Question

Let $A=\left[\begin{array}{cc}\cos \left(\theta \right)& -\sin \left(\theta \right)\\ \sin \left(\theta \right)& \cos \left(\theta \right)\end{array}\right]$ find the value of $A^{-50}$ at $\theta =\frac{\pi }{12}$.

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

Answer 1

The correct option is B

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

The explanation for the correct option

The given matrix, $A=\left[\begin{array}{cc}\cos \left(\theta \right)& -\sin \left(\theta \right)\\ \sin \left(\theta \right)& \cos \left(\theta \right)\end{array}\right]$.

The given matrix is an example of rotation matrix.

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

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

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

Put $\theta =\frac{\pi }{12}$.

$$\begin{gathered} \Rightarrow A^{-50}=\left[\begin{array}{cc}\cos \left(50\times \frac{\mathrm{\pi }}{12}\right)& \sin \left(50\times \frac{\mathrm{\pi }}{12}\right)\\ -\sin \left(50\times \frac{\mathrm{\pi }}{12}\right)& \cos \left(50\times \frac{\mathrm{\pi }}{12}\right)\end{array}\right] \\ \Rightarrow A^{-50}=\left[\begin{array}{cc}\cos \left(\frac{25\mathrm{\pi }}{6}\right)& \sin \left(\frac{25\mathrm{\pi }}{6}\right)\\ -\sin \left(\frac{25\mathrm{\pi }}{6}\right)& \cos \left(\frac{25\mathrm{\pi }}{6}\right)\end{array}\right] \\ \Rightarrow A^{-50}=\left[\begin{array}{cc}\cos \left(4\mathrm{\pi }+\frac{\mathrm{\pi }}{6}\right)& \sin \left(4\mathrm{\pi }+\frac{\mathrm{\pi }}{6}\right)\\ -\sin \left(4\mathrm{\pi }+\frac{\mathrm{\pi }}{6}\right)& \cos \left(4\mathrm{\pi }+\frac{\mathrm{\pi }}{6}\right)\end{array}\right] \\ \Rightarrow A^{-50}=\left[\begin{array}{cc}\cos \left(\frac{\mathrm{\pi }}{6}\right)& \sin \left(\frac{\mathrm{\pi }}{6}\right)\\ -\sin \left(\frac{\mathrm{\pi }}{6}\right)& \cos \left(\frac{\mathrm{\pi }}{6}\right)\end{array}\right]\left[\because \sin \left(4\pi +\theta \right)=\sin \left(\theta \right),\cos \left(4\pi +\theta \right)=\cos \left(\theta \right)\right] \\ \Rightarrow A^{-50}=\left[\begin{array}{cc}\frac{\sqrt{3}}{2}& \frac{1}{2}\\ -\frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\right]\left[\because \sin \left(\frac{\pi }{6}\right)=\frac{1}{2},\cos \left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}\right] \end{gathered}$$

Therefore, $A^{-50}=\left[\begin{array}{cc}\frac{\sqrt{3}}{2}& \frac{1}{2}\\ -\frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\right]$.

Hence, (B) is the correct option.