\(A=\begin{bmatrix}2&3&4\\5&6&-10\\8&9&4\end{bmatrix}\)

∵ 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 \(=\begin{bmatrix}\frac{-19}{14}&\frac{-2}{7}&\frac{9}{14}\\\frac{25}{21}&\frac{2}{7}&\frac{-10}{21}\\\frac{1}{28}&\frac{-1}{14}&\frac{1}{28}\end{bmatrix}\)