Question

Let $A=\left[\begin{array}{cc}3& 2\\ 7& 5\end{array}\right]\mathrm{and}B=\left[\begin{array}{cc}6& 7\\ 8& 9\end{array}\right].\mathrm{Find}{\left(AB\right)}^{-1}$

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ A=\left[\begin{array}{cc}3& 2\\ 7& 5\end{array}\right] \\ B=\left[\begin{array}{cc}6& 7\\ 8& 9\end{array}\right] \\ AB=\left[\begin{array}{cc}34& 39\\ 82& 94\end{array}\right] \\ \mathrm{Now}, \\ \left|AB\right|=-2 \\ \mathrm{Since},\left|AB\right|\ne 0 \\ \mathrm{Hence},AB\mathrm{is}\mathrm{invertible}.\mathrm{Let}{C}_{ij}\mathrm{be}\mathrm{the}\mathrm{cofactor}\mathrm{of}{a}_{in}\mathrm{in}AB=\left[a_{ij}\right] \\ {C}_{11}=94,C_{12}=-82,C_{21}=-39\mathrm{and}{C}_{22}=34 \\ \mathrm{adj}\left(AB\right)={\left[\begin{array}{cc}94& -82\\ -39& 34\end{array}\right]}^T=\left[\begin{array}{cc}94& -39\\ -82& 34\end{array}\right] \\ \therefore {\left(AB\right)}^{-1}=-\frac{1}{2}\left[\begin{array}{cc}94& -39\\ -82& 34\end{array}\right]=\left[\begin{array}{cc}-47& \frac{39}{2}\\ 41& -17\end{array}\right] \end{gathered}$$