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प्रश्न
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}2 & - 1 & 3 \\ 1 & 2 & 4 \\ 3 & 1 & 1\end{bmatrix}\]
उत्तर
\[A = \begin{bmatrix}2 & - 1 & 3 \\ 1 & 2 & 4 \\ 3 & 1 & 1\end{bmatrix}\]
We know
\[A = IA \]
\[ \Rightarrow \begin{bmatrix}2 & - 1 & 3 \\ 1 & 2 & 4 \\ 3 & 1 & 1\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} A\]
\[ \Rightarrow \begin{bmatrix}1 & - \frac{1}{2} & \frac{3}{2} \\ 1 & 2 & 4 \\ 3 & 1 & 1\end{bmatrix} = \begin{bmatrix}\frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}A \left[\text{ Applying }R_1 \to \frac{1}{2} R_1 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & - \frac{1}{2} & \frac{3}{2} \\ 0 & \frac{5}{2} & \frac{5}{2} \\ 0 & \frac{5}{2} & \frac{- 7}{2}\end{bmatrix} = \begin{bmatrix}\frac{1}{2} & 0 & 0 \\ - \frac{1}{2} & 1 & 0 \\ - \frac{3}{2} & 0 & 1\end{bmatrix} A \left[\text{ Applying }R_2 \to R_2 - R_1\text{ and }R_3 \to R_3 - 3 R_1 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & - \frac{1}{2} & \frac{3}{2} \\ 0 & 1 & 1 \\ 0 & \frac{5}{2} & \frac{- 7}{2}\end{bmatrix} = \begin{bmatrix}\frac{1}{2} & 0 & 0 \\ - \frac{1}{5} & \frac{2}{5} & 0 \\ - \frac{3}{2} & 0 & 1\end{bmatrix} A \left[\text{ Applying }R_2 \to \frac{2}{5} R_2 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & - 6\end{bmatrix} = \begin{bmatrix}\frac{2}{5} & \frac{1}{5} & 0 \\ - \frac{1}{5} & \frac{2}{5} & 0 \\ - 1 & - 1 & 1\end{bmatrix}A \left[\text{ Applying }R_1 \to R_1 + \frac{1}{2} R_2\text{ and }R_3 \to R_3 - \frac{5}{2} R_2 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}\frac{2}{5} & \frac{1}{5} & 0 \\ - \frac{1}{5} & \frac{2}{5} & 0 \\ \frac{1}{6} & \frac{1}{6} & - \frac{1}{6}\end{bmatrix} A \left[\text{ Applying }R_3 \to - \frac{1}{6} R_2 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}\frac{1}{15} & \frac{- 2}{15} & \frac{1}{3} \\ - \frac{11}{30} & \frac{7}{30} & \frac{1}{6} \\ \frac{1}{6} & \frac{1}{6} & - \frac{1}{6}\end{bmatrix} A \left[\text{ Applying }R_2 \to R_2 - R_3\text{ and }R_1 \to R_1 - 2 R_3 \right]\]
\[ \Rightarrow A^{- 1} = - \frac{1}{30}\begin{bmatrix}- 2 & 4 & - 10 \\ 11 & - 7 & - 5 \\ - 5 & - 5 & 5\end{bmatrix}\]
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