Find the inverse of matrix \(A\) by elementary row transformations, where \(A=\begin{bmatrix}2&-3\\-1&2\end{bmatrix}\)

Answer: \(A^{-1}=\begin{bmatrix}2&3\\1&2\end{bmatrix}\)

\(|A|=4-3=1\ne0\), so \(A^{-1}\) exists. Consider \(AA^{-1}=I\), i.e. \(\begin{bmatrix}2&-3\\-1&2\end{bmatrix}A^{-1}=\begin{bmatrix}1&0\\0&1\end{bmatrix}\).

\(R_1\to R_1+R_2\): \(\begin{bmatrix}1&-1\\-1&2\end{bmatrix}A^{-1}=\begin{bmatrix}1&1\\0&1\end{bmatrix}\)

\(R_2\to R_2+R_1\): \(\begin{bmatrix}1&-1\\0&1\end{bmatrix}A^{-1}=\begin{bmatrix}1&1\\1&2\end{bmatrix}\)

\(R_1\to R_1+R_2\): \(\begin{bmatrix}1&0\\0&1\end{bmatrix}A^{-1}=\begin{bmatrix}2&3\\1&2\end{bmatrix}\)

Hence \(A^{-1}=\begin{bmatrix}2&3\\1&2\end{bmatrix}\). (Check: \(AA^{-1}=\begin{bmatrix}4-3&6-6\\-2+2&-3+4\end{bmatrix}=I\).)