# Solve this

Question:

If $\left[\begin{array}{cc}\cos \frac{2 \pi}{7} & -\sin \frac{2 \pi}{7} \\ \sin \frac{2 \pi}{7} & \cos \frac{2 \pi}{7}\end{array}\right]^{k}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$, then the least positive integral value of $k$ is

(a) 3

(b) 4

(c) 6

(d) 7

Solution:

(d) 7

Here,

$A=\left[\begin{array}{cc}\cos \frac{2 \pi}{7} & -\sin \frac{2 \pi}{7} \\ \sin \frac{2 \pi}{7} & \cos \frac{2 \pi}{7}\end{array}\right]$

$\Rightarrow A^{2}=A \times A$

$\Rightarrow A^{2}=\left[\begin{array}{cc}\cos \frac{2 \pi}{7} & -\sin \frac{2 \pi}{7} \\ \sin \frac{2 \pi}{7} & \cos \frac{2 \pi}{7}\end{array}\right]\left[\begin{array}{cc}\cos \frac{2 \pi}{7} & -\sin \frac{2 \pi}{7} \\ \sin \frac{2 \pi}{7} & \cos \frac{2 \pi}{7}\end{array}\right]$

$\Rightarrow A^{2}=\left[\begin{array}{cc}\cos ^{2} \frac{2 \pi}{7}-\sin ^{2} \frac{2 \pi}{7} & \left(-2 \cos \frac{2 \pi}{7} \sin \frac{2 \pi}{7}\right) \\ 2 \cos \frac{2 \pi}{7} \sin \frac{2 \pi}{7} & \cos ^{2} \frac{2 \pi}{7}-\sin ^{2} \frac{2 \pi}{7}\end{array}\right]$

$\Rightarrow A^{2}=\left[\begin{array}{cc}\cos \frac{4 \pi}{7} & -\sin \frac{4 \pi}{7} \\ \sin \frac{4 \pi}{7} & \cos \frac{4 \pi}{7}\end{array}\right]$    $\left[\begin{array}{c}\because \quad \cos ^{2} \theta-\sin ^{2} \theta=\cos 2 \theta \\ 2 \sin \theta \cos \theta=\sin 2 \theta\end{array}\right]$

$\Rightarrow A^{3}=A^{2} \times A$

$\Rightarrow A^{3}=\left[\begin{array}{cc}\cos \frac{4 \pi}{7} & -\sin \frac{4 \pi}{7} \\ \sin \frac{4 \pi}{7} & \cos \frac{4 \pi}{7}\end{array}\right]\left[\begin{array}{cc}\cos \frac{2 \pi}{7} & -\sin \frac{2 \pi}{7} \\ \sin \frac{2 \pi}{7} & \cos \frac{2 \pi}{7}\end{array}\right]$

$\Rightarrow A^{3}=\left[\begin{array}{ll}\left(\cos \frac{4 \pi}{7} \cos \frac{2 \pi}{7}-\sin \frac{4 \pi}{7} \sin \frac{2 \pi}{7}\right) & \left(-\cos \frac{4 \pi}{7} \sin \frac{2 \pi}{7}-\sin \frac{4 \pi}{7} \cos \frac{2 \pi}{7}\right) \\ \left(\sin \frac{4 \pi}{7} \cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7} \sin \frac{2 \pi}{7}\right) & \left(-\sin \frac{2 \pi}{7} \sin \frac{4 \pi}{7}+\cos \frac{4 \pi}{7} \cos \frac{2 \pi}{7}\right)\end{array}\right]$

$\Rightarrow A^{3}=\left[\begin{array}{cc}\cos \frac{6 \pi}{7} & -\sin \frac{6 \pi}{7} \\ \sin \frac{6 \pi}{7} & \cos \frac{6 \pi}{7}\end{array}\right]$         $\left[\begin{array}{c}\because \cos (A+B)=\cos A \cos B-\sin A \sin B \\ \sin (A+B)=\sin A \cos B+\cos A \sin B\end{array}\right]$

Now we check if the pattern is same for $k=6$.

Here,

$A^{6}=A^{3} \cdot A^{3}$

$\Rightarrow A^{6}=\left[\begin{array}{cc}\cos \frac{6 \pi}{7} & -\sin \frac{6 \pi}{7} \\ \sin \frac{6 \pi}{7} & \cos \frac{6 \pi}{7}\end{array}\right]\left[\begin{array}{cc}\cos \frac{6 \pi}{7} & -\sin \frac{6 \pi}{7} \\ \sin \frac{6 \pi}{7} & \cos \frac{6 \pi}{7}\end{array}\right]$

$\Rightarrow A^{6}=\left[\begin{array}{cc}\cos \frac{12 \pi}{7} & -\sin \frac{12 \pi}{7} \\ \sin \frac{12 \pi}{7} & \cos \frac{12 \pi}{7}\end{array}\right]$

Now, we check if the pattern is same for $k=7$. Here,

$A^{7}=A^{6} \times A$

$\Rightarrow A^{7}=\left[\begin{array}{cc}\cos \frac{12 \pi}{7} & -\sin \frac{12 \pi}{7} \\ \sin \frac{12 \pi}{7} & \cos \frac{12 \pi}{7}\end{array}\right]\left[\begin{array}{cc}\cos \frac{2 \pi}{7} & -\sin \frac{2 \pi}{7} \\ \sin \frac{2 \pi}{7} & \cos \frac{2 \pi}{7}\end{array}\right]$

$\Rightarrow A^{7}=\left[\begin{array}{cc}\cos \frac{14 \pi}{7} & -\sin \frac{14 \pi}{7} \\ \sin \frac{14 \pi}{7} & \cos \frac{14 \pi}{7}\end{array}\right]$

$\Rightarrow A^{7}=\left[\begin{array}{cc}\cos 2 \pi & -\sin 2 \pi \\ \sin 2 \pi & \cos 2 \pi\end{array}\right] \quad\left[\because \frac{14 \pi}{7}=2 \pi\right]$

$=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

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