If $\omega$ is the 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 the 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

(b)
$\ \left| \begin{matrix}
1 & \omega & \omega^{2} \
\omega & \omega^{2} & 1 \
\omega^{2} & 1 & \omega
\end{matrix} ight|$
Applying $C_{1} ightarrow C_{1} + C_{2} + C_{3}$
$= \left| \begin{matrix}
1 + \omega + \omega^{2} & \omega & \omega^{2} \
1 + \omega + \omega^{2} & \omega^{2} & 1 \
1 + \omega + \omega^{2} & 1 & \omega
\end{matrix} ight|$ $(\because 1 + \omega + \omega^{2} = 0)$
$= \left| \begin{matrix}
0 & \omega & \omega^{2} \
0 & \omega^{2} & 1 \
0 & 1 & \omega
\end{matrix} ight|$
$= 0$

Q. If $\omega$ is the 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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