Question

If $A=\left[\begin{array}{cc}\alpha & 2\\ 2& \alpha \end{array}\right]$ and $|A^3|=27$, then $\alpha$ is equal to

• A. $\pm 1$
• B. $\pm 2$
• C. $\pm \sqrt{7}$
• D. $\pm \sqrt{5}$

Answer 1

Answer: (C)

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