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JEE Main & Advanced Mathematics Complex Numbers and Quadratic Equations Question Bank De Moivre's theorem and Roots of unity

  • question_answer
    Let \[\Delta =\left| \,\begin{matrix}    1 & \omega  & 2{{\omega }^{2}}  \\    2 & 2{{\omega }^{2}} & 4{{\omega }^{3}}  \\    3 & 3{{\omega }^{3}} & 6{{\omega }^{4}}  \\ \end{matrix}\, \right|\] where \[\omega \] is the cube root of unity, then

    A) \[\Delta =0\]

    B) \[\Delta =1\]

    C) \[\Delta =2\]

    D) \[\Delta =3\]

    Correct Answer: A

    Solution :

    We have \[\Delta =\left| \begin{matrix}    1 & \omega  & 2{{\omega }^{2}}  \\    2 & 2{{\omega }^{2}} & 4{{\omega }^{3}}  \\    3 & 3{{\omega }^{3}} & 6{{\omega }^{4}}  \\ \end{matrix} \right|=2\omega \left| \begin{matrix}    1 & \omega  & \omega   \\    2 & 2{{\omega }^{2}} & 2{{\omega }^{2}}  \\    3 & 3{{\omega }^{3}} & 3{{\omega }^{3}}  \\ \end{matrix} \right|=0\]


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