If $n = 3k\ and\ 1\ \omega,\ \omega^{2}$ are the cube roots of unity, then $\mathrm{\Delta} = \left| \begin{matrix} 1 & \omega^{n} & \omega^{2n} \ \omega^{2n} & 1 & \omega^{n} \ \omega^{n} & \omega^{2n} & 1 \end{matrix} ight|$ has the value

Question

If $n = 3k\ and\ 1\ \omega,\ \omega^{2}$ are the cube roots of unity, then $\mathrm{\Delta} = \left| \begin{matrix} 1 & \omega^{n} & \omega^{2n} \ \omega^{2n} & 1 & \omega^{n} \ \omega^{n} & \omega^{2n} & 1 \end{matrix} ight|$ has the value

Dr. Abubacker · Accepted Answer

(a)
Given that, $\mathrm{\Delta} = \left| \begin{matrix}
1 & \omega^{n} & \omega^{2n} \
\omega^{2n} & 1 & \omega^{n} \
\omega^{n} & \omega^{2n} & 1
\end{matrix} ight|$
Applying $C_{1} ightarrow C_{1} + C_{2} + C_{3}$
$= \left| \begin{matrix}
1 + \omega^{n} + \omega^{2n} & \omega^{n} & \omega^{2n} \
1 + \omega^{n} + \omega^{2n} & 1 & \omega^{n} \
1 + \omega^{n} + \omega^{2n} & \omega^{2n} & 1
\end{matrix} ight|$
$= \left| \begin{matrix}
0 & \omega^{n} & \omega^{2n} \
0 & 1 & \omega^{n} \
0 & \omega^{2n} & 1
\end{matrix} ight|$
$(\because\$If $n$ multiple of 3, then $1 + \omega^{n} + \omega^{2n} = 0$)
$= 0$

Q. If $n = 3k\ and\ 1\ \omega,\ \omega^{2}$ are the cube roots of unity, then $\mathrm{\Delta} = \left| \begin{matrix} 1 & \omega^{n} & \omega^{2n} \\ \omega^{2n} & 1 & \omega^{n} \\ \omega^{n} & \omega^{2n} & 1 \end{matrix} \right|$ has the value

Explanation

Reviewed by Dr. Abubacker

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