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JEE Main & Advanced JEE Main Solved Paper-2017

  • question_answer
    Let \[\omega \]be a complex number such that  \[2\omega +1=z\]where \[z=\sqrt{-3}.\]If \[\left| \begin{matrix}    1 & 1 & 1  \\    1 & -{{\omega }^{2}}-1 & {{\omega }^{2}}  \\    1 & {{\omega }^{2}} & {{\omega }^{7}}  \\ \end{matrix} \right|=3k,\] then k is equal to:-       JEE Main Solved Paper-2017

    A)  1                                            

    B)  \[-z\]

    C)  z                                            

    D)  -1

    Correct Answer: B

    Solution :

     Here \[\omega \]is complex cube root of unity                 \[{{R}_{1}}\to {{R}_{1}}+{{R}_{2}}+{{R}_{3}}\]                 \[=\left| \begin{matrix}    3 & 0 & 0  \\    1 & -{{\omega }^{2}}-1 & {{\omega }^{2}}  \\    1 & {{\omega }^{2}} & \omega   \\ \end{matrix} \right|\]                 \[=3(-1-\omega -\omega )=-3z\Rightarrow k=-z\]


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