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JCECE Engineering JCECE Engineering Solved Paper-2006

  • question_answer
    If \[(\omega \ne 1)\] is a cubic root of unity, then\[\left| \begin{matrix}    1 & 1+1+{{\omega }^{2}} & {{\omega }^{2}}  \\    1-i & -1 & {{\omega }^{2}}-1  \\    -i & -1+\omega -i & -1  \\ \end{matrix} \right|\]equals:

    A) \[zero\]                                               

    B) \[1\]

    C) \[i\]                                       

    D) \[\omega \]

    Correct Answer: D

    Solution :

    Applying \[{{R}_{1}}\to {{R}_{1}}+{{R}_{3}}\], we obtain                 \[\left| \begin{matrix}    1-i & {{\omega }^{2}}+\omega  & {{\omega }^{2}}-1  \\    1-i & -1 & {{\omega }^{2}}-1  \\    -i & -1+\omega -i & -1  \\ \end{matrix} \right|=0\] \[\because \]\[{{\omega }^{2}}+\omega =-1\]which \[{{R}_{1}}\] and \[{{R}_{2}}\] become identical.


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