Question

If $A=\left[\begin{array}{cc}\cos \left(2\pi /33\right)& \sin \left(2\pi /33\right)\\ -\sin \left(2\pi /33\right)& \cos \left(2\pi /33\right)\end{array}\right],$ then $A^{2017}=?$

• A. $A$
• B. $A^2$
• C. $A^4$
• D. $I$

Answer 1

The correct option is C $A^4$

Given

$A=\left(\begin{array}{cc}cos(\frac{2\pi }{33}& sin(\frac{2\pi }{33}\\ -sin\left(\frac{2\pi }{33}\right)& cos\left(\frac{2\pi }{33}\right)\end{array}\right)$

$\therefore A^{2017}=A^{2013}.A^4-\left(1\right)$

$A^{2013}=\left(\begin{array}{cc}cos\left(2013\right)\left(\frac{2\pi }{33}\right)& sin\left(2013\right)\left(\frac{2\pi }{33}\right)\\ -sin\left(2013\right)\left(\frac{2\pi }{33}\right)& cos\left(2013\right)\left(\frac{2\pi }{33}\right)\end{array}\right)=\left(\begin{array}{cc}cos132\pi & sin132\pi \\ -sin132\pi & cos132\pi \end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=I$

eqs,(1)$⟹A^{2017}=A^{2013}.A^4=I.A^4=A^4$