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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2011

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

    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 :

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


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