Dear student,

if A is a non-singular square matrix of order n,then we know that
$A\left(adjA\right)=\left|A\right|{I}_{n}$
$⇒A\left(adjA\right)={\left[\begin{array}{cccccc}\left|A\right|& 0& 0& & & 0\\ 0& \left|A\right|& 0& & & 0\\ 0& 0& \left|A\right|& & & 0\\ & & & & & \\ & & & & & \\ 0& 0& 0& & & \left|A\right|\end{array}\right]}_{n*n}$

as we know that $\left|AB\right|=\left|A\right|\left|B\right|$
$\left|A\right|\left|adjA\right|={\left|A\right|}^{n}$
$\left|adjA\right|={\left|A\right|}^{n-1}$
now if we replace A by adjA, we get
$\left|adj\left(adjA\right)\right|={\left|adjA\right|}^{n-1}$
$\left|adj\left(adjA\right)\right|=\left({\left|A\right|}^{n-1}{\right)}^{n-1}$
$\left|adj\left(adjA\right)\right|={\left|A\right|}^{\left(n-1{\right)}^{2}}$

enjoy

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