Let A= $\begin{array}{cc} 2 & 5 \\ 1 & 3 \end{array}$ We can write A = IA i.e. $\begin{array}{cc} 2 & 5 \\ 1 & 3 \end{array}$ = $\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}$ A Applying $R_1$ → $\frac{1}{2} R_1$ , we get $\begin{array}{cc} 1 & \frac{5}{2} \\ 1 & 3 \end{array}$ = $\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & 1 \end{array}$ A Applying $R_2$ → $R_2 - R_1$ , gives $\begin{array}{cc} 1 & \frac{5}{2} \\ 0 & \frac{1}{2} \end{array}$ = $\begin{array}{cc} \frac{1}{2} & 0 \\ -\frac{1}{2} & 1 \end{array}$ A Applying , $R_1$ → $R_1 - 5 R_2$ we obtain $\begin{array}{cc} 1 & 0 \\ 0 & \frac{1}{2} \end{array}$ = $\begin{array}{cc} 3 & -5 \\ -\frac{1}{2} & 1 \end{array}$ A Applying $R_2$ → $2 R_2$ , gives $\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}$ = $[able 3,-5;-1,2]$ A Therefore, $A^{-1}$ = $\begin{array}{cc} 3 & -5 \\ -1 & 2 \end{array}$