Let Q = begin{bmatrix} cos frac{pi}{4} & -sin | vedclass

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(C) Given $Q = \begin{bmatrix} \cos \frac{\pi}{4} & -\sin \frac{\pi}{4} \ \sin \frac{\pi}{4} & \cos \frac{\pi}{4} \end{bmatrix}$ and $x = \begin{bmat

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$\begin{bmatrix} -1 \ 0 \end{bmatrix}$

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(C) Given $Q = \begin{bmatrix} \cos \frac{\pi}{4} & -\sin \frac{\pi}{4} \ \sin \frac{\pi}{4} & \cos \frac{\pi}{4} \end{bmatrix}$ and $x = \begin{bmatrix} \frac{1}{\sqrt{2}} \ \frac{1}{\sqrt{2}} \end{bmatrix}$.We know that the matrix $Q(	heta) = \begin{bmatrix} \cos 	heta & -\sin 	heta \ \sin 	heta & \cos 	heta \end{bmatrix}$ represents a rotation matrix.By the property of rotation matrices,$Q^{n}(	heta) = Q(n	heta)$.Therefore,$Q^{3} = Q\left(3 	imes \frac{\pi}{4}ight) = Q\left(\frac{3\pi}{4}ight) = \begin{bmatrix} \cos \frac{3\pi}{4} & -\sin \frac{3\pi}{4} \ \sin \frac{3\pi}{4} & \cos \frac{3\pi}{4} \end{bmatrix}$.Evaluating the trigonometric values: $\cos \frac{3\pi}{4} = -\frac{1}{\sqrt{2}}$ and $\sin \frac{3\pi}{4} = \frac{1}{\sqrt{2}}$.So,$Q^{3} = \begin{bmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix}$.Now,calculate $Q^{3}x = \begin{bmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{2}} \ \frac{1}{\sqrt{2}} \end{bmatrix}$.$Q^{3}x = \begin{bmatrix} (-\frac{1}{\sqrt{2}} 	imes \frac{1}{\sqrt{2}}) + (-\frac{1}{\sqrt{2}} 	imes \frac{1}{\sqrt{2}}) \ (\frac{1}{\sqrt{2}} 	imes \frac{1}{\sqrt{2}}) + (-\frac{1}{\sqrt{2}} 	imes \frac{1}{\sqrt{2}}) \end{bmatrix} = \begin{bmatrix} -\frac{1}{2} - \frac{1}{2} \ \frac{1}{2} - \frac{1}{2} \end{bmatrix} = \begin{bmatrix} -1 \ 0 \end{bmatrix}$.

Let $Q = \begin{bmatrix} \cos \frac{\pi}{4} & -\sin \frac{\pi}{4} \\ \sin \frac{\pi}{4} & \cos \frac{\pi}{4} \end{bmatrix}$ and $x = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{bmatrix}$. Then $Q^{3} x$ is equal to:

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