I was deepening about figure transformations described by matrices, and I encountered an issue that I wasn't able to solve by myself.
In similar shapes you have a constant linear factor k s.t. $$k = \frac{A'B'}{AB}$$ where apices means the image of the point for the similitude: $$T(A) = A'\,or\,\,T(x_A, y_A) = (x'_A, y'_A)$$ Where T represents the transformation function.
Now, the similitude can be described like: $$ \begin{pmatrix} x'\\ y'\\ \end{pmatrix}= \begin{pmatrix} a, b\\ -b, a\\ \end{pmatrix} * \begin{pmatrix} x\\ y\\ \end{pmatrix} + \begin{pmatrix} e\\ f\\ \end{pmatrix} $$
In this case, it is easy to see that the linear factor is the square root of the surface factor, which is the determinant of the matrix: $$ k^2 = \begin{vmatrix}a,b\\-b,a\\\end{vmatrix} $$
This can be easily proved, but in my resource it was written (without explanation or proof) that this is a particular case of a more general one:
$$\frac{Surface(S')}{Surface(S)} = \begin{vmatrix}a,b\\c,d\\\end{vmatrix}$$
Where Surface is an hypotetically surface calcolator function and S and S' are the two surfaces, with $$ T(S)=S'$$ while a, b, c, d are the coefficient of the affinity defined as: $$ \begin{pmatrix} x'\\ y'\\ \end{pmatrix}= \begin{pmatrix} a, b\\ c, d\\ \end{pmatrix} * \begin{pmatrix} x\\ y\\ \end{pmatrix} + \begin{pmatrix} e\\ f\\ \end{pmatrix} $$
My question is:
Is it true? A complete answer would contain a clear proof, maybe also with an intuitive one. If this is not the case, a counter-example would be nice.
If $A$, $B$, $C$ are three points in the plane, then you can check that the signed area of triangle $ABC$ is given by: $$ 2\ Area(ABC)=|A,B|+|B,C|+|C,A|, \quad\hbox{where:}\quad |A,B|=\det\pmatrix{x_A & x_B\\ y_A & y_B},\ \hbox{and so on.} $$ If $A'=TA$ etc., where $T$ is some matrix, it follows that: $$ \begin{align} 2\ Area(A'B'C')&=|A',B'|+|B',C'|+|C',A'|\\ &=|TA,TB|+|TB,TC|+|TC,TA|\\ &=|T||A,B|+|T||B,C|+|T||C,A|\\ &=2\ |T|\ Area(ABC). \end{align} $$ As any polygon can be decomposed into triangles, this completes the proof.