Why does adding a row to another in an augmented matrix preserves the solution set? I get why multiplying one equation by a constant, and swapping equations, preserves the solution set(s) in a system of equations. 
But what I can't wrap my head around is why adding two rows in an augmented matrix preserves the solution set. Is there a proof/simply reasoning for this? 
Each row in my matrix corresponds to a linear equation. In such a matrix, 
$$
\left[
\begin {matrix}
1 & -2 & \;\;\;1 & \;\;\;0 \\
0  & \;\;\;1 &-4 & \;\;\;4 \\
0 &-3 &\;13 &-9 \\
\end {matrix}
\right]
$$
row one would be $x_1 + -2x_2 + x_3 = 0 $, and so on.
 A: I'm assuming you have a matrix equation of the form $Ax = b$. A row of the augmented matrix corresponds to a linear equation with $x_1, x_2, \ldots, x_n$ as the unknowns.
Let the two rows you are adding be $[p_1~p_2~\cdots~p_n~c]$ and $[q_1~q_2~\cdots~q_n~d]$, where the $p_i$ and $q_i$ are from $A$, and $c$ and $d$ from $b$. These correspond to the equations:
$$p_1 x_1 + p_2 x_2 + \cdots + p_n x_n = c \qquad (1)$$
$$q_1 x_1 + q_2 x_2 + \cdots + q_n x_n = d  \qquad (2)$$
If you add $k$ times the first row to the second, you get a new second row $[kp_1 + q_1~kp_2 + q_2~\cdots~kp_n + q_n~kc + d]$, which corresponds to this equation:
$$ (kp_1 + q_1) x_1 + (kp_2 + q_2) x_2 + \cdots + (kp_n + q_n) x_n = kc + d \qquad (2^\prime)$$
The hardest part about this is justifying what exactly you need to check. We must show that $x$ satisfies $(1)$ and $(2)$ if and only if it satisfies $(1)$ and $(2^\prime)$. Checking that these are true just relies on a bunch of distributing and whatever the name for this theorem is: $u = v \textrm{ and } x = y \implies u + x = v + y$.
Forward direction:
Assume $(1)$ and $(2)$ are true. Then $kp_1 x_1 + kp_2 x_2 + \cdots + kp_n x_n = kc$, because we can multiply both sides of an equation by a constant. Then, we can add $(2)$ to this, getting: $(kp_1 + q_1) x_1 + (kp_2 + q_2) x_2 + \cdots + (kp_n + q_n) x_n = kc + d$, which is exactly what we wanted to show, i.e. equation $(2^\prime)$.
Reverse direction is similar, see if you can work out the details.
A: In principal adding rows will change the solution set. The equations you add might increase the constraint on the solution set, making the solution set smaller. If the original matrix is just $[1,1\mid 0]$ and you add the row $[0,1\mid 0]$ then your solution space goes from matrices of the form $(x,-x)$ down to $(0,0)$.
On the other hand, if you're adding rows that are just linear combinations of the other rows, that would not change the solution set, since they don't add any more constraints.
