Question

$\mathrm{If}{A}^{-1}=\left[\begin{array}{ccc}3& -1& 1\\ -15& 6& -5\\ 5& -2& 2\end{array}\right]\mathrm{and}B=\left[\begin{array}{ccc}1& 2& -2\\ -1& 3& 0\\ 0& -2& 1\end{array}\right],\mathrm{find}{\left(AB\right)}^{-1}.$

Answer 1

We know that (AB)⁻¹ = B⁻¹ A⁻¹.

$$\begin{gathered} B=\left[\begin{array}{ccc}1& 2& -2\\ -1& 3& 0\\ 0& -2& 1\end{array}\right] \\ {B}^{-1}=\frac{1}{\left|B\right|}\mathrm{Adj}.B \\ \mathrm{Now}, \\ \left|B\right|=\left|\begin{array}{ccc}1& 2& -2\\ -1& 3& 0\\ 0& -2& 1\end{array}\right| \\ =1\left(3+0\right)+1\left(2-4\right) \\ =1 \\ \mathrm{Now},\mathrm{to}\mathrm{find}\mathrm{Adj}.B \\ \begin{array}{ccc}{B}_{11}={\left(-1\right)}^{1+1}\left(3\right)=3 \\ {B}_{12}={\left(-1\right)}^{1+2}\left(-1\right)=1 \\ {B}_{13}={\left(-1\right)}^{1+3}\left(2\right)=2& B_{21}={\left(-1\right)}^{2+1}\left(2-4\right)=2 \\ {B}_{22}={\left(-1\right)}^{2+2}\left(1\right)=1 \\ {B}_{23}={\left(-1\right)}^{2+3}\left(-2\right)=2& B_{31}={\left(-1\right)}^{3+1}\left(6\right)=6 \\ {B}_{32}={\left(-1\right)}^{3+2}\left(-2\right)=2 \\ {B}_{33}={\left(-1\right)}^{3+3}\left(3+2\right)=5\end{array} \\ \mathrm{Therefore}, \\ \mathrm{Adj}.B=\left[\begin{array}{ccc}3& 2& 6\\ 1& 1& 2\\ 2& 2& 5\end{array}\right] \\ \mathrm{Thus}, \\ {B}^{-1}=\left[\begin{array}{ccc}3& 2& 6\\ 1& 1& 2\\ 2& 2& 5\end{array}\right]. \\ {\left(AB\right)}^{-1}=B^{-1}{A}^{-1} \\ =\left[\begin{array}{ccc}3& 2& 6\\ 1& 1& 2\\ 2& 2& 5\end{array}\right]\left[\begin{array}{ccc}3& -1& 1\\ -15& 6& -5\\ 5& -2& 2\end{array}\right] \\ =\left[\begin{array}{ccc}9-30+30& -3+12-12& 3-10+12\\ 3-15+10& -1+6-4& 1-5+4\\ 6-30+25& -2+12-10& 2-10+10\end{array}\right] \\ =\left[\begin{array}{ccc}9& -3& 5\\ -2& 1& 0\\ 1& 0& 2\end{array}\right] \\ \mathrm{Hence},{\left(AB\right)}^{-1}=\left[\begin{array}{ccc}9& -3& 5\\ -2& 1& 0\\ 1& 0& 2\end{array}\right] \end{gathered}$$