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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2008

  • question_answer
    If the matrix M, is given by \[{{M}_{r}}=\left[ \begin{matrix}    r & r-1  \\    r-1 & r  \\ \end{matrix} \right],r=1,2,3,....,\]then the value of\[\det ({{M}_{1}})+\det ({{M}_{2}})+....+\det ({{M}_{2008}})\] is

    A)  2007                     

    B)         2008

    C)  \[{{(2008)}^{2}}\]        

    D)         \[{{(2007)}^{2}}\]

    E)  \[2009\]

    Correct Answer: C

    Solution :

    Given, \[{{M}_{r}}=\left[ \begin{matrix}    r & r-1  \\    r-1 & r  \\ \end{matrix} \right]\] \[\therefore \]\[\det ({{M}_{r}})={{r}^{2}}-{{(r-1)}^{2}}=2r-1\] \[\therefore \]\[\det ({{M}_{1}})+\det ({{M}_{2}})+....+\det ({{M}_{2008}})\] \[=1+3+5+...+4015\] \[=\frac{2008}{2}[2+(2008-1)2]\] \[=2008(2008)={{(2008)}^{2}}\]


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