Question

$ifA=\left[\begin{array}{cc}ab& b^2\\ -a^2& -ab\end{array}\right],thenobtain{|A|}^2$

Answer 1

$A^2=\left[\begin{array}{cc}ab& b^2\\ {-a}^2& -ab\end{array}\right]\left[\begin{array}{cc}ab& b^2\\ {-a}^2& -ab\end{array}\right]=\left[\begin{array}{cc}{a}^2{b}^2-a^2{b}^2& a{b}^2-a{b}^2\\ -a^{3}b+a^{3}b& -a^2{b}^2+a^2{b}^2\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

$\therefore |A^2|=0$