Prove the following

Question:

If $\mathrm{A}=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right]$, then the matrix $\mathrm{A}^{-50}$

when $\theta=\frac{\pi}{12}$, is equal to :

  1. $\left[\begin{array}{cc}\frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right]$

  2. $\left[\begin{array}{cc}\frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & \frac{1}{2}\end{array}\right]$

  3. $\left[\begin{array}{cc}\frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2}\end{array}\right]$

  4. $\left[\begin{array}{cc}\frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right]$


Correct Option: 1

Solution:

Here, $\mathrm{AA}^{\mathrm{T}}=\mathrm{I}$

$\Rightarrow A^{-1}=A^{\mathrm{T}}=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$

Also, $\mathrm{A}^{-\mathrm{n}}=\left[\begin{array}{cc}\cos (\mathrm{n} \theta) & \sin (\mathrm{n} \theta) \\ -\sin (\mathrm{n} \theta) & \cos (\mathrm{n} \theta)\end{array}\right]$

$\therefore \mathrm{A}^{-50}=\left[\begin{array}{cc}\cos (50) \theta & \sin (50) \theta \\ -\sin (50) \theta & \cos (50) \theta\end{array}\right]$

$=\left[\begin{array}{cc}\frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right]$

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