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J & K CET Engineering J and K - CET Engineering Solved Paper-2007

  • question_answer
    \[A=\left[ \begin{matrix}    \cos \alpha  & -\sin \alpha  & 0  \\    \sin \alpha  & \cos \alpha  & 0  \\    0 & 0 & 1  \\ \end{matrix} \right],\] then \[{{A}^{-1}}\] is

    A)  \[A\]

    B)  \[-A\]

    C)  \[adj\,(A)\]

    D)  \[-adj\,(A)\]

    Correct Answer: C

    Solution :

    \[A=\left[ \begin{matrix}    \cos \,\alpha  & -\sin \alpha  & 0  \\    \sin \,\alpha  & \cos \,\alpha  & 0  \\    0 & 0 & 1  \\ \end{matrix} \right]\] \[\therefore \]  \[|A|=1.({{\cos }^{2}}\,\alpha +{{\sin }^{2}}\alpha )=1\] Now, \[{{A}^{-1}}=\frac{1}{|A|}\,adj\,(A)=adj\,(A)\]


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