Question

$A=\left[\begin{array}{ccc}\cos \frac{\pi }{4}& \sin \frac{\pi }{4}& 0\\ -\sin \frac{\pi }{4}& \cos \frac{\pi }{4}& 0\\ 0& 0& 1\end{array}\right]$
$B=\left[\begin{array}{ccc}1& 0& 0\\ 0& \cos \frac{\pi }{3}& \sin \frac{\pi }{3}\\ 0& -\sin \frac{\pi }{3}& \cos \frac{\pi }{3}\end{array}\right]$
$C=\left[\begin{array}{ccc}\cos \frac{\pi }{6}& 0& \sin \frac{\pi }{6}\\ 0& 1& 0\\ -\sin \frac{\pi }{6}& \cos \frac{\pi }{6}& 0\end{array}\right]$
$D=\left[\begin{array}{ccc}\cos \frac{\pi }{2}& \sin \frac{\pi }{2}& 0\\ -\sin \frac{\pi }{2}& \cos \frac{\pi }{2}& 0\\ 0& 0& 1\end{array}\right]$

• A. $A^{2020}=I$
• B. $B^{2020}=I$
• C. $A{D}^{2019}=I$
• D. $B^{2022}=I$

Answer 1

Answer: (D)

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