Let A be a n x n matrix such that $\left|\mathrm{A}\right|=2$. If the determinant of the matrix $\mathrm{Adj}\left(2\mathrm{Adj}\left(2{\mathrm{A}}^{-1}\right)\right)\text{ is }{2}^{84}$, then $n$ is equal to ______

$\left|\mathrm{A}\right|=2,\mathrm{A}\text{ is }\mathrm{n}\times \mathrm{n}\text{ matrix, }$

$\left|\mathrm{Adj}\left(2\mathrm{Adj}\left(2{\mathrm{A}}^{-1}\right)\right)\right|={\left|2\mathrm{Adj}\left(2{\mathrm{A}}^{-1}\right)\right|}^{\mathrm{n}-1}$

$\begin{array}{c}\left|\mathrm{Adj}\left(2\mathrm{Adj}\left(2{\mathrm{A}}^{-1}\right)\right)\right|=2^{\mathrm{n}(\mathrm{n}-1)}{\left|\mathrm{Adj}\left(2{\mathrm{A}}^{-1}\right)\right|}^{\mathrm{n}-1}\end{array}$
$\begin{array}{c}\left|\mathrm{Adj}\left(2\mathrm{Adj}\left(2{\mathrm{A}}^{-1}\right)\right)\right|=2^{\mathrm{n}(\mathrm{n}-1)}{\left(2{\mathrm{A}}^{-1}\right)}^{(\mathrm{n}-1{)}^2}\\ =2^{\mathrm{n}(\mathrm{n}-1)}\cdot 2^{\mathrm{n}(\mathrm{n}-1{)}^2}\cdot \frac{1}{|\mathrm{A}{|}^{(\mathrm{n}-1{)}^2}}\\ =2^{n(\mathrm{n}-1)+\mathrm{n}(\mathrm{n}-1{)}^2-(\mathrm{n}-1{)}^2}\\ =2^{(\mathrm{n}-1)\left(\mathrm{n}+{\mathrm{n}}^2-\mathrm{n}-\mathrm{n}+1\right)=2^{(\mathrm{n}-1)\left({\mathrm{n}}^2-\mathrm{n}+1\right)}}=2^{84}\end{array}$
So $\mathrm{n}=5(\mathrm{n}\in \mathrm{N})$