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