Step1:
Given
$A=\begin{bmatrix}1 & -3\\-2 & 6\end{bmatrix}$
In order to use elementary row operation we may write A=IA.
$\Rightarrow \begin{bmatrix}1 & -3\\-2 & 6\end{bmatrix}=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}A$
Step2:
Apply $R_1=2R_1+R_2$
$\Rightarrow \begin{bmatrix}0 & 0\\-2 & 6\end{bmatrix}=\begin{bmatrix}2 & 1\\0 & 1\end{bmatrix}A$
In first row of the LHS each element is zero hence $A^{-1}$ does not exist.