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J & K CET Engineering J and K - CET Engineering Solved Paper-2006

  • question_answer
    If \[A\left[ \begin{matrix}    a & b & 0  \\    -b & a & 0  \\    0 & 0 & 1  \\ \end{matrix} \right],\] where \[{{a}^{2}}+{{b}^{2}}=1,\] then adj  is equal to                      

    A)  \[{{A}^{-1}}\]           

    B)  \[{{A}^{T}}\]

    C)  \[A\]             

    D)  \[-A\] (Here, \[{{A}^{T}}\] is the transpose of A) 

    Correct Answer: A

    Solution :

    Given,  \[A=\left[ \begin{matrix}    a & b & 0  \\    -b & a & 0  \\    0 & 0 & 1  \\ \end{matrix} \right]\] \[\Rightarrow \] \[|A|={{a}^{2}}+{{b}^{2}}=1\] \[\therefore \]  \[{{A}^{-1}}=\frac{1}{|A|}adj\,(A)=adj\,(A)\]


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