find the

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

We have,

$|A|=15-14=1$

Now,

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

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

$\therefore A^{-1}=\frac{1}{|A|} \cdot \operatorname{adj} A=\left[\begin{array}{rr}5 & -7 \\ -2 & 3\end{array}\right]$

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

We have

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

$\therefore \operatorname{adj} B=\left[\begin{array}{rr}9 & -8 \\ -7 & 6\end{array}\right]$

$\therefore B^{-1}=\frac{1}{|B|} \operatorname{adj} B=-\frac{1}{2}\left[\begin{array}{rr}9 & -8 \\ -7 & 6\end{array}\right]=\left[\begin{array}{rr}-\frac{9}{2} & 4 \\ \frac{7}{2} & -3\end{array}\right]$

Now,

$B^{-1} A^{-1}=\left[\begin{array}{cc}-\frac{9}{2} & 4 \\ \frac{7}{2} & -3\end{array}\right]\left[\begin{array}{rr}5 & -7 \\ -2 & 3\end{array}\right]$

$=\left[\begin{array}{cc}-\frac{45}{2}-8 & \frac{63}{2}+12 \\ \frac{35}{2}+6 & -\frac{49}{2}-9\end{array}\right]=\left[\begin{array}{cc}-\frac{61}{2} & \frac{87}{2} \\ \frac{47}{2} & -\frac{67}{2}\end{array}\right]$  ...(1)

Then,

$\begin{aligned} A B &=\left[\begin{array}{ll}3 & 7 \\ 2 & 5\end{array}\right]\left[\begin{array}{lr}6 & 8 \\ 7 & 9\end{array}\right] \\ &=\left[\begin{array}{ll}18+49 & 24+63 \\ 12+35 & 16+45\end{array}\right] \\ &=\left[\begin{array}{ll}67 & 87 \\ 47 & 61\end{array}\right] \end{aligned}$

Therefore,we have $|A B|=67 \times 61-87 \times 47=4087-4089=-2$.

Also,

$\operatorname{adj}(A B)=\left[\begin{array}{rr}61 & -87 \\ -47 & 67\end{array}\right]$

$\therefore(A B)^{-1}=\frac{1}{|A B|} a d j(A B)=-\frac{1}{2}\left[\begin{array}{ll}61 & -87 \\ -47 & 67\end{array}\right]$

$=\left[\begin{array}{cc}-\frac{61}{2} & \frac{87}{2} \\ \frac{47}{2} & -\frac{67}{2}\end{array}\right] \ldots(2)$

From (1) and (2), we have:

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

Hence, the given result is proved.