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

Answer 1

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

Given, the matrix
$A=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]$
$A^T=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$
$\therefore A{A}^T=I$ so $A$ is a orthogonal matrix.
$\Rightarrow A^{-1}=A^T\Rightarrow A^{-50}=(A^{50}{)}^T$
Now finding $A^2$ and then $A^3$,
$A^2=\left[\begin{array}{cc}\cos 2\theta & -\sin 2\theta \\ \sin 2\theta & \cos 2\theta \end{array}\right]$

$A^3=A^2.A=\left[\begin{array}{cc}\cos 3\theta & -\sin 3\theta \\ \sin 3\theta & \cos 3\theta \end{array}\right]$

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

Now evaluating $50\theta =50.\frac{\pi }{12}=\frac{25\pi }{6}=4\pi +\frac{\pi }{6}$

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

$\Rightarrow A^{50}=\left[\begin{array}{cc}\cos \frac{\pi }{6}& -\sin \frac{\pi }{6}\\ \sin \frac{\pi }{6}& \cos \frac{\pi }{6}\end{array}\right]$

$\Rightarrow A^{50}=\left[\begin{array}{cc}\frac{\sqrt{3}}{2}& -\frac{1}{2}\\ \frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\right]\Rightarrow (A^{50}{)}^T=\left[\begin{array}{cc}\frac{\sqrt{3}}{2}& \frac{1}{2}\\ -\frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\right]$
Hence,
$A^{-50}=\left[\begin{array}{cc}\frac{\sqrt{3}}{2}& \frac{1}{2}\\ -\frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\right]$