Transformation and Matrices - Points and Vectors Right question, I am stuck. We have been working on matrices and I think I understand them, however I have no idea how to apply these to this transformation question.
Consider the points O = (0; 0; 0), A = (1; 3; 3) and B = (-2; -6; 4)
together with the transformation which carries out a scaling to increase
distances in the x direction by a factor of two, decrease those in the
y - direction by a factor of three and leave distances in the z - direction
unchanged. It also leaves the point A unchanged. What is the matrix
associated with this transformation? Use the matrix to find the images
of the following
(a) the point O
(b) the point A
(c) the point B
(d) the vector A->B
(e) the vector O->B

 A: If the matrix you're looking for is a 3x3 matrix, then no such matrix exists, since
$A = x + 3y + z$, so $MA = Mx + 3My + 3Mz = (2,1,3) \neq A$.
But since you make the distinction between points and vectors, are you maybe working in projective space?
Then take the matrix
$
N = \begin{bmatrix}
1 & 0 & 0 & A_x \\
0 & 1 & 0 & A_y \\
0 & 0 & 1 & A_z \\
0 & 0 & 0 & 1 \\
\end{bmatrix}
$
and let
$M = NSN^{-1}$, 
where $S$ is the diagonal matrix with entries $(2,\frac{1}{3},1,1)$.
A: The above mentioned transformation can be written as an affine mapping in $R^3$:
$$
\begin{bmatrix}
x' \\ y' \\ z' \\ 
\end{bmatrix}
= 
\begin{bmatrix} 
2 & 0 & 0 \\ 0 & 1/3 & 0 \\ 0 & 0 & 1 \\ 
\end{bmatrix}
\begin{bmatrix}
x \\ y \\ z \\ 
\end{bmatrix}
+
\begin{bmatrix}
-1 \\ 2 \\ 0 \\ 
\end{bmatrix}
$$
It can also be written as a linear mapping in $R^4$ by introducing homogenous coordinates:
$$
\begin{bmatrix}
x' \\ y' \\ z' \\ w' 
\end{bmatrix}
= 
\begin{bmatrix} 
2 & 0 & 0 & -1 \\ 0 & 1/3 & 0 & 2 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 
\end{bmatrix}
\begin{bmatrix}
x \\ y \\ z \\ 1 
\end{bmatrix}
$$
Using any one of these equations we get for the images (...)' of (...):
(a) O' = (-1, 2, 0)
(b) A' = (1, 3, 3) = A
(c) B' = (-5, 0, 4)
(d) (A -> B)' = B' - A' = (-6, -3, 1)
(e) (O -> B)' = B' - O' = (-4, -2, 4) 

