Question

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

Answer 1

We know that

$(AB{)}^{-1}=B^{-1}{A}^{-1}$ .....(1)

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

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

Here, $|B|=1(3-0)+1(2-4)+0$

$\Rightarrow |B|=3-2=1$

Since, $|B|\ne 0$

Hence, $B^{-1}$ exists.

$B^{-1}=\frac{adjB}{|A|}$ and $adjB=C^T$

$C_{11}=(-1{)}^{1+1}\left|\begin{array}{cc}3& 0\\ -2& 1\end{array}\right|$

$\Rightarrow C_{11}=3-0=3$

$C_{12}=(-1{)}^{1+2}\left|\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right|$

$\Rightarrow C_{12}=-(-1-0)=1$

$C_{13}=(-1{)}^{1+3}\left|\begin{array}{cc}-1& 3\\ 0& -2\end{array}\right|$

$\Rightarrow C_{13}=2-0=2$

$C_{21}=(-1{)}^{2+1}\left|\begin{array}{cc}2& -2\\ -2& 1\end{array}\right|$

$\Rightarrow C_{21}=-(2-4)=2$

$C_{22}=(-1{)}^{2+2}\left|\begin{array}{cc}1& -2\\ 0& 1\end{array}\right|$

$\Rightarrow C_{22}=1-0=1$

$C_{23}=(-1{)}^{2+3}\left|\begin{array}{cc}1& 2\\ 0& -2\end{array}\right|$

$\Rightarrow C_{23}=-(-2-0)=2$

$C_{31}=(-1{)}^{3+1}\left|\begin{array}{cc}2& -2\\ 3& 0\end{array}\right|$

$\Rightarrow C_{31}=0+6=6$

$C_{32}=(-1{)}^{3+2}\left|\begin{array}{cc}1& -2\\ -1& 0\end{array}\right|$

$\Rightarrow C_{32}=-(0-2)=2$

$C_{33}=(-1{)}^{3+3}\left|\begin{array}{cc}1& 2\\ -1& 3\end{array}\right|$

$\Rightarrow C_{33}=3+2=5$

Hence, the co-factor matrix is $C=\left[\begin{array}{ccc}3& 1& 2\\ 2& 1& 2\\ 6& 2& 5\end{array}\right]$

$\Rightarrow adjB=C^T=\left[\begin{array}{ccc}3& 2& 6\\ 1& 1& 2\\ 2& 2& 5\end{array}\right]$

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

Substituting the values in (1), we get

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

$(AB{)}^{-1}=\left[\begin{array}{ccc}9-30+30& -3+12-12& 3-10+12\\ 3-15+10& -1+6+4& 1-5+4\\ 6-30+25& -2+12-10& 2-10+10\end{array}\right]$

$(AB{)}^{-1}=\left[\begin{array}{ccc}9& -3& 5\\ -2& 9& 0\\ 1& 0& 2\end{array}\right]$