Question

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

${\left(AB\right)}^{-1}=?$

Answer 1

$B=\left[\begin{array}{ccc}1& 2& -2\\ -1& 3& 0\\ 0& -2& 1\end{array}\right]$

$B^{-1}=\left[\begin{array}{ccc}3& 1& 2\\ 2& 1& 2\\ 6& 2& 5\end{array}\right]$

We can find $B^{-1}$ by first creating a matrix of minors, then of co factors, and finally transposing it before dividing it by its determinant

${\left(AB\right)}^{-1}=B^{-1}{A}^{-1}$

$=\left[\begin{array}{ccc}3& 1& 2\\ 2& 1& 2\\ 6& 2& 5\end{array}\right]\left[\begin{array}{ccc}3& -1& 1\\ -15& 6& -5\\ 5& -2& 2\end{array}\right]$

$=\left[\begin{array}{ccc}3\times 3+1\times -15+2\times 5& 3\times -1+1\times 6+2\times -2& 3\times 1+1\times -5+2\times 2\\ 2\times 3+1\times -15+2\times 5& 2\times -1+1\times 6+2\times -2& 2\times 1+1\times -5+2\times 2\\ 6\times 3+2\times -15+5\times 5& 6\times -1+2\times 6+5\times -2& 6\times 1+2\times -5+5\times 2\end{array}\right]$

${\left(AB\right)}^{-1}=\left[\begin{array}{ccc}4& -1& 2\\ 1& 0& 1\\ 13& -4& 6\end{array}\right]$