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)

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}$