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If $A=\left[\begin{array}{cc}\alpha & 2 \\ 2 & \alpha\end{array}\right]$ and $\left|A^{3}\right|=125$, then $\alpha$ is equal to
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Verified Answer
The correct answer is:
$\pm 3$
Given, $\quad A=\left[\begin{array}{ll}\alpha & 2 \\ 2 & \alpha\end{array}\right]$
Now, $\quad|A|=\left|\begin{array}{cc}\alpha & 2 \\ 2 & \alpha\end{array}\right|=\alpha^{2}-4...(i)$
Also, given,
$\begin{array}{ll}
& \left|A^{3}\right|=125 \\
\Rightarrow \quad & |A|^{3}=125=(5)^{3} \quad\left(\because\left|A^{n}\right|=|A|^{n}\right) \\
\therefore & |A|=5
\end{array}$
On putting this value in Eq. (i), we get
$\begin{aligned}
5 &=\alpha^{2}-4 \\
\Rightarrow \quad \alpha^{2} &=9=(3)^{2} \\
\Rightarrow \quad \alpha &=\pm 3
\end{aligned}$
Now, $\quad|A|=\left|\begin{array}{cc}\alpha & 2 \\ 2 & \alpha\end{array}\right|=\alpha^{2}-4...(i)$
Also, given,
$\begin{array}{ll}
& \left|A^{3}\right|=125 \\
\Rightarrow \quad & |A|^{3}=125=(5)^{3} \quad\left(\because\left|A^{n}\right|=|A|^{n}\right) \\
\therefore & |A|=5
\end{array}$
On putting this value in Eq. (i), we get
$\begin{aligned}
5 &=\alpha^{2}-4 \\
\Rightarrow \quad \alpha^{2} &=9=(3)^{2} \\
\Rightarrow \quad \alpha &=\pm 3
\end{aligned}$
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