Answer
See below
Work Step by Step
The given system can be written as $x′ = Ax$,
where $A=\begin{bmatrix}
-7 & -4\\
18 & 11
\end{bmatrix}$
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
$p(\lambda)=\begin{bmatrix}
-7-\lambda & -4\\
18 & 11-\lambda
\end{bmatrix}=(\lambda-5)(\lambda+1)$
Hence, A is nondefective by Corollary 7.2.10. The eigenvectors are easily computed:
$\lambda_1=5: v=r(1,-3) \\
\lambda_2=-1: v=s(2,-3)\\$
Set $S = \begin{bmatrix}1 &2\\
-3 & -3
\end{bmatrix}$
then from Theorem 7.3.4, $S−1AS = diag(5,-1)$
so that the system becomes:
$A^3=SD^3S^{-1}=S\begin{bmatrix}
125 & 0\\
0 & -1
\end{bmatrix}S^{-1}=S\begin{bmatrix}
-127 & -84\\
378 & 251
\end{bmatrix}$
$A^5=SD^5S^{-1}=S\begin{bmatrix}
3125 & 0\\
0 & -1
\end{bmatrix}S^{-1}=S\begin{bmatrix}
-3127 & -2084\\
9378 & 6251
\end{bmatrix}$
