Choose the correct option for the matrices gi | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$B^{2022}=I$

Vedclass · Answer

(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 $\sin(674\pi) = 0$,$B^{2022} = I$.

Similar Questions

If $A = \begin{bmatrix} 2 & 2 \\ -3 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$,then $(B^{-1}A^{-1})^{-1} = $

If $A = \begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix}$, then $A^3 - A^2$ is equal to

Choose the correct statement regarding matrices.

Which one of the following is not true?

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,

Vedclass Test Series

Exam Paper Generator

Online Exam Module