જો P = begin{bmatrix} cos frac{pi}{4} & -sin | vedclass

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(C) આપેલ છે કે $P = \begin{bmatrix} \cos \frac{\pi}{4} & -\sin \frac{\pi}{4} \ \sin \frac{\pi}{4} & \cos \frac{\pi}{4} \end{bmatrix} = \begin{bmatrix

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

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(C) આપેલ છે કે $P = \begin{bmatrix} \cos \frac{\pi}{4} & -\sin \frac{\pi}{4} \ \sin \frac{\pi}{4} & \cos \frac{\pi}{4} \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{bmatrix}$.અહીં $P$ એ રોટેશન મેટ્રિક્સ $R_{	heta}$ છે જ્યાં $	heta = \frac{\pi}{4}$.રોટેશન મેટ્રિક્સનો ગુણધર્મ છે કે $R_{	heta}^n = R_{n	heta}$.તેથી, $P^3 = R_{3 	imes \frac{\pi}{4}} = R_{\frac{3\pi}{4}} = \begin{bmatrix} \cos \frac{3\pi}{4} & -\sin \frac{3\pi}{4} \ \sin \frac{3\pi}{4} & \cos \frac{3\pi}{4} \end{bmatrix} = \begin{bmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix}$.હવે, $P^3 X = \begin{bmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{2}} \ \frac{1}{\sqrt{2}} \end{bmatrix}$.$P^3 X = \begin{bmatrix} (-\frac{1}{\sqrt{2}} 	imes \frac{1}{\sqrt{2}}) + (-\frac{1}{\sqrt{2}} 	imes \frac{1}{\sqrt{2}}) \ (\frac{1}{\sqrt{2}} 	imes \frac{1}{\sqrt{2}}) + (-\frac{1}{\sqrt{2}} 	imes \frac{1}{\sqrt{2}}) \end{bmatrix} = \begin{bmatrix} -\frac{1}{2} - \frac{1}{2} \ \frac{1}{2} - \frac{1}{2} \end{bmatrix} = \begin{bmatrix} -1 \ 0 \end{bmatrix}$.

જો $P = \begin{bmatrix} \cos \frac{\pi}{4} & -\sin \frac{\pi}{4} \\ \sin \frac{\pi}{4} & \cos \frac{\pi}{4} \end{bmatrix}$ અને $X = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{bmatrix}$ હોય, તો $P^3 X$ ની કિંમત શોધો.

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