Any geometric transformation can be written like a linear system of matrices or like a system like:

$$$ \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \cdot \begin{pmatrix} x \\ y \end{pmatrix} + \begin{pmatrix} u \\ v \end{pmatrix} \Leftrightarrow \ \vec{x}=A\cdot\vec{x}+\vec{b} $$$

Next, we are going to give a scheme of classification that depends on the previous equations system.

In the scheme is interpreted that the central symmetry is a rotation of $$180^\circ$$.

This classification is only valid when:

- In the axial symmetry, the symmetry axis is one of the axes of coordinates.
- In the central symmetry or in the rotation, the center is the origin of coordinates.

In the cases that this is not satisfied, the scheme previously given is false. To do a general classification scheme a little more elaborated, then mathematical concepts would be needed.

Let the equations system be $$$ \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ -1 & 1 \end{pmatrix} \cdot \begin{pmatrix} x \\ y \end{pmatrix}$$$

We are going to see which kind of transformation it is. To start, we are going to calculate:

$$$\det (A)=2-(-1)=2+1=3>0$$$

Therefore, it is a question of a direct transformation. Besides, since its determinant is different from $$1$$, the transformation is a similarity, therefore, the above mentioned transformation trebles the distances between the points and the lengths of the segments of the plane.

Given the following system $$$ \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} x \\ y \end{pmatrix}$$$

say, what type of transformation is it?

Since $$\det (A) = -1$$, the transformation is inverse and the only inverse transformation is the axial symmetry. Therefore this is transformation symmetry of the $$y$$-coordinate axis.

Finally, we are going to give an example of draft. Consider the following equations system

$$$ \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} \dfrac{1}{2} & \dfrac{\sqrt{3}}{2} \\ -\dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \end{pmatrix} \cdot \begin{pmatrix} x \\ y \end{pmatrix}$$$

then the determinant of the system is:

$$$\det(a)=\dfrac{1}{2}\cdot \dfrac{1}{2} + \dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt{3}}{2}= \dfrac{1}{4}+\dfrac{3}{4}=1>0$$$

Therefore, it is a direct transformation. Also, as $$det (A) = 1$$, is an isometric transformation and finally, as we have no term $$b$$, it might be a rotation.