Linear Independence easy question I have these vectors $B = \{u, v, w\}$ with
$$u = (-1, 1, -1),\, v = (19, 10, -9),\, w = (-1, x, y)$$
And i want to prove that these vectors are linearly independent. 
I have no problem to proove that three vectors without unknown variables are linearly indepedent but i have difficulties in this that i have two unknown variables $(x, y)$.
I find that $\det(B)$ is $-28x - 29y -1$ but i do not know how this helps.
The actual question is to prove that B forms a basis in R3
PS. New to Linear Algebra, dont shoot !
 A: Good question. In fact, the vectors are linearly independent whenever the determinant $-28x -29y -1$ is not equal to zero. So, the answer depends on $x$ and $y$.
How I know this: The determinant can be thought of as a function of $m$ vectors in $m$-dim'l space, in which case one of the properties that characterizes it is that if one of its inputs is a linear combination of some or all of the others, then the function will evaluate to zero. Also, the function will evaluate to zero only in this case. Hence, a set of $m$ vectors in an $m$-dim'l space will be linearly independent if and only if their determinant is nonzero. See this section of wiki article for more details, specifically where it says that properties 1, 7, and 8 completely characterize the determinant.
Back to your specific question. Set $$-28x - 29y - 1 = 0$$ and rewrite it as $$y = - \frac{28}{29}x  -\frac{1}{29}.$$ Your three vectors are linearly independent when $x,y$ lie away from this line, which you can easily plot now that it's written in slope-intercept form. And since you want to know when these vectors form a basis for $V := \mathbb{R}^3$, and since you know that any linearly independent set of 3 vectors in $V$ forms a basis, you know that these vectors form a basis when $x,y$ lie away from the line given above.
Hope that helps.
