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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2000

  • question_answer
    The  value   of  the  determinant\[\left| \begin{matrix}    1 & {{\omega }^{2}} & {{\omega }^{5}}  \\    {{\omega }^{2}} & 1 & {{\omega }^{4}}  \\    {{\omega }^{5}} & {{\omega }^{4}} & 1  \\ \end{matrix} \right|\]where \[\omega \]is an  imaginary cube root of unity is:

    A)  \[{{(1-\omega )}^{2}}\]               

    B)  3

    C)  \[-3\]                                   

    D)  zero

    Correct Answer: B

    Solution :

     We have \[\left| \begin{matrix}    1 & {{\omega }^{2}} & {{\omega }^{5}}  \\    {{\omega }^{3}} & 1 & {{\omega }^{4}}  \\    {{\omega }^{5}} & {{\omega }^{4}} & 1  \\ \end{matrix} \right|=\left| \begin{matrix}    1 & 1 & {{\omega }^{2}}  \\    1 & 1 & {{\omega }^{2}}  \\    {{\omega }^{2}} & \omega  & 1  \\ \end{matrix} \right|\] \[=2{{\omega }^{3}}-{{\omega }^{2}}-{{\omega }^{4}}=2-{{\omega }^{2}}-\omega \] \[=2-(\omega +{{\omega }^{2}})\] \[=2-(-1)=3\]


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