Answer
See below
Work Step by Step
The given system can be written as $x′ = Ax$,
where $A=\begin{bmatrix}
-12 & -7\\
16 & 10
\end{bmatrix}$
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
$p(λ) = \begin{bmatrix}
-12-\lambda & -7\\
16 & 10-\lambda
\end{bmatrix}=(-12-\lambda)(10-\lambda)+112=(\lambda-2)(\lambda+4)$
Hence, A is nondefective by Corollary 7.2.10. The eigenvectors are easily computed:
$\lambda_1=2: v=r(1,-2)\\
\lambda_2=-4: v=s(7,-8)$
Set $S = \begin{bmatrix}
1 & 7\\
-2 & -8
\end{bmatrix}$
then from Theorem 7.3.4, $S−1AS = diag(2, -4)$
so that the system becomes:
$\begin{bmatrix}
y_1'\\
y_2'
\end{bmatrix}=\begin{bmatrix}
2 & 0\\
0 & -4
\end{bmatrix}\begin{bmatrix}
y_1\\
y_2
\end{bmatrix}$
Hence, $y_1'=2y_1\\
y_2'=-4y_2$
Both of these equations can be integrated to obtain $y_1(t)=C_1e^{2t}\\
y_2(t)=C_2e^{-4t}$
Return to the original variables, we have:
$x=Sy=\begin{bmatrix}
1 & 7\\
-2 & -8
\end{bmatrix}\begin{bmatrix}
C_1e^{2t}\\
C_2e^{-4t}
\end{bmatrix} \\
\rightarrow x_1(t)=C_1e^{2t}+7C_2e^{-4t}\\
x_2(t)=-2C_1e^{2t}-8C_2e^{-4t}$
