JEE Main & Advanced Sample Paper JEE Main Sample Paper-10

  • question_answer
    Consider \[A=\left[ \begin{matrix}    \cos \theta  & \sin \theta   \\    -\sin \theta  & \cos \theta   \\ \end{matrix} \right],\]then the value of\[\underset{x\to \infty }{\mathop{\lim }}\,\frac{{{A}^{n}}}{n}\] (where \[\theta \in R\]) is equal to

    A)  10                         

    B)  zero matrix            

    C)  symmetric matrix     

    D)  4

    Correct Answer: B

    Solution :

    Idea \[\because \]Let \[A=[{{a}_{i\,j}}]\]be \[{{a}_{m\times n}}\]matrix and k be any number, then \[{{k}_{A}}={{[k{{a}_{i\,j}}]}_{m\times n}}\]and use formula such as\[\underset{x\to \infty }{\mathop{\lim }}\,\frac{\cos n\theta }{n}=\frac{\sin \theta }{n}=0\] Given that \[A=\left[ \begin{matrix}    \cos \theta  & \sin \theta   \\    -\sin \theta  & \cos \theta   \\ \end{matrix} \right]\] \[{{A}^{n}}=\left[ \begin{matrix}    \cos n\theta  & \sin n\theta   \\    -\sin n\theta  & \cos n\theta   \\ \end{matrix} \right]\] \[\frac{{{A}^{n}}}{n}=\left[ \begin{matrix}    \frac{\cos n\theta }{n} & \frac{\sin n\theta }{n}  \\    \frac{-\sin n\theta }{n} & \frac{\cos n\theta }{n}  \\ \end{matrix} \right]\] \[\underset{x\to \infty }{\mathop{\lim }}\,\frac{{{A}^{n}}}{n}=\left[ \begin{matrix}    \underset{x\to \infty }{\mathop{\lim }}\,\frac{\cos n\theta }{n} & \underset{x\to \infty }{\mathop{\lim }}\,\frac{\sin n\theta }{n}  \\    \underset{x\to \infty }{\mathop{\lim }}\,\frac{-\sin n\theta }{n} & \underset{x\to \infty }{\mathop{\lim }}\,\frac{\cos n\theta }{n}  \\ \end{matrix} \right]\] \[=\left[ \begin{matrix}    0 & 0  \\    0 & 0  \\ \end{matrix} \right]=\]zero matrix TEST Edge Properties of scalar multiplication and addition of matrices based questions are asked. These questions are solved with the help of properties of matrices.


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