The least positive integer n such that left(b | vedclass

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(B) Let $A = \begin{bmatrix} \cos \frac{\pi}{4} & \sin \frac{\pi}{4} \ -\sin \frac{\pi}{4} & \cos \frac{\pi}{4} \end{bmatrix} = \begin{bmatrix} \frac

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$8$

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(B) Let $A = \begin{bmatrix} \cos \frac{\pi}{4} & \sin \frac{\pi}{4} \ -\sin \frac{\pi}{4} & \cos \frac{\pi}{4} \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{bmatrix}$.This is a rotation matrix $R_{	heta} = \begin{bmatrix} \cos 	heta & \sin 	heta \ -\sin 	heta & \cos 	heta \end{bmatrix}$ with $	heta = \frac{\pi}{4}$.By the property of rotation matrices, $A^n = R_{n	heta} = \begin{bmatrix} \cos(n	heta) & \sin(n	heta) \ -\sin(n	heta) & \cos(n	heta) \end{bmatrix}$.We want $A^n = I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$.This implies $\cos(n	heta) = 1$ and $\sin(n	heta) = 0$.This occurs when $n	heta = 2k\pi$ for some integer $k$.Substituting $	heta = \frac{\pi}{4}$, we get $n \cdot \frac{\pi}{4} = 2k\pi$.$n = 8k$.For the least positive integer $n$, we set $k = 1$, which gives $n = 8$.

The least positive integer $n$ such that $\left(\begin{array}{cc}\cos \frac{\pi}{4} & \sin \frac{\pi}{4} \\ -\sin \frac{\pi}{4} & \cos \frac{\pi}{4}\end{array}\right)^{n}$ is an identity matrix of order $2$ is

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