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. 

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.
