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(D) The rotation matrix $R_{	heta}$ is given by $\left[\begin{array}{cc}\cos 	heta & \sin 	heta \ -\sin 	heta & \cos 	heta\end{array}ight]$.Fo

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$B^{2022}=I$

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(D) The rotation matrix $R_{	heta}$ is given by $\left[\begin{array}{cc}\cos 	heta & \sin 	heta \ -\sin 	heta & \cos 	heta\end{array}ight]$.For matrix $A$,$	heta = \frac{\pi}{4}$. Thus,$A^n = \left[\begin{array}{ccc}\cos \frac{n\pi}{4} & \sin \frac{n\pi}{4} & 0 \ -\sin \frac{n\pi}{4} & \cos \frac{n\pi}{4} & 0 \ 0 & 0 & 1\end{array}ight]$.For $A^{2020}$,$n=2020$,so $\frac{2020\pi}{4} = 505\pi$. Since $\cos(505\pi) = -1$ and $\sin(505\pi) = 0$,$A^{2020} 
eq I$.For matrix $D$,$	heta = \frac{\pi}{2}$. Thus,$D^n = \left[\begin{array}{ccc}\cos \frac{n\pi}{2} & \sin \frac{n\pi}{2} & 0 \ -\sin \frac{n\pi}{2} & \cos \frac{n\pi}{2} & 0 \ 0 & 0 & 1\end{array}ight]$.For $D^{2019}$,$n=2019$,$\frac{2019\pi}{2} = 1009.5\pi$,so $D^{2019} 
eq I$.For matrix $B$,$	heta = \frac{\pi}{3}$. Thus,$B^n = \left[\begin{array}{ccc}1 & 0 & 0 \ 0 & \cos \frac{n\pi}{3} & \sin \frac{n\pi}{3} \ 0 & -\sin \frac{n\pi}{3} & \cos \frac{n\pi}{3}\end{array}ight]$.For $B^{2022}$,$n=2022$,$\frac{2022\pi}{3} = 674\pi$. Since $\cos(674\pi) = 1$ and $\sin(674\pi) = 0$,$B^{2022} = I$.

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