For the matrix $A=\left(\begin{array}{rrrr} 2 & 1 & 5 & 0\\ 1&2&4& -3\\ 1 & 1 & 3 & -1 \end{array}\right)$ I needed to find a range vector $\mathbf{b}$ such that $b_1=5$, $b_2=2$.
I'm not very clear on what algebraic characteristics this vector should have, so as to allow $A\mathbf{x}=\mathbf{b}$ to have solutions. I understand geometrically that all the equations should provide at least non-contradictory inputs.
I used a few steps of elimination by pivoting on $[A|\mathbf{b}]$: $\begin{align*} R_1&:=R_1/2;\\ R_2&:=-R_1;\\ R_3&:=-R_1; \end{align*}$
At this point I noticed that $R_2$ was just 3 times $R_3$ (if only the matrix $A$, not vector b is considered), so I proceeded to solve $-.5=3(b_3-2.5) \implies b_3=2\frac{1}{3}$, which extended this relationship to the rows of vector $\mathbf{b}$. Even though this would mean the third equation was denoting a parallel space, it just seemed easy to solve the equality. When testing with a calculator 2 1/3 for b3 does work and several random values didn't. However, most importantly, I'm not clear about the linear algebra behind this. Please explain this simply and concisely.
Because I was unsure if that's the right way to solve this, I also proceeded with elimination to reach the form: $\left(\begin{array}{rrrr|r} 1&0&2&1&2 \frac{2}{3}\\ 0&1&1&-2 & -\frac{1}{3} \end{array}\right)$
That didn't seem helpful because the only variable I was concerned with, b3, disappeared as the third row became 0.
I apologize for the lengthy post because the question is rather simple, but I was trying to be detailed to help the reader understand what I'm talking about.