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

(A) Given $A = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}$.
$|A| = (3 \times 5) - (7 \times 2) = 15 - 14 = 1$.
$adj(A) = \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}$.
$A^{-1} = \frac{1}{|A|} adj(A) = \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}$.
Given $B = \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix}$.
$|B| = (6 \times 9) - (8 \times 7) = 54 - 56 = -2$.
$adj(B) = \begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix}$.
$B^{-1} = \frac{1}{|B|} adj(B) = -\frac{1}{2} \begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix} = \begin{bmatrix} -\frac{9}{2} & 4 \\ \frac{7}{2} & -3 \end{bmatrix}$.
Now,$B^{-1} A^{-1} = \begin{bmatrix} -\frac{9}{2} & 4 \\ \frac{7}{2} & -3 \end{bmatrix} \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix} = \begin{bmatrix} -\frac{45}{2} - 8 & \frac{63}{2} + 12 \\ \frac{35}{2} + 6 & -\frac{49}{2} - 9 \end{bmatrix} = \begin{bmatrix} -\frac{61}{2} & \frac{87}{2} \\ \frac{47}{2} & -\frac{67}{2} \end{bmatrix} \dots (1)$.
Now,$AB = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix} \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix} = \begin{bmatrix} 18+49 & 24+63 \\ 12+35 & 16+45 \end{bmatrix} = \begin{bmatrix} 67 & 87 \\ 47 & 61 \end{bmatrix}$.
$|AB| = (67 \times 61) - (87 \times 47) = 4087 - 4089 = -2$.
$adj(AB) = \begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix}$.
$(AB)^{-1} = \frac{1}{|AB|} adj(AB) = -\frac{1}{2} \begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix} = \begin{bmatrix} -\frac{61}{2} & \frac{87}{2} \\ \frac{47}{2} & -\frac{67}{2} \end{bmatrix} \dots (2)$.
From $(1)$ and $(2)$,$(AB)^{-1} = B^{-1} A^{-1}$ is verified.