Question

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

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

Answer 1

The correct option is C $\pm \sqrt{7}$

$A=\left[\begin{array}{cc}\alpha & 2\\ 2& \alpha \end{array}\right]$
$\Rightarrow |A|={\alpha }^2-4$
$\Rightarrow |A{|}^3=({\alpha }^2-4{)}^3$
$\text{Also,}|A^3|=27\Rightarrow |A^3|=27$
$\therefore ({\alpha }^2-4{)}^3=27$
$\Rightarrow ({\alpha }^2-4{)}^3-(3{)}^3=0$
$\Rightarrow ({\alpha }^2-4-3)\left[\right({\alpha }^2-4{)}^2+3^2+3({\alpha }^2-4)]=0$
$\Rightarrow ({\alpha }^2-7)({\alpha }^4-5{\alpha }^2+13)=0$
$\Rightarrow \alpha =\pm \sqrt{7}$