Let n geq 2 be an integer. A = begin{bmatrix} | vedclass

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(A) The matrix $A$ represents a rotation matrix in $3D$ space about the $z$-axis by an angle $	heta = \frac{2\pi}{n}$.By the property of rotation mat

Vedclass · Accepted Answer

$A^{n} = I$ and $A^{n-1} 
eq I$

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(A) The matrix $A$ represents a rotation matrix in $3D$ space about the $z$-axis by an angle $	heta = \frac{2\pi}{n}$.By the property of rotation matrices, $A^k = \begin{bmatrix} \cos (k	heta) & \sin (k	heta) & 0 \ -\sin (k	heta) & \cos (k	heta) & 0 \ 0 & 0 & 1 \end{bmatrix}$.For $A^n$, we have $k = n$, so $k	heta = n 	imes \frac{2\pi}{n} = 2\pi$.Thus, $A^n = \begin{bmatrix} \cos (2\pi) & \sin (2\pi) & 0 \ -\sin (2\pi) & \cos (2\pi) & 0 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} = I$.For $A^{n-1}$, the angle is $(n-1) 	imes \frac{2\pi}{n} = 2\pi - \frac{2\pi}{n}$.Since $n \geq 2$, $\frac{2\pi}{n}$ is not a multiple of $2\pi$, so $A^{n-1} 
eq I$.Therefore, the correct option is $A$.

Let $n \geq 2$ be an integer. $A = \begin{bmatrix} \cos (2 \pi / n) & \sin (2 \pi / n) & 0 \\ -\sin (2 \pi / n) & \cos (2 \pi / n) & 0 \\ 0 & 0 & 1 \end{bmatrix}$ and $I$ is the identity matrix of order $3$. Then,

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