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If $p=\left(\begin{array}{cc}\cos \frac{\pi}{4} & -\sin \frac{\pi}{4} \\ \sin \frac{\pi}{4} & \cos \frac{\pi}{4}\end{array}\right)$ and $X=\left(\begin{array}{c}\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}\end{array}\right)$. Then,
$P^{3} X$ is equal to
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$P^{3} X$ is equal to
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Verified Answer
The correct answer is:
$\left(\begin{array}{r}-1 \\ 0\end{array}\right)$
Given, $P=\left(\begin{array}{cc}\cos \frac{\pi}{4} & -\sin \frac{\pi}{4} \\ \sin \frac{\pi}{4} & \cos \frac{\pi}{4}\end{array}\right)=\left(\begin{array}{cc}\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{array}\right)$
$\Rightarrow \quad P=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1 & -1 \\ 1 & 1\end{array}\right)$
$\begin{aligned} \text { Now, } P^{2} &=P \cdot P=\frac{1}{2}\left(\begin{array}{cc}1 & -1 \\ 1 & 1\end{array}\right)\left(\begin{array}{cc}1 & -1 \\ 1 & 1\end{array}\right) \\ &=\frac{1}{2}\left(\begin{array}{cc}1-1 & -1 & -1 \\ 1+1 & -1+1\end{array}\right) \\ &=\frac{1}{2}\left(\begin{array}{cc}0 & -2 \\ 2 & 0\end{array}\right)=\left(\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right) \end{aligned}$
$\begin{aligned} P^{3} &=P \cdot P^{2}=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1 & -1 \\ 1 & 1\end{array}\right) \cdot\left(\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right) \\ &=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}0-1 & -1-0 \\ 0+1 & -1+0\end{array}\right) \\ &=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}-1 & -1 \\ 1 & -1\end{array}\right) \end{aligned}$
Also, given $X=\left(\begin{array}{c}1 / \sqrt{2} \\ \frac{1}{\sqrt{2}}\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}{l}1 \\ 1\end{array}\right)$
$\begin{aligned} P^{3} X &=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}-1 & -1 \\ 1 & -1\end{array}\right) \cdot \frac{1}{\sqrt{2}}\left(\begin{array}{l}1 \\ 1\end{array}\right) \\ &=\frac{1}{2}\left(\begin{array}{c}-1-1 \\ 1-1\end{array}\right)=\frac{1}{2}\left(\begin{array}{c}-2 \\ 0\end{array}\right)=\left(\begin{array}{c}-1 \\ 0\end{array}\right) \end{aligned}$
$\Rightarrow \quad P=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1 & -1 \\ 1 & 1\end{array}\right)$
$\begin{aligned} \text { Now, } P^{2} &=P \cdot P=\frac{1}{2}\left(\begin{array}{cc}1 & -1 \\ 1 & 1\end{array}\right)\left(\begin{array}{cc}1 & -1 \\ 1 & 1\end{array}\right) \\ &=\frac{1}{2}\left(\begin{array}{cc}1-1 & -1 & -1 \\ 1+1 & -1+1\end{array}\right) \\ &=\frac{1}{2}\left(\begin{array}{cc}0 & -2 \\ 2 & 0\end{array}\right)=\left(\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right) \end{aligned}$
$\begin{aligned} P^{3} &=P \cdot P^{2}=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1 & -1 \\ 1 & 1\end{array}\right) \cdot\left(\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right) \\ &=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}0-1 & -1-0 \\ 0+1 & -1+0\end{array}\right) \\ &=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}-1 & -1 \\ 1 & -1\end{array}\right) \end{aligned}$
Also, given $X=\left(\begin{array}{c}1 / \sqrt{2} \\ \frac{1}{\sqrt{2}}\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}{l}1 \\ 1\end{array}\right)$
$\begin{aligned} P^{3} X &=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}-1 & -1 \\ 1 & -1\end{array}\right) \cdot \frac{1}{\sqrt{2}}\left(\begin{array}{l}1 \\ 1\end{array}\right) \\ &=\frac{1}{2}\left(\begin{array}{c}-1-1 \\ 1-1\end{array}\right)=\frac{1}{2}\left(\begin{array}{c}-2 \\ 0\end{array}\right)=\left(\begin{array}{c}-1 \\ 0\end{array}\right) \end{aligned}$
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