Let $\left[\begin{array}{ccc}\begin{array}{c}2\\ 1\\ 0\end{array}& \begin{array}{c}1\\ 2\\ -1\end{array}& \begin{array}{c}0\\ -1\\ 2\end{array}\end{array}\right]$. If $\left|adj(adj\left(adj2A)\right)\right|=(16{)}^n$, then $n$ is equal to

Given that $A=\left[\begin{array}{ccc}2& 1& 0\\ 1& 2& -1\\ 0& -1& 2\end{array}\right]$ and $\left|adj\left(adj\left(adj2A\right)\right)\right|={16}^n$.

We know that $\left|adjA\right|={\left|A\right|}^{n-1}$ and $\left|adj\left(adjA\right)\right|={\left|A\right|}^{{\left(n-1\right)}^2}$

Similarly, $\left|adj\left(adj\left(adjA\right)\right)\right|={\left|A\right|}^{{\left(n-1\right)}^3}$ where $n$ is the order of the square matrix.

Let us find $\left|A\right|$.

$\left|A\right|=2\left(2\times 2-\left(-1\right)\left(-1\right)\right)-1\left(1\times 2-0\left(-1\right)\right)+0\left(1\times \left(-1\right)-0\times 2\right)$

$=2\left(3\right)-\left(2\right)+0=4$

$\Rightarrow {\left|2A\right|}^{{\left(n-1\right)}^3}={\left|2A\right|}^{{\left(3-1\right)}^3}$

$={\left|2A\right|}^8={\left(2^8\right)}^3{\left|A\right|}^8$

$\Rightarrow 2^{24}\times 4^8={16}^n$

$\Rightarrow {16}^6\times {16}^4={16}^n$

$\Rightarrow {16}^{10}={16}^n$

$\Rightarrow n=10$

Therefore, the value of $n$ is $10$.