vela said:
Sorry, I assumed you were familiar with row-echelon form jargon. A pivot is the first non-zero entry in a row. In this problem, the -2 at the start of the first row is a pivot, so x1 would not be free. Since there are no other non-zero rows, the remaining variables are free.
To add to what Ray said, this isn't the only way to decide which variables are free and which aren't. It's just that once you have the system in row-echelon form, it's a straightforward way to choose.
Your description vs. mine would be a bit clearer if we had different numbers in row 1, so say we have
\left| \begin{array}{cccc}<br />
1&2&0&0\\ <br />
0&0&0&0 \\<br />
0&0&0&0 \end{array} \right|,
corresponding to listing the variables as x_1, x_2, x_3 from left to right.
By some definitions, that makes x_1 a "leading" variable and x_2 free (as well as x_3). That would regard x_2 as an independent variable and would correspond to writing x_1 = -2 x_2. Obviously, we can let x_2 have any value under the Sun and can then determine a corresponding unique value of x_1.
However, there is nothing sacred about the ordering x_1, x_2, x_3. We could re-order them as x_2, x_1, x_3,, giving the augmented system in which columns 1 and 2 are swapped:
\left| \begin{array}{cccc}<br />
2&1&0&0\\<br />
0&0&0&0 \\<br />
0&0&0&0<br />
\end{array} \right| .
Now x_2 is the "leading" variable and x_1 is free. This corresponds to writing
x_2 = -(1/2)x_1. We can let x_1 be anything and then determine the value of x_2.
So, which variables are free depends on how you order the columns; most well-written numerical linear algebra codes allow for column interchanges in order to decrease roundoff errors in finite-wordlength calculations, so would not decide the free variables ahead of time.
RGV