Given the following two graphics I need to matrix A so that T(x) = Ax.
What I see is that 0 is a fixed point at the origin after the transformation, so I don't need to worry about it when finding matrix A.
I also found that:
b* = 2[0 1/2]T
a* = 2[1/2 0]T
c* = 3[1/3 1/3]T
Now I don't know how I can use this to find the matrix A. What is especially confusing to me is that a and b have swapped places so I don't know how it will effect the matrix A.
Sorry for the formatting.
On
We have $$T(3,3)= (1,1)\;\;\;{\rm and}\;\;\;T(2,1)= (0,1)$$ so $$T(1,0) = (-{1\over 3},{2\over 3}) \;\;\;{\rm and}\;\;\; T(1,0) = ({2\over 3},-{1\over 3})$$
and thus $$T= \begin{pmatrix}-{1\over 3}&{2\over 3}\\ {2\over 3}&-{1\over 3} \end{pmatrix} $$
Edit $$T(3,0) =3T(2,1)-T(3,3) = (0,3)-(1,1) = (-1,2)$$ now we divide this by 3 and thus $$T(1,0) = (-{1\over 3},{2\over 3}) $$ Similary for $T(0,1)$
Look at $T^{-1}$. As the columns of $A^{-1}$ you can choose the vectors 1. going from $0$ to $a$ and 2. going from $0$ to $b$. The vector from $0$ to $a$ is $(1,2)^T$ and the other one is $(2,1)^T$, so we get
$$A^{-1}=\begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}.$$
Inverting yields
$$A=\frac{1}{3}\begin{pmatrix} -1 & 2 \\ 2 & -1 \end{pmatrix}.$$
Our transformation map is then
$$T: \mathbb{R}^2 \longrightarrow \mathbb{R}^2, \ x \longmapsto Ax = \frac{1}{3}\begin{pmatrix} -1 & 2 \\ 2 & -1 \end{pmatrix}x.$$