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प्रश्न
If \[A = \begin{bmatrix}- 1 & 2 & 0 \\ - 1 & 1 & 1 \\ 0 & 1 & 0\end{bmatrix}\] , show that \[A^2 = A^{- 1} .\]
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उत्तर
\[\text{ We have,} A = \begin{bmatrix} - 1 & 2 & 0\\- 1 & 1 & 1\\ 0 & 1 & 0 \end{bmatrix}\]
Now,
\[ A^2 = \begin{bmatrix} - 1 & 2 & 0\\ - 1 & 1 & 1\\ 0 & 1 & 0 \end{bmatrix}\begin{bmatrix} - 1 & 2 & 0\\ - 1 & 1 & 1\\ 0 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 - 2 + 0 & - 2 + 2 + 0 & 0 + 2 + 0\\1 - 1 + 0 & - 2 + 1 + 1 & 0 + 1 + 0\\ 0 - 1 + 0 & 0 + 1 + 0 & 0 + 1 + 0 \end{bmatrix} = \begin{bmatrix} - 1 & 0 & 2\\ 0 & 0 & 1\\ - 1 & 1 & 1 \end{bmatrix}\]
\[\text{ and }A^2 \times A = \begin{bmatrix} - 1 & 0 & 2\\ 0 & 0 & 1\\ - 1 & 1 & 1 \end{bmatrix}\begin{bmatrix} - 1 & 2 & 0\\ - 1 & 1 & 1\\ 0 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 + 0 + 0 & - 2 + 0 + 2 & 0\\0 + 0 + 0 & 0 + 0 + 1 & 0\\ 0 & - 2 + 1 + 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} = I_3 [\text{ Identity matrix of order 3 }]\]
\[ \Rightarrow A^2 \times A = I_3 \]
\[ \Rightarrow A^2 = A^{- 1}\]
Now,
\[ A^2 = \begin{bmatrix} - 1 & 2 & 0\\ - 1 & 1 & 1\\ 0 & 1 & 0 \end{bmatrix}\begin{bmatrix} - 1 & 2 & 0\\ - 1 & 1 & 1\\ 0 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 - 2 + 0 & - 2 + 2 + 0 & 0 + 2 + 0\\1 - 1 + 0 & - 2 + 1 + 1 & 0 + 1 + 0\\ 0 - 1 + 0 & 0 + 1 + 0 & 0 + 1 + 0 \end{bmatrix} = \begin{bmatrix} - 1 & 0 & 2\\ 0 & 0 & 1\\ - 1 & 1 & 1 \end{bmatrix}\]
\[\text{ and }A^2 \times A = \begin{bmatrix} - 1 & 0 & 2\\ 0 & 0 & 1\\ - 1 & 1 & 1 \end{bmatrix}\begin{bmatrix} - 1 & 2 & 0\\ - 1 & 1 & 1\\ 0 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 + 0 + 0 & - 2 + 0 + 2 & 0\\0 + 0 + 0 & 0 + 0 + 1 & 0\\ 0 & - 2 + 1 + 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} = I_3 [\text{ Identity matrix of order 3 }]\]
\[ \Rightarrow A^2 \times A = I_3 \]
\[ \Rightarrow A^2 = A^{- 1}\]
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