Question

If $A=\begin{bmatrix} {\alpha}&{2} \\\ {2}&{\alpha} \\\ \end{bmatrix}$ $| A^3|= 27$then $\alpha$ is equal to

• A. $\pm 1$
• B. $\pm2$
• C. $\pm \sqrt {7}$
• D. $\pm \sqrt {5}$

Answer 1

Answer: (C)

$\pm \sqrt {7}$

Answer 2

We have,
$A=\begin{bmatrix}\alpha & 2 \\\2 & \alpha\end{bmatrix}$
$\Rightarrow |A| =\begin{vmatrix}\alpha & 2 \\\ 2 & \alpha \end{vmatrix}=\alpha^{2}-4$
Now, $\left|A^{3}\right|=27$
$\Rightarrow |A|^{3}=27$
$|A^{n}|=|A|^{n}]$
$\Rightarrow \left(\alpha^{2}-4\right)^{3}=27$
$[\because |A|=\alpha^{2}-4 ]$
$\Rightarrow \alpha^{2}-4=3$
$\Rightarrow \alpha^{2}=7$
$\Rightarrow \alpha=\pm \sqrt{7}$