Question

If  \[\begin{bmatrix}\cos\frac{2\pi}{7} & - \sin\frac{2\pi}{7} \\ \sin\frac{2\pi}{7} & \cos\frac{2\pi}{7}\end{bmatrix}^k = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\] then the least positive integral value of k is _____________.

विकल्प

• 3

• 4

• 6

• 7

Answer 1

 7

\[Here, \] 
\[A = \begin{bmatrix}\cos \frac{2\pi}{7} & - \sin\frac{2\pi}{7} \\ \sin\frac{2\pi}{7} & \cos\frac{2\pi}{7}\end{bmatrix}\]

\[ \Rightarrow A^2 = A \times A\]

\[ \Rightarrow A^2 = \begin{bmatrix}\cos \frac{2\pi}{7} & - \sin\frac{2\pi}{7} \\ \sin\frac{2\pi}{7} & \cos\frac{2\pi}{7}\end{bmatrix} \begin{bmatrix}\cos \frac{2\pi}{7} & - \sin\frac{2\pi}{7} \\ \sin\frac{2\pi}{7} & \cos\frac{2\pi}{7}\end{bmatrix}\]

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

`⇒ A^2 =[[cos  (4π   )/7   -sin (4π  )/7   ] , [ sin (4π  )/7      cos (4π)/7] ]`     `[[∵   cos^2 θ - sin^2 θ  = cos 2 θ ],[ 2 sin θ  cos θ  = sin θ ]]`

\[ \Rightarrow A^3 = A^2 \times A\]

\[ \Rightarrow A^3 = \begin{bmatrix}\cos\frac{4\pi}{7} & - \sin\frac{4\pi}{7} \\ \sin\frac{4\pi}{7} & \cos\frac{4\pi}{7}\end{bmatrix} \begin{bmatrix}\cos \frac{2\pi}{7} & - \sin\frac{2\pi}{7} \\ \sin\frac{2\pi}{7} & \cos\frac{2\pi}{7}\end{bmatrix}\]
\[ \Rightarrow A^3 = \begin{bmatrix}\left( \cos \frac{4\pi}{7}\cos\frac{2\pi}{7} - \sin\frac{4\pi}{7}\sin\frac{2\pi}{7} \right) & \left( - \cos\frac{4\pi}{7}\sin\frac{2\pi}{7} - \sin\frac{4\pi}{7}\cos\frac{2\pi}{7} \right) \\ \left( \sin\frac{4\pi}{7}\cos\frac{2\pi}{7} + \cos\frac{4\pi}{7}\sin\frac{2\pi}{7} \right) & \left( - \sin\frac{2\pi}{7}\sin\frac{4\pi}{7} + \cos\frac{4\pi}{7}\cos\frac{2\pi}{7} \right)\end{bmatrix}\]
  `⇒ A^2 =[[cos  (6π   )/7   -sin (6π  )/7   ] , [ sin (6π  )/7      cos (6π)/7] ]`    `[[∵   cos(A+B) = cos A cos B - sin A sin B ],[ sin (A+B) =sin A  cos B   + cos A sin  B ]]`

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

\[A^6 = A^3 . A^3 \]
\[ \Rightarrow A^6 = \begin{bmatrix}\cos \frac{6\pi}{7} & - \sin\frac{6\pi}{7} \\ \sin\frac{6\pi}{7} & \cos\frac{6\pi}{7}\end{bmatrix} \begin{bmatrix}\cos \frac{6\pi}{7} & - \sin\frac{6\pi}{7} \\ \sin\frac{6\pi}{7} & \cos\frac{6\pi}{7}\end{bmatrix}\]
\[ \Rightarrow A^6 = \begin{bmatrix}\cos \frac{12\pi}{7} & - \sin\frac{12\pi}{7} \\ \sin \frac{12\pi}{7} & \cos \frac{12\pi}{7}\end{bmatrix}\]

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

\[A^7 = A^6 \times A\]
\[ \Rightarrow A^7 = \begin{bmatrix}\cos \frac{12\pi}{7} & - \sin\frac{12\pi}{7} \\ \sin \frac{12\pi}{7} & \cos \frac{12\pi}{7}\end{bmatrix} \begin{bmatrix}\cos\frac{2\pi}{7} & - \sin\frac{2\pi}{7} \\ \sin\frac{2\pi}{7} & \cos\frac{2\pi}{7}\end{bmatrix}\]
\[ \Rightarrow A^7 = \begin{bmatrix}\cos \frac{14\pi}{7} & - \sin\frac{14\pi}{7} \\ \sin \frac{14\pi}{7} & \cos \frac{14\pi}{7}\end{bmatrix}\]
\[ \Rightarrow A^7 = \begin{bmatrix}\cos 2\pi & - \sin2\pi \\ \sin 2\pi & \cos 2\pi\end{bmatrix} \left[ \because \frac{14\pi}{7} = 2\pi \right]\]
\[ = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\]

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