Answer
$A^4=\begin{bmatrix}
226 & -525 \\
90 & -209\\ \end{bmatrix}$
Work Step by Step
\[ P = \begin{bmatrix}
5 & 7 \\
2 & 3\\ \end{bmatrix}, D=\begin{bmatrix}
2 & 0\\
0 & 1\\ \end{bmatrix}\]
First, remember that since the given matrix $D$ is diagonal, then $A^k=PD^kP^{-1}$ where $D^k$ is rather simple to solve and $P$ is invertible. To use the above equation we need to find $P^{-1}$:
\[ P = \begin{bmatrix}
a & b \\
c & d\\ \end{bmatrix}, P^{-1}=\frac{1}{ad-bc}\begin{bmatrix}
d & -b\\
-c & a\\ \end{bmatrix}\]
Using the formula for the inverse of a 2x2 matrix (shown above), calculate $P^{-1}$:
\[P^{-1}=\frac{1}{(5)(3)-(7)(2)}\begin{bmatrix}
3 & -7\\
-2 & 5\\ \end{bmatrix}=\begin{bmatrix}
3 & -7\\
-2 & 5\\\end{bmatrix}\]
Then, to the equation $A^k=PD^kP^{-1}$ to solve for $A^4$:
$A^4=PD^4P^{-1}$
$A^4=\begin{bmatrix}
5 & 7 \\
2 & 3\\ \end{bmatrix} \begin{bmatrix}
2 & 0\\
0 & 1\\ \end{bmatrix}^4 \begin{bmatrix}
3 & -7\\
-2 & 5\\\end{bmatrix}$
Remember that for any matrix $M$ with zero entries in all positions except the main diagonal, $M^k$ is equal to $M$ with each value on the main diagonal raised to the power of $k$. This strategy is used to go from the line of work above to the line of work below.
$A^4=\begin{bmatrix}
5 & 7 \\
2 & 3\\ \end{bmatrix} \begin{bmatrix}
2^4 & 0\\
0 & 1^4\\ \end{bmatrix} \begin{bmatrix}
3 & -7\\
-2 & 5\\\end{bmatrix}$
$A^4=\begin{bmatrix}
5 & 7 \\
2 & 3\\ \end{bmatrix} \begin{bmatrix}
16 & 0\\
0 & 1\\ \end{bmatrix} \begin{bmatrix}
3 & -7\\
-2 & 5\\\end{bmatrix}$
Use matrix multiplication to simplify:
$A^4=\begin{bmatrix}
(5)(16)+(7)(0) & (5)(0)+(7)(1) \\
(2)(16)+(3)(0) & (2)(0)+(3)(1)\\ \end{bmatrix} \begin{bmatrix}
3 & -7\\
-2 & 5\\\end{bmatrix}$
$A^4=\begin{bmatrix}
80 & 7 \\
32 & 3\\ \end{bmatrix} \begin{bmatrix}
3 & -7\\
-2 & 5\\\end{bmatrix}$
$A^4=\begin{bmatrix}
(80)(3)+(7)(-2) & (80)(-7)+(7)(5) \\
(32)(3)+(3)(-2) & (32)(-7)+(3)(5)\\ \end{bmatrix}$
$A^4=\begin{bmatrix}
226 & -525 \\
90 & -209\\ \end{bmatrix}$
