Question

If $A=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]$, then the matrix $A^{-50}$ when $\theta =\frac{\pi }{12}$, is equal to

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

Answer 1

The correct option is A $\left[\begin{array}{cc}\frac{\sqrt{3}}{2}& \frac{1}{2}\\ -\frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\right]$

Here, $A{A}^T=1$
$\Rightarrow A^{-1}=A^T=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$
Also, $A^{-n}=\left[\begin{array}{cc}\cos \left(n\theta \right)& \sin \left(n\theta \right)\\ -\sin \left(n\theta \right)& \cos \left(n\theta \right)\end{array}\right]$
$\therefore A^{-50}=\left[\begin{array}{cc}\cos \left(50\right)\theta & \sin \left(50\right)\theta \\ -\sin \left(50\right)\theta & \cos \left(50\right)\theta \end{array}\right]$

$\therefore A^{-50}=\left[\begin{array}{cc}\cos \left(50\right)\frac{\pi }{12}& \sin \left(50\right)\frac{\pi }{12}\\ -\sin \left(50\right)\frac{\pi }{12}& \cos \left(50\right)\frac{\pi }{12}\end{array}\right]$

$50\times \frac{\pi }{12}=4\left(360\right)+60$

$\therefore A^{-50}=\left[\begin{array}{cc}\cos \left(4\right(360)+60)& \sin \left(4\right(360)+60)\\ -\sin \left(4\right(360)+60)& \cos \left(4\right(360)+60)\end{array}\right]$

$\therefore A^{-50}=\left[\begin{array}{cc}\cos \left(60\right)& \sin \left(60\right)\\ -\sin \left(60\right)& \cos \left(60\right)\end{array}\right]$
$\left[\begin{array}{cc}\frac{\sqrt{3}}{2}& \frac{1}{2}\\ -\frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\right]$.