Question

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

a) 3
b) 4
c) 6
d) 7​

Answer 1

ANSWER:

• The least positive integral value of k is 7.

GIVEN:

${\left[ \begin{array}{c c} \cos \dfrac{2 \pi}{7} & - \sin \dfrac{2 \pi}{7} \\ \\ \sin \dfrac{2 \pi }{7} & \cos \dfrac{2 \pi}{7} \end{array}\right]}^{k} = \left[ \begin{array}{c c} 1 & 0\\ 0 & 1 \end{array}\right]$

TO FIND:

• The least positive integral value of k.

EXPLANATION:

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

Let cos 2π/7 = x and sin 2π/7 = y.

$\left[ \begin{array}{c c} x & - y\\ \\ y& x \end{array}\right]\left[ \begin{array}{c c} x & - y\\ \\ y& x \end{array}\right]$

$\left[ \begin{array}{c c} {x}^{2} - {y}^{2} & - xy - xy\\ \\ xy + xy& - {y}^{2} + {x}^{2} \end{array}\right]$

$\left[ \begin{array}{c c} {x}^{2} - {y}^{2} & -2xy\\ \\ \\ 2xy& {x}^{2}- {y}^{2} \end{array}\right]$

x² - y² = cos² 2π/7 - sin² 2π/7

cos² x - sin² x = cos 2x

x² - y² = cos 2(2π/7)

2xy = 2 cos 2π/7 sin 2π/7

2 sin x cos x = sin 2x

2xy = sin 2(2π/7)

$\left[ \begin{array}{c c} cos \ 2 \left(\dfrac{2 \pi}{7} \right) & -sin \ 2\left(\dfrac{2 \pi}{7} \right)\\ \\ \\ sin \ 2\left(\dfrac{2 \pi}{7} \right)& cos \ 2\left(\dfrac{2 \pi}{7} \right) \end{array}\right]$

Let cos 2(2π/7) = a and sin 2(2π/7) = b.

$\left[ \begin{array}{c c} a & - b\\ \\ b& a \end{array}\right] \left[ \begin{array}{c c} x & - y\\ \\ y& x \end{array}\right]$

$\left[ \begin{array}{c c} ax - by & - ay - bx\\ \\ bx + ay & - by + ax \end{array}\right]$

$\left[ \begin{array}{c c} ax - by & - (ay + bx)\\ \\ bx + ay & ax - by \end{array}\right]$

(ax - by) = cos 2(2π/7) cos 2π/7 - sin 2(2π/7) sin 2π/7

Cos x cos y - sin x sin y = cos (x + y)

(ax - by) = cos ( 2(2π/7) + 2π/7)

(ax - by) = cos 3(2π/7)

ay + bx = cos 2(2π/7) sin 2π/7 + sin 2(2π/7) cos 2π/7

Cos x sin y + sin x cos y = sin (x + y)

(ay + bx) = sin (2(2π/7) + 2π/7)

(ay + bx) = sin 3(2π/7)

$\left[ \begin{array}{c c} cos \ 3 \left(\dfrac{2 \pi}{7} \right) & -sin \ 3\left(\dfrac{2 \pi}{7} \right)\\ \\ \\ sin \ 3\left(\dfrac{2 \pi}{7} \right)& cos \ 3\left(\dfrac{2 \pi}{7} \right) \end{array}\right]$

Let 2π/7 = θ

$Let\ A= \left[ \begin{array}{c c} \cos \ \theta & - \sin\ \theta \\ \\ \sin\ \theta & \cos\ \theta \end{array}\right]$

$Let\ A^2= \left[ \begin{array}{c c} \cos \ 2\theta & - \sin\ 2\theta \\ \\ \sin\ 2\theta & \cos\ 2\theta \end{array}\right]$

$Let\ A^3= \left[ \begin{array}{c c} \cos \ 3\theta & - \sin\ 3\theta \\ \\ \sin\ 3\theta & \cos\ 3\theta \end{array}\right]$

Similarly for

$Let\ A^7= \left[ \begin{array}{c c} \cos \ 7\theta & - \sin\ 7\theta \\ \\ \sin\ 7\theta & \cos\ 7\theta \end{array}\right]$

Substitute θ = 2π/7

$\left[ \begin{array}{c c} cos \ 7 \left(\dfrac{2 \pi}{7} \right) & -sin \ 7\left(\dfrac{2 \pi}{7} \right)\\ \\ \\ sin \ 7\left(\dfrac{2 \pi}{7} \right)& cos \ 7\left(\dfrac{2 \pi}{7} \right) \end{array}\right]$

$\left[ \begin{array}{c c} cos \ 2 \pi & -sin \ 2 \pi\\ \\ \\ sin \ 2 \pi& cos \ 2 \pi \end{array}\right]$

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

${\left[ \begin{array}{c c} \cos \dfrac{2 \pi}{7} & - \sin \dfrac{2 \pi}{7} \\ \\ \sin \dfrac{2 \pi }{7} & \cos \dfrac{2 \pi}{7} \end{array}\right]}^7 = \left[ \begin{array}{c c} 1 & 0\\ \\ 0 & 1 \end{array}\right]$