Coloquially, the geometric transformations are the geometric operation/s that let us create a new figure from one previously given. This new figure is called the homologous of the original one. We can classify the above mentioned transformations under two big groups:

- Direct: if the double preserves the orientation of the original.

- Inverse: if the homologous has the opposite sense.

We can also classify the geometric transformations depending on the form of the homologous with regard to the original. In this case, we have three big groups:

- Isometric: the homologous preserves the distances and the angles. This group, is also called movements in the plane.

- Isomorphic: the homologous preserves the form and the angles. Therefore, proportionality exists between the sides of the homologous and the original.

- Anamorphic: it changes the form of the original figure. In this cell, these transformations are not going to be explained.

Formally, the geometric transformations are the linear applications $$\varphi: \mathbb{R}^2 \rightarrow \mathbb{R}^2$$. Let $$(e_1,e_2)$$ ube an orthonormal basis (orthogonal of module $$1$$) of $$\mathbb{r}^2$$. As the geometric transformations are linear applications, then we can represent them by means of a two-dimensional system of linear equations. So, let $$\vec{x}=(x_1,x_2)$$ be any vector of $$E$$ and let $$\vec{x'}=(x'_1,x'_2)$$ be the vector transformed by means of the geometric transformation. Then, these two vectors satisfy the following equation:

$$$ \begin{pmatrix} x'_1 \\ x'_2 \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \cdot \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} + \begin{pmatrix} b_1 \\ b_2 \end{pmatrix} $$$

where the matrix $$A= \begin{pmatrix} a& b \\ c & d \end{pmatrix}$$ s the matrix that represents how they change the vectors of the basis regarding the transformation.

That is, in the first column there is the new components of the first vector of the basis and in the second one the components of the second basic vector. Besides, the vector $$\vec{b}=(b_1,b_2)^{T}$$ shows us how the origin of the changes of coordinates by means of the transformation.

Therefore, thanks to this algebraic formulation of the geometric transformations, we can re-formulate the previous classifications using only the matrix associated with the transformation. Let's remember that, in the first classification we had that:

- Direct transformation: If it preserves the orientation and this will happen if and only if $$\det (A)> 0$$.
- Inverse transformation: If it inverted the orientation, and this happens if and only if $$\det (A) <0$$.

Therefore, by means of the sign of the determinant of the matrix associated with the transformation, we are able to know whether or not it preserves its orientation.

On the other hand, the second classification that we have done of the geometric transformations, told us that:

- Isometric: It preserves the angles and the distances. This fact is equivalent to say that $$\det (A) = \pm 1$$.
- Isomorphic: It preserves the angles and its form, existing a proportionality reason between the sides of the original and of the homologous. This is equivalent to saying that $$det (A) =\pm K \ $$ and $$ \ K\neq 1$$, and that the distances that existed in the original figure turn out to be multiplied by the factor $$|K|$$. Therefore, $$K$$ is the similarity ratio between two figures.
- Anamorphic: They cannot be represented by a matrix since they do not preserve neither the angles nor the proportions. Therefore, this kind of applications are non linear and we are not going to study them.

To finish, we are going to give an example of classification of transformations. Given the system$$$ \begin{pmatrix} x'_1 \\ x'_2 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} \cdot \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} + \begin{pmatrix} 1 \\ 1 \end{pmatrix} $$$

we are going to classify it under the two classifications given in advance. First we need to calculate the determinant of the matrix associated with the transformation. Then, $$\det(A)=2\cdot2-1\cdot1=3$$, with that we can see that the transformation is direct, since its determinant is positive, and it is an isomorphic transformation, since the determinant is $$3$$. Therefore, the figures where we apply this transformation will turn out to be multiplied by a ratio of similarity equal to $$3$$.