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प्रश्न
If w is a complex cube root of unity, show that
`([[1 w w^2],[w w^2 1],[w^2 1 w]]+[[w w^2 1],[w^2 1 w],[w w^2 1]])[[1],[w],[w^2]]=[[0],[0],[0]]`
उत्तर
\[Here, \]
\[LHS = \left( \begin{bmatrix}1 & w & w^2 \\ w & w^2 & 1 \\ w^2 & 1 & w\end{bmatrix} + \begin{bmatrix}w & w^2 & 1 \\ w^2 & 1 & w \\ w & w^2 & 1\end{bmatrix} \right)\begin{bmatrix}1 \\ w \\ w^2\end{bmatrix}\]
\[ = \begin{bmatrix}1 + w & w + w^2 & w^2 + 1 \\ w + w^2 & w^2 + 1 & 1 + w \\ w^2 + w & 1 + w^2 & w + 1\end{bmatrix}\begin{bmatrix}1 \\ w \\ w^2\end{bmatrix}\]
\[ = \begin{bmatrix}- w^2 & - 1 & - w \\ - 1 & - w & - w^2 \\ - 1 & - w & - w^2\end{bmatrix}\begin{bmatrix}1 \\ w \\ w^2\end{bmatrix} \] `(∴ 1 + w +w^2 = 0 and w^3 =1 )`
\[ = \begin{bmatrix}- w^2 - w - w^3 \\ - 1 - w^2 - w^4 \\ - 1 - w^2 - w^4\end{bmatrix}\]
\[ = \begin{bmatrix}- w\left( 1 + w + w^2 \right) \\ - 1 - w^2 - w^3 w \\ - 1 - w^2 - w^3 w\end{bmatrix}\]
`[ (-w xx 0) , (-1 -w -w) ,( -1 -w^2 -w)]` ` (∵ 1 + w + w^2 = 0 and w^3 = 1)`
\[ = \begin{bmatrix}0 \\ - 0 \\ - 0\end{bmatrix}\]
\[ = \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}\]
\[ \therefore \left( \begin{bmatrix}1 & w & w^2 \\ w & w^2 & 1 \\ w^2 & 1 & w\end{bmatrix} + \begin{bmatrix}w & w^2 & 1 \\ w^2 & 1 & w \\ w & w^2 & 1\end{bmatrix} \right)\begin{bmatrix}1 \\ w \\ w^2\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}\]
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