If  $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$(where  $bc\ne 0$) satisfies the equation  $x^2+k=0$ then  $(\left|A\right|=ad-bc)$

 $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$
 $A^2=\left[\begin{array}{cc}{a}^2+bc& ab+db\\ ac+cd& bc+d^2\end{array}\right]$
As $A$ satisfies  $x^2+k=0$, we have  $A^2+KI=0$
 $$\begin{gathered} \left[\begin{array}{cc}{a}^2+bc+K& (a+d)b\\ (a+d)c& (bc+d^2+K)\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \\ {a}^2+bc+K=0(a+d)b=0 \\ (a+d)c=0bc+d^2+K=0 \end{gathered}$$
 As  $bc\ne 0,b\ne 0c\ne 0$
 $a+d=0$
Also  $K=-(a^2+bc)$
 $K=-\left|A\right|$