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# Find $A^{-1}$ if it exists by using elementary transformations where A=$\begin{bmatrix}6 & -3\\-2 & 1\end{bmatrix}$

$\begin{bmatrix}6 & -3\\-2 & 1\end{bmatrix}=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}A$
Step 1: Apply $R_1=\frac{1}{6}R_1$
$\begin{bmatrix}1 & \frac{-1}{2}\\-2 & 1\end{bmatrix}=\begin{bmatrix}\frac{1}{6} & 0\\0 & 1\end{bmatrix}A$
Step 2: Apply $R_2\rightarrow R_2+2R_1$
$\begin{bmatrix}1 & \frac{-1}{2}\\0 & 0\end{bmatrix}=\begin{bmatrix}\frac{1}{6} & 0\\\frac{1}{3} & 1\end{bmatrix}A$
Step 3: Since in the matrix on LHS all elements of second row are zero.Therefore $A^{-1}$ does not exists.