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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2008

  • question_answer
    Let\[A=\left[ \begin{matrix}    {{\alpha }^{2}} & 5  \\    5 & -\alpha   \\ \end{matrix} \right]\]and\[|{{A}^{10}}\text{ }\!\!|\!\!\text{ }=1024,\]then a is equal to

    A)  2                                            

    B)  \[-2\]

    C)  3                            

    D)         \[-3\]

    E)  \[-5\]

    Correct Answer: D

    Solution :

    Since,        \[A=\left[ \begin{matrix}    {{\alpha }^{2}} & 5  \\    5 & -\alpha   \\ \end{matrix} \right]\] Now,     \[|A|=-{{a}^{3}}-25\] Also,     \[|{{A}^{10}}|=1024\] \[\Rightarrow \]               \[|{{A}^{10}}|={{2}^{10}}\] \[\Rightarrow \]               \[{{(-{{\alpha }^{3}}-25)}^{10}}={{2}^{10}}\] \[\Rightarrow \]               \[-{{\alpha }^{3}}-25=2\] \[\Rightarrow \]               \[-{{\alpha }^{3}}=27\] \[\Rightarrow \]               \[\alpha =-3\]


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