If $\begin{bmatrix}\cos\frac{2π}{7}&-\sin\frac{2π}{7}\\\sin\frac{2π}{7}&\cos\frac{2π}{7}\end{bmatrix}^k=\begin{bmatrix}1&0\\0&1\end{bmatrix}$, then the least positive integral value of k, is
Answer & explanation
Correct answer: option 4
We have,
$\begin{bmatrix}\cos\frac{2π}{7}&-\sin\frac{2π}{7}\\\sin\frac{2π}{7}&\cos\frac{2π}{7}\end{bmatrix}^2$
$=\begin{bmatrix}\cos\frac{2π}{7}&-\sin\frac{2π}{7}\\2\sin\frac{2π}{7}&\cos\frac{2π}{7}\end{bmatrix}\begin{bmatrix}\cos\frac{2π}{7}&-\sin\frac{2π}{7}\\\sin\frac{2π}{7}&\cos\frac{2π}{7}\end{bmatrix}$
$=\begin{bmatrix}\cos^2\frac{2π}{7}-\sin^2\frac{2π}{7}&-2\sin\frac{2π}{7}\cos\frac{2π}{7}\\2\sin\frac{2π}{7}\cos\frac{2π}{7}&-\sin^2\frac{2π}{7}+\cos^2\frac{2π}{7}\end{bmatrix}$
$=\begin{bmatrix}\cos\frac{4π}{7}&-\sin\frac{4π}{7}\\\sin\frac{4π}{7}&\cos\frac{4π}{7}\end{bmatrix}$
Continuing in this manner, we get
$\begin{bmatrix}\cos\frac{2π}{7}&-\sin\frac{2π}{7}\\\sin\frac{2π}{7}&\cos\frac{2π}{7}\end{bmatrix}^7=\begin{bmatrix}\cos 2π&-\sin 2π\\\sin 2π&\cos 2π\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
Hence, the least value of k is 7.