A) \[x=0\]
B) \[x=1\]
C) \[x=-1\]
D) None of these
Correct Answer: A
Solution :
\[\left| \,\begin{matrix} x+{{\omega }^{2}} & \omega & 1 \\ \omega & {{\omega }^{2}} & 1+x \\ 1 & x+\omega & {{\omega }^{2}} \\ \end{matrix}\, \right|\] = 0 Check at \[x=0,\] we get \[\left| \,\begin{matrix} {{\omega }^{2}} & \omega & 1 \\ \omega & {{\omega }^{2}} & 1 \\ 1 & \omega & {{\omega }^{2}} \\ \end{matrix}\, \right|\] = \[{{\omega }^{2}}({{\omega }^{4}}-\omega )-\omega ({{\omega }^{3}}-1)+1({{\omega }^{2}}-{{\omega }^{2}})\] = \[{{\omega }^{2}}(\omega -\omega )-\omega \,(1-1)+0=0\] Or \[\Delta =\left| \,\begin{matrix} 1+\omega +{{\omega }^{2}}+x & \omega & 1 \\ 1+\omega +{{\omega }^{2}}+x & {{\omega }^{2}} & 1+x \\ 1+\omega +{{\omega }^{2}}+x & x+\omega & {{\omega }^{2}} \\ \end{matrix}\, \right|\] by \[{{C}_{1}}\to {{C}_{1}}+{{C}_{2}}+{{C}_{3}}\] \[=\,\left| \,\begin{matrix} x & \omega & 1 \\ x & {{\omega }^{2}} & 1+x \\ x & x+\omega & {{\omega }^{2}} \\ \end{matrix}\, \right|\], (\[\because \,1+\omega +{{\omega }^{2}}=0\]) = 0, if \[x=0\].
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