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JEE Main & Advanced Mathematics Determinants & Matrices Question Bank Special types of matrices, Transpose, Adjoint and Inverse of matrices

  • question_answer
    If \[P=\left[ \begin{matrix}    \frac{\sqrt{3}}{2} & \frac{1}{2}  \\    -\frac{1}{2} & \frac{\sqrt{3}}{2}  \\ \end{matrix} \right],\,A=\left[ \begin{matrix}    1 & 1  \\    0 & 1  \\ \end{matrix} \right]\] and \[Q=PA{{P}^{T}}\], then \[P({{Q}^{2005}}){{P}^{T}}\] equal to [IIT Screening 2005]

    A) \[\left[ \begin{matrix}    1 & 2005  \\    0 & 1  \\ \end{matrix} \right]\]

    B) \[\left[ \begin{matrix}    \sqrt{3}/2 & 2005  \\    1 & 0  \\ \end{matrix} \right]\]

    C) \[\left[ \begin{matrix}    1 & 2005  \\    \sqrt{3}/2 & 1  \\ \end{matrix} \right]\]

    D) \[\left[ \begin{matrix}    1 & \sqrt{3}/2  \\    0 & 2005  \\ \end{matrix} \right]\]

    Correct Answer: A

    Solution :

    If   \[Q=PA{{P}^{T}}\]      \[{{P}^{T}}Q=A{{P}^{T}}\], \[(\text{as}\,\,P{{P}^{T}}=I)\] \[{{P}^{T}}{{Q}^{2005}}P=A{{P}^{T}}{{Q}^{2004}}P\]        \[={{A}^{2}}{{P}^{T}}{{Q}^{2003}}P\]\[={{A}^{3}}{{P}^{T}}{{Q}^{2002}}P\]\[={{A}^{2004}}{{P}^{T}}(QP)\] \[={{A}^{2004}}{{P}^{T}}(PA)\]\[(Q=PA{{P}^{T}}\Rightarrow QP=PA)\]\[={{A}^{2005}}\] Þ \[{{A}^{2005}}=\left[ \begin{matrix}    1 & 2005  \\    0 & 1  \\ \end{matrix} \right]\].


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