Question

Let $P = \begin{pmatrix}cos \frac{\pi}{4}&-sin \frac{\pi}{4}\\\ sin \frac{\pi}{4}&cos; \frac{\pi}{4}\end{pmatrix}$ and $X = \begin{pmatrix}\frac{1}{\sqrt{2}}\\\ \frac{1}{\sqrt{2}}\end{pmatrix}$. Then $P^{3}X$ is equal to

• A. $\binom{0}{1}$
• B. $\begin{pmatrix}\frac{-1}{\sqrt{2}}\\\ \frac{1}{\sqrt{2}}\end{pmatrix}$
• C. $\binom{-1}{0}$
• D. $\begin{pmatrix}-\frac{1}{\sqrt{2}}\\\ -\frac{1}{\sqrt{2}}\end{pmatrix}$

Answer 1

Answer: (C)

$\binom{-1}{0}$

Answer 2

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