1. ## homogeneus matrix question..

$
\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
$
\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: $3x_1-x_4=0\Longrightarrow x_4=3x_1$ (if you want the first variable to be the free one), or $x_1=\frac{1}{3}x_4$ , if you want the fourth variable to be the free one.
But this time is better to put $x_4$ as function of $x_1$ since then row 2 gives you $x_6$ as function of the first one, and row 4 gives $x_2$ as function of $x_4=3x_1$ , 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