\(A=\begin{bmatrix}2&3&4\\5&6&-10\\8&9&4\end{bmatrix}\)
∵ A = IA
⇒ \(\begin{bmatrix}2&3&4\\5&6&-10\\8&9&4\end{bmatrix}\) \(=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}A\)
R2 → 2R2 - 5R1
Applying R3 → R3 - 4R1
\(\begin{bmatrix}2&3&4\\0&-3&-40\\0&-3&-12\end{bmatrix}\)\(=\begin{bmatrix}1&0&0\\-5&2&0\\-4&0&1\end{bmatrix}A\)
Applying R3 → R3 - R2
\(\begin{bmatrix}2&3&4\\0&-3&-40\\0&0&28\end{bmatrix}\) \(=\begin{bmatrix}1&0&0\\-5&2&0\\1&-2&1\end{bmatrix}A\)
Applying R3 → R3/28
\(\begin{bmatrix}2&3&4\\0&-3&-40\\0&0&1\end{bmatrix}\) \(=\begin{bmatrix}1&0&0\\-5&2&0\\\frac{1}{28}&\frac{-1}{14}&\frac{1}{28}\end{bmatrix}A\)
Applying R1 → R1 - 4R3
R2 → R2 + 40R3
\(\begin{bmatrix}2&3&0\\0&-3&0\\0&0&1\end{bmatrix}\) \(=\begin{bmatrix}\frac{6}{7}&\frac{2}{7}&\frac{-1}{7}\\\frac{-25}{7}&\frac{-6}{7}&\frac{10}{7}\\\frac{1}{28}&\frac{-1}{14}&\frac{1}{28}\end{bmatrix}A\)
Applying R1 → R1 + R2
\(\begin{bmatrix}2&0&0\\0&-3&0\\0&0&1\end{bmatrix}\) \(=\begin{bmatrix}\frac{-19}{7}&\frac{-4}{7}&\frac{9}{7}\\\frac{-25}{7}&\frac{-6}{7}&\frac{10}{7}\\\frac{1}{28}&\frac{-1}{14}&\frac{1}{28}\end{bmatrix}A\)
Applying R1 → R1/2, R2 → R2/-3
\(\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\) \(=\begin{bmatrix}\frac{-19}{14}&\frac{-2}{7}&\frac{9}{14}\\\frac{25}{21}&\frac{2}{7}&\frac{-10}{21}\\\frac{1}{28}&\frac{-1}{14}&\frac{1}{28}\end{bmatrix}A\)
We obtain A-1A = I
Therefore, A-1 \(=\begin{bmatrix}\frac{-19}{14}&\frac{-2}{7}&\frac{9}{14}\\\frac{25}{21}&\frac{2}{7}&\frac{-10}{21}\\\frac{1}{28}&\frac{-1}{14}&\frac{1}{28}\end{bmatrix}\)