Answer
$\begin{bmatrix}
4-3\cdot2^k&12\cdot2^k-12\\
1-2^k&4\cdot2^k-3
\end{bmatrix}$
Work Step by Step
The matrices in the equation as written correspond to $A=PDP^{-1}$.
$A^k=PD^kP^{-1}=\begin{bmatrix}
3&4\\1&1
\end{bmatrix}\begin{bmatrix}
2^k&0\\0&1
\end{bmatrix}\begin{bmatrix}
-1&4\\1&-3
\end{bmatrix}=\begin{bmatrix}
4-3\cdot2^k&12\cdot2^k-12\\
1-2^k&4\cdot2^k-3
\end{bmatrix}$
