Question

If $A_{3\times 3}$ and det $A=6$, then det $\left(2adjA\right)$ :

• A. $8$
• B. $48$
• C. $288$
• D. $1$

Answer 1

The correct option is C $288$

$det\left(KA\right)=K^{n}det\left(A\right)$
$det\left(adjA\right)={\left|A\right|}^{n-1}$
using the above 2
thus $\left|2adjA\right|=2^3\left|adjA\right|=2^3{\left|A\right|}^2$
$=2^3\times 6^2$
$=288$