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

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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