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VIT Engineering VIT Engineering Solved Paper-2010

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

    A)  \[\det \,({{A}^{-1}})\]

    B)  \[\det \,({{B}^{-1}})\]

    C)  \[\det \,(A)\]

    D)  \[\det \,(B)\]

    Correct Answer: C

    Solution :

    \[\det \,({{B}^{-1}}AB)\] \[=\det \,({{B}^{-1}})\,detA\,\det B\] \[=\det \,({{B}^{-1}})\,\cdot \det B\,\cdot \det A\] \[=\det \,({{B}^{-1}}B)\,\det A\] \[=\det I\cdot \det A\] \[=1\cdot \det A=\det A\]


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