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

A) $x^{3}$
B) $2x^{3}$
C) $3x^{3}$
D) None of these
Explanation

(d)
$\left| \begin{matrix}
x + 1 & \omega & \omega^{2} \\
\omega & x + \omega^{2} & 1 \\
\omega^{2} & 1 & x + \omega
\end{matrix} \right|$
$= \ \left| \begin{matrix}
x + 1 + \omega + \omega^{2} & \omega & \omega^{2} \\
\omega + x + \omega^{2} + 1 & x + \omega^{2} & 1 \\
\omega^{2} + 1 + x + \omega & 1 & x + \omega
\end{matrix} \right|$
$(C_{1} \rightarrow C_{1} + C_{2} + C_{3})$
$= x\left| \begin{matrix}
1 & \omega & \omega^{2} \\
1 & x + \omega^{2} & 1 \\
1 & 1 & x + \omega
\end{matrix} \right|\ \ \ \ (\because 1 + \omega + \omega^{2} = 0)$
$= x\left| \begin{matrix}
1 & \omega & \omega^{2} \\
0 & x + \omega^{2} - \omega & 1 - \omega^{2} \\
0 & 1 - \omega & x + \omega - \omega^{2}
\end{matrix} \right|$
$(R_{2} \rightarrow R_{2} - R_{1},R_{3} \rightarrow R_{3} - R_{1})$
$= x\lbrack\left( x + \omega^{2} - \omega \right)\left( x + \omega - \omega^{2} \right) - (1 - \omega)\left( 1 - \omega^{2} \right)\rbrack$
$= x\lbrack x + 3 - 3\rbrack$
$= x^{2}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.