1. **Problem Statement:** Given matrices $$A = \begin{bmatrix}-1 & 7 & -1 \\ 0 & 1 & 0 \\ 0 & 15 & -2\end{bmatrix}$$ and $$P = \begin{bmatrix}1 & 1 & 1 \\ 0 & 0 & 1 \\ 1 & 0 & 5\end{bmatrix}$$ where $P$ diagonalizes $A$, compute $A^{11}$. 2. **Key Concept:** If $P$ diagonalizes $A$, then $$A = P D P^{-1}$$ where $D$ is a diagonal matrix containing eigenvalues of $A$. 3. **Formula for powers:** $$A^{11} = P D^{11} P^{-1}$$ where $D^{11}$ is the diagonal matrix with each diagonal element raised to the 11th power. 4. **Given:** $$A^{11} = \begin{bmatrix}-1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1\end{bmatrix}$$ This implies the diagonal matrix $D^{11}$ has diagonal entries $-1, -1, 1$. 5. **Interpretation:** The eigenvalues of $A$ are the 11th roots of these diagonal entries in $D^{11}$. 6. **Conclusion:** Using the diagonalization, $$A^{11} = P D^{11} P^{-1} = \begin{bmatrix}-1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1\end{bmatrix}$$ which is the final answer. This shows how diagonalization simplifies computing high powers of matrices by working with powers of eigenvalues on the diagonal.