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प्रश्न
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}- 1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{bmatrix}\]
उत्तर
\[\text{ Let }A = \begin{bmatrix}- 1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{bmatrix} . \]
To find inverse, first write A = IA .
\[i . e . , \begin{bmatrix}- 1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}A\]
\[ \Rightarrow \begin{bmatrix}1 & - 1 & - 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{bmatrix} = \begin{bmatrix}- 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}A \left[\text{ Applying }R_1 \to \left( - 1 \right) R_1 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & - 1 & - 2 \\ 0 & 3 & 5 \\ 0 & 4 & 7\end{bmatrix} = \begin{bmatrix}- 1 & 0 & 0 \\ 1 & 1 & 0 \\ 3 & 0 & 1\end{bmatrix}A \left[\text{ Applying }R_2 \to R_2 - R_1\text{ and }R_3 \to R_3 - 3 R_1 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & - 1 & - 2 \\ 0 & 1 & \frac{5}{3} \\ 0 & 4 & 7\end{bmatrix} = \begin{bmatrix}- 1 & 0 & 0 \\ \frac{1}{3} & \frac{1}{3} & 0 \\ 3 & 0 & 1\end{bmatrix}A \left[\text{ Applying }R_2 \to \frac{1}{3} R_2 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & \frac{- 1}{3} \\ 0 & 1 & \frac{5}{3} \\ 0 & 0 & \frac{1}{3}\end{bmatrix} = \begin{bmatrix}\frac{- 2}{3} & \frac{1}{3} & 0 \\ \frac{1}{3} & \frac{1}{3} & 0 \\ \frac{5}{3} & - \frac{4}{3} & 1\end{bmatrix}A \left[\text{ Applying }R_3 \to R_3 - 4 R_2\text{ and }R_1 \to R_1 + R_2 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & \frac{- 1}{3} \\ 0 & 1 & \frac{5}{3} \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}\frac{- 2}{3} & \frac{1}{3} & 0 \\ \frac{1}{3} & \frac{1}{3} & 0 \\ 5 & - 4 & 3\end{bmatrix}A \left[\text{ Applying }R_3 \to 3 R_3 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}1 & - 1 & 1 \\ - 8 & 7 & - 5 \\ 5 & - 4 & 3\end{bmatrix}A \left[\text{ Applying }R_2 \to R_2 - \frac{5}{3} R_3\text{ and }R_1 \to R_1 + \frac{1}{3} R_3 \right]\]
\[\text{ Hence, }A^{- 1} = \begin{bmatrix}1 & - 1 & 1 \\ - 8 & 7 & - 5 \\ 5 & - 4 & 3\end{bmatrix} .\]
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