A) -1
B) 1
C) 0
D) \[\omega \]
Correct Answer: C
Solution :
Let \[A=\left| \begin{matrix} 1 & \omega & {{\omega }^{2}} \\ \omega & {{\omega }^{2}} & 1 \\ {{\omega }^{2}} & 1 & \omega \\ \end{matrix} \right|\] \[=\left| \begin{matrix} 1+\omega +{{\omega }^{2}} & \omega & {{\omega }^{2}} \\ 1+\omega +{{\omega }^{2}} & {{\omega }^{2}} & 1 \\ 1+\omega +{{\omega }^{2}} & 1 & \omega \\ \end{matrix} \right|\] \[{{C}_{1}}\to {{C}_{1}}+{{C}_{2}}+{{C}_{3}}\] \[=\left| \begin{matrix} 0 & \omega & {{\omega }^{2}} \\ 0 & {{\omega }^{2}} & 1 \\ 0 & 1 & \omega \\ \end{matrix} \right|\] \[[\because \,\,1+\omega +{{\omega }^{2}}=0]\] = 0
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