Let A= $\left[\begin{array}{cc}2& 5\\ 1& 3\end{array}\right]$
We can write A = IA
i.e. $\left[\begin{array}{cc}2& 5\\ 1& 3\end{array}\right]$ = $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ A
Applying ${\mathit{R}}_1$ → $\frac{1}{2}{\mathit{R}}_1$ , we get
$\left[\begin{array}{cc}1& \frac{5}{2}\\ 1& 3\end{array}\right]$ = $\left[\begin{array}{cc}\frac{1}{2}& 0\\ 0& 1\end{array}\right]$ A
Applying ${\mathit{R}}_2$ → ${\mathit{R}}_2-{\mathit{R}}_1$ , gives
$\left[\begin{array}{cc}1& \frac{5}{2}\\ 0& \frac{1}{2}\end{array}\right]$ = $\left[\begin{array}{cc}\frac{1}{2}& 0\\ -\frac{1}{2}& 1\end{array}\right]$ A
Applying , ${\mathit{R}}_1$ → ${\mathit{R}}_1-5{\mathit{R}}_2$ we obtain
$\left[\begin{array}{cc}1& 0\\ 0& \frac{1}{2}\end{array}\right]$ = $\left[\begin{array}{cc}3& -5\\ -\frac{1}{2}& 1\end{array}\right]$ A
Applying ${\mathit{R}}_2$ → $2{\mathit{R}}_2$ , gives
$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ = $\left[\mathrm{able}3,-5;-1,2\right]$ A
Therefore, ${\mathit{A}}^{-1}$ = $\left[\begin{array}{cc}3& -5\\ -1& 2\end{array}\right]$