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  • How to solve this problem- - If , then the matrix A-50when , is equal to: - Matrices and Determinants - JEE Main

If A = \begin{bmatrix} \cos\theta & \sin\theta\\ \sin\theta & \cos\theta \end{bmatrix} , then the matrix A-50 when \theta =\frac{\pi}{12} , is equal to:

  • Option 1)

    \begin{bmatrix} \frac{1}{2} &-\frac{\sqrt3}{2} \\ \frac{\sqrt3}{2} & \frac{1}{2} \end{bmatrix}

  • Option 2)

    \begin{bmatrix} \frac{\sqrt3}{2} &\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt3}{2} \end{bmatrix}

  • Option 3)

    \begin{bmatrix} \frac{\sqrt3}{2} &\frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt3}{2} \end{bmatrix}

  • Option 4)

    \begin{bmatrix} \frac{1}{2} &\frac{\sqrt3}{2} \\ -\frac{\sqrt3}{2} & \frac{1}{2} \end{bmatrix}

Answers (1)

best_answer

 

Inverse of a matrix -

 A^{-1}=\frac{1}{\left | A \right |}\cdot adjA

-

 

A=\begin{bmatrix} \cos \left ( \theta \right ) & -\sin \left ( \theta \right )\\ \sin \left ( \theta \right )& \cos \left ( \theta \right ) \end{bmatrix}

A^{-1}=\frac{1}{\left | A \right |}adjA=\frac{1}{1}\begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}

\left ( A^{-1} \right )^{2}=\begin{bmatrix} \cos\theta & \sin \theta \\ - \sin \theta & \cos\theta \end{bmatrix}\begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}=\begin{bmatrix} \cos \left ( 2\theta \right ) &\sin \left ( 2\theta \right ) \\ -\sin \left ( 2\theta \right ) & \sin \left ( 2\theta \right ) \end{bmatrix}

\left ( A^{-1} \right )^{3}=\begin{bmatrix} \cos \left (3 \theta \right ) & \sin \left (3 \theta \right )\\ - \sin \left ( 3\theta \right )&\sin \left ( 3\theta \right ) \end{bmatrix}

\left ( A^{-1} \right )^{50}=\begin{bmatrix} \cos \left (50 \theta \right ) & \sin \left (50 \theta \right )\\ - \sin \left ( 50\theta \right )&\sin \left ( 50\theta \right ) \end{bmatrix}

when \theta =\frac{\pi }{12}

\therefore A^{-50}=\begin{bmatrix} \frac{\sqrt{3}}{2} &\frac{1}{2} \\ -\frac{1}{2}& \frac{\sqrt{3}}{2} \end{bmatrix}

or

Here, AA^{T}=I\Rightarrow A^{-1}=A^{T}

\Rightarrow A^{-1}=\begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}

Also , A^{-n}=\begin{bmatrix} \cos \left ( n\theta \right ) & \sin \left (n \theta \right )\\ -\sin \left ( n\theta \right )& \cos \left (n \theta \right ) \end{bmatrix}


Option 1)

\begin{bmatrix} \frac{1}{2} &-\frac{\sqrt3}{2} \\ \frac{\sqrt3}{2} & \frac{1}{2} \end{bmatrix}

Option 2)

\begin{bmatrix} \frac{\sqrt3}{2} &\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt3}{2} \end{bmatrix}

Option 3)

\begin{bmatrix} \frac{\sqrt3}{2} &\frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt3}{2} \end{bmatrix}

Option 4)

\begin{bmatrix} \frac{1}{2} &\frac{\sqrt3}{2} \\ -\frac{\sqrt3}{2} & \frac{1}{2} \end{bmatrix}

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