Question

Find $(AB{)}^{-1}$, if $A=\left[\begin{array}{ccc}1& 2& 3\\ 1& -2& -3\end{array}\right],B=\left[\begin{array}{cc}1& -1\\ 1& 2\\ 1& -2\end{array}\right]$

Answer 1

We have, $A=\left[\begin{array}{ccc}1& 2& 3\\ 1& -2& -3\end{array}\right]$ and $B=\left[\begin{array}{cc}1& -1\\ 1& 2\\ 1& -2\end{array}\right]$
Now, $AB=\left[\begin{array}{ccc}1& 2& 3\\ 1& -2& -3\end{array}\right]\left[\begin{array}{cc}1& -1\\ 1& 2\\ 1& -2\end{array}\right]$
$AB=\left[\begin{array}{cc}1+2+3& -1+4-6\\ 1-2-3& -1-4+6\end{array}\right]$
$\therefore AB=\left[\begin{array}{cc}6& -3\\ -4& 1\end{array}\right]$
Now, we get $M_{11}=1,A_{11}=(-1{)}^{1+1}(1)=1$
$M_{12}=-4,A_{12}=(-1{)}^{1+2}(-4)=4$
$M_{21}=-3,A_{21}=(-1{)}^{2+1}(-3)=3$
$M_{22}=6,A_{22}=(-1{)}^{2+2}(6)=6$
$\therefore \text{adj}\left(AB\right)={\left[\begin{array}{cc}1& 4\\ 3& 6\end{array}\right]}^T=\left[\begin{array}{cc}1& 3\\ 4& 6\end{array}\right]$
and $|AB|=\left|\begin{array}{cc}1& 3\\ 4& 6\end{array}\right|=6-12=-6\ne 0$
$\therefore$ Using $A^{-1}=\frac{1}{|A|}\text{adj}A$
$\therefore (AB{)}^{-1}=-\frac{1}{6}\left[\begin{array}{cc}1& 3\\ 4& 6\end{array}\right]$ which is the required inverse.