If $\omega\$is a complex cube root of unity, then $\left| \begin{matrix} 1 & \omega & \omega^{2}\ \ \omega & \omega^{2} & 1 \ \omega^{2} & 1 & \omega \end{matrix} ight|$ is equal to

Question

If $\omega\$is a complex cube root of unity, then $\left| \begin{matrix} 1 & \omega & \omega^{2}\ \ \omega & \omega^{2} & 1 \ \omega^{2} & 1 & \omega \end{matrix} ight|$ is equal to

Dr. Abubacker · Accepted Answer

(c)
Let $A = \left| \begin{matrix}
1 & \omega & \omega^{2}\  \
\omega & \omega^{2} & 1 \
\omega^{2} & 1 & \omega
\end{matrix} ight| = \left| \begin{matrix}
1 + \omega + \omega^{2} & \omega & \omega^{2}\  \
1 + \omega + \omega^{2} & \omega^{2} & 1 \
1 + \omega + \omega^{2} & 1 & \omega
\end{matrix} ight|$
[$C_{1} ightarrow C_{1} + C_{2} + C_{3}$
$= \left| \begin{matrix}
0 & \omega & \omega^{2}\  \
0 & \omega^{2} & 1 \
o & 1 & \omega
\end{matrix} ight| = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \lbrack\because 1 + \omega + \omega^{2} = 0brack$

Q. If $\omega\$is a complex cube root of unity, then $\left| \begin{matrix} 1 & \omega & \omega^{2}\ \\ \omega & \omega^{2} & 1 \\ \omega^{2} & 1 & \omega \end{matrix} \right|$ is equal to

Explanation

Reviewed by Dr. Abubacker

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