Question: Determine the conditions on $b\_1$, $b\_2$, $b\_3$, if any, in order to guarantee that the linear system is consistent.

$x + 3y − 2z = b\_1$ 

$5x − 5y + 4z = b\_2$ 

$−3x + 11y − 8z = b\_3$.

Consider the system of equations

$x + y + 2z = a $

$x + z = b$

$2x + y + 3z = c$. 

Show that in order for this system to be consistent, $a$, $b$, and $c$ must satisfy $c=a+b$.

I need to determine the conditions of the $b$'s. I tried to solve this by using Gauss elimination to convert it into echelon form, but that seems unnecessarily cumbersome to solve this. I was hoping there would be an easier way to do this as well as for the the second part it seems they need the same concept which I don't know. Thank you in advance.

Answer: 
In general, the condition you are looking for is known (more often than not) as the [Rouché-Capelli theorem](https://en.m.wikipedia.org/wiki/Rouch%C3%A9%E2%80%93Capelli_theorem).

You construct the matrix of the coefficients of the equations $A$; in the first case:
$$ A = \begin{bmatrix}
1 &3& -2\\
5 &-5&4\\
-3&11&-8
\end{bmatrix}.$$

Then, you impose that the rank of $A$ is equal to the rank of the augmented matrix you get by adding the right-hand sife of each equation to each row of $A$; in the first case:
$$ [A|b] = \begin{bmatrix}
1 &3& -2&b\_1\\
5 &-5&4&b\_2\\
-3&11&-8&b\_3
\end{bmatrix}.$$

If $A$ and $[A|b]$ have the same rank, the system has a solution.
Supplementing [this other answer](https://math.stackexchange.com/a/3214716/265466), a different way of characterizing the equal-rank condition is that $(b\_1,b\_2,b\_3)^T$ must lie in the column space of the coefficient matrix $A$ for there to be any solution at all, so what you’re really looking for is a description of this space. Row-reducing $A^T$ (equivalently, *column*-reducing $A$) will generate a “nice” basis, in the sense that the basis vectors have a simple, regular structure. In this case, $$\begin{bmatrix}1&5&-3\\3&-5&11\\-2&4&-8\end{bmatrix} \to \begin{bmatrix}1&0&2\\0&1&-1\\0&0&0\end{bmatrix},$$ so a basis for $C(A)$ consists of $(1,0,2)^T$ and $(0,1,-1)^T$. In other words, every vector in $C(A)$ is of the form $b\_1(1,0,2)^T+b\_2(0,1,-1)^T = (b\_1,b\_2,2b\_1-b\_2)^T$, which gives you the condition $b\_3=2b\_1-b\_2$. Similarly, in the second problem you can compute the basis $\{(1,0,1)^T,(0,1,1)^T\}$, from which $c=a+b$.