Consider a matrix
\overline{A}=\left|\begin{matrix} 2&6 \\ 3&4 \end{matrix} \right|
It is required to find its inverse.
Attach a unit matrix of 2 × 2 and perform the operations as shown as follows:
\left|\begin{matrix} 2&6 \\ 3&4 \end{matrix} \right| \left|\begin{matrix} 1&0 \\ 0&1 \end{matrix} \right| → \frac{R_{1}}{2} \left|\begin{matrix} 1&3 \\ 3&4 \end{matrix} \right|\left|\begin{matrix} \frac{1}{2}&0 \\ 0&1 \end{matrix} \right| → R_{2} \ – \ 3R_{1} \left|\begin{matrix} 1&3 \\ 0&-5 \end{matrix} \right|\left|\begin{matrix} \frac{1}{2}&0 \\\\ \frac{-3}{2}&1 \end{matrix} \right|→ R_{1}+ \frac{5}{3}R_{2} \left|\begin{matrix} 1&0 \\ 0&-5 \end{matrix} \right|\left|\begin{matrix} \frac{-2}{5} & \frac{3}{5} \\\\ \frac{-3}{2}&1 \end{matrix} \right|→ R_{2} \ – \frac{1}{5}\left|\begin{matrix} 1&0 \\ 0&1 \end{matrix} \right| \left|\begin{matrix} \frac{-2}{5} & \frac{3}{5} \\\\ \frac{3}{10}& \frac{-1}{5} \end{matrix} \right|Thus, the inverse is
\overline{A}^{-1}= \left|\begin{matrix} \frac{-2}{5} & \frac{3}{5} \\\\ \frac{3}{10}&\frac{-1}{5} \end{matrix} \right|