If $A=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right]$, then the matrix $A^{-50}$ when $\theta=\frac{\pi}{12}$, is equal to

(C) $\left[\begin{array}{cc}\frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right]$

(A) $\left[\begin{array}{cc}\frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2}\end{array}\right]$

(B) $\left[\begin{array}{ll}\frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right]$

(D) $\left[\begin{array}{cc}\frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & \frac{1}{2}\end{array}\right]$

JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Matrices

A = [cos θ sin θ - sin θ cos θ]

, then the matrix

A^{- 50}

when

θ = \frac{π}{12}

, is equal to

Answer: (C)

The matrix is a rotation by

θ

, so

A^{n}

is a rotation by

n θ

and

A^{- 50}

is a rotation by

- 50 θ

. With

θ = \frac{π}{12}

- 50 θ = - \frac{25 π}{6}

, which differs from

- \frac{π}{6}

by a whole multiple of

2 π

. So the matrix is the rotation by

- \frac{π}{6}

, namely

[cos \frac{π}{6} - sin \frac{π}{6} sin \frac{π}{6} cos \frac{π}{6}] = [\frac{3}{2} - \frac{1}{2} \frac{1}{2} \frac{3}{2}]

JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Single correct option