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If $\mathrm{A}=\left[\begin{array}{cc}\alpha & 2 \\ 2 & \alpha\end{array}\right]$ and $\operatorname{det}\left(\mathrm{A}^{3}\right)=125$, then $\alpha$ is equal to
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$\pm 3$
$\quad A=\left[\begin{array}{ll}\alpha & 2 \\ 2 & \alpha\end{array}\right], \operatorname{det}\left(A^{3}\right)=125$
$\left|A^{3}\right|=125 \Rightarrow|A|=5$
$\therefore \alpha 2-4=5 \Rightarrow \alpha^{2}=9 \Rightarrow \alpha=\pm 3$
$\left|A^{3}\right|=125 \Rightarrow|A|=5$
$\therefore \alpha 2-4=5 \Rightarrow \alpha^{2}=9 \Rightarrow \alpha=\pm 3$
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