If \(A=\left[\begin{array}{ccc}2 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -1\end{array}\right]\), then \(A^{4} A^{-1}=\)

Understand that \(\mathrm{A}\) is a diagonal matrix
\(\therefore A^n=\left[\begin{array}{ccc}a_{11}^n & 0 & 0 \\ 0 & a_{22}^n & 0 \\ 0 & 0 & a_{33}^n\end{array}\right] \Rightarrow A^4=\left[\begin{array}{ccc}16 & 0 & 0 \\ 0 & 16 & 0 \\ 0 & 0 & 1\end{array}\right]\)
Here \(|A|=\left[\left.\begin{array}{ccc}2 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -1\end{array} \quad \right\rvert\,=2(2)=4\right.\)
\(\operatorname{adj} A=\left[\begin{array}{ccc}2 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -4\end{array}\right] \Rightarrow A^{-1}\) \(=\frac{1}{4}\left[\begin{array}{ccc}2 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -4\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \\ 0 & \frac{-1}{2} & 0 \\ 0 & 0 & -1\end{array}\right]\)
\(A^4 A^{-1}=\left[\begin{array}{ccc}16 & 0 & 0 \\ 0 & 16 & 0 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \\ 0 & \frac{-1}{2} & 0 \\ 0 & 0 & -1\end{array}\right]\)
\(=\left[\begin{array}{ccc}8 & 0 & 0 \\ 0 & -8 & 0 \\ 0 & 0 & -1\end{array}\right]\)