<s>[INST] Find $b_3$ in a $3\times 1$ range vector 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.
 [/INST] You start with the Augmented Matrix
$$A=\left(\begin{array}{rrrr|r}
2 & 1 & 5 & 0&5\\
1&2&4& -3&2\\
1 & 1 & 3 & -1&b_3
\end{array}\right)$$
and you want to find a value of $b_3$ for which the matrix corresponds to a system with a solution; that is, you want to make sure that after you row reduce, you don't have a line which has all $0$s before the vertical bar equal to something with a nonzero thing after it (that would correspond to an equation that reads "$0=\text{something nonzero}$", which is bad).
If we divide the first row by $2$, we get (avoid mixed fractions! They lead to errors)
$$\left(\begin{array}{rrrr|r}
1 & \frac{1}{2} & \frac{5}2 & 0&\frac{5}{2}\\
1&2&4& -3&2\\
1 & 1 & 3 & -1&b_3
\end{array}\right)$$
Then you seem to say that you subtracted Row 1 from both Row 2 and Row 3; if you do that, we get:
$$\left(\begin{array}{rrrr|r}
1 & \frac{1}{2} & \frac{5}{2} & 0 & \frac{5}{2}\\
0 & \frac{3}{2} & \frac{3}{2} & -3 & -\frac{1}{2}\\
0 & \frac{1}{2} & \frac{1}{2} & -1 & b_3-\frac{5}{2}
\end{array}\right).$$
(Aside: it would have been easier, and avoided fractions, to first exchange rows 1 and 2).
Your next observation was that if you divide the second row by $3$, you get something whose first four entries equal the third row:
$$\left(\begin{array}{rrrr|r}
1 & \frac{1}{2} & \frac{5}{2} & 0 & \frac{5}{2}\\
0 & \frac{1}{2} & \frac{1}{2} & -1 & -\frac{1}{6}\\
0 & \frac{1}{2} & \frac{1}{2} & -1 & b_3-\frac{5}{2}
\end{array}\right).$$
Now, if we subract the second row from the third, we would get:
$$\left(\begin{array}{rrrr|r}
1 & \frac{1}{2} & \frac{5}{2} & 0 & \frac{5}{2}\\
0 & \frac{1}{2} & \frac{1}{2} & -1 & -\frac{1}{6}\\
0 & 0 & 0 & 0 & b_3-\frac{5}{2}+\frac{1}{6}
\end{array}\right).$$
Now remember: for the system to have a solution, we cannot have a row that has $0\ 0\ 0\ 0|x$ with $x\neq 0$. So that final row tells us that we must have:
$$b_3 - \frac{5}{2}+\frac{1}{6} = 0.$$
This tells us what $b_3$ needs to be:
$$b_3 = \frac{5}{2}-\frac{1}{6} = \frac{14}{6}=\frac{7}{3}.$$
This is the same answer you got. And indeed, what we did above shows that $b_3=\frac{7}{3}$ is the only value that can work: any other value for $b_3$ would lead to a system that is inconsistent, and so has no solutions.
Explanation. From the perspective of systems of linear equations, all we are doing is transforming the system into simpler and simpler systems, just like we usually do with row reduction.  The key is that the third equation is a linear combination of the first two: the first equation reads
$$2x+y+5z + 0w = b_1$$
and the second reads
$$x+2y+4z-3w = b_2;$$
if you add them, we get
$$3x + 3y + 9z - 3w = (b_1+b_2),$$
and you can note that the third equation is
$$x+y+3z-w = b_3.$$
If the system is consistent, then we must have
$$\begin{align*}
\frac{1}{3}(b_1+b_2) &= \frac{1}{3}(3x+3y+9z-3w)\\
&= x+y+3z-w\\
&= b_3,
\end{align*}$$
which is why we have $b_3 = \frac{7}{3} = \frac{1}{3}(5+2) = \frac{1}{3}(b_1+b_2)$. The linear dependency between the first two rows of $A$ and the third forces the same linear dependency between the first two entries of $\mathbf{b}$ and the third. The row reduction just takes the guessing and luck out of figuring out the linear dependencies between the rows.
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