Question about describing the range of linear transformations So say I have some linear transformation $T(\vec{x}) = A\vec{x}$
So we say
$$
\left[
\begin{array}{ccc} 
a_1 & \dots & a_n\\
a_1 & \ddots & \vdots\\
a_1 & \dots & a_n\\
\end{array}
\right]
\times
\left[
\begin{array}{c}
x_1 \\
\vdots \\
x_n \\
\end{array}
\right]
=
\left[
\begin{array}{c}
b_1 \\
\vdots \\
b_n \\
\end{array}
\right]
$$
So my question is, if we take the following augmented matrix: 
$$
\left[
\begin{array}{ccc|c} 
a_1 & \dots & a_n & b_1\\
a_1 & \ddots & \vdots & \vdots\\
a_1 & \dots & a_n & b_n\\
\end{array}
\right]
$$
If we were to put the above matrix into reduced row echelon form, (assuming we are given some numbers to plug into matrix A), while leaving the b vector in terms of $b_1, b_2, ... b_n$, the right side of the matrix would end up being just a bunch of combinations of the terms $b_1, b_2, ... b_n$. 
My question is, what does this resulting vector tell us about the range of this function.
I am interested in any usefulness that this new vector has, or any information we can glean from it.
 A: The question of whether $A\mathbf{x}=\mathbf{b}$ has a solution is equivalent to asking whether $\mathbf{b}$ lies in the columnspace of $A$. 
Elementary row operations do not respect the columns space; however, every time you row-reduce the augmented matrix, you have a new system $A'\mathbf{x}=\mathbf{b}'$, and what you know is that $A\mathbf{x}=\mathbf{b}$ has a solution if and only if $A'\mathbf{x}=\mathbf{b}'$ has a solution, and in fact it has the same solutions. That means that if $\mathbf{b'}$ is in the column space of $A$, and you can write
$$\mathbf{b'} = \alpha_1 A'_1 + \cdots + \alpha_k A'_k,$$
(where $A'_i$ is the $i$th column of $A'$, and $\alpha_i$ are scalars), then you will also have
$$\mathbf{b} = \alpha_1A_1 + \cdots + \alpha_kA_k$$
(where $A_i$ is the $i$th column of $A$), with the same scalars. And conversely.
The point of going all the way to the reduced row-echelon form is that it makes it very easy to see what the solutions to $A'\mathbf{x}=\mathbf{b}'$ are; using those solutions you can express the original $\mathbf{b}$ as a linear combination of the columns of the original $A$.
(What is happening "behind the scenes" is that each time you do an elementary row operation, you are "really" performing an invertible linear transformation $E$, so that you are going from asking "is $\mathbf{b}$ in the range of $A$?" to asking "is $E\mathbf{b}$ in the range of $EA$?"; because $E$ is invertible, you can go back and forth from one question to the other).
(I'm not 100% sure that I am answering what you are asking; I hope I'm not too far off the mark)
