Let [ array ll 2 1 0 array array ll 1 2 - 1 array array ll 0 - 1 2...

Given that A = \left[\begin{cases}2 & 1 & 0 \\ 1 & 2 & - 1 \\ 0 & - 1 & 2\end{cases}\right] and \left|adj\left(adj\left(adj2A\right)\right)\right| = {16}^{n}$.</p><p>We know that \left|adjA\right| = {\left|A\right|}^{n - 1} and$\left|adj\left(adjA\right)\right| = {\left|A\right|}^{{\left(n - 1\right)}^{2}}$</p><p>Similarly, \left|adj\left(adj\left(adjA\right)\right)\right| = {\left|A\right|}^{{\left(n - 1\right)}^{3}}$ where $n$ is the order of the square matrix.

Let us find \left|A\right|$.</p><p>\left|A\right| = 2\left(2 \times 2 - \left(- 1\right)\left(- 1\right)\right) - 1\left(1 \times 2 - 0\left(- 1\right)\right) + 0\left(1 \times \left(- 1\right) - 0 \times 2\right)$

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

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

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

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

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

$⇒ {16}^{10} = {16}^{n}$

$⇒ n = 10$

Therefore, the value of $n$ is $10$.