TITLE: Given a reduced row exhelon form of a $4 \times 4$ matrix and two columns, how do you find the other two columns?
QUESTION [5 upvotes]: I am given the following :

Let $A$ be a $4 \times 4$ matrix with RREF given by: 
  $$
U = \begin{bmatrix}
1 & 0 & 2 & 1 \\
0 & 1 & 1 & 4 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{bmatrix}.
$$
  If 
  $$
a_1 = \begin{bmatrix} -3 \\ 5 \\ 2 \\ 1 \end{bmatrix} \text{ and } a_2 = \begin{bmatrix} 4 \\ -3 \\ 7 \\ -1 \end{bmatrix},  
$$
  find $a_3$ and $a_4$. [Note: $a_1,a_2,a_3,a_4$ are the columns of $A$.]

This is in the section with column, row, and null space. I don't know how those apply no this problem though... I have absolutely no idea how to do this. Can anyone help point me in the right direction?

REPLY [5 votes]: The (homogeneous) linear relations which hold between the columns of the RREF also hold among the columns of the original matrix.
So your final two columns are precisely $2a_1+a_2$ and $a_1+4a_2$.
$$A = \begin{bmatrix} a_1 & a_2 & a_3 & a_4 \end{bmatrix} = \begin{bmatrix} a_1 & a_2 & 2a_1+a_2 & a_1+4a_2 \end{bmatrix} = \begin{bmatrix} -3 & 4 & -2 & 13 \\ 5 & -3 & 7 & 7 \\ 2 & 7 & 11 & 30 \\ 1 & -1 & 3 & -3 \end{bmatrix}$$ 
As you preform row operation, the columns are never mixed. So if column 3 is 2 times  column 1 plus column 2 in the RREF, it must have been that way in the original  matrix. 
This means that given the RREF and a list of pivot columns (the columns corresponding to leading non-zero row entries) one can easily reconstruct the original matrix. 
This also explains why the pivot columns form a basis for the column space.