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