If $A=\left[\begin{array}{cc}3 & 1 \\ 2 & -3\end{array}\right]$, then find |adj $A \mid$.
$|A|=\left|\begin{array}{cc}3 & 1 \\ 2 & -3\end{array}\right|=-11$
$\therefore|\operatorname{adj} A|=|A|^{n-1}=(-11)^{2-1}=-11$
Leave a comment
All Study Material
