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Find the Inverse by Using Elementary Row Transformations: ⎡ ⎢ ⎣ 2 − 1 4 4 0 7 3 − 2 7 ⎤ ⎥ ⎦ - Mathematics

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प्रश्न

Find the inverse by using elementary row transformations:

\[\begin{bmatrix}2 & - 1 & 4 \\ 4 & 0 & 7 \\ 3 & - 2 & 7\end{bmatrix}\]

योग

उत्तर

\[A = \begin{bmatrix}2 & - 1 & 4 \\ 4 & 0 & 2 \\ 3 & - 2 & 7\end{bmatrix}\]
We know
\[A = IA \]
\[ \Rightarrow \begin{bmatrix}2 & - 1 & 4 \\ 4 & 0 & 2 \\ 3 & - 2 & 7\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} A\]
\[ \Rightarrow \begin{bmatrix}1 & - \frac{1}{2} & 2 \\ 4 & 0 & 2 \\ 3 & - 2 & 7\end{bmatrix} = \begin{bmatrix}\frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} A \left[\text{ Applying } R_1 \to \frac{1}{2} R_1 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & - \frac{1}{2} & 2 \\ 0 & 2 & - 6 \\ 0 & - \frac{1}{2} & 1\end{bmatrix} = \begin{bmatrix}\frac{1}{2} & 0 & 0 \\ - 2 & 1 & 0 \\ \frac{- 3}{2} & 0 & 1\end{bmatrix}A \left[\text{ Applying }R_2 \to R_2 - 4 R_1\text{ and }R_3 \to R_3 - 3 R_1 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & - \frac{1}{2} & 2 \\ 0 & 1 & - 3 \\ 0 & - \frac{1}{2} & 1\end{bmatrix} = \begin{bmatrix}\frac{1}{2} & 0 & 0 \\ - 1 & \frac{1}{2} & 0 \\ \frac{- 3}{2} & 0 & 1\end{bmatrix} A \left[\text{ Applying }R_2 \to \frac{1}{2} R_2 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & \frac{1}{2} \\ 0 & 1 & - 3 \\ 0 & 0 & - \frac{1}{2}\end{bmatrix} = \begin{bmatrix}0 & \frac{1}{4} & 0 \\ - 1 & \frac{1}{2} & 0 \\ - 2 & \frac{1}{4} & 1\end{bmatrix} A \left[\text{ Applying }R_1 \to R_1 + \frac{1}{2} R_2\text{ and }R_3 \to R_3 + \frac{1}{2} R_2 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & \frac{1}{2} \\ 0 & 1 & - 3 \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}0 & \frac{1}{4} & 0 \\ - 1 & \frac{1}{2} & 0 \\ 4 & - \frac{1}{2} & - 2\end{bmatrix} A \left[\text{ Applying }R_3 \to - 2 R_3 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}- 2 & \frac{1}{2} & 1 \\ 11 & - 1 & - 6 \\ 4 & - \frac{1}{2} & - 2\end{bmatrix} A \left[\text{ Applying }R_1 \to R_1 - \frac{1}{2} R_3\text{ and }R_2 \to R_2 + 3 R_3 \right]\]
\[ \Rightarrow A^{- 1} = \begin{bmatrix}- 2 & \frac{1}{2} & 1 \\ 11 & - 1 & - 6 \\ 4 & - \frac{1}{2} & - 2\end{bmatrix} \]

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अध्याय 7: Adjoint and Inverse of a Matrix - Exercise 7.2 [पृष्ठ ३४]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 7 Adjoint and Inverse of a Matrix
Exercise 7.2 | Q 13 | पृष्ठ ३४

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