Question

If ${\left[\begin{array}{cc}\cos \frac{2\mathrm{\pi }}{7}& -\sin \frac{2\mathrm{\pi }}{7}\\ \sin \frac{2\mathrm{\pi }}{7}& \cos \frac{2\mathrm{\pi }}{7}\end{array}\right]}^k=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$, then the least positive integral value of k is
(a) 3
(b) 4
(c) 6
(d) 7

Answer 1

(d) 7


$$\begin{gathered} \mathrm{Here}, \\ A=\left[\begin{array}{cc}\cos \frac{2\mathrm{\pi }}{7}& -\sin \frac{2\mathrm{\pi }}{7}\\ \sin \frac{2\mathrm{\pi }}{7}& \cos \frac{2\mathrm{\pi }}{7}\end{array}\right] \\ \Rightarrow A^2=A\times A \\ \Rightarrow A^2=\left[\begin{array}{cc}\cos \frac{2\mathrm{\pi }}{7}& -\sin \frac{2\mathrm{\pi }}{7}\\ \sin \frac{2\mathrm{\pi }}{7}& \cos \frac{2\mathrm{\pi }}{7}\end{array}\right]\left[\begin{array}{cc}\cos \frac{2\mathrm{\pi }}{7}& -\sin \frac{2\mathrm{\pi }}{7}\\ \sin \frac{2\mathrm{\pi }}{7}& \cos \frac{2\mathrm{\pi }}{7}\end{array}\right] \\ \Rightarrow A^2=\left[\begin{array}{cc}{\cos }^2\frac{2\mathrm{\pi }}{7}-{\sin }^2\frac{2\mathrm{\pi }}{7}& \left(-2\cos \frac{2\mathrm{\pi }}{7}\sin \frac{2\mathrm{\pi }}{7}\right)\\ 2\cos \frac{2\mathrm{\pi }}{7}\sin \frac{2\mathrm{\pi }}{7}& {\cos }^2\frac{2\mathrm{\pi }}{7}-{\sin }^2\frac{2\mathrm{\pi }}{7}\end{array}\right] \\ \Rightarrow A^2=\left[\begin{array}{cc}\cos \frac{4\mathrm{\pi }}{7}& -\sin \frac{4\mathrm{\pi }}{7}\\ \sin \frac{4\mathrm{\pi }}{7}& \cos \frac{4\mathrm{\pi }}{7}\end{array}\right]\left[\begin{array}{c}\because {\cos }^2\theta -{\sin }^2\theta =\cos 2\theta \\ 2\sin \theta \cos \theta =\sin 2\theta \end{array}\right] \\ \Rightarrow A^3=A^2\times A \\ \Rightarrow A^3=\left[\begin{array}{cc}\cos \frac{4\mathrm{\pi }}{7}& -\sin \frac{4\mathrm{\pi }}{7}\\ \sin \frac{4\mathrm{\pi }}{7}& \cos \frac{4\mathrm{\pi }}{7}\end{array}\right]\left[\begin{array}{cc}\cos \frac{2\mathrm{\pi }}{7}& -\sin \frac{2\mathrm{\pi }}{7}\\ \sin \frac{2\mathrm{\pi }}{7}& \cos \frac{2\mathrm{\pi }}{7}\end{array}\right] \\ \Rightarrow A^3=\left[\begin{array}{cc}\left(\cos \frac{4\mathrm{\pi }}{7}\cos \frac{2\mathrm{\pi }}{7}-\sin \frac{4\mathrm{\pi }}{7}\sin \frac{2\mathrm{\pi }}{7}\right)& \left(-\cos \frac{4\mathrm{\pi }}{7}\sin \frac{2\mathrm{\pi }}{7}-\sin \frac{4\mathrm{\pi }}{7}\cos \frac{2\mathrm{\pi }}{7}\right)\\ \left(\sin \frac{4\mathrm{\pi }}{7}\cos \frac{2\mathrm{\pi }}{7}+\cos \frac{4\mathrm{\pi }}{7}\sin \frac{2\mathrm{\pi }}{7}\right)& \left(-\sin \frac{2\mathrm{\pi }}{7}\sin \frac{4\mathrm{\pi }}{7}+\cos \frac{4\mathrm{\pi }}{7}\cos \frac{2\mathrm{\pi }}{7}\right)\end{array}\right] \\ \Rightarrow A^3=\left[\begin{array}{cc}\cos \frac{6\mathrm{\pi }}{7}& -\sin \frac{6\mathrm{\pi }}{7}\\ \sin \frac{6\mathrm{\pi }}{7}& \cos \frac{6\mathrm{\pi }}{7}\end{array}\right]\left[\begin{array}{c}\because \cos \left(A+B\right)=\cos A\cos B-\sin A\sin B\\ \sin \left(A+B\right)=\sin A\cos B+\cos A\sin B\end{array}\right] \end{gathered}$$

Now we check if the pattern is same for k = 6.
Here,
$$\begin{gathered} A^6=A^3.A^3 \\ \Rightarrow A^6=\left[\begin{array}{cc}\cos \frac{6\mathrm{\pi }}{7}& -\sin \frac{6\mathrm{\pi }}{7}\\ \sin \frac{6\mathrm{\pi }}{7}& \cos \frac{6\mathrm{\pi }}{7}\end{array}\right]\left[\begin{array}{cc}\cos \frac{6\mathrm{\pi }}{7}& -\sin \frac{6\mathrm{\pi }}{7}\\ \sin \frac{6\mathrm{\pi }}{7}& \cos \frac{6\mathrm{\pi }}{7}\end{array}\right] \\ \Rightarrow A^6=\left[\begin{array}{cc}\cos \frac{12\mathrm{\pi }}{7}& -\sin \frac{12\mathrm{\pi }}{7}\\ \sin \frac{12\mathrm{\pi }}{7}& \cos \frac{12\mathrm{\pi }}{7}\end{array}\right] \end{gathered}$$

Now, we check if the pattern is same for k = 7.
Here,
$$\begin{gathered} A^7=A^6\times A \\ \Rightarrow A^7=\left[\begin{array}{cc}\cos \frac{12\mathrm{\pi }}{7}& -\sin \frac{12\mathrm{\pi }}{7}\\ \sin \frac{12\mathrm{\pi }}{7}& \cos \frac{12\mathrm{\pi }}{7}\end{array}\right]\left[\begin{array}{cc}\cos \frac{2\mathrm{\pi }}{7}& -\sin \frac{2\mathrm{\pi }}{7}\\ \sin \frac{2\mathrm{\pi }}{7}& \cos \frac{2\mathrm{\pi }}{7}\end{array}\right] \\ \Rightarrow A^7=\left[\begin{array}{cc}\cos \frac{14\mathrm{\pi }}{7}& -\sin \frac{14\mathrm{\pi }}{7}\\ \sin \frac{14\mathrm{\pi }}{7}& \cos \frac{14\mathrm{\pi }}{7}\end{array}\right] \\ \Rightarrow A^7=\left[\begin{array}{cc}\cos 2\mathrm{\pi }& -\sin 2\mathrm{\pi }\\ \sin 2\mathrm{\pi }& \cos 2\mathrm{\pi }\end{array}\right]\left[\because \frac{14\mathrm{\pi }}{7}=2\mathrm{\pi }\right] \\ =\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \end{gathered}$$

So, the least positive integral value of k is 7.