Let A be a 3 × 3 invertible matrix. If adj 24 A = adj 3 adj 2 A | , then A 2 is equal to

We know that $\left|adjA\right|={\left|A\right|}^{n-1}$ 

$\left|KA\right|=K^n\left|A\right|$

where $n$ is order of matrix

Given 

$\left|\mathrm{adj}\left(24A\right)\right|=\left|\mathrm{adj}3\left(\mathrm{adj}2A\right)\right|$

$\Rightarrow {\left|24A\right|}^2={\left|3adj\left(2A\right)\right|}^2$

$\Rightarrow \left|24A\right|=\left|3adj\left(2A\right)\right|$

$\Rightarrow {24}^3\left|A\right|=3^3\left|adj2A\right|$ $\Rightarrow {24}^3\left|A\right|=3^3\left|2A\right|\left|2A\right|$

$\Rightarrow {24}^3\left|A\right|=3^3{2}^3{2}^3\left|A\right|\left|A\right|$ $$

$\Rightarrow \left|A\right|=\frac{{24}^3}{2^3{2}^3{3}^3}$ $=\frac{{\left(2^3·3\right)}^3}{2^6{3}^3}$ $=8$

$\Rightarrow {\left|A\right|}^2=64$