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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2005

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

    A)  zero                                     

    B)  1

    C)  \[i\]                                      

    D)  \[\omega \]

    Correct Answer: A

    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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