Question

If ${\left[\begin{array}{cc}\cos \frac{2\pi }{3}& -\sin \frac{2\pi }{3}\\ \sin \frac{2\pi }{3}& \cos \frac{2\pi }{3}\end{array}\right]}^k=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$, then least value of $k$ equals $(k\ne 0)$

• A. $1$
• B. $2$
• C. $-1$
• D. $3$

Answer 1

Answer: (D)

$3$

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

where $t=\frac{2\pi }{3}$

$\therefore 3t=2\pi$

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

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

$=\left[\begin{array}{cc}\cos 2t& -\sin 2t\\ \sin 2t& \cos 2t\end{array}\right]$

$\Rightarrow A^2.A=A^3=\left[\begin{array}{cc}\cos 2t& -\sin 2t\\ \sin 2t& \cos 2t\end{array}\right]\times \left[\begin{array}{cc}\cos t& -\sin t\\ \sin t& \cos t\end{array}\right]$

$\Rightarrow A^3=\left[\begin{array}{cc}\cos 3t& -\sin 3t\\ \sin 3t& \cos 3t\end{array}\right]=\left[\begin{array}{cc}\cos 2\pi & -\sin 2\pi \\ \sin 2\pi & \cos 2\pi \end{array}\right]$

$\Rightarrow A^3=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$\therefore$ The least value of $k=3$