Question

If $A=\left[\begin{array}{cc}3& 2\\ 7& 5\end{array}\right]$ and $B=\left[\begin{array}{cc}6& 7\\ 8& 9\end{array}\right]$, verify that $(AB{)}^{-1}=B^{-1}{A}^{-1}$

Answer 1

$\Rightarrow A=\left[\begin{array}{cc}3& 2\\ 7& 5\end{array}\right]B=\left[\begin{array}{cc}6& 7\\ 8& 9\end{array}\right]$

Now, $AB=\left[\begin{array}{cc}3& 2\\ 7& 5\end{array}\right]\times \left[\begin{array}{cc}6& 7\\ 8& 9\end{array}\right]$

$=\left[\begin{array}{cc}3\times 6+2\times 8& 3\times 7+9\times 2\\ 7\times 6+5\times 8& 7\times 7+5\times 9\end{array}\right]$

$=\left[\begin{array}{cc}34& 39\\ 82& 94\end{array}\right]$

If $X$ is a matrix of $2\times 2$ order i.e,

$\Rightarrow X=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

Then the inverse of $X$ is found by following formulae.

$\Rightarrow X^{-1}=\frac{1}{\left|X\right|}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$

Where $\left|X\right|,$ is the determinant of matrix $X$

Similarly, ${\left(AB\right)}^{-1}=\frac{1}{\left|AB\right|}\left[\begin{array}{cc}94& -39\\ -82& 34\end{array}\right]$

Now, $\left|AB\right|=\left[\begin{array}{cc}34& 39\\ 82& 94\end{array}\right]$

$=[(34\times 94)-(39\times 82)]$

$=-2$

$\therefore {\left(AB\right)}^{-1}=\frac{1}{-2}\left[\begin{array}{cc}94& -39\\ -82& 34\end{array}\right]$

Again to verify ${\left(AB\right)}^{-1}={\left(B\right)}^{-1}{\left(A\right)}^{-1}:$

$\Rightarrow B^{-1}=\frac{1}{\left|B\right|}\left[\begin{array}{cc}9& -7\\ -8& 6\end{array}\right]$

$=\frac{1}{[(6\times 9)-(7\times 8)]}\left[\begin{array}{cc}9& -7\\ -8& 6\end{array}\right]$

$=\frac{1}{-2}\left[\begin{array}{cc}9& -7\\ -8& 6\end{array}\right]$

$\Rightarrow A^{-1}=\frac{1}{\left|A\right|}\left[\begin{array}{cc}5& -2\\ -7& 3\end{array}\right]$

$=\frac{1}{[(3\times 5)-(7\times 2)]}\left[\begin{array}{cc}5& -2\\ -7& 3\end{array}\right]$

$=\left[\begin{array}{cc}5& -2\\ -7& 3\end{array}\right]$

$\Rightarrow B^{-1}{A}^{-1}=\frac{1}{-2}\left[\begin{array}{cc}9& -7\\ -8& 6\end{array}\right]\times \left[\begin{array}{cc}5& -2\\ -7& 3\end{array}\right]$

$=\frac{1}{-2}\left[\frac{[(9\times 5)+(-7\times -7)][(9\times (-2))+((-7)\times 3)]}{\left[((-8)\times 5)(6\times (-7))\right][((-8)\times (-2))+(6\times 3)]}\right]$

Hence, the answer is ${\left(AB\right)}^{-1}=\frac{1}{-2}\left[\begin{array}{cc}94& -39\\ -82& 34\end{array}\right]={\left(B\right)}^{-1}{\left(A\right)}^{-1}.$