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JEE Main & Advanced Mathematics Determinants & Matrices Question Bank Mock Test - Matrices

  • question_answer
    If A and B are square matrices of the same order and A is nonsingular, then for a positive integer n, \[{{({{A}^{-1}}BA)}^{n}}\] is equal

    A)  \[{{A}^{-n}}{{B}^{n}}{{A}^{n}}\]

    B)  \[{{A}^{n}}{{B}^{n}}{{A}^{-n}}\]

    C)  \[{{A}^{-1}}{{B}^{n}}A\]   

    D)  \[n({{A}^{-1}}BA)\]

    Correct Answer: C

    Solution :

    [c] \[{{({{A}^{-1}}BA)}^{2}}=({{A}^{-1}}BA)({{A}^{-1}}BA)\] \[={{A}^{-1}}B(A{{A}^{-1}})BA\] \[={{A}^{-1}}BIBA={{A}^{-1}}{{B}^{2}}A\] \[\Rightarrow {{({{A}^{-1}}BA)}^{3}}=({{A}^{-1}}{{B}^{2}}A)({{A}^{-1}}BA)\] \[={{A}^{-1}}{{B}^{2}}(A{{A}^{-1}})BA\] \[={{A}^{-1}}{{B}^{2}}IBA\] \[={{A}^{-1}}{{B}^{3}}A\] and so on \[\Rightarrow {{({{A}^{-1}}BA)}^{n}}={{A}^{-1}}{{B}^{n}}A\]


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