# How do rotational matrices work? [duplicate]

I am confuse on the how exactly rotational matrices work. So I understand that you can rotate a point around the x, y and z axis but if asked how you find a single matrix that will show the same rotation if you were to rotate it along the x, y and z axis in that order. Any help would be appreciated!

• This is totally unclear. Rotate a matrix around the $x,y$ and $z$ axes? Mar 18, 2014 at 10:07

Wikipedia has a decent explantion: http://en.wikipedia.org/wiki/Transformation_matrix#Rotation

To understand how they get their equations let's consider a 2D case where you have a point $(x,y)$ and you want to rotate to $\phi$ degrees clockwise to get $(x',y')$. If we convert to polar coorindates then this would be really simple so lets do that.

$$x=r\cos(\theta)$$ $$y=r\sin(\theta)$$

now rotate by $\phi$ degress and covert back to Cartesian coordiantes:

$$x' = r\cos{(\theta-\phi)}$$ $$=r\cos(\theta)\cos(-\phi)-r\sin(\theta)\sin(-\phi)$$ $$\therefore x' =x\cos(\phi)+y\sin(\phi)$$

similarly $$y'=-x\sin(\phi)+y\sin(\phi)$$

so the idea of the rotation matrix is to just summarise those two equations in to one matrix equation:

$$\begin{pmatrix}x'\\y'\end{pmatrix} = \begin{pmatrix}\cos \phi & \sin \phi \\ -\sin \phi & \cos \phi \end{pmatrix} \begin{pmatrix}x\\y\end{pmatrix}$$

Which if you multiply out give you exactly the rotation equations we derived by rotation in polar coordinates.

The nice thing about matrices as representations for linear mappings is that you can combine (concatenate) mappings simply by multiplying their respective matrices. A rotation around the $x$-Axis in $\mathbb{R}^3$ is specified by the matrix $$M_x(\varphi) = \begin{pmatrix}1 & 0 & 0 \\ 0 & cos \varphi & -\sin \varphi \\ 0 & \sin \varphi & \cos \varphi \end{pmatrix} \text{,}$$ a rotation around the $y$-Axis is specified by $$M_y(\vartheta) = \begin{pmatrix}\cos\vartheta & 0 & -\sin\vartheta \\ 0 & 1 & 0 \\ \sin\vartheta & 0 & \cos \vartheta \end{pmatrix} \text{,}$$ and a rotation around the $z$-Axis by $$M_z(\rho) = \begin{pmatrix}cos \rho & -\sin \rho & 0\\ \sin \rho & \cos \rho & 0 \\ 0 & 0 & 1\end{pmatrix} \text{.}$$ To find the matrix representing the mapping which first rotates around the $x$-Axis, then around the $y$-Axis and finally around the $z$-Axis, just multiply the three matrices, i.e. compute $$M(\varphi,\vartheta,\rho) = M_z(\varphi)M_y(\vartheta)M_x(\rho) \text{.}$$ Note that it is the rightmost matrix in such a product that is applied first. This is because such products are associative, i.e. $$M \mathbf{x} = (M_x M_y M_z) \mathbf{x} = M_x (M_y (M_z \mathbf{x})) \text{.}$$

• Thank you very much! that is exactly what I needed! Mar 18, 2014 at 10:02