If $\omega$ be a complex cube root of unity, then $\left| \begin{matrix} 1 & \omega & - \omega^{2}/2 \ 1 & 1 & 1 \ 1 & - 1 & 0 \end{matrix} ight|$ is equal to

Question

If $\omega$ be a complex cube root of unity, then $\left| \begin{matrix} 1 & \omega & - \omega^{2}/2 \ 1 & 1 & 1 \ 1 & - 1 & 0 \end{matrix} ight|$ is equal to

Dr. Abubacker · Accepted Answer

(a)
$\left| \begin{matrix}
1 & \omega & - \omega^{2}/2 \
1 & 1 & 1 \
1 & - 1 & 0
\end{matrix} ight| = - \frac{1}{2}\left| \begin{matrix}
1 & \omega & \omega^{2} \
1 & 1 & - 2 \
1 & - 1 & 0
\end{matrix} ight|$
$= - \frac{1}{2}\left| \begin{matrix}
1 + \omega + \omega^{2} & \omega & \omega^{2} \
0 & 1 & - 2 \
0 & - 1 & 0
\end{matrix} ight|$ ${(C}_{1} ightarrow C_{1} + C_{2} + C_{3})$
$= - \frac{1}{2}\left| \begin{matrix}
0 & \omega & \omega^{2} \
0 & 1 & - 2 \
0 & - 1 & 0
\end{matrix} ight|$ $(\because 1 + \omega + \omega^{2} = 0)$
$= 0$

Q. If $\omega$ be a complex cube root of unity, then $\left| \begin{matrix} 1 & \omega & - \omega^{2}/2 \\ 1 & 1 & 1 \\ 1 & - 1 & 0 \end{matrix} \right|$ is equal to

Explanation

Reviewed by Dr. Abubacker

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