CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2006

  • question_answer
    The value of the determinant, \[\left| \begin{matrix}    \sqrt{13}+\sqrt{3} & 2\sqrt{5} & \sqrt{5}  \\    15+\sqrt{26} & 5 & \sqrt{10}  \\    3+\sqrt{65} & \sqrt{15} & 5  \\ \end{matrix} \right|\]is:

    A)  \[5(\sqrt{6}-5)\]             

    B)         \[5\sqrt{3}(\sqrt{6}-5)\]

    C)  \[\sqrt{5}(\sqrt{6}-\sqrt{3})\]

    D)         \[\sqrt{2}(\sqrt{7}-\sqrt{5})\]

    E)  \[3(\sqrt{5}-\sqrt{2})\]

    Correct Answer: B

    Solution :

    \[\left| \begin{matrix}    \sqrt{13}+\sqrt{3} & 2\sqrt{5} & \sqrt{5}  \\    \sqrt{15}+\sqrt{26} & 5 & \sqrt{10}  \\    3+\sqrt{65} & \sqrt{15} & 5  \\ \end{matrix} \right|\] \[=\left| \begin{matrix}    \sqrt{13} & 2\sqrt{5} & \sqrt{5}  \\    \sqrt{26} & 5 & \sqrt{10}  \\    \sqrt{65} & \sqrt{15} & 5  \\ \end{matrix} \right|+\left| \begin{matrix}    \sqrt{3} & 2\sqrt{5} & \sqrt{5}  \\    \sqrt{15} & 5 & \sqrt{10}  \\    3 & \sqrt{15} & 5  \\ \end{matrix} \right|\] \[=\sqrt{13}.\sqrt{5}.\sqrt{5}\left| \begin{matrix}    1 & 2 & 1  \\    \sqrt{2} & \sqrt{5} & \sqrt{2}  \\    \sqrt{5} & \sqrt{3} & \sqrt{5}  \\ \end{matrix} \right|\]                 \[\sqrt{3}.\sqrt{5}.\sqrt{5}\left| \begin{matrix}    1 & 2 & 1  \\    \sqrt{5} & \sqrt{5} & \sqrt{2}  \\    \sqrt{3} & \sqrt{3} & \sqrt{5}  \\ \end{matrix} \right|\] \[=0+5\sqrt{3}\left| \begin{matrix}    -1 & 2 & 1  \\    0 & \sqrt{5} & \sqrt{2}  \\    0 & \sqrt{3} & \sqrt{5}  \\ \end{matrix} \right|=5\sqrt{3}(\sqrt{6}-5)\]


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