If $1,\omega,\omega^{2}$are the cube roots of unity, then $\mathrm{\Delta} = \left| \begin{matrix} 1 & \omega^{n} & \omega^{2n} \ \omega^{n} & \omega^{2n} & 1 \ \omega^{2n} & 1 & \omega^{n} \end{matrix} ight|$ is equal to

Question

If $1,\omega,\omega^{2}$are the cube roots of unity, then $\mathrm{\Delta} = \left| \begin{matrix} 1 & \omega^{n} & \omega^{2n} \ \omega^{n} & \omega^{2n} & 1 \ \omega^{2n} & 1 & \omega^{n} \end{matrix} ight|$ is equal to

Dr. Abubacker · Accepted Answer

(a)
Given, $\mathrm{\Delta} = \left| \begin{matrix}
1 & \omega^{n} & \omega^{2n} \
\omega^{n} & \omega^{2n} & 1 \
\omega^{2n} & 1 & \omega^{n}
\end{matrix} ight|$
$= 1\left( \omega^{3n} - 1 ight) - \omega^{n}\left( \omega^{2n} - \omega^{2n} ight) + \omega^{2n}(\omega^{n} - \omega^{4n})$
$= 1(1 - 1) - 0 + \omega^{2n}\left( \omega^{n} - \omega^{n} ight)\ \ \ \ \ \ \lbrack\because\omega^{3} - 1brack$
$= 0$

Q. If $1,\omega,\omega^{2}$are the cube roots of unity, then $\mathrm{\Delta} = \left| \begin{matrix} 1 & \omega^{n} & \omega^{2n} \\ \omega^{n} & \omega^{2n} & 1 \\ \omega^{2n} & 1 & \omega^{n} \end{matrix} \right|$ is equal to

Explanation

Reviewed by Dr. Abubacker

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