\(A=\begin{bmatrix}4&-2\\3&1\end{bmatrix}\)
We have A = IA
\(\begin{bmatrix}4&-2\\3&1\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}A\)
⇒ \(\begin{bmatrix}4&-2\\0&10\end{bmatrix}=\begin{bmatrix}1&0\\-3&4\end{bmatrix}A\) (By applying R2 → 4R2 - 3R1)
⇒ \(\begin{bmatrix}1&-1/2\\0&1\end{bmatrix}=\begin{bmatrix}1/4&0\\-3/10&2/5\end{bmatrix}A\) (By applying R1 → R1/4 and R2 → R2/10)
⇒ \(\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}1/10&1/5\\-3/10&2/5\end{bmatrix}A\) (By applying R1 → R1 + 1/2 R2)
⇒ \(\begin{bmatrix}1&0\\0&1\end{bmatrix}=\frac1{10}\begin{bmatrix}1&2\\-3&4\end{bmatrix}A\)
We get I = A-1A
∴ A-1 \(=\frac1{10}\begin{bmatrix}1&2\\-3&4\end{bmatrix}.\)