Let $A$ and $B$ be two $3\times 3$ matrices such that $AB=I$ and $\left|A\right|=\frac{1}{8}$ then $\left|adj\left(Badj\left(2A\right)\right)\right|$ is equal to

Given, $AB=I$ and $\left|A\right|=\frac{1}{8}$

Now by using property $\mathrm{adj}\left(PQ\right)=\left(adjQ\right)\left(adjP\right)$ we get, $\left|adj\left(B·adj2A\right)\right|=\left|adj\left(adj2A\right)\right|\left|adjB\right| ...\left(\mathrm{i}\right)$

Now $\left|\mathrm{adj}\left(\mathrm{adj}2A\right)\right|={\left|\mathrm{adj}2A\right|}^2$ $={\left|2A\right|}^4$

$={\left|2^3\right|}^4{\left|A\right|}^4={\left(2^3\right)}^4{\left|A\right|}^4=8^4{\left|A\right|}^4 ...\left(\mathrm{ii}\right)$

{Property used $\left|adjA\right|={\left|A\right|}^{n-1}$ and $\left|kA\right|=k^n\left|A\right|$ where $n$ is order of matrix}

Now $\left|adjB\right|={\left|B\right|}^2 .............\left(iii\right)$

Now putting the value of equation $\left(\mathrm{ii}\right) \& \left(\mathrm{iii}\right)$ in eq $\left(\mathrm{i}\right)$

we get $\left|adj\left(B·adj2A\right)\right|=8^4{\left|A\right|}^4{\left|B\right|}^2$ 

Now using$\left|A\right|=\frac{1}{\left|B\right|}$as $AB=I$, we get

$\left|adj\left(B·adj2A\right)\right|=8^4{\left|A\right|}^4\times \frac{1}{{\left|A\right|}^2}$  $=8^4\times {\left|A\right|}^2=8^4\times \frac{1}{8^2}=64$