A) \[-4\]
B) \[{{\omega }^{2}}-4\]
C) \[{{\omega }^{2}}\]
D) 4
Correct Answer: B
Solution :
Given, \[{{\omega }^{3}}=1\] and \[1+\omega +{{\omega }^{2}}=0\] \[=\left| \begin{matrix} 1 & {{\omega }^{2}} & 1-{{\omega }^{4}} \\ \omega & 1 & 1+{{\omega }^{2}} \\ 1 & \omega & {{\omega }^{2}} \\ \end{matrix} \right|=\left| \begin{matrix} 1 & {{\omega }^{2}} & 1-\omega \\ \omega & 1 & 1+{{\omega }^{2}} \\ 1 & \omega & {{\omega }^{2}} \\ \end{matrix} \right|\] Expand with respect to \[{{R}_{1}}\] \[=({{\omega }^{2}}-\omega -{{\omega }^{2}})-{{\omega }^{2}}({{\omega }^{3}}-1-{{\omega }^{2}})\] \[+(1-\omega )\,({{\omega }^{2}}-1)\] \[=({{\omega }^{2}}-\omega -1)-{{\omega }^{2}}(1-1-{{\omega }^{2}})\] \[+(1-\omega )\,({{\omega }^{2}}-1)\] \[={{\omega }^{2}}-\omega -1+{{\omega }^{4}}+{{\omega }^{2}}-1-{{\omega }^{3}}+\omega \] \[=2{{\omega }^{2}}-2+\omega -1=2{{\omega }^{2}}+\omega -3\] \[={{\omega }^{2}}+(\omega +{{\omega }^{2}})-3\] \[={{\omega }^{2}}-1-3={{\omega }^{2}}-4\]
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