Question

If the adjoint of the matrix
$A=\left[\begin{array}{ccc}1& 2& 3\\ 0& 5& 0\\ 2& 4& 3\end{array}\right]$ is $\left[\begin{array}{ccc}15& a& -15\\ 0& -3& 0\\ -15& 0& b\end{array}\right]$, find the value of $ab$?

Answer 1

If $C$ be the matrix of cofactors of the element is $|A|$, then
$C=\left[\begin{array}{ccc}{C}_{11}& C_{12}& C_{13}\\ C_{21}& C_{22}& C_{23}\\ C_{31}& C_{32}& C_{33}\end{array}\right]$
Here $C_{11}=\left|\begin{array}{cc}5& 0\\ 4& 3\end{array}\right|=15;C_{12}=-\left|\begin{array}{cc}0& 0\\ 2& 3\end{array}\right|=0;C_{13}=\left|\begin{array}{cc}0& 5\\ 2& 4\end{array}\right|=-10$
$C_{21}=-\left|\begin{array}{cc}2& 3\\ 4& 3\end{array}\right|=6;C_{22}=\left|\begin{array}{cc}1& 3\\ 2& 3\end{array}\right|=0;C_{23}=\left|\begin{array}{cc}1& 2\\ 2& 4\end{array}\right|=-10$
$C_{31}=-\left|\begin{array}{cc}2& 3\\ 5& 0\end{array}\right|=-15;C_{32}=-\left|\begin{array}{cc}1& 3\\ 0& 0\end{array}\right|=0;C_{33}=\left|\begin{array}{cc}1& 2\\ 0& 5\end{array}\right|=5$
$\therefore C=\left[\begin{array}{ccc}15& 0& -10\\ 6& -3& 0\\ -15& 0& 5\end{array}\right]$
$\therefore Adj\left(A\right)=\text{transpose of C}=C^{\prime }=\left[\begin{array}{ccc}15& 6& -15\\ 0& -3& 0\\ -15& 0& 5\end{array}\right]$
$\Rightarrow ab=30$