System of equations from given solution I'm stuck at part b) of the following problem:
a) Find a $2\times 3$ system $Ax = b$ whose complete solution is
$$\vec x = \begin{bmatrix}
1\\
2\\
0
\end{bmatrix} + w \begin{bmatrix}
1\\
3\\
1
\end{bmatrix}.$$
b) Find a $3\times 3$ system with these solutions exactly when $b_1 +b_2 = b_3$.
(Linear Algebra and Its Applications by Gilbert Strang - Chapter 2.2 problem 10)

For part a) I arrived at the following solution:
$$\begin{bmatrix}
1 & 0 & -1\\
0 & 1 & -3
\end{bmatrix} \vec x = \begin{bmatrix}
1\\
2
\end{bmatrix} \quad \quad (R \vec x = \vec b)$$
by "reading of" the entries for $x_1$ and $x_2$ from the nullspace $w\begin{bmatrix}
1\\
3\\
1
\end{bmatrix}$ and putting them into a row reduced echelon form matrix $R$ with switched signs and by again reading of the values for $b_1$ and $b_2$ from the particular solution $\begin{bmatrix}
1\\
2\\
0
\end{bmatrix}$.
For part b) I thought that I could do the same:
The matrix  $R=\begin{bmatrix}
1 & 0 &-1\\
0 & 1 &-3\\
0 & 0 &0
\end{bmatrix}$ has $N(R) = w\begin{bmatrix}
1\\
3\\
0
\end{bmatrix}$. Now I wanted to solve for $\vec b$:
$$\begin{bmatrix}
1 & 0 &-1\\
0 & 1 &-3\\
0 & 0 &0
\end{bmatrix} \vec x =\begin{bmatrix}
b_1\\
b_2\\
b_3
\end{bmatrix} $$
$$\begin{bmatrix}
1 & 0 &-1\\
0 & 1 &-3\\
0 & 0 &0
\end{bmatrix} \vec x =\begin{bmatrix}
b_1\\
b_2\\
b_1+b_2
\end{bmatrix} $$
Since $\vec x = \begin{bmatrix}
1\\
2\\
0
\end{bmatrix}$ is a particular solution: $b_1 = 1$,$b_2 = 2$.
However: $b_1 + b_2 = 3$ which makes the system inconsistent. Is there a system with the given solution which is consistent? And if there is one, where did I make a mistake?
 A: Suppose the matrix $A=\begin{bmatrix}\mathbf{3u} & \mathbf{-(u+v)} & 3\mathbf{v}\end{bmatrix}$, where $\bf{u}$ and $\bf{v}$ are $3 \times 1$ column vectors. The reason for starting with this $A$ is to make sure that the null space of $A$ is spanned by $(1,3,1)$ (as required by the problem). (Note: of course we want to make sure that $\bf{u}$ and $\bf{v}$ are independent so that null space is $1-$dimensional.)
Now we want $(1,2,0)$ to be a solution for $Ax=\begin{bmatrix}b_1\\b_2\\b_1+b_2\end{bmatrix}$. This means we want
$$1(\mathbf{3u})-2 \mathbf{(u+v)}+0\mathbf{(3v)}=\begin{bmatrix}b_1\\b_2\\b_1+b_2\end{bmatrix}.$$
This is equivalent to
$$\mathbf{u}-2 \mathbf{v}=\begin{bmatrix}b_1\\b_2\\b_1+b_2\end{bmatrix}.$$
So for example, if we have
$$\mathbf{u}=\begin{bmatrix}b_1\\b_2\\0\end{bmatrix},$$
then we can have
$$\mathbf{v}=\begin{bmatrix}0\\0\\-\frac{b_1+b_2}{2}\end{bmatrix}.$$
Now you have your $A$. If you assume that at least one of $b_1$ or $b_2$ is non-zero, and $b_1+b_2 \neq 0$, then the conditions are (easily) satisfied.
Your error: is that while you are row reducing the matrix $A$ but you are not applying the row operations to the right hand side vector $\bf{b}$.
