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