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Question
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}3 & - 3 & 4 \\ 2 & - 3 & 4 \\ 0 & - 1 & 1\end{bmatrix}\]
Solution
\[A = \begin{bmatrix}3 & - 3 & 4 \\ 2 & - 3 & 4 \\ 0 & - 1 & 1\end{bmatrix}\]
We know
\[A = IA \]
\[ \Rightarrow \begin{bmatrix}3 & - 3 & 4 \\ 2 & - 3 & 4 \\ 0 & - 1 & 1\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} A\]
\[ \Rightarrow \begin{bmatrix}1 & - 1 & \frac{4}{3} \\ 2 & - 3 & 4 \\ 0 & - 1 & 1\end{bmatrix} = \begin{bmatrix}\frac{1}{3} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} A \left[\text{ Applying }R_1 \to \frac{1}{3} R_1 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & - 1 & \frac{4}{3} \\ 0 & - 1 & \frac{4}{3} \\ 0 & - 1 & 1\end{bmatrix} = \begin{bmatrix}\frac{1}{3} & 0 & 0 \\ \frac{- 2}{3} & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} A \left[\text{ Applying }R_2 \to R_2 - 2 R_1 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & - 1 & \frac{4}{3} \\ 0 & 1 & \frac{- 4}{3} \\ 0 & - 1 & 1\end{bmatrix} = \begin{bmatrix}\frac{1}{3} & 0 & 0 \\ \frac{2}{3} & - 1 & 0 \\ 0 & 0 & 1\end{bmatrix} A \left[\text{ Applying }R_2 \to - R_2 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & \frac{- 4}{3} \\ 0 & 0 & \frac{- 1}{3}\end{bmatrix} = \begin{bmatrix}1 & - 1 & 0 \\ \frac{2}{3} & - 1 & 0 \\ \frac{2}{3} & - 1 & 1\end{bmatrix} A \left[\text{ Applying }R_1 \to R_1 + R_2\text{ and }R_3 \to R_3 + R_2 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & \frac{- 4}{3} \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}1 & - 1 & 0 \\ \frac{2}{3} & - 1 & 0 \\ - 2 & 3 & - 3\end{bmatrix} A \left[\text{ Applying }R_3 \to - 3 R_3 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}1 & - 1 & 0 \\ - 2 & 3 & - 4 \\ - 2 & 3 & - 3\end{bmatrix} A \left[\text{ Applying }R_2 \to R_2 + \frac{4}{3} R_3 \right]\]
\[ \therefore A^{- 1} = \begin{bmatrix}1 & - 1 & 0 \\ - 2 & 3 & - 4 \\ - 2 & 3 & - 3\end{bmatrix}\]
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