Question

Let $Q=\left(\begin{array}{cc}\cos \frac{\pi }{4}& -\sin \frac{\pi }{4}\\ \sin \frac{\pi }{4}& \cos \frac{\pi }{4}\end{array}\right)$ and
$x=\left(\begin{array}{c}\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\end{array}\right)$ then$Q^{3}x$ is equal to

• A. $\left(\begin{array}{c}0\\ 1\end{array}\right)$
• B. $\left(\begin{array}{c}-\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\end{array}\right)$
• C. $\left(\begin{array}{c}-1\\ 0\end{array}\right)$
• D. $\left(\begin{array}{c}-\frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}}\end{array}\right)$

Answer 1

The correct option is C $\left(\begin{array}{c}-1\\ 0\end{array}\right)$

Assuming $\theta =\frac{\pi }{4}$
$Q=\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)$
$\Rightarrow Q^2=\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)\Rightarrow Q^2=\left(\begin{array}{cc}\cos 2\theta & -\sin 2\theta \\ \sin 2\theta & \cos 2\theta \end{array}\right)$

Similarly,
$\Rightarrow Q^3=\left(\begin{array}{cc}\cos 3\theta & -\sin 3\theta \\ \sin 3\theta & \cos 3\theta \end{array}\right)\Rightarrow Q^3=\left(\begin{array}{cc}\cos \frac{3\pi }{4}& -\sin \frac{3\pi }{4}\\ \sin \frac{3\pi }{4}& \cos \frac{3\pi }{4}\end{array}\right)\Rightarrow Q^{3}x=\left(\begin{array}{cc}-\frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\end{array}\right)\left(\begin{array}{c}\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\end{array}\right)\Rightarrow Q^{3}x=\left(\begin{array}{c}-1\\ 0\end{array}\right)$