Find the transformation matrix given a set of points The map $A:ℝ^3→ℝ^3$ satisfies:
$A(2,−1,2)=(−2,−9,−30)$
$A(−1,2,2)=(−23,12,−9)$
$A(2,2,−1)=(13,6,−6)$
So if I understand it correctly there is a matrix $A$ such that
$ [A] \begin{bmatrix}2\\-1\\2\end{bmatrix} = \begin{bmatrix}-2\\-9\\-30\end{bmatrix}$
and
$ [A] \begin{bmatrix}-1\\2\\2\end{bmatrix} = \begin{bmatrix}-23\\12\\-9\end{bmatrix}$
and
$ [A] \begin{bmatrix}2\\2\\-1\end{bmatrix} = \begin{bmatrix}13\\6\\-6\end{bmatrix}$
I'm not sure where to start. 
 A: The ordinary way to find matrix of a linear transformation according to a given ordered basis is: calculate the image of elements of the basis so the coefficients of the first element will be the first column  of the matrix and so on. That is you have find the matrix.  
A: Notice the vectors you have listed are linearly independent and call this basis $\alpha$. Thus you have a basis for $\mathbb{R}^3$. Find the transformation matrix with respect to $\alpha = \mathbb{R}^3$ and you have the matrix for $T$. That is you will have to solve the system:
$$[A]_{\alpha}^{\alpha}[v]_{\alpha} = [T(v_1),  T(v_2), T(v_3)]_{\alpha}$$
And then row reduce and solve for $A$. This can be done by ($v_i$ are column vectors):
$$A[v_1 \ v_2 \ v_3] = [A(v_1) \ A(v_2) \ A(v_3)]$$
And now solve for $A$ by $Av_i = [A(v_i)]_{\alpha}$, you have both sides of the equation
A: An answer for noobs.
You said you had this:
(1)
$ [A] \begin{bmatrix}2\\-1\\2\end{bmatrix} = \begin{bmatrix}-2\\-9\\-30\end{bmatrix}$
(2)
$ [A] \begin{bmatrix}-1\\2\\2\end{bmatrix} = \begin{bmatrix}-23\\12\\-9\end{bmatrix}$
(3)
$ [A] \begin{bmatrix}2\\2\\-1\end{bmatrix} = \begin{bmatrix}13\\6\\-6\end{bmatrix}$

$ A = \begin{bmatrix}A_{00} A_{01} A_{02}\\ A_{10} A_{11} A_{12} \\ A_{20} A_{21} A_{22} \end{bmatrix}$
So, the first equation is actually:
$ 2A_{00}-1A_{01}+2A_{02}+0A_{10}+0A_{11}+0A_{12}+0A_{20}+0A_{21}+0A_{22}=-2$
$ 0A_{00}+0A_{01}+0A_{02}+2A_{10}-1A_{11}+2A_{12}+0A_{20}+0A_{21}+0A_{22}=-9$
$ 0A_{00}+0A_{01}+0A_{02}+0A_{10}+0A_{11}+0A_{12}+2A_{20}-1A_{21}+2A_{22}=-30$
Doing this for all 3 equations, you get the matrix M in:
$ M \begin{bmatrix}A_{00} \\ A_{01} \\ A_{02}\\ A_{10} \\ A_{11} \\ A_{12} \\ A_{20} \\ A_{21} \\ A_{22} \end{bmatrix} = \begin{bmatrix}-2 \\ -9 \\ -30 \\ -23 \\ 12 \\ -9 \\ 13 \\ 6 \\ -6 \end{bmatrix}$
Then solve for vector A.
