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?
$\endgroup$1 Answer
$\begingroup$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.
$\endgroup$ 2
