For clarification: For a line that is represented in the Euclidean plane as: $l: a_1x_1 + a_2x_2 + a_3 = 0$ We can represent it using parallel coordinates by using the transformation $$ A =\begin{bmatrix} 0 & d &0 \\ 0&0 &-1 \\ 1 & 1& 0 \end{bmatrix}$$
Wherein $d$ is the distance between the $x_1$ and $x_2$ axes. This transformation gives us a point in parallel coordinates (a duality). Given such a point, we can go back to a line in the Euclidean plane using the transformation $$ B =\begin{bmatrix} -1 & 1 &0 \\ 0&0 &-d \\ d & 0& 0 \end{bmatrix}$$ .
The question is, after a finite amount of transormations $A \rightarrow B \rightarrow ... \rightarrow B$, will I get back to the line I started with. In other words, is $A \cdot B \cdot ... \cdot B = I_3$? What about if I start with a point and not a line.
My thinking is this: First off, $d = 1$ or else it'll never happen because of determinant reasons. Then, I calculated the matrix $AB$ ($d=1$ and it doesn't seem to me like it will ever be $I_3$, but I'm not sure how to prove it or if I'm on the right track. I also think I'm mistaken somewhere..
Any help would be appreciated!
I am changing my answer because it appears you are talking about a given $B$, rather than trying to undo $A$ as I originally supposed.
In this case, unless $d = \pm 1$, it will never be $I_3$. Because $\det A = -d$ and $\det B = d^2$, we see that the determinant of any finite product of them is $\pm d^k$ for some $k$. As $\det I_3 = 1$, the only possible way to get $I_3$ as a product is to have either $d = 1$ and have an even number of $A$s in the product, or have $d = -1$ (in which case there is no resriction on the number of $A$s).
Of course, this doesn't say it is possible for $d = \pm 1$, just that it is impossible for any other value of $d$.