Question

If $A=\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right)$ where $\theta =\frac{2\pi }{19}$ then $A^{2017}=?$

• A. $A$
• B. $A^3$
• C. $A^5$
• D. $I$

Answer 1

The correct option is B $A^3$

Given

$A=\left(\begin{array}{cc}cos\theta & sin\theta \\ -sin\theta & cos\theta \end{array}\right)$

with$\theta =\frac{2\pi }{19}$

$\therefore A^{2017}=A^{2014}.A^3-\left(1\right)$

$A^{2014}=\left(\begin{array}{cc}cos\left(2014\right)\theta & sin\left(2014\right)\theta \\ -sin\left(2014\right)\theta & cos\left(2014\right)\theta \end{array}\right)=\left(\begin{array}{cc}cos32\pi & sin32\pi \\ -sin32\pi & cos32\pi \end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=I$

eqs,(1)$⟹A^{2017}=A^{2014}.A^3=I.A^3=A^3$