Let \(A = \begin{bmatrix} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 0 & 0 & 1\end{bmatrix}\).
What is \(\det(A^{6}\))?
The answer is \(64\).
Since \(A\) is upper triangular, \(\det(A) = 1 \cdot 2 \cdot 1 = 2\). Then, \(\det(A^6) = \det(A)^6 = 2^6 = 64\).