Suppose we are solving a system of linear equation and arrive at the following (augmented) matrix:
$ \begin{bmatrix} 1& 6 & 0 & 3 & 0 & 0\\ 0& 0 & 1 & -4 & 0 & 5\\ 0& 0 & 0 & 0 & 1 & 7 \end{bmatrix} $
The question that arises is then: Which variables are basic and which are free? I know how to pick them up: You look for pivot and non-pivot columns. But what I wonder is: Why is it that non-pivot columns are those containing free variables and pivot columns are those containing the basic variables?
You seem to draw your example from here. Now, in that example, it seems quite easy to reverse the roles, and have $x_1$, $x_3$, and $x_5$ as free variables. However, this is not always the case.
First, consider a regular system, for example
$$A = \begin{bmatrix} 1 & 2 & 3 \\ & 4 & 5 \\ & & 6 \end{bmatrix}.$$
We cannot really choose anything here, right? We have $3$ pivot columns, and $3$ basic variables, while we have no non-pivot columns and no free variables.
Another example, where we do have some free variables, would be something like this:
$$A = \begin{bmatrix} 1 & 2 & 3 & 4 & 5 \\ & & & 6 & 7 \end{bmatrix}.$$
Obviously, we have only $2$ basic variables, and $3$ free ones. Pivot columns correspond to $x_1$, and $x_4$, which is $2$.
If you think about it, the number of pivot columns corresponds to the rank of the system, which is also the number of basic variables. Equivalently, the number of non-pivot columns corresponds to the number of free variables. That's why it makes sense to use that correspondence for a general rule, even though sometimes other choices would also do well.