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

Question

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

Dr. Abubacker · Accepted Answer

(d)
$\left| \begin{matrix}
x + 1 & \omega & \omega^{2} \
\omega & x + \omega^{2} & 1 \
\omega^{2} & 1 & x + \omega
\end{matrix} ight|$
$= \left| \begin{matrix}
x + 1 + \omega + \omega^{2} & \omega & \omega^{2} \
x + 1 + \omega + \omega^{2} & x + \omega^{2} & 1 \
x + 1 + \omega + \omega^{2} & 1 & x + \omega
\end{matrix} ight|$ $C_{1} ightarrow C_{1} + C_{2} + C_{3})$
$= x\left| \begin{matrix}
1 & \omega & \omega^{2} \
1 & x + \omega^{2} & 1 \
1 & 1 & x + \omega
\end{matrix} ight|$ $\ (\because 1 + \omega + \omega^{2} = 0$
$= x\lbrack 1\{\left( x + \omega^{2} ight)(x + \omega) - 1\} + \omega\{ 1 -$($\ x + \omega)\} + \omega^{2}\{ 1 - \left( x + \omega^{2} ight)\}brack$
$= x\lbrack(x^{2} + \omega x + \omega^{2}x + \omega^{3} - 1 + \omega - \omega x - \omega^{2} + \omega^{2} - \omega^{2}x - \omega^{4}brack$
${= x}^{3}$ $\ (\because\omega^{3} = 1)$

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

Explanation

Reviewed by Dr. Abubacker

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