Alternative matrix representation for translation The ''usual" way to write translation for $\textbf{v}\in \mathbb{R}^2$ is with the following $3\times3$ matrix
$$  \left( {\begin{array}{ccc}
   1 & 0 & x_{0} \\
   0 & 1 & y_{0} \\
   0 & 0 & 1 \\
  \end{array} } \right) \cdot \left( {\begin{array}{c}
   x \\
   y \\
   1 \\
  \end{array} } \right)= \left( {\begin{array}{c} x+x_{0}\\y+y_{0}\\1\end{array}}\right)$$
But also it seems to be possible to write this transformation with the following $2\times2$  matrix:
$$  \left( {\begin{array}{cc}
   1 & \frac{x_{0}}{y} \\
   \frac{y_{0}}{x} & 1 \\
   \end{array} } \right) \cdot \left( {\begin{array}{c}
   x \\
   y \\
    \end{array} } \right)= \left( {\begin{array}{c} x+x_{0}\\y+y_{0}\end{array}}\right)$$
I didn't find a source using this as a version of translation, what are the problems associated with this representation?  

I want to answer my own question in response to the answer given above. In the answer Marc van Leeuwen says that ''matrix representing linear transformation should not depend on the input" 
Why exactly? What properties of linear transformations prevent that?If we find a transformation that depends on the input but satisfies $A(\alpha \textbf{v}+\beta \textbf{u})=\alpha A(\textbf{v})+\beta A(\textbf{u})$ isn't that automatically a linear transformation? 
If we try it on the above transformation we see where the problem comes from:
Let $\textbf{u} = \left( {\begin{array}{c} u^1\\u^2\end{array}}\right)$ and  $\textbf{v} = \left( {\begin{array}{c} v^1\\v^2\end{array}}\right)$ Then from the definition of $A$ we have:
$$A(\alpha \textbf{v}+\beta \textbf{u})=A\cdot \Bigg( {\begin{array}{c} \alpha u^{1}+\beta v^{1} \\ \alpha u^{2}+\beta v^{2}  \end{array}} \Bigg) = \Bigg( \begin{array}{cc} 1&\frac{x_{0}}{\alpha u^2+\beta v^2}\\
\frac{y_{0}}{\alpha u^1 + \beta v^1} &1 \end{array} \Bigg)\cdot \Bigg( {\begin{array}{c} \alpha u^{1}+\beta v^{1} \\ \alpha u^{2}+\beta v^{2}  \end{array}} \Bigg) $$
$$=\Bigg( {\begin{array}{c} \alpha u^{1}+\beta v^{1} + x_{0} \\ \alpha u^{2}+\beta v^{2} + y_{0} \end{array}} \Bigg)$$
On the other hand 
$$\alpha A(\textbf{v})+\beta A(\textbf{u}) = \alpha \cdot \Bigg( \begin{array}{cc} 1&\frac{x_{0}}{ u^2}\\
\frac{y_{0}}{u^1} &1 \end{array} \Bigg)\cdot \Bigg( {\begin{array}{c} u^{1} \\  u^{2}  \end{array}} \Bigg)+\beta \cdot \Bigg( \begin{array}{cc} 1&\frac{x_{0}}{ v^2}\\
\frac{y_{0}}{v^1} &1 \end{array} \Bigg)\cdot \Bigg( {\begin{array}{c} v^{1} \\  v^{2}  \end{array}} \Bigg)$$
$$=\Bigg( {\begin{array}{c} \alpha u^{1}+\beta v^{1} + (\alpha + \beta)x_{0} \\ \alpha u^{2}+\beta v^{2} + (\alpha + \beta)y_{0} \end{array}} \Bigg)$$
So we proved that $A(\alpha \textbf{v}+\beta \textbf{u})\neq \alpha A(\textbf{v})+\beta A(\textbf{u})$ thus the transformation provided by the matrix $A$ is not linear.
