A) \[x=1\]
B) \[x=\omega \]
C) \[x={{\omega }^{2}}\]
D) \[x=0\]
Correct Answer: D
Solution :
Given that \[\left| \begin{matrix} x+1 & \omega & {{\omega }^{2}} \\ \omega & x+{{\omega }^{2}} & 1 \\ {{\omega }^{2}} & 1 & x+\omega \\ \end{matrix} \right|=0\] Applying transformation\[{{R}_{1}}\to {{R}_{1}}+{{R}_{2}}+{{R}_{3}}\], we get \[x\left| \begin{matrix} 1 & 1 & 1 \\ \omega & x+{{\omega }^{2}} & 1 \\ {{\omega }^{2}} & 1 & x+\omega \\ \end{matrix} \right|=0\] Þ\[(x+{{\omega }^{2}})(x+\omega )-1+{{\omega }^{2}}-\omega (x+\omega )+\omega \] \[-{{\omega }^{2}}(x+{{\omega }^{2}})=0\] Þ \[{{x}^{2}}=0\]Þ\[x=0\] Trick: Putting \[x=0,\] we get \[\left| \begin{matrix} 1 & \omega & {{\omega }^{2}} \\ \omega & {{\omega }^{2}} & 1 \\ {{\omega }^{2}} & 1 & \omega \\ \end{matrix} \right|=0\]
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