For the following pairs of matrices

Solution:

(i) We have, $A=\left[\begin{array}{ll}3 & 2 \\ 7 & 5\end{array}\right]$ and $B=\left[\begin{array}{ll}4 & 6 \\ 3 & 2\end{array}\right]$

$\therefore A B=\left[\begin{array}{ll}18 & 22 \\ 43 & 52\end{array}\right]$

$|A B|=-10$

Since, $|A B| \neq 0$

Hence, $A B$ is invertible. Let $C_{i j}$ be the cofactor of $a_{i n}$ in $A B=\left[a_{i j}\right]$

$C_{11}=52, C_{12}=-43, C_{21}=-22$ and $C_{22}=18$

$\operatorname{adj}(A B)=\left[\begin{array}{cc}52 & -43 \\ -22 & 18\end{array}\right]^{T}=\left[\begin{array}{cc}52 & -22 \\ -43 & 18\end{array}\right]$

$\therefore(A B)^{-1}=-\frac{1}{10}\left[\begin{array}{cc}52 & -22 \\ -43 & 18\end{array}\right]$          .....(1)

Now, $B=\left[\begin{array}{ll}4 & 6 \\ 3 & 2\end{array}\right]$

$|B|=-10$

Since, $|B| \neq 0$

Hence, $B$ is invertible. Let $C_{i j}$ be the cofactor of $a_{i n}$ in $B=\left[a_{i j}\right]$

$C_{11}=2, C_{12}=-3, C_{21}=-6$ and $C_{22}=4$

$\operatorname{adj} B=\left[\begin{array}{cc}2 & -3 \\ -6 & 4\end{array}\right]^{T}=\left[\begin{array}{cc}2 & -6 \\ -3 & 4\end{array}\right]$

$\therefore B^{-1}=-\frac{1}{10}\left[\begin{array}{cc}2 & -6 \\ -3 & 4\end{array}\right]$

$|A|=1$

Since, $|A| \neq 0$

Hence, $A$ is invertible. Let $C_{i j}$ be the cofactor of $a_{i n}$ in $A=\left[a_{i j}\right]$

$C_{11}=5, C_{12}=-7, C_{21}=-2$ and $C_{22}=3$

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

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

Now, $B^{-1} A^{-1}=-\frac{1}{10}\left[\begin{array}{cc}52 & -22 \\ -43 & 18\end{array}\right]$         .....(2)

From eq. (1) and (2), we have

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

Hence verified.

(ii) We have, $A=\left[\begin{array}{ll}2 & 1 \\ 5 & 3\end{array}\right]$ and $B=\left[\begin{array}{ll}4 & 5 \\ 3 & 4\end{array}\right]$

$\therefore A B=\left[\begin{array}{ll}11 & 14 \\ 29 & 37\end{array}\right]$

Now,

$|A B|=1$

Since, $|A B| \neq 0$

Hence, $A B$ is invertible. Let $C_{i j}$ be the cofactor of $a_{i n}$ in $A B=\left[a_{i j}\right]$

$C_{11}=37, C_{12}=-29, C_{21}=-14$ and $C_{22}=11$

$\operatorname{adj}(A B)=\left[\begin{array}{cc}37 & -14 \\ -29 & 11\end{array}\right]$

$\therefore(A B)^{-1}=\left[\begin{array}{cc}37 & -14 \\ -29 & 11\end{array}\right]$            ....(1)

$|B|=1$

Since, $|B| \neq 0$

Hence, $B$ is invertible. Let $C_{i j}$ be the cofactor of $a_{i n}$ in $B=\left[a_{i j}\right]$

$C_{11}=4, C_{12}=-3, C_{21}=-5$ and $C_{22}=4$

$\operatorname{adj} B=\left[\begin{array}{cc}4 & -5 \\ -3 & 4\end{array}\right]$

$\therefore B^{-1}=\left[\begin{array}{cc}4 & -5 \\ -3 & 4\end{array}\right]$

$|A|=1$

Since, $|A| \neq 0$

Hence, $A$ is invertible. Let $C_{i j}$ be the cofactor of $a_{i n}$ in $A=\left[a_{i j}\right]$

$C_{11}=3, C_{12}=-5, C_{21}=-1$ and $C_{22}=2$

$\operatorname{adj} A=\left[\begin{array}{cc}3 & -1 \\ -5 & 2\end{array}\right]$

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

Now, $B^{-1} A^{-1}=\left[\begin{array}{cc}37 & -14 \\ -29 & 11\end{array}\right]$              ....(2)

From eq. (1) and (2), we have

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

Hence verified.