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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2008

  • question_answer
    If \[\omega \ne 1\]is a cube root of unity, then the value of \[\left| \begin{matrix}    1+2{{\omega }^{100}}+{{\omega }^{200}}  \\    1  \\    \omega   \\ \end{matrix}\begin{matrix}    {{\omega }^{2}}  \\    1+{{\omega }^{100}}+2{{\omega }^{200}}  \\    {{\omega }^{2}}  \\ \end{matrix}\begin{matrix}    1  \\    \omega   \\    1+{{\omega }^{100}}+2{{\omega }^{200}}  \\ \end{matrix} \right|\] is equal to

    A) \[0\]                                    

    B) \[1\]                    

    C) \[\omega \]                      

    D) \[{{\omega }^{2}}\]

    E) \[1+\omega \]

    Correct Answer: A

    Solution :

    Let\[\Delta =\left| \begin{matrix}    1+2{{\omega }^{100}}+{{\omega }^{200}} & {{\omega }^{2}}  \\    1 & 1+{{\omega }^{100}}+2{{\omega }^{200}}  \\    \omega  & {{\omega }^{2}}  \\ \end{matrix} \right.\] \[\left. \begin{matrix}    1  \\    \omega   \\    2+{{\omega }^{100}}+{{\omega }^{200}}  \\ \end{matrix} \right|\] \[=\left| \begin{matrix}    1+2\omega +{{\omega }^{2}} & {{\omega }^{2}} & 1  \\    1 & 1+\omega +2{{\omega }^{2}} & \omega   \\    \omega  & {{\omega }^{2}} & 2+\omega +{{\omega }^{2}}  \\ \end{matrix} \right|\] \[=\left| \begin{matrix}    \omega  & {{\omega }^{2}} & 1  \\    1 & {{\omega }^{2}} & \omega   \\    \omega  & {{\omega }^{2}} & 1  \\ \end{matrix} \right|\] \[=0\]                           (\[\because \]Rows\[{{R}_{1}}\]and\[{{R}_{3}}\]are identical)


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