Q.If $F(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}$ and $[F(x)]^2 = F(kx)$, then the value of $k$ is: (A) $1$ (B) $2$ (C) $0$ (D) $-2$

Matrix Rotation Power

A rotation matrix turns every vector in the plane through a fixed angle. So what happens when you apply it again and again? Applying a rotation of $\theta$ twice is just a rotation of $2\theta$; three times, $3\theta$; and so on. Matrix power is exactly this idea written algebraically: $A^n$ means "apply the transformation $A$ a total of $n$ times."

The intuition

Multiplying a vector by a matrix $A$ transforms it once. Multiplying by $A$ again transforms the result once more. Hence

with the conventions $A^1 = A$ and $A^0 = I$ (the identity), just as $x^0 = 1$ for numbers.

The rotation case

The cleanest example is the rotation matrix through angle $\theta$ (counterclockwise):

Because stacking two rotations adds their angles,

Proving it by induction

This is a classic exam result, proved by mathematical induction on $n$.

• Base case ($n = 1$): $R_\theta^{\,1} = R_\theta = R_{1\cdot\theta}$, true.
• Inductive step: assume $R_\theta^{\,k} = R_{k\theta}$. Then

Multiplying the two matrices and using the addition formulas

$\cos k\theta\cos\theta - \sin k\theta\sin\theta = \cos(k+1)\theta, \qquad \sin k\theta\cos\theta + \cos k\theta\sin\theta = \sin(k+1)\theta,$ …