<s>[INST] Friedberg Linear algebra exercise 3.4.6 Let the reduced row echelon form of $A$ be 
\begin{pmatrix} 
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{pmatrix}
Determine $A$ if the first, third, and sixth colum of $A$ are 
\begin{pmatrix} 
1 \\
-2 \\
-1 \\
3 
\end{pmatrix}
\begin{pmatrix} 
-1 \\
1 \\
2 \\
-4 
\end{pmatrix}
\begin{pmatrix} 
3 \\
-9 \\
2 \\
5 
\end{pmatrix}

Using (d), I can easily find the first and fourth column of $A$ (for example, the second column of $A$ is the first column of $A$ multiplied by $-3$, similarly I can compute the 4th column), but the 5th column cannot be found in this way. Could you give some help?  
 [/INST] Let $B$ the reduced row echelon form of $A$ and let $a_1,a_2,a_3,a_4,a_5,a_6$ and $b_1,b_2,b_3,b_4,b_5,b_6$ be the ordered columns of $A$ and $B$ respectively.
It is easy to see that $b_2=-3b_1$, so, by $(\textrm d)$ we conclude that also
$$a_2=-3a_1=\begin{pmatrix} -3\\6\\3\\-9 \end{pmatrix}$$
Similarly, the fourth column of $B$, $b_4$, is equal to $4b_1+3b_3$. Hence, the fourth column of $A$ is
$$a_4=4a_1+3a_3=4 \begin{pmatrix} 1\\-2\\-1\\3 \end{pmatrix} + 3 \begin{pmatrix} -1\\1\\2\\-4 \end{pmatrix} = \begin{pmatrix} 1\\-5\\2\\0 \end{pmatrix}$$
Finally, to get the fifth column of $ A $ note that $b_5=5b_1+2b_3-b_6$, that is
$$a_5=\begin{pmatrix} 0\\1\\-3\\2 \end{pmatrix}$$
Therefore, we already have all the columns of $ A $ and then it would look
$$A=\begin{pmatrix}
1&-3&-1&1&0&3 \\
-2&6&1&-5&1&-9 \\
-1&3&2&2&-3&2 \\
3&-9&-4&0&2&5 
\end{pmatrix}$$
</s>