If $(\omega
eq 1)$ is a cubic root of unity, then $\left| \begin{matrix} 1 & 1 + i + \omega^{2} & \omega^{2} \ 1 - i & - 1 & \omega^{2} - 1 \ - i & - 1 + \omega - i & - 1 \end{matrix} ight|$

Question

If $(\omega 
eq 1)$ is a cubic root of unity, then $\left| \begin{matrix} 1 & 1 + i + \omega^{2} & \omega^{2} \ 1 - i & - 1 & \omega^{2} - 1 \ - i & - 1 + \omega - i & - 1 \end{matrix} ight|$

Dr. Abubacker · Accepted Answer

(a)
Applying $R_{1} ightarrow R_{1} + R_{3},$we get
$\left| \begin{matrix}
1 - i & \omega^{2} + \omega & \omega^{2} - 1 \
1 - i & - 1 & \omega^{2} - 1 \
 - i & - 1 + \omega - i & - 1
\end{matrix} ight|$=0
$\lbrack	herefore\omega^{2} + \omega = - 1$, so $R_{1}$and $R_{2}$become identical]

Q. If $(\omega \neq 1)$ is a cubic root of unity, then $\left| \begin{matrix} 1 & 1 + i + \omega^{2} & \omega^{2} \\ 1 - i & - 1 & \omega^{2} - 1 \\ - i & - 1 + \omega - i & - 1 \end{matrix} \right|$

Explanation

Reviewed by Dr. Abubacker

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