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JEE Main & Advanced AIEEE Solved Paper-2003

  • question_answer
    If \[1,\,\omega ,\,{{\omega }^{2}}\]are the cube roots of unity, then                 \[\Delta =\left| \begin{matrix}    1 & {{\omega }^{n}} & {{\omega }^{2n}}  \\    {{\omega }^{n}} & {{\omega }^{2n}} & 1  \\    {{\omega }^{2n}} & 1 & {{\omega }^{n}}  \\ \end{matrix} \right|\] is equal to     AIEEE  Solved  Paper-2003

    A)                         0                             

    B) 1                             

    C) \[\omega \]                       

    D)       \[{{\omega }^{2}}\]

    Correct Answer: A

    Solution :

    Given, \[\Delta =\left| \begin{matrix}    1 & {{\omega }^{n}} & {{\omega }^{2n}}  \\    {{\omega }^{n}} & {{\omega }^{2n}} & 1  \\    {{\omega }^{2n}} & 1 & {{\omega }^{n}}  \\ \end{matrix} \right|\] \[=1\,({{\omega }^{3n}}-1)-{{\omega }^{n}}({{\omega }^{2n}}-{{\omega }^{2n}})+{{\omega }^{2n}}({{\omega }^{n}}-{{\omega }^{4n}})\] \[=1(1-1)-0+{{\omega }^{2n}}({{\omega }^{n}}-{{\omega }^{n}})=0\]          \[(\because {{\omega }^{3}}=1)\]


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