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Question
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{bmatrix}\]
Solution
\[A = \begin{bmatrix}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{bmatrix}\]
We know
\[A = IA \]
\[ \Rightarrow \begin{bmatrix}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} A\]
\[ \Rightarrow \begin{bmatrix}0 & 1 & 2 \\ - 2 & 1 & 2 \\ 3 & 1 & 1\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & - 1 \\ 0 & 0 & 1\end{bmatrix} A \left[\text{ Applying }R_2 \to R_2 - R_3 \right]\]
\[ \Rightarrow \begin{bmatrix}- 3 & 0 & 1 \\ - 2 & 1 & 2 \\ 3 & 1 & 1\end{bmatrix} = \begin{bmatrix}1 & 0 & - 1 \\ 0 & 1 & - 1 \\ 0 & 0 & 1\end{bmatrix} A \left[\text{ Applying }R_1 \to R_1 - R_3 \right]\]
\[ \Rightarrow \begin{bmatrix}- 3 & 0 & 1 \\ - 2 & 1 & 2 \\ 0 & 1 & 2\end{bmatrix} = \begin{bmatrix}1 & 0 & - 1 \\ 0 & 1 & - 1 \\ 1 & 0 & 0\end{bmatrix} A \left[\text{ Applying }R_3 \to R_3 + R_1 \right]\]
\[ \Rightarrow \begin{bmatrix}- 3 & 0 & 1 \\ 0 & 3 & 4 \\ 0 & 1 & 2\end{bmatrix} = \begin{bmatrix}1 & 0 & - 1 \\ - 2 & 3 & - 1 \\ 1 & 0 & 0\end{bmatrix} A \left[\text{ Applying }R_2 \to 3 R_2 - 2 R_1 \right]\]
\[ \Rightarrow \begin{bmatrix}- 3 & 0 & 1 \\ 0 & 3 & 4 \\ 0 & - 2 & - 2\end{bmatrix} = \begin{bmatrix}1 & 0 & - 1 \\ - 2 & 3 & - 1 \\ 3 & - 3 & 1\end{bmatrix} A \left[\text{ Applying }R_3 \to R_3 - R_2 \right]\]
\[ \Rightarrow \begin{bmatrix}- 3 & 0 & 1 \\ 0 & 3 & 4 \\ 0 & 1 & 1\end{bmatrix} = \begin{bmatrix}1 & 0 & - 1 \\ - 2 & 3 & - 1 \\ \frac{- 3}{2} & \frac{3}{2} & \frac{- 1}{2}\end{bmatrix} A \left[\text{ Applying }R_3 \to - \frac{1}{2} R_3 \right]\]
\[ \Rightarrow \begin{bmatrix}- 3 & 0 & 1 \\ 0 & - 1 & 0 \\ 0 & 1 & 1\end{bmatrix} = \begin{bmatrix}1 & 0 & - 1 \\ 4 & - 3 & 1 \\ \frac{- 3}{2} & \frac{3}{2} & \frac{- 1}{2}\end{bmatrix} A \left[ \text{ Applying }R_2 \to R_2 - 4 R_3 \right]\]
\[ \Rightarrow \begin{bmatrix}- 3 & 0 & 1 \\ 0 & - 1 & 0 \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}1 & 0 & - 1 \\ 4 & - 3 & 1 \\ \frac{5}{2} & \frac{- 3}{2} & \frac{1}{2}\end{bmatrix} A \left[ \text{ Applying }R_3 \to R_3 + R_2 \right]\]
\[ \Rightarrow \begin{bmatrix}- 3 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}- \frac{3}{2} & \frac{3}{2} & - \frac{3}{2} \\ 4 & - 3 & 1 \\ \frac{5}{2} & \frac{- 3}{2} & \frac{1}{2}\end{bmatrix} A \left[\text{ Applying }R_1 \to R_1 - R_3 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}\frac{1}{2} & \frac{- 1}{2} & \frac{1}{2} \\ 4 & - 3 & 1 \\ \frac{5}{2} & \frac{- 3}{2} & \frac{1}{2}\end{bmatrix} A \left[\text{ Applying }R_1 \to \frac{- 1}{3} R_1 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}\frac{1}{2} & \frac{- 1}{2} & \frac{1}{2} \\ - 4 & 3 & - 1 \\ \frac{5}{2} & \frac{- 3}{2} & \frac{1}{2}\end{bmatrix} A \left[\text{ Applying }R_2 \to - R_2 \right]\]
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