Choose the correct option about the matrices given below \(\begin{aligned} & A=\left[\begin{array}{ccc}\cos \frac{\pi}{4} & \sin \frac{\pi}{4} & 0 \ -\sin \frac{\pi}{4} & \cos \frac{\pi}{4} & 0 \ 0 & 0 & 1\end{array}\right] \ & B=\left[\begin{array}{ccc}1 & 0 & 0 \ 0 & \cos \frac{\pi}{3} &…

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Choose the correct option about the matrices given below $\begin{aligned} & A=\left[\begin{array}{ccc}\cos \frac{\pi}{4} & \sin \frac{\pi}{4} & 0 \ -\sin \frac{\pi}{4} & \cos \frac{\pi}{4} & 0 \ 0 & 0 & 1\end{array}\right] \ & B=\left[\begin{array}{ccc}1 & 0 & 0 \ 0 & \cos \frac{\pi}{3} & \sin \frac{\pi}{3} \ 0 & -\sin \frac{\pi}{3} & \cos \frac{\pi}{3}\end{array}\right] \ & C=\left[\begin{array}{ccc}\cos \frac{\pi}{6} & 0 & \sin \frac{\pi}{6} \ 0 & 1 & 0 \ -\sin \frac{\pi}{6} & \cos \frac{\pi}{6} & 0\end{array}\right]\end{aligned}$ $D=\left[\begin{array}{ccc} \cos \frac{\pi}{2} & \sin \frac{\pi}{2} & 0 \ -\sin \frac{\pi}{2} & \cos \frac{\pi}{2} & 0 \ 0 & 0 & 1 \end{array}\right]$

Aicharya · Accepted Answer

Given matrix $\begin{aligned} & A=\left[\begin{array}{ccc} \cos \frac{\pi}{4} & \sin \frac{\pi}{4} & 0 \ -\sin \frac{\pi}{4} & \cos \frac{\pi}{4} & 0 \ 0 & 0 & 1 \end{array}\right] \ & \Rightarrow \quad A^2=\left[\begin{array}{ccc} \cos \frac{\pi}{2} & \sin \frac{\pi}{2} & 0 \ -\sin \frac{\pi}{2} & \cos \frac{\pi}{2} & 0 \ 0 & 0 & 1 \end{array}\right] \ \end{aligned}$ $\begin{array}{ll} \therefore & A^2=D=\left[\begin{array}{ccc} \cos \frac{\pi}{2} & \sin \frac{\pi}{2} & 0 \ -\sin \frac{\pi}{2} & \cos \frac{\pi}{2} & 0 \ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 0 & 1 & 0 \ -1 & 0 & 0 \ 0 & 0 & 1 \end{array}\right] \ \therefore & A^4=\left[\begin{array}{ccc} -1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & 1 \end{array}\right]=D^2 \Rightarrow A^8=\left[\begin{array}{lll} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{array}\right] \ \therefore & A^{2020}=\left(A^{505}\right)^4=\left(\left(A^8\right)^{63} \cdot A\right)^4=\left(I^{63} \cdot A\right)^4=A^4 \neq I \end{array}$ Similarly, $D^4=I$, so $D^{2019} \neq I$ Now, $B=\left[\begin{array}{ccc}1 & 0 & 0 \ 0 & \cos \frac{\pi}{3} & \sin \frac{\pi}{3} \ 0 & -\sin \frac{\pi}{3} & \cos \frac{\pi}{3}\end{array}\right]$ $\Rightarrow \quad B^2=\left[\begin{array}{ccc} 1 & 0 & 0 \ 0 & \cos \frac{2 \pi}{3} & \sin \frac{2 \pi}{3} \ 0 & -\sin \frac{2 \pi}{3} & \cos \frac{2 \pi}{3} \end{array}\right]$ $\begin{aligned} & \Rightarrow B^4=\left[\begin{array}{ccc} 1 & 0 & 0 \ 0 & \cos \frac{4 \pi}{3} & \sin \frac{4 \pi}{3} \ 0 & -\sin \frac{4 \pi}{3} & \cos \frac{4 \pi}{3} \end{array}\right] \ & \Rightarrow B^6=\left[\begin{array}{ccc} 1 & 0 & 0 \ 0 & \cos \frac{6 \pi}{3} & \sin \frac{6 \pi}{3} \ 0 & -\sin \frac{6 \pi}{3} & \cos \frac{6 \pi}{3} \end{array}\right]=I \ & \therefore B^{2020} \neq I \text { but } B^{2022}=\left(B^6\right)^{337}=I^{337}=I \end{aligned}$

Choose the correct option about the matrices given below \(\begin{aligned} & A=\left[\begin{array}{ccc}\cos…

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