Question

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

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

Answer 1

Answer: (C)

Given, $P=\left(\begin{array}{cc}\cos \frac{\pi }{4}& -\sin \frac{\pi }{4}\\ \sin \frac{\pi }{4}& \cos \frac{\pi }{4}\end{array}\right)=\left(\begin{array}{cc}\frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right)$
$\Rightarrow P=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1& -1\\ 1& 1\end{array}\right)$
Now, $P^2=P\cdot P=\frac{1}{2}\left(\begin{array}{cc}1& -1\\ 1& 1\end{array}\right)\left(\begin{array}{cc}1& -1\\ 1& 1\end{array}\right)$
$=\frac{1}{2}\left(\begin{array}{cc}1-1& -1-1\\ 1+1& -1+1\end{array}\right)$
$=\frac{1}{2}\left(\begin{array}{cc}0& -2\\ 2& 0\end{array}\right)=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$
$P^3=P\cdot P^2=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1& -1\\ 1& 1\end{array}\right)\cdot \left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$
$=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}0-1& -1-0\\ 0+1& -1+0\end{array}\right)$
$=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}-1& -1\\ 1& -1\end{array}\right)$
Also, given
$X=\left(\begin{array}{c}1/\sqrt{2}\\ \frac{1}{\sqrt{2}}\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\ 1\end{array}\right)$
$\therefore P^{3}X=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}-1& -1\\ 1& -1\end{array}\right)\cdot \frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\ 1\end{array}\right)$
$=\frac{1}{2}\left(\begin{array}{c}-1-1\\ 1-1\end{array}\right)=\frac{1}{2}\left(\begin{array}{c}-2\\ 0\end{array}\right)=\left(\begin{array}{c}-1\\ 0\end{array}\right)$