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JEE Main & Advanced Mathematics Determinants & Matrices Question Bank Self Evaluation Test - Matrices

  • question_answer
    If \[P\left[ \begin{matrix}    \cos (\pi /6) & \sin (\pi /6)  \\    -\sin (\pi /6) & \cos (\pi /6)  \\ \end{matrix} \right],A=\left[ \begin{matrix}    1 & 1  \\    0 & 1  \\ \end{matrix} \right]\] and \[Q=PAP'\] then \[P'{{Q}^{2007}}P\] is equal to

    A) \[\left[ \begin{matrix}    1 & 2007  \\    0 & 1  \\ \end{matrix} \right]\]

    B) \[\left[ \begin{matrix}    1 & \sqrt{3}/2  \\    0 & 2007  \\ \end{matrix} \right]\]

    C) \[\left[ \begin{matrix}    \sqrt{3}/2 & 2007  \\    0 & 1  \\ \end{matrix} \right]\]

    D) \[\left[ \begin{matrix}    \sqrt{3}/2 & -1/2  \\    1 & 2007  \\ \end{matrix} \right]\]

    Correct Answer: A

    Solution :

    [a] Note that \[P'={{P}^{-1}}\] Now, \[Q=PAP'=PA{{P}^{-1}}\] \[\Rightarrow {{Q}^{2007}}=P{{A}^{2007}}{{P}^{-1}}\] \[\therefore P'\,{{Q}^{2007}}P={{P}^{-1}}(P{{A}^{2007}}{{P}^{-}}^{1})P\]             \[={{A}^{2007}}={{(I+B)}^{2007}}\] Where \[B=\left[ \begin{matrix}    0 & 1  \\    0 & 0  \\ \end{matrix} \right]\]. As \[{{B}^{2}}=0\], we get \[{{B}^{r}}=0\forall r\ge 2\]. Thus, by binomial theorem, \[{{A}^{2007}}=I+2007B=\left[ \begin{matrix}    1 & 2007  \\    0 & 1  \\ \end{matrix} \right]\]


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