# If $A=\begin{bmatrix}\alpha&2\\2&\alpha\end{bmatrix}$ and $\mid A^3\mid=125$ then the value of $\alpha$ is

$(a)\;\pm 1\qquad(b)\;\pm 2\qquad(c)\;\pm 3\qquad(d)\;\pm 5$

$A=\begin{vmatrix}\alpha&2\\2 &\alpha\end{vmatrix}$
$\mid A^3\mid=125$
$\Rightarrow \mid A\mid=\alpha^2-4$
$\Rightarrow \mid A\mid^3=125$
$(\alpha^2-4)^3=125$
$(\alpha^2-4)^3=5^3$
$\alpha^2-4=5$
$\alpha^2=9$
$\alpha=\pm 3$
Hence (c) is the correct answer.