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Question
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}1 & 3 & - 2 \\ - 3 & 0 & - 1 \\ 2 & 1 & 0\end{bmatrix}\]
Solution
\[A = \begin{bmatrix}1 & 3 & - 2 \\ - 3 & 0 & - 1 \\ 2 & 1 & 0\end{bmatrix}\]
We know
\[A = IA \]
\[ \Rightarrow \begin{bmatrix}1 & 3 & - 2 \\ - 3 & 0 & - 1 \\ 2 & 1 & 0\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} A\]
\[ \Rightarrow \begin{bmatrix}1 & 3 & - 2 \\ 0 & 9 & - 7 \\ 0 & - 5 & 4\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 \\ 3 & 1 & 0 \\ - 2 & 0 & 1\end{bmatrix} A \left[\text{ Applying }R_2 \to R_2 + 3 R_1\text{ and }R_3 \to R_3 - 2 R_1 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 3 & - 2 \\ 0 & 1 & - \frac{7}{9} \\ 0 & - 5 & 4\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 \\ \frac{1}{3} & \frac{1}{9} & 0 \\ - 2 & 0 & 1\end{bmatrix} A \left[\text{ Applying }R_2 \to \frac{1}{9} R_2 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & \frac{1}{3} \\ 0 & 1 & - \frac{7}{9} \\ 0 & 0 & \frac{1}{9}\end{bmatrix} = \begin{bmatrix}0 & - \frac{1}{3} & 0 \\ \frac{1}{3} & \frac{1}{9} & 0 \\ - \frac{1}{3} & \frac{5}{9} & 1\end{bmatrix} A \left[\text{ Applying }R_1 \to R_1 - 3 R_2\text{ and }R_3 \to R_3 + 5 R_2 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & \frac{1}{3} \\ 0 & 1 & - \frac{7}{9} \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}0 & - \frac{1}{3} & 0 \\ \frac{1}{3} & \frac{1}{9} & 0 \\ - 3 & 5 & 9\end{bmatrix} A \left[\text{ Applying }R_3 \to 9 R_3 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}1 & - 2 & - 3 \\ - 2 & 4 & 7 \\ - 3 & 5 & 9\end{bmatrix} A \left[\text{ Applying }R_2 \to R_2 + \frac{7}{9} R_3\text{ and }R_1 \to R_1 - \frac{1}{3} R_3 \right]\]
\[ \Rightarrow A^{- 1} = \begin{bmatrix}1 & - 2 & - 3 \\ - 2 & 4 & 7 \\ - 3 & 5 & 9\end{bmatrix} \]
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