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  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    If B is a non-singular matrix and A is a square matrix such that\[{{B}^{-1}}AB\]exists, then det\[({{B}^{-1}}AB)\]is equal to

    A)  det\[({{A}^{-1}})\]

    B)                         det\[({{B}^{-1}})\]

    C)  \[\det \,(B)\]                   

    D)  \[\det \,(A)\]

    E)  det \[(A{{B}^{-1}})\]

    Correct Answer: D

    Solution :

    Given,     \[A={{B}^{-1}}AB\] \[\Rightarrow \]               \[BA=AB\] \[\therefore \]  \[\det ({{B}^{-1}}AB)=\det ({{B}^{-1}}BA)\]                                 \[=\det (A)\]


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