If $A=\frac{1}{5!6!7!}\left[\begin{array}{lll}5!& 6!& 7!\\ 6!& 7!& 8!\\ 7!& 8!& 9!\end{array}\right]$, then $\left|adj\left(adj\left(2A\right)\right)\right|$  is equal to

Given,

$A=\frac{1}{5!6!7!}\left[\begin{array}{lll}5!& 6!& 7!\\ 6!& 7!& 8!\\ 7!& 8!& 9!\end{array}\right]$

$\Rightarrow \left|A\right|=\frac{1}{5!6!7!}\left|\begin{array}{lll}5!& 6!& 7!\\ 6!& 7!& 8!\\ 7!& 8!& 9!\end{array}\right|$

$\Rightarrow \left|A\right|=\frac{1}{5!6!7!}5!\times 6!\times 7!\left|\begin{array}{lll}1& 6& 42\\ 1& 7& 56\\ 1& 8& 72\end{array}\right|$

$\Rightarrow \left|A\right|=2$

Now using the property of adjoint of matrix we get,

$\left|adj\left(adj\left(A\right)\right)\right|={\left|A\right|}^{(n-1{)}^2} \& \left|kA\right|=k^n\left|A\right|$, where $n$ is order of square matrix,

Now using the above two formula we get,

$\left|adj\left(adj\left(2A\right)\right)\right|={\left|2A\right|}^4=2^{12}{\left|A\right|}^4$

$\Rightarrow \left|adj\left(adj\left(2A\right)\right)\right|=2^{12}·2^4$

$\Rightarrow \left|adj\left(adj\left(2A\right)\right)\right|=2^{16}$