Q.If $f(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}$, then prove that $f(x + y) = f(x) \cdot f(y)$

Rotation Matrix Composition

Hold a book, rotate it so the cover faces you, then tilt it to look down at the spine. The final orientation depends on which turn you do first. That is the whole idea: composing rotations means applying one after another, and in 3D the order matters.

Each rotation is a matrix, and doing two in sequence corresponds to multiplying their matrices. The subtlety is that matrix multiplication is generally not commutative: $AB \neq BA$.

The Rule

Read right to left: a rotation acts on a column vector as $R\mathbf{v}$, so applying $R_1$ then $R_2$ gives $R_2(R_1\mathbf{v}) = (R_2 R_1)\mathbf{v}$. The rotation done first sits on the right.

A 2D Example

In the plane, a counter-clockwise rotation by angle $\theta$ is

Here $R(60^\circ)\,R(30^\circ) = R(90^\circ)$: the angles simply add. In 2D rotations do commute, because they all share the same axis (the axis pointing out of the plane).

Why 3D Is Different

In 3D, rotations are about different axes, and swapping the order changes the result. Use the standard right-handed matrices

and track the vector $\mathbf{v} = (0,0,1)$.

Order 1 — $R_x$ first, then $R_y$ (product $R_y R_x$):

• $R_x(90^\circ)\mathbf{v} = (0,-1,0)$ — the point moves onto the $y$-axis.
• $R_y$ leaves the $y$-axis fixed, so the result stays $(0,-1,0)$.

Order 2 — $R_y$ first, then $R_x$ (product $R_x R_y$):

• $R_y(90^\circ)\mathbf{v} = (1,0,0)$ — the point moves onto the $x$-axis.
• $R_x$ leaves the $x$-axis fixed, so the result stays $(1,0,0)$.