Given a matrix A=

```
1   2   2
3   1   4
4   3   6

```

find a basis for the column space of A. Then express the 3rd column of A in that basis.

My first thought is to use RREF and than get the matrix from the augmented matrix

```
1 0 1.2 |0
0 1 .4  |0
0 0  0  |0 

```

from here I get confused/most likely wrong. Since leading 1s are in row 1 and 2 the basis of this matrix would than be vector {[1, 0, 1.2] and [0, 1, .4]} with X1=-1.2X3 and X2=-.4X3. With X3 being arbitrary. If I made a mistake anywhere let me know, I do not know how to finish this problem.