Given a matrix and its RREF, find missing columns We are given the matrix $$A=\begin{pmatrix}
-1&|&-6&1&2&| \\ 
2&c_2&4&3&0&c_6 \\ 
3&|&4&-1&1&|\\ 
0&|&-4&1&1&| 
\end{pmatrix}$$
$$C=\begin{pmatrix}
\color{red}{1}&0&2&0&0&0\\ 
0&\color{red}{1}&-2&0&0&2\\ 
0&0&0&\color{red}{1}&0&1 \\ 
0&0&0&0&\color{red}{1}&-1
\end{pmatrix}$$
where $C$ is the $rref(A)$.
I am asked to find the columns $c_2, c_6$ in the origin matrix $A$, but I have no idea how to do this. The only thing I have done is colored the pivots in. How do I proceed for here?
 A: To reverse engineer the echelon form we must go back, step by step how the matrix $A$ has been reduced. Evidently we replace the values of the "missing" columns by indeterminates. We know that the first row of $C$ is a linear combination of the rows of $A$ but we don't know which were the coefficients, so call them  $x,y,z$ and $u$ so we have $\begin{pmatrix} x & y & z & u\end{pmatrix} A = C[1]$ giving the system of equations
$$
\begin{cases}
-x+2y+3z = 1 \\ xa+yc+ze+ug  = 0 \\ -6x+4y+4z-5u = 2 \\
      x+3y-z+u = 0 \\ 2x+z+u = 0 \\ xb+yd+zj+uh =0
\end{cases}
$$
If we keep away the second and last equation (which don't learn us anything) we can solve the coefficients $x, y, z$  and $u$, namely $x = -1/9, y = 1/9, z= 2/9, u = 0$. Moreover we well retain the discarded relations as $ -a/9+y/9+2e/9 = 0 $ and $-b/9+d/9+2j/9 = 0$. This completes the first step.
For the second step we note that the second row of $C$ is a combination of the last three rows with unknown coefficients $x, y, z$ . Just like in the first step this will give us five equations in which we discard two and use the others to solve for $x, y, z$ and substitute these values into the discarded equations.
After having performed step 3 and step 4 we end up with eight "discarded" equations in eight unknown. Solving this system will give you the desired solution.
