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Question
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}1 & 2 & 0 \\ 2 & 3 & - 1 \\ 1 & - 1 & 3\end{bmatrix}\]
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Solution
\[A = \begin{bmatrix}1 & 2 & 0 \\ 2 & 3 & - 1 \\ 1 & - 1 & 3\end{bmatrix}\]
We know
\[A = IA \]
\[ \Rightarrow \begin{bmatrix}1 & 2 & 0 \\ 2 & 3 & - 1 \\ 1 & - 1 & 3\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} A\]
\[ \Rightarrow \begin{bmatrix}1 & 2 & 0 \\ 0 & - 1 & - 1 \\ 0 & - 3 & 3\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 \\ - 2 & 1 & 0 \\ - 1 & 0 & 1\end{bmatrix} A \left[\text{ Applying }R_2 \to R_2 - 2 R_1\text{ and }R_3 \to R_3 - R_1 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 2 & 0 \\ 0 & 1 & 1 \\ 0 & - 3 & 3\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 \\ 2 & - 1 & 0 \\ - 1 & 0 & 1\end{bmatrix} A \left[\text{ Applying }R_2 \to - R_2 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & - 2 \\ 0 & 1 & 1 \\ 0 & 0 & 6\end{bmatrix} = \begin{bmatrix}- 3 & 2 & 0 \\ 2 & - 1 & 0 \\ 5 & - 3 & 1\end{bmatrix} A \left[\text{ Applying }R_1 \to R_1 - 2 R_2\text{ and }R_3 \to R_3 + 3 R_2 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & - 2 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}- 3 & 2 & 0 \\ 2 & - 1 & 0 \\ \frac{5}{6} & - \frac{1}{2} & \frac{1}{6}\end{bmatrix} A \left[\text{ Applying }R_3 \to \frac{1}{6} R_3 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}- \frac{4}{3} & 1 & \frac{1}{3} \\ \frac{7}{6} & - \frac{1}{2} & \frac{- 1}{6} \\ \frac{5}{6} & - \frac{1}{2} & \frac{1}{6}\end{bmatrix}A \left[\text{ Applying }R_1 \to R_1 + 2 R_3\text{ and }R_2 \to R_2 - R_3 \right]\]
\[ \therefore A^{- 1} = \begin{bmatrix}- \frac{4}{3} & 1 & \frac{1}{3} \\ \frac{7}{6} & - \frac{1}{2} & \frac{- 1}{6} \\ \frac{5}{6} & - \frac{1}{2} & \frac{1}{6}\end{bmatrix}\]
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