Find the inverse of each of the matrices, if it exists.

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

We know that A = IA

$\therefore\left[\begin{array}{ll}2 & 3 \\ 5 & 7\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] A$

$\Rightarrow\left[\begin{array}{ll}1 & \frac{3}{2} \\ 5 & 7\end{array}\right]=\left[\begin{array}{ll}\frac{1}{2} & 0 \\ 0 & 1\end{array}\right] A \quad\left(\mathrm{R}_{1} \rightarrow \frac{1}{2} \mathrm{R}_{1}\right)$

$\Rightarrow\left[\begin{array}{lr}1 & \frac{3}{2} \\ 0 & -\frac{1}{2}\end{array}\right]=\left[\begin{array}{ll}\frac{1}{2} & 0 \\ -\frac{5}{2} & 1\end{array}\right] A \quad\left(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-5 \mathrm{R}_{1}\right)$

$\Rightarrow\left[\begin{array}{cc}1 & 0 \\ 0 & -\frac{1}{2}\end{array}\right]=\left[\begin{array}{cc}-7 & 3 \\ -\frac{5}{2} & 1\end{array}\right] A \quad\left(\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+3 \mathrm{R}_{2}\right)$

$\Rightarrow\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}-7 & 3 \\ 5 & -2\end{array}\right] A \quad\left(\mathrm{R}_{2} \rightarrow-2 \mathrm{R}_{1}\right)$

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