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Question:

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

Solution:

Given :

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

$B=\left[\begin{array}{ll}6 & 7 \\ 8 & 9\end{array}\right]$

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

Now,

$|A B|=-2$

Since, $|A B| \neq 0$

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

$C_{11}=94, C_{12}=-82, C_{21}=-39$ and $C_{22}=34$

$\operatorname{adj}(A B)=\left[\begin{array}{cc}94 & -82 \\ -39 & 34\end{array}\right]^{T}=\left[\begin{array}{cc}94 & -39 \\ -82 & 34\end{array}\right]$

$\therefore(A B)^{-1}=-\frac{1}{2}\left[\begin{array}{cc}94 & -39 \\ -82 & 34\end{array}\right]=\left[\begin{array}{cc}-47 & \frac{39}{2} \\ 41 & -17\end{array}\right]$

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