2. Solved Papers
  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2011

  • question_answer
    If\[A=\left[ \begin{matrix}    2 & 1  \\    0 & x  \\ \end{matrix} \right]\]and\[{{A}^{-1}}=\left[ \begin{matrix}    \frac{1}{2} & \frac{1}{6}  \\    0, & \frac{1}{x}  \\ \end{matrix} \right]\],then the value of\[x\]is equal to

    A)  \[-3\]                                   

    B) \[3\]     

    C)  \[-2\]                   

    D)        \[6\]

    E)  \[-6\]

    Correct Answer: A

    Solution :

    \[A=\left[ \begin{matrix}    2 & 1  \\    0 & x  \\ \end{matrix} \right]{{A}^{-1}}=\left[ \begin{matrix}    1/2 & 1/6  \\    0 & 1/x  \\ \end{matrix} \right]\] \[\Rightarrow \] \[|A|=2x\] And adj \[(A)=\left[ \begin{matrix}    x & -1  \\    0 & 2  \\ \end{matrix} \right]\] Now, \[{{A}^{-1}}=\frac{adj(A)}{|A|}\] \[\Rightarrow \]               \[\left[ \begin{matrix}    1/2 & 1/6  \\    0 & 1/x  \\ \end{matrix} \right]=\frac{1}{2x}\left[ \begin{matrix}    x & -1  \\    0 & 2  \\ \end{matrix} \right]\]                                 \[=\left[ \begin{matrix}    1/2 & -1/2x  \\    0 & 1/x  \\ \end{matrix} \right]\] On comparing, \[-1/2x=1/6\] \[\Rightarrow \]               \[x=-3\]


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