Canonical matrix of a transformation 
Let $T: \mathbb{R}^2 \to \mathbb{R}^3$ and $T(-2,3)=(-1,0,1)$ and $T(1,2)=(0,-1,0)$.
Obtain the canonical matrix of $T$ and the transformation $T(x,y)$.

How I obtain the canonical matrix of a transformation?
This is the "official answer": It's hard to see but maybe it gives a light on what was done. But I couldn't understand it.

 A: The canonical matrix is a matrix whose columns are the images of the basis elements under the transformation. In your case, the basis is {$(-2,3),(1,2)$}. Can you see how to set it up?
EDIT According to the comment by Zev Chonoles, the canonical representation of a transformation is one in which the basis vectors are {$(1,0),(0,1)$} ( or, I would guess, $E_i =\delta_{ij}$. In this case, we must first do a change of basis from {$(-2,3),(1,2)$} to {$(1,0),(0,1)$} (assuming, I guess, the bases are ordered ). We then find the change-of-basis matrix and then describe the transformation bases on this.
The matrix $M:=(-2,3)^t(1, 2)^t$ , where $(a,b)^t$ is the transpose of the vector $(a,b)$  takes the basis {$(1,0),(0, 1)$} to {$(-2 3 ),(1 , 2)$} , and the matrix $M^{-1}$ goes in the opposite direction. Notice that $$ T( 1, 0) = T(M^{-1}( 2, 3)=M^{-1}T(2, 3) $$. Can you see how to use this to find  $T(1, 0), T(0,1)$?
A: The canonical basis of the vector space $\mathbb{R}^n$ is the set of vectors $e_{1},\ldots,e_{n}\in\mathbb{R}^n$ defined by
$$e_{1}=\underbrace{(1,0,\ldots,0)}_{n\text{ terms}},\qquad e_{2}=\underbrace{(0,1,\ldots,0)}_{n\text{ terms}},\qquad\ldots,\qquad e_{n}=\underbrace{(0,0,\ldots,1)}_{n\text{ terms}}$$
Given a linear transformation $T:\mathbb{R}^n\to\mathbb{R}^m$, I would say the canonical matrix for $T$ is the $m\times n$ matrix formed by
$$\left[\begin{array}{c|c|c|c}
 &  &  &   \\
T(e_1) & T(e_2) & \cdots & T(e_n)\\
 &  &  &  
\end{array}\right]$$
Thus, for example,
$$\underbrace{\left[\begin{array}{c|c|c|c}
 &  &  &   \\
T(e_1) & T(e_2) & \cdots & T(e_n)\\
 &  &  &  
\end{array}\right]}_{\text{canonical matrix for }T}\underbrace{\left[\begin{array}{c|c|c|c}
1 \\
\vdots\\
0
\end{array}\right]}_{e_1}=T(e_1)$$
Therefore, your task is to use the facts
$$T\begin{bmatrix}
-2\\
3
\end{bmatrix}=\begin{bmatrix}
-1\\
0\\
1
\end{bmatrix}\qquad\qquad T\begin{bmatrix}
1\\
2
\end{bmatrix}=\begin{bmatrix}
0\\
1\\
0
\end{bmatrix}$$
to compute 
$$T\begin{bmatrix}
1\\
0
\end{bmatrix}=\begin{bmatrix}
?\\
?\\
?
\end{bmatrix}\qquad\qquad T\begin{bmatrix}
0\\
1
\end{bmatrix}=\begin{bmatrix}
?\\
?\\
?
\end{bmatrix}$$
How to do that? Here is an example. If there existed real numbers $a$ and $b$ (hint: they exist!) such that
$$\begin{bmatrix}
1\\
0
\end{bmatrix}=\begin{bmatrix}
-2a+b\\
3a+2b
\end{bmatrix}=a\begin{bmatrix}
-2\\
3
\end{bmatrix}+b\begin{bmatrix}
1\\
2
\end{bmatrix}$$
then you could use the fact that $T$ is a linear transformation to conclude that
$$T\begin{bmatrix}
1\\
0
\end{bmatrix}=T\left(a\begin{bmatrix}
-2\\
3
\end{bmatrix}+b\begin{bmatrix}
1\\
2
\end{bmatrix}\right)=a\cdot T\begin{bmatrix}
-2\\
3
\end{bmatrix}+b\cdot T\begin{bmatrix}
1\\
2
\end{bmatrix}=a\begin{bmatrix}
-1\\
0\\
1
\end{bmatrix}+b\begin{bmatrix}
0\\
1\\
0
\end{bmatrix}=\begin{bmatrix}
-a\\
b\\
a
\end{bmatrix}$$
