concatenating translation and rotation matrix - what am I doing wrong? I'm teaching myself linear algebra trying to write a 3D script. Mostly relying on Khan Academy for revising math itself, and supplementing with 3D graphics blogs, websites, etc. 
I got as far as using matrix multiplication to perform compound rotations (say in 3 axes, depending on 3D package's convention for rotation order), but now I'm trying to understand concatenating a rotation and a translation.  
I did this example in 2D (3D homogeneous space) for simplicity of graphing.
Working from Wikipedia (Translation: Matrix representation [see disclaimer at bottom]) I understand I need to convert my vectors to homogeneous space with w = 1, and multiply by the translation matrix T. 
This works for only translation (T) and for rotation*translation (R * T). But produces garbage for T * R. As you can see in my graph and calculations, I would expect A' = TRA somewhere around [0.3,6] point but it is not.  
What am I missing? I'd appreciate pointing me in the right way.

(^In afterthought: I labelled operations in the order of reading, so "A - transl/rot" is technically actually: R * T * A )

Many Thanks.
 A: If you want to rotate and translate a 2D point $p=[x,y]^T$, then this can be written $$ q = Rp + t $$ where $t=[t_x,t_y]^T$ and $R\in SO(2)$ as a $2\times 2$ matrix.
To put this as a matrix multiplication, notice that 
$$
\underbrace{
\begin{bmatrix}
q \\ 1
\end{bmatrix}}_{q_h}
=
\underbrace{
\begin{bmatrix}
R & t \\ 0 & 1
\end{bmatrix}}_{M}
\underbrace{
\begin{bmatrix}
p \\ 1
\end{bmatrix}}_{p_h}
$$
simply as a fact of linear algebra. This can be interpreted as moving $p$ to homogeneous coordinates $p_h$, and then constructing $M$ by appending $t$ to it (after moving $t$ to homogeneous coordinates). 
From a geometric perspective, why must we place zeros underneath $R$ (which is the solution you discovered)? It is because we need the last element of $q_h$ (say $a$) to be $1$, else we would be moving (scaling) $q$ by $a$ in real space! 
A: There are deeper points here that are worth noting: 


*

*Translation is not a linear transformation. Hence it cannot be written as a multiplication by a  matrix (unless one uses clever tricks).

*Rotating and then translating will give different results to translating and then rotating.
Proof that translation is not linear: Define $T(x)$ as the transformation that translates the point $x$ by a displacement $b$. Then $T(x) = x + b$ for any point $x$. Now suppose that $T$ is a linear transformation. Then by definition we must have $T(x+y) = T(x) + T(y)$ for all $x$ and $y$. Now $T(x+y) = x+y + b$ but $T(x) + T(y) = x+b + y+b = x+y + 2b$ so we have equality for all $x$ and $y$ if and only if $b=0$.
Reference that $T(x)$ is a linear transformation (on a finite-dimensioned vector space) if and only if $T(x) = Ax$ for some matrix $A$: Wolfram and it is also a common student exercise.
