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