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?