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