How to obtain a column of a matrix representing a homogeneous linear system by the value of the adjacent column? A homogeneous linear system of equations has a coefficient matrix A which
is row equivalent to the following matrix R in reduced echelon form:
\begin{bmatrix}
  1 & 2 & 0 & 3 & 0 &5\\
  0 & 0 & 1 &4&0&2 \\
  0 & 0 & 0&0&1 &-3 
\end{bmatrix}
(a) Describe the solution set in parametric vector form.
(b) Suppose the first column of A is:
\begin{bmatrix}
  2\\
  3 \\
  5
\end{bmatrix}
What is the second column of A?
(c) If, in addition to the above, the third column of A is:
\begin{bmatrix}
  3\\
  -1 \\
  4
\end{bmatrix}
what is the fourth column of A? Hint: Use special solutions of Ax = 0
obtained by setting one free variable to 1, and the others to 0. Each
solution of Ax = 0 gives a linear combination of the columns of A that
is 0. The special solutions make it easy to solve for the free columns
in terms of the pivot columns.
I already did part (a) but I am unsure on how to get the columns of A. Could I just obtain them through performing row operations on R, since they are row equivalent? And how can I use special solutions to solve for the fourth column?
 A: In part (a), we find that every solution to $A\vec{x}=\vec{O}$ is of the form
\begin{align*}
\vec{x}
&= \left[\begin{array}{r}
x_{1} \\
x_{2} \\
x_{3} \\
x_{4} \\
x_{5} \\
x_{6}
\end{array}\right] 
= \left[\begin{array}{r}
-2 \, x_{2} - 3 \, x_{4} - 5 \, x_{6} \\
x_{2} \\
-4 \, x_{4} - 2 \, x_{6} \\
x_{4} \\
3 \, x_{6} \\
x_{6}
\end{array}\right] \\
&= x_2\left[\begin{array}{r}
-2 \\
1 \\
0 \\
0 \\
0 \\
0
\end{array}\right]+x_4\left[\begin{array}{r}
-3 \\
0 \\
-4 \\
1 \\
0 \\
0
\end{array}\right]+x_6\left[\begin{array}{r}
-5 \\
0 \\
-2 \\
0 \\
3 \\
1
\end{array}\right]
\end{align*}
This means that the three vectors
\begin{align*}
\vec{v}_1 &= \left\langle-2,\,1,\,0,\,0,\,0,\,0\right\rangle &
\vec{v}_2 &= \left\langle-3,\,0,\,-4,\,1,\,0,\,0\right\rangle & 
\vec{v}_3 &= \left\langle-3,\,0,\,-4,\,1,\,0,\,0\right\rangle
\end{align*}
form a basis of $\operatorname{Null}(A)$.
In particular, the vector $\vec{v}_1=\left\langle-2,\,1,\,0,\,0,\,0,\,0\right\rangle$ satisfies $A\vec{v}_1=\vec{O}$. This means that
$$
(-2)\cdot\vec{a}_1+1\cdot\vec{a}_2+0\cdot\vec{a}_3+0\cdot\vec{a}_4+0\cdot\vec{a}_5+0\cdot\vec{a}_6=\vec{O}
$$
where $\{\vec{a}_1, \vec{a}_2, \vec{a}_3, \vec{a}_4, \vec{a}_5, \vec{a}_6 \}$ are the columns of $A$. This gives $\vec{a}_2=2\cdot\vec{a}_1$. 
So, if the first column of $A$ is $\vec{a}_1=\left\langle2,\,3,\,5\right\rangle$, then the second column of $A$ is $\vec{a}_2=2\cdot\vec{a}_1=\left\langle4,\,6,\,10\right\rangle$.
Also note that the vector $\vec{v}_2=\left\langle-3,\,0,\,-4,\,1,\,0,\,0\right\rangle$ satisfies $A\vec{v}_2=\vec{O}$. This means that
$$
3\cdot\vec{a}_1+0\cdot\vec{a}_2+(-4)\cdot\vec{a}_3+1\cdot\vec{a}_4+0\cdot\vec{a}_5+0\cdot\vec{a}_6=\vec{O}
$$
Solving for $\vec{a}_4$ gives
$$
\vec{a}_4 = -3\cdot\vec{a}_1+4\cdot\vec{a}_3
$$
So, if $\vec{a}_1=\left\langle2,\,3,\,5\right\rangle$ and $\vec{a}_3=\left\langle3,\,-1,\,4\right\rangle$, then
$$
\vec{a}_4 
= -3\cdot\vec{a}_1+4\cdot\vec{a}_3
= -3\cdot\left\langle2,\,3,\,5\right\rangle+4\cdot\left\langle3,\,-1,\,4\right\rangle
= \left\langle6,\,-13,\,1\right\rangle
$$
