Answer
$x_1(t)=-C_1e^{-t}+C_3e^{2t}\\
x_2(t)=C_1e^{-t}+C_2e^{2t} \\
x_3(t)=-C_1e^{-t}+(C_2-C_3)e^{2t}$
Work Step by Step
The given system can be written as $x′ = Ax$,
where $A=\begin{bmatrix}
1 & 1 & -1\\
1 & 1 & 1\\
-1 & 1 & 1
\end{bmatrix}$
The transformed system is $y′ = (S−1AS)y$ where $x = Sy$
To determine S, we need the eigenvalues and eigenvectors of A. The characteristic polynomial of A is
$p(\lambda)=\begin{bmatrix}
1-\lambda & 1 & -1\\
1 & 1-\lambda & 1\\
-1 & 1 & 1-\lambda
\end{bmatrix}=-\lambda^3+3\lambda^2-4$
Hence, A is nondefective by Corollary 7.2.10. The eigenvectors are easily computed:
\lambda_1=-1: v=r(-1,1,-1)$
\lambda_2=2: v=s(0,1,1)\\
\lambda_3=2: v=t(1,0,-1) \\
$
Set $S = \begin{bmatrix}
-1 & 0 & 1\\
1 & 1 & 0\\
-1 & 1 & -1
\end{bmatrix}$
then from Theorem 7.3.4, $S−1AS = diag(-1,2,2)$
so that the system becomes:
$\begin{bmatrix}
y_1'\\
y_2' \\
y_3'
\end{bmatrix}=\begin{bmatrix}
-1 & 0 & 0\\
0 & 2 & 0\\
0 & 0 & 2
\end{bmatrix}\begin{bmatrix}
y_1\\
y_2 \\
y_3
\end{bmatrix}$
Hence, $y_1'=-y_1\\
y_2'=2y_2 \\
y_3'=2y_3$
Both of these equations can be integrated to obtain $y_1(t)=C_1e^{-t}\\
y_2(t)=C_2e^{2t}\\
y_3(t)=C_3e^{2t}$
Return to the original variables, we have:
$x=Sy=\begin{bmatrix}
-1 & & 1\\
1 & 1 & 0\\
-1 & 1 & 1
\end{bmatrix}\begin{bmatrix}
C_1e^{-t}\\
C_2e^{2t}\\
C_3e^{2t}
\end{bmatrix} \\
\rightarrow x_1(t)=-C_1e^{-t}+C_3e^{2t}\\
x_2(t)=C_1e^{-t}+C_2e^{2t} \\
x_3(t)=-C_1e^{-t}+(C_2-C_3)e^{2t}$
