Question

If $\theta =\frac{\pi }{12}$ and
A=$\left[\begin{array}{cc}\cos \theta & \sin \theta \\ \sin \theta & \cos \theta \end{array}\right]$ then det $\left(A^6\right)$ =

• A. $\frac{27}{64}$
• B. $\frac{3}{2}$
• C. -1
• D. $\frac{9}{16}$

Answer 1

The correct option is A $\frac{27}{64}$

$A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ \sin \theta & \cos \theta \end{array}\right]$

$A^2=A\times A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ \sin \theta & \cos \theta \end{array}\right]\times \left[\begin{array}{cc}\cos \theta & \sin \theta \\ \sin \theta & \cos \theta \end{array}\right]$

$=\left[\begin{array}{cc}{\cos }^2\theta {\sin }^2\theta & \cos \theta \sin \theta +\sin \theta \cos \theta \\ \sin \theta \cos \theta +\sin \theta \cos \theta & {\cos }^2\theta +{\sin }^2\theta \end{array}\right]$

$=\left[\begin{array}{cc}1& {\sin }^2\theta \\ {\sin }^2\theta & 1\end{array}\right]$

As $\theta =\frac{\pi }{12}$

$\therefore A^2=\left[\begin{array}{cc}1& \sin \left(\frac{2\pi }{12}\right)\\ \sin \left(\frac{2\pi }{12}\right)& 1\end{array}\right]=\left[\begin{array}{cc}1& \sin \frac{\pi }{6}\\ \sin \frac{\pi }{6}& 1\end{array}\right]=\left[\begin{array}{cc}1& \frac{1}{2}\\ \frac{1}{2}& 1\end{array}\right]$

$\therefore A^4=A^2\times A^2=\left[\begin{array}{cc}1& \frac{1}{2}\\ \frac{1}{2}& 1\end{array}\right]\left[\begin{array}{cc}1& \frac{1}{2}\\ \frac{1}{2}& 1\end{array}\right]=\left[\begin{array}{cc}1+\frac{1}{4}& \frac{1}{2}+\frac{1}{2}\\ \frac{1}{2}+\frac{1}{2}& \frac{1}{4}+1\end{array}\right]=\left[\begin{array}{cc}\frac{5}{4}& +1\\ 1& \frac{5}{4}\end{array}\right]$

Hence $A^6=A^4\times A^2=\left[\begin{array}{cc}\frac{5}{4}& 1\\ 1& \frac{5}{4}\end{array}\right]\left[\begin{array}{cc}1& \frac{1}{2}\\ \frac{1}{2}& 1\end{array}\right]$

$=\left[\begin{array}{cc}\frac{5}{4}+\frac{1}{2}& \frac{5}{8}+1\\ 1+\frac{5}{8}& \frac{1}{2}+\frac{5}{4}\end{array}\right]=\left[\begin{array}{cc}\frac{7}{4}& \frac{13}{8}\\ \frac{13}{8}& \frac{7}{4}\end{array}\right]$

$=\frac{7}{4}\times \frac{7}{4}-\frac{13}{8}\times \frac{13}{8}=\frac{27}{64}$