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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2005

  • question_answer
    If \[A=\left[ \begin{matrix}    3 & 5  \\    2 & 0  \\ \end{matrix} \right]\] and \[B=\left[ \begin{matrix}    1 & 17  \\    0 & -10  \\ \end{matrix} \right]\], then \[|AB|\] is equal to :

    A)  80                                         

    B)  100

    C)  - 110                                    

    D)  92

    Correct Answer: B

    Solution :

    \[A=\left[ \begin{matrix}    3 & 5  \\    2 & 0  \\ \end{matrix} \right]\] and \[B=\left[ \begin{matrix}    1 & 17  \\    0 & -10  \\ \end{matrix} \right]\] \[\therefore \]  \[AB=\left[ \begin{matrix}    3 & 5  \\    2 & 0  \\ \end{matrix} \right]\,\,\left[ \begin{matrix}    1 & 17  \\    0 & -10  \\ \end{matrix} \right]\]                 \[=\left[ \begin{matrix}    3+0 & 51-50  \\    2+0 & 34-0  \\ \end{matrix} \right]=\left[ \begin{matrix}    3 & 1  \\    2 & 34  \\ \end{matrix} \right]\] \[\Rightarrow \]               \[|AB|=\left| \begin{matrix}    1 & 1  \\    2 & 34  \\ \end{matrix} \right|\]                 = 102 - 2                 = 100


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