2. Solved Papers
  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    If\[A=\left| \begin{matrix}    1 & 0 & 0  \\    x & 1 & 0  \\    x & x & 1  \\ \end{matrix} \right|\]and\[I=\left| \begin{matrix}    1 & 0 & 0  \\    0 & 1 & 0  \\    0 & 0 & 1  \\ \end{matrix} \right|,\]then \[{{A}^{3}}-4{{A}^{2}}+3A+I\]is equal to

    A)  \[3I\]                                   

    B)  \[I\]

    C)  \[-I\]                    

    D)         \[-2I\]

    E)  \[2I\]

    Correct Answer: B

    Solution :

    Given, \[A=\left| \begin{matrix}    1 & 0 & 0  \\    x & 1 & 0  \\    x & x & 1  \\ \end{matrix} \right|\Rightarrow A=1\] \[\therefore \] \[{{A}^{3}}-4{{A}^{2}}+3A+I={{(1)}^{3}}-4{{(1)}^{3}}+3(1)+I\]                 \[=I\]


You need to login to perform this action.
You will be redirected in 3 sec spinner