If $A={\left[a_{ij}\right]}_{4\times 4}$ such that $a_{ij}=\{\begin{array}{l}2wheni=j\\ 0wheni\ne j\end{array}$ then  $\left\{\frac{\det \left(adj\left(adjA\right)\right)}{7}\right\}$ is [When$\{.\}$  represents fractional part function]

Let  $A=\left[\begin{array}{cccc}2& 0& 0& 0\\ 0& 2& 0& 0\\ 0& 0& 2& 0\\ 0& 0& 0& 2\end{array}\right]$
 $\left|A\right|=2^4=16$
 $\det \left(adj\left(adjA\right)\right)=\left|adj\left(adjA\right)\right|$
 $\Rightarrow {\left|A\right|}^{3^2}={\left|A\right|}^9$
 $\Rightarrow {\left(2^4\right)}^9={\left(2^3\right)}^{12}\Rightarrow {(1+7)}^{12}$
 $\Rightarrow 1{+}^{12}{C}_1\left(7\right){+}^{12}{C}_2{\left(7\right)}^2+\mathrm{.....}$
 $\therefore \frac{\det \left(adj\left(adjA\right)\right)}{7}=\frac{1}{7}+\text{positive\hspace{0.17em}integer}$
 $\frac{\det \left(adj\left(adjA\right)\right)}{7}=\frac{1}{7}$