TITLE: How to check if a set of vectors is a basis
QUESTION [31 upvotes]: OK, I am having a real problem with this and I am desperate. 
I have a set of vectors. {(1,0.-1), (2,5,1), (0,-4,3)}
How do I check is this is a basis for $R^3?$
My text says a basis B for a vector space V is a linearly independent subset of V that generates V. OK then. I need to see if these vectors are linearly independent, yes? 
If that is so, then for these to be linearly independent the following must be true: 
$a_1v_1 + a_2v_2 + ... + a_nv_n $ ≠ 0 for any scalars $a_i$
Is this the case or not? 
If it is, then I just have to see if 
$a_1(1,0.-1)+ a_2(2,5,1)+ a_3(0,-4,3)$ = 0 
or 
$a_1 + 2a_2 + 0a_3 = 0$ 
$0a_1 + 5a_2 - 4a_3 = 0$ 
$-a_1 + a_2  + 3a_3 = 0$ 
has a solution. 
Adding these equations up I get $8a_2 - a_3 = 0$ or $a_3 = 8a_2$ 
so $5a_2 - 32a_2 = 0$ which gets me $a_2 = 0$ and that implies $a_1 = 0 $and$ a_3=0$ as well. 
So they are all linearly dependent and thus they are not a basis for $R^3$. 
Something tells me that this is wrong. But I am having a hell of a time figuring this stuff out. Please someone help, and I ask: pretend I am the dumbest student you ever met.

REPLY [8 votes]: The easiest way to check whether a given set $\{(a,b,c),(d,e,f),(p,q,r)\} $of three vectors are linearly independent in $\Bbb R^3$ is to find the determinant of the matrix, $$\begin{bmatrix}
a & b & c\\
d & e& f\\
p&q&r
\end{bmatrix}$$ is zero or not.
If the determinant is zero then the set is linearly dependent else i.e. determinant is nonzero it is linearly independent. Which is the same as saying that the system of linear equations, $$\begin{bmatrix}a&b&c\\d&e&f\\p&q&r\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$$ has only the trivial solution.