Is this the way to build a translation matrix? I am thinking of passive transformations.
I have read the idea that a translation rule can be expressed as a matrix by adding a new dimension. Can this dimension be precisely the distance (d) between the two origins of the bases or systems of coordinates?
For example, imagine that a non-primed frame measures that a house is 10 m away. A primed frame puts its origin 3 m to the right and would logically measure that the house is 7 m away.
So I have thought that the direct and inverse transformations would be as follows:
$$\begin{array}{l}\left( {\begin{array}{*{20}{c}}{x'}\\{d'}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&{ - 1}\\0&{ - 1}\end{array}} \right)\left( {\begin{array}{*{20}{c}}x\\d\end{array}} \right) \to \\x' = x - d = 10 - 3 = 7\\d' =  - d =  - 3\end{array}
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$$
$$\begin{array}{l}\left( {\begin{array}{*{20}{c}}x\\d\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&{ - 1}\\0&{ - 1}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{x'}\\{d'}\end{array}} \right) \to \\x = x' - d' = 7 - ( - 3) = 10\\d =  - d' =  - ( - 3) = 3\end{array}
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$$
Is this the way to do it?
To be noted that the transformation matrix is its own inverse.
 A: On the real line $\mathbb R^1$, shifting the origin $d$ units to the right (i.e., in the positive direction) subtracts $d$ from the coordinate of a point: $$x'=x-d.\tag1$$ The conventional way to express this in matrix form is to introduce homogeneous coordinates and write $$\pmatrix{x'\\1} = \pmatrix{1&-d\\0&1}\pmatrix{x\\1}.\tag2$$ What you’ve effectively done is to choose a different way to divvy up the terms in (1) between the $2\times2$ “transformation matrix” and the $2\times1$ “coordinate vector” (and also to negate the second coordinate, which I don’t quite understand). Algebraically, both produce the same value for $x'$, so when looked at in isolation, there’s nothing wrong with your idea per se. If I understood your construction correctly, the $2\times2$ matrix for any translation is the same—the translation distance is encoded elsewhere. Compared to the conventional approach, you’ve “crossed the beams:” the $2\times2$ matrix encodes the mathematical operations to be performed to effect a translation, but the distance is encoded in what appears to be a coordinate vector. The latter is ad hoc: different translations require different vectors. By contrast, the conventional method packs all of the information needed to effect the translation into the $2\times2$ transformation matrix; the coordinate vector is independent of the translation distance (and more generally, is independent of the transformation to be applied to it). This hints at some potential difficulties with your approach.  
The power of the conventional representation is that composition of transformations translates into multiplication of their respective associated matrices. It’s not immediately clear to me how one would combine a sequence of translations into one in your formulation, but it seems like it’s going to be more involved. Similarly, the matrix of the inverse of a transformation is just the inverse of the transformation’s matrix. As you’ve noted elsewhere, for a translation matrix this amounts to negating the upper-left element.  
What is meant by “adding a dimension” in the conventional construction is that we identify the real line $\mathbb R^1$ with the line $y=1$ in $\mathbb R^2$. That is, the point with coordinate $x$ on the real line is identified with the point $(x,1)^T$ on the Cartesian plane. Linear transformations of the line (which are just mutiplying the coordinates by some constant $a$) obviously correspond to a subset of linear transformations of the plane—scaling in $x$ and reflection in the $y$-axis. In addition, a translation along the line $y=1$ can be effected by a shear transformation of the plane parallel to the $x$-axis, which is also a linear transformation. Compositions of linear transformations are also linear, so we can represent any affine transformation $x\mapsto ax+b$ of the real line as a linear transformation of the plane. Moreover, we can build it up from simpler transformations simply by multiplying the corresponding matrices together. Without going into detail, by going a step further and identifying the point with coordinate $x$ on the real line with the line in $\mathbb R^2$ that passes through the origin and $(x,1)^T$ (so that for $k\ne0$, $(x,1)^T$ and $(kx,k)^T$ represent the same point on the real line) allows even more types of nonlinear transformations of the line to be identified with linear transformations of the plane.  
This uniformity of representation of various important classes of transformations is another advantage of the conventional approach. I’m not sure that your formulation lends itself to incorporating other types of transformation as conveniently. It’s certainly something to explore.  
Finally, this idea of “adding a dimension” by embedding $\mathbb R^1$ into $\mathbb R^2$ has an obvious generalization to $\mathbb R^n$: we identify it with a certain hyperplane of $\mathbb R^{n+1}$. We always add but one dimension to “linearize” those non-linear transformations. With your construction, it looks to me like generalizing it to higher-dimensional spaces requires doubling the dimensions. A general translation can have a component along each coordinate axis, so the “coordinate” vector needs $2n$ elements: the original coordinates of the point and all of the offsets.
A: A translation in 2D cannot be defined by a $2 \times 2$ matrix, but can with a $3 \times 3$ matrix:
$$\begin{bmatrix}
x'\\ y' \\ 1\\
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 & T_x \\
0 & 1 & T_y \\
0 & 0 & 1 \\
\end{bmatrix}
\begin{bmatrix}
x\\ y \\ 1\\
\end{bmatrix}$$
Where $T_x, T_y$ is the translation vector.
You can easy do the same for 3D, 4D, etc.
