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

Let $A=\left[\begin{array}{lr}6 & -3 \\ -2 & 1\end{array}\right]$

We know that $A=I A$

$\therefore\left[\begin{array}{cr}6 & -3 \\ -2 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] A$

$\Rightarrow\left[\begin{array}{cr}1 & -\frac{1}{2} \\ -2 & 1\end{array}\right]=\left[\begin{array}{cc}\frac{1}{6} & 0 \\ 0 & 1\end{array}\right] A \quad\left(\mathrm{R}_{1} \rightarrow \frac{1}{6} \mathrm{R}_{1}\right)$

$\Rightarrow\left[\begin{array}{cc}1 & -\frac{1}{2} \\ 0 & 0\end{array}\right]=\left[\begin{array}{cc}\frac{1}{6} & 0 \\ \frac{1}{3} & 1\end{array}\right] A \quad\left(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}+2 \mathrm{R}_{1}\right)$

Now, in the above equation, we can see all the zeros in the second row of the matrix on the L.H.S.

Therefore, $A^{-1}$ does not exist.