Question

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

Answer 1

$A=\left[\begin{array}{ccc}1& 2& 3\\ 1& -2& -3\end{array}\right],B=\left[\begin{array}{cc}1& -1\\ 1& 2\\ 1& -2\end{array}\right]$

$AB=\left[\begin{array}{ccc}1& 2& 3\\ 1& -2& -3\end{array}\right],B=\left[\begin{array}{cc}1& -1\\ 1& 2\\ 1& -2\end{array}\right]=\left[\begin{array}{cc}6& -3\\ -4& 1\end{array}\right]$

det $\left(AB\right)=6-12=-6.$

$(AB{)}^{-1}=\frac{1}{det\left(AB\right)}\left[\begin{array}{cc}1& 3\\ 4& 6\end{array}\right]=\frac{-1}{6}\left[\begin{array}{cc}1& 3\\ 4& 6\end{array}\right]$

$\therefore (AB{)}^{-1}=\left[\begin{array}{cc}\frac{-1}{6}& \frac{-1}{2}\\ \frac{-2}{3}& 6\end{array}\right]$