Answer
See below
Work Step by Step
Matrices are inverses of one another if
$AA^{-1}=A^{-1}A=I_n$
hence
$AA^{-1}=\begin{bmatrix}
a &c\\c&d
\end{bmatrix}.\frac{1}{ad-bc}\begin{bmatrix}
d&-b\\-c& a
\end{bmatrix}=\frac{1}{ad-bc}\begin{bmatrix}
ad-bc & -ab+ab\\cd-cd & -bc+ad
\end{bmatrix}=\frac{1}{ad-bc}\begin{bmatrix}
ad-bc &0\\0&ad-bc
\end{bmatrix}=\begin{bmatrix}
1& 0\\ 0 &1
\end{bmatrix}$
$A^{-1}A=\frac{1}{ad-bc}\begin{bmatrix}
d&-b\\-c & a
\end{bmatrix}.\begin{bmatrix}
a &c\\c&d
\end{bmatrix}=\frac{1}{ad-bc}\begin{bmatrix}
ad-bc & 0\\0 &ad-bc
\end{bmatrix}=\begin{bmatrix}
1 & 0\\
0& 1
\end{bmatrix}$
It shows that the given matrices are inverses of one another.
