Given:$A=\begin{bmatrix}6 & -3\\-2 &1\end{bmatrix}$
Step 1: In order to use row elementary transformation we write as A=IA.
$\begin{bmatrix}6 & -3\\-2& 1\end{bmatrix}=\begin{bmatrix}1 & 0\\0&1\end{bmatrix}A$
Step 2: Applying $R_1\leftrightarrow R_2$
$\begin{bmatrix}-2 & 1\\6& -3\end{bmatrix}=\begin{bmatrix}1 & 0\\0&1\end{bmatrix}A$
Step 3: Applying $R_2\Rightarrow R_2+3R_1$
$\begin{bmatrix}-2 & 1\\6+3(-2)& -3+3(1)\end{bmatrix}=\begin{bmatrix}0 & 1\\1+3(0)&0+3(1)\end{bmatrix}A$
$\begin{bmatrix}-2 & 1\\0& 0\end{bmatrix}=\begin{bmatrix}0 & 1\\1&3\end{bmatrix}A$
In the second row of LHS elements are zero .
Hence $A^{-1}$ does not exist.