Why composite transformations are multiplied to the right side? I have seen that many composite transformations have the later transformation multiplied to the right side of the matrix. Say I have matrix an existing transformation matrix $\mathbf{M}$ and then other transformation matrices $S$ for scaling, $T$ for translation, $R$ for rotation. I want to do a rotation first then translation then scaling, I have to do this: $\mathbf{M}(R)(T)(S)=\mathbf{M'}$. Now the final matrix $\mathbf{M'}$ has the composite transformations in the sequence that I could use it to transform vectors with $\mathbf{M'}\vec{x}=\vec{x'}$.
But the question is why are transformations multiplied to the right side of the matrices and how can I prove that it has to be multiplied to the right side? What would the meaning be if I had multiplied them to the left side instead as $(S)(T)(R)\mathbf{M}=\mathbf{M'}$? 
Sometimes, I have this tendency to multiply to the left because in row elementary operations, the matrices are multiplied to the left.
 A: Sometime one has to left-multiply, sometimes one has to right-multiply. This really depends.
Prerequisites:
You are performing scaling, rotation, and translation. So let us assume we have linear point transformation of the general form:
$$\mathtt T = \left[ \begin{array}{cc} s\mathtt R & \mathbf t  \\ \mathtt O& 1\end{array} \right]$$
which first rotates a point by $\mathtt R$ , then scales  it by $s$ and then adds the translation $\mathbf t$:
$$\mathtt T \cdot \left( \begin{array}{c} \mathbf x  \\  1\end{array} \right)   = \left[ \begin{array}{c} s(\mathtt R\cdot \mathbf x) + \mathbf t  \\  1\end{array} \right]$$
(Note that rotation and scaling commutes: $s(\mathtt R\cdot \mathbf x)=\mathtt R(s\cdot \mathbf x)$)
From now on we will assume that all points $\mathbf y$ are homogenous points ($\mathbf y= (\mathbf x,  1)^\top $).
Mind the reference frames: In order to make it clear whether you need a left or right multiplication, it is important to highlight in which reference frame your points are!
Let us assume, we have points $\mathbf y_a$ in reference frame $a$, and you want to transform them into reference frame $b$, you do 
$$ \mathbf y_b = \mathtt T_{ba} \mathbf y_a$$
where $\mathtt T_{ba}$ is a transformation to $b$ from $a$. Note that the indices must match!
Now, let us look at a more complicated example. One might be interested in:
$$\mathbf y_a = \mathtt T_{ab}\mathtt T_{bc}\mathtt T_{cd}\mathbf y_d$$
Further, let's assume that we receive the poses in order (First $\mathtt T_{ab}$, then $\mathtt T_{bc}$...).
We would calculate in an algorithm:
$\mathtt T_{ai} := \mathtt T_{ab}$ 
(thus, $i=b$)
$\mathtt T_{ai} := \mathtt T_{ai}\cdot \mathtt T_{bc}$ 
(now, $i=c$)
$\mathtt T_{ai} := \mathtt T_{ai}\cdot \mathtt T_{cd}$ 
($i=d$)
Thus, we right-multiplied and $\mathtt T_{ai}$ means now $\mathtt T_{ad}$, the transformation from $d$ to $a$. Finally, we can transform our points:
$$\mathbf y_a :=  \mathtt T_{ad} \mathbf y_d $$ 
However, if one really wants to left-multiply, this is possible too! Note that $\mathtt T_{ia}=\mathtt T_{ai}^{-1}.$ Thus, we can do:
$\mathtt T_{ia} := \mathtt T_{ab}^{-1}$ 
($i=b$)
$\mathtt T_{ia} :=  \mathtt T_{bc}^{-1} \mathtt T_{ia}$
($i=c$)
$\mathtt T_{ia} := \mathtt T_{cd}^{-1}\mathtt T_{ia}$ 
($i=d$)
Thus, we have $\mathtt T_{ia} = \mathtt T_{da}$, and therefore we can transfrom the point from $d$ to $a$ using the inverse:
$$\mathbf y_a :=  \mathtt T_{da}^{-1} \mathbf y_d $$ 
