How does a row of zeros make a free variable in linear systems of equations? I don't understand how a row of zeros gives a free variable when solving systems of linear equations. Here's an example matrix and let us say that we're trying to solve Ax=0:
$$\left[
        \begin{matrix}
        2 & -3 & 0 \\
        3 & 5 & 0 \\
        0 & 1 & 0 \\
        \end{matrix}
\right]$$
This makes sense to me - you have a COLUMN of numbers corresponding to the number of times the variable $x_3$ is used for each equation and since they are all zeros, $x_3$ could have been any real number because we never get to manipulate our equations to determine a value for it. Hence, it is a free variable as it is not subject to the constraints of the equations.
$$\left[
        \begin{matrix}
        2 & -3 & 5 \\
        0 & 5 & 1 \\
        0 & 0 & 0 \\
        \end{matrix}
\right]$$
Now for this second example, $x_3$ would still be a free variable. Why is this so? $x_3$ is being used in the other two equations where you could certainly come up with a finite set of answers for this variable rather than saying "It could've been anything!" right?
Also is it entirely arbitrary that $x_3$ is the free variable or could it be decided that $x_1$ or $x_2$ is free instead?
Could someone explain to me in a more layman or simplified form on why a row of zeros magically makes a free variable?
Please help me :(

How many $\langle x,y,z\rangle$ satisfy both $x + 5y + 3z = 0$ and $2x-3y+z =0$?

 Infinitely many: $x + 5y + 3z = 0$ and $2x-3y+z =0$ are non-parallel planes; as such, it should be obvious that they intersect in a line, and there is no unique solution to this system.

What does this question have to do with $\begin{bmatrix}1&5&3 \\ 2&-3&1 \\ 0&0&0\end{bmatrix}$?

 The above is the exact same problem as finding the solutions to $\begin{bmatrix}1&5&3 \\ 2&-3&1 \\ 0&0&0\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}0\\ 0\\ 0\end{bmatrix}.$
 
 The key to realize here is that if a system of equations is going to have finite solutions, you need a certain number of equations. Having a row of zeroes means you have one less equation than necessary to have finite solutions, and hence you have an infinite set of solutions.

It is indeed entirely arbitrary that $x_3$ is the free variable. Remember, we're just choosing variable names at our own convenience - if we let $\mathbf{x}=\begin{bmatrix}x_1\\ x_3 \\x_2\end{bmatrix}$, then $x_2$ would be our free variable. But because it makes sense to name the coordinates of $\mathbf{x}$ in numerical order, it's much easier to remember and to write $\mathbf{x}=\begin{bmatrix}x_1\\ x_2 \\x_3\end{bmatrix}$ instead.
