Are these vectors linearly independent? 3 vectors. I have this vector set S = {(2,-1,2), (1,0,3), (3,-2,1)}. Are they linearly independent?
$$(0, 0, 0) = c_1(2,-1,2) + c_2(1,0,3) + c_3(3,-2,1)$$
Here are the corresponding equations:
$$2c_1 + c_2 + 3c_3= 0$$
$$-c_1 - 2c_3 = 0$$
$$2c_1 + + 3c_2 + c_3 = 0$$
->
$$\begin{bmatrix} 2 & 1 & 3 & 0 \\ -1 & 0 & -2 & 0 \\ 2 & 3 & 1 & 0 \end{bmatrix}$$
$$ -> \begin{bmatrix} 2 & 1 & 3 & 0 \\ 0 & 0.5 & 0.5 & 0 \\ 0 & 2 & -2 & 0 \end{bmatrix}$$
$$-> \begin{bmatrix} 1 & 0.5 & 1.5 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$
Let $c_3 = t, c_2 = t, c_1 = -2t$
So, there are infinite non-trivial solutions and as a result the vectors are linearly dependent. Is this right?
 A: Yes, your work is correct. 
You may as well check the determinant of your matrix to see if it is zero in which case the vectors are dependent.
A: Personally, I wouldn't use matrices for a simple set of equations like these.  You have 
$2c_1+ c_2+ 3c_3= 0$
$-c_1- 2c_3= 0$
$2c_1+ 3c_2+ c_3= 0$
From the second equation $c_1= -2c_3$.  So the first equation is $-4c_3+ c_2+ 3c_3= c_2- c_3= 0$ and the second is 
$-4c_3+ 3c_2+ c_3= 3c_2- 3c_3= 0$.  
Both of those equations give $c_2= c_3$ so, yes, there are an infinite number of solutions and these vectors are dependent.
A: I can't comment because I don't have enough rep. But I just want to say that you have shown these vectors to be linearly dependent. At the end of your question you write indepdenent. This might just be a typo, but it is worth making sure. 
These vectors are linearly dependent. Your working proves there are infinitely many choices of $c_1, c_2, c_3$ which form solutions to the equation. Just one such solution would imply these vectors are linearly dependent. 
