Question

If $A=\left(\begin{array}{cc}a& 3\\ 3& a\end{array}\right)$ and $det{A}^3=4096$ then the value of $a$ equals

• A. $\pm 3$
• B. $\pm 16$
• C. $\pm 5$
• D. $\pm 9$

Answer 1

Answer: (C)

$\pm 5$

concept $det\left(A^n\right)={\left(detA\right)}^n...\left(1\right)$
Now $A=\left\{\begin{array}{cc}a& 3\\ 3& a\end{array}\right\}$
$\therefore \left|A\right|=a^2-9$
given $det{A}^3=4096$
$\Rightarrow {\left(detA\right)}^3=4096$
${(a^2-9)}^3={\left(16\right)}^3\Rightarrow a^2-9=16$
$\therefore a=\pm 5$