Question

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

Answer 1

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

Inverse of $AB$ :

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

$\Rightarrow AB=\left[\begin{array}{cc}18+49& 24+63\\ 12+35& 16+45\end{array}\right]$

$\Rightarrow AB=\left[\begin{array}{cc}67& 87\\ 47& 61\end{array}\right]$

Now, $|AB|=4087-4089=-2$

Since, $|AB|\ne 0$

Hence, $(AB{)}^{-1}$ exists.

$(AB{)}^{-1}=\frac{adj\left(AB\right)}{|AB|}$

Now, we will find $adj\left(AB\right)$

For this , we will find co-factors of each element of $AB$.

$C_{11}=(-1{)}^{1+1}61=61$

$C_{12}=(-1{)}^{1+2}47=-47$

$C_{21}=(-1{)}^{2+1}87=-87$

$C_{22}=(-1{)}^{1+1}67=67$

Hence, the cofactor matrix is $\left[\begin{array}{cc}61& -47\\ -87& 67\end{array}\right]$

$adjAB=C^T=\left[\begin{array}{cc}61& -87\\ -47& 67\end{array}\right]$

$\Rightarrow (AB{)}^{-1}=\frac{adj\left(AB\right)}{|AB|}=\frac{1}{-2}\left[\begin{array}{cc}61& -87\\ -47& 67\end{array}\right]$

Inverse of $A$ :

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

$|A|=15-14=1$

Since, $|A|\ne 0$

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

$A^{-1}=\frac{adjA}{|A|}$

Now, we will find $adjA$

For this , we will find co-factors of each element of $A$.

$C_{11}=(-1{)}^{1+1}5=5$

$C_{12}=(-1{)}^{1+2}2=-2$

$C_{21}=(-1{)}^{2+1}7=-7$

$C_{22}=(-1{)}^{1+1}3=3$

Hence, the cofactor matrix is $\left[\begin{array}{cc}5& -2\\ -7& 3\end{array}\right]$

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

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

Inverse of $B$ :

We have $A=\left[\begin{array}{cc}6& 8\\ 7& 9\end{array}\right]$

$|B|=54-56=-2$

Since, $|B|\ne 0$

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

$B^{-1}=\frac{adjB}{|B|}$

Now, we will find $adjB$

For this , we will find co-factors of each element of $B$.

$C_{11}=(-1{)}^{1+1}9=9$

$C_{12}=(-1{)}^{1+2}7=-7$

$C_{21}=(-1{)}^{2+1}8=-8$

$C_{22}=(-1{)}^{1+1}6=6$

Hence, the cofactor matrix is $\left[\begin{array}{cc}9& -7\\ -8& 6\end{array}\right]$

$adjB=C^T=\left[\begin{array}{cc}9& -8\\ -7& 6\end{array}\right]$

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

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

$=\frac{1}{-2}\left[\begin{array}{cc}45+16& -63-24\\ -35-12& 49+18\end{array}\right]$

$\Rightarrow B^{-1}{A}^{-1}=\frac{1}{-2}\left[\begin{array}{cc}61& -87\\ -47& 67\end{array}\right]$

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