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प्रश्न
पर्याय
\[\begin{bmatrix}x^{- 1} & 0 & 0 \\ 0 & y^{- 1} & 0 \\ 0 & 0 & z^{- 1}\end{bmatrix}\]
\[xyz \begin{bmatrix}x^{- 1} & 0 & 0 \\ 0 & y^{- 1} & 0 \\ 0 & 0 & z^{- 1}\end{bmatrix}\]
\[\frac{1}{xyz}\begin{bmatrix}x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{bmatrix}\]
\[\frac{1}{xyz} \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\]
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उत्तर
\[\begin{bmatrix}x^{- 1} & 0 & 0 \\ 0 & y^{- 1} & 0 \\ 0 & 0 & z^{- 1}\end{bmatrix}\]
\[A = IA\]
\[ \Rightarrow \begin{bmatrix}x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}A\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}x^{- 1} & 0 & 0 \\ 0 & y^{- 1} & 0 \\ 0 & 0 & z^{- 1}\end{bmatrix} A .................\left[\text{ Applying }R_1 = \frac{1}{x} R_1 , R_2 = \frac{1}{y} R_2\text{ and }R_3 = \frac{1}{z} R_3 \right]\]
\[ \Rightarrow A^{- 1} = \begin{bmatrix}x^{- 1} & 0 & 0 \\ 0 & y^{- 1} & 0 \\ 0 & 0 & z^{- 1}\end{bmatrix}\]
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