Let A = m n p q ,   d = A ≠ 0 and A - d Adj A = 0 . Then

Given, $A=\left[\begin{array}{ll}m& n\\ p& q\end{array}\right], \left|A-d\left(adjA\right)\right|=0$

Here, $d=\left|A\right|=mq-np$

$\Rightarrow \left|A-d\left(adjA\right)\right|=\left|\left[\begin{array}{cc}m& n\\ p& q\end{array}\right]-d\left[\begin{array}{cc}q& -n\\ -p& m\end{array}\right]\right|=0$

$\Rightarrow \left|\begin{array}{cc}m-qd& n\left(1+d\right)\\ p\left(1+d\right)& q-md\end{array}\right|=0$

$\Rightarrow \left(m-qd\right)\left(q-md\right)-np{\left(1+d\right)}^2=0$

$\Rightarrow mq-m^{2}d-q^{2}d+mq{d}^2-np\left(1+2d+d^2\right)=0$

$\Rightarrow mq-m^{2}d-q^{2}d+mq{d}^2-np-2npd-np{d}^2=0$

$\Rightarrow \left(mq-np\right)+d^2\left(mq-np\right)-d\left(m^2+q^2+2np\right)=0$

$\Rightarrow (mq-np)(1+d^2)-d(m+q{)}^2=0$

$\Rightarrow 1+d^2=\frac{d(m+q{)}^2}{(mq-np)}$

$\Rightarrow 1+d^2=\frac{(mq-np)(m+q{)}^2}{(mq-np)} \left[\because d=mq-np\right]$

$\Rightarrow {\left(1+d\right)}^2={\left(m+q\right)}^2$

Hence, option $\left(A\right)$ is correct.