1. homogeneus matrix question..

$\displaystyle \begin{pmatrix} 3 &0 &0 &-1 &0 &0 \\ 2& 0& 0 & 0 & 0 &-4 \\ 8&2 &0 &-3 &-1 &0 \\ 0& 1& 0& -1& 0& 0\\ 0& 0 &1 &0 &-1 &0 \end{pmatrix}$
i got here 5 equations
and 6 variables

how to know who is the free variable
without making a triangle of zeros
on the bottom
?

2. Originally Posted by transgalactic
$\displaystyle \begin{pmatrix} 3 &0 &0 &-1 &0 &0 \\ 2& 0& 0 & 0 & 0 &-4 \\ 8&2 &0 &-3 &-1 &0 \\ 0& 1& 0& -1& 0& 0\\ 0& 0 &1 &0 &-1 &0 \end{pmatrix}$
i got here 5 equations
and 6 variables

how to know who is the free variable
without making a triangle of zeros
on the bottom
?

From the matrix there are several "natural" candidates: take row 1, for instance: $\displaystyle 3x_1-x_4=0\Longrightarrow x_4=3x_1$ (if you want the first variable to be the free one), or $\displaystyle x_1=\frac{1}{3}x_4$ , if you want the fourth variable to be the free one.
But this time is better to put $\displaystyle x_4$ as function of $\displaystyle x_1$ since then row 2 gives you $\displaystyle x_6$ as function of the first one, and row 4 gives $\displaystyle x_2$ as function of $\displaystyle x_4=3x_1$ , so as function of the first one, and then 3 row gives you the fifth and...etc.

In fact, and after checking quickly the matrix, the above seems to be the best option, and by far. But it is always better to bring your matrix to echelon form

Tonio