Let the reduced row echelon form of A = $\begin{bmatrix}1&-3&0&4&0&5\\ 0&0&1&3&0&2\\ 0&0&0&0&1&-1\\ 0&0&0&0&0&0\end{bmatrix}$. Determine A if ... Let the reduced row echelon form of $A$ be $R$ = $\begin{bmatrix}1&-3&0&4&0&5\\ 0&0&1&3&0&2\\ 0&0&0&0&1&-1\\ 0&0&0&0&0&0\end{bmatrix}$.
Determine A if the ﬁrst, third, and sixth columns of A are $\begin{bmatrix} 1\\-2\\-1\\3\end{bmatrix}$, $\begin{bmatrix} -1\\1\\2\\-4\end{bmatrix}$, and $\begin{bmatrix} 3\\-9\\2\\5\end{bmatrix}$, respectively.
Given this information, I know second column of $A$ is $-3$ times the first column (by looking at the first two columns of $R$). How can I solve for the remaining two columns?
$A$ = $\begin{bmatrix}
  1 & -3 & -1 & ? & ? & 3  \\
 -2 & 6  & 1  & ? & ? & -9 \\
 -1 & 3  & 2  & ? & ? & 2  \\
  3 & -9 & -4 & ? & ? & 5
\end{bmatrix}$.

Let $C_{i}$ be the $i$th column. By observing $R$ we notice that $C_{4} = 3C_{3} + 4C_{1}$ and $C_{5} = -C_{6} + 2C_{3} + 5C_{1}$. 
The idea is to find a linear combination of the other columns which each 
 have only one non-zero entry ,to then express the values in the column that we're evaluating. You did right for column $2$.
