+3c_2e^{2t}-c_3e^{-3t}
\end{bmatrix}$
