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

Question

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

Aicharya · Accepted Answer

Understand that $\mathrm{A}$ is a diagonal matrix $	herefore A^{n}=\left[\begin{array}{ccc} a_{11}^{n} & 0 & 0 \ 0 & a_{22}^{n} & 0 \ 0 & 0 & a_{33}^{n} \end{array}ight] \Rightarrow A^{4}=\left[\begin{array}{ccc} 16 & 0 & 0 \ 0 & 16 & 0 \ 0 & 0 & 1 \end{array}ight]$ $\begin{aligned} 	ext { Here }|A| &=\left[\begin{array}{ccc}2 & 0 & 0 \ 0 & -2 & 0 \ 0 & 0 & -1\end{array} \mid=2(2)=4ight.\ \operatorname{adj} A &=\left[\begin{array}{ccc}2 & 0 & 0 \ 0 & -2 & 0 \ 0 & 0 & -4\end{array}ight] \Rightarrow A^{-1}=\frac{1}{4}\left[\begin{array}{ccc}2 & 0 & 0 \ 0 & -2 & 0 \ 0 & 0 & -4\end{array}ight]=\left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \ 0 & \frac{-1}{2} & 0 \ 0 & 0 & -1\end{array}ight] \ A^{4} A^{-1} &=\left[\begin{array}{ccc}16 & 0 & 0 \ 0 & 16 & 0 \ 0 & 0 & 1\end{array}ight]\left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \ 0 & \frac{-1}{2} & 0 \ 0 & 0 & -1\end{array}ight] \ &=\left[\begin{array}{ccc}8 & 0 & 0 \ 0 & -8 & 0 \ 0 & 0 & -1\end{array}ight] \end{aligned}$

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

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