Is thi set of vectors, $\{(2, 1), (3, 2), (1, 2)\}$, is linearly dependent or independent? Given a set of vectors 
S = $\left\{
\begin{bmatrix} 2 \\ 1 \end{bmatrix},
\begin{bmatrix} 3 \\ 2 \end{bmatrix},
\begin{bmatrix} 1 \\ 2 \end{bmatrix}
\right\}
 $
Find out if the vectors are linearly dependent or independent 
I know that for a set of vectors to be linearly dependent, they must satisfy the below equation:  
$$c_1v_1 + c_2v_2 ... c_nv_n = \mathbf 0 $$  
such that not all $c_i$ are zero.
So, I decided to apply Gauss Elimination and I got the following equation:
$
c_1\begin{bmatrix} 2 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} 3 \\ 2 \end{bmatrix} + c_3 \begin{bmatrix} 1 \\2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0\end{bmatrix}
$  
And, needless to say, I get an under-determined system of equations below:
$$2c_1 + 3c_2 + c3 = 0$$
$$c_1 + 2c_2 + 2c_3 = 0$$  
And after solving, I get this:
$$c_1 = 4c_3$$
$$c_2 = -3c_3$$
So, $c_3$ is the free variable.  
Assuming $c_3$ is non zero, the vectors are linearly dependent. If it is zero, vectors are linearly dependent.  
How can it be that a free variable decides whether vectors are linearly dependent or not ? Shouldn't it a 100% yes or no answer that does not fluctuate depending on values of constants?
 A: The dimension of the vector space $\mathbb R^2$ is $2$ hence every set of vectors with cardinal greater or equal $3$ is linearly dependant.
A: You say: 

Assuming $c_3$ is non zero, the vectors are linearly dependent. If it is zero, vectors are linearly dependent.

This is not quite correct. A better way to say this is: As we can choose $c_3$ to be nonzero (for example, $c_3=1$ and then by the above equatoins $c_1=4$, $c_2=-3$), the vectors are linearly dependant.
If there were no possibility to choose $c_3\ne0$, i.e. if we had necessarily that $c_3=0$ (which we don't!), then the vectors would be linearly independant.
A: The answer does not fluctuate; you’ve misunderstood the definition of linear independence. A set $\{v_1,\dots,v_n\}$ of vectors is linearly dependent if and only if there is at least one set of coefficients $\{c_1,\dots,c_n\}$ such that
$$c_1v_1+\ldots c_nv_n=0$$
and at least one of the coefficients is non-zero. It is linearly independent if and only if it is not linearly dependent. Your calculations show that $4v_1-3v_2+v_3=0$, with coefficients $4,-3$, and $1$; it’s certainly true that at least one of these is non-zero(!), so your set of vectors is linearly dependent.
Note that it’s always true that $0v_1+\ldots 0v_n=0$; that’s never at issue and does not make the set of vectors linearly dependent. The question is whether there is any other set of coefficients that makes the sum $0$. If there is, the set of vectors is linearly dependent; if not, it’s linearly independent.
A: When you have vectors like $\displaystyle\left[\begin{array}{c}1\\2\end{array}\right]=1\mathbf{i}+2\mathbf{j}$, they live on the plane where there are essentially two different directions: left/right ($\mathbf{i}$) and up/down ($\mathbf{j}$).
Every other direction can be made out of combinations of left/right and up/down; e.g. $1\mathbf{i}+2\mathbf{j}$.
Now for a set to be linearly independent means that altogether, they all describe essentially different directions. A set that isn't linearly independent is the set $\{\mathbf{i},\mathbf{j},3\mathbf{i}-2\mathbf{j}\}$. The reason being because $3\mathbf{i}-2\mathbf{j}$ isn't essentially different from the others because it can be made out of $\mathbf{i}$ and $\mathbf{j}$. At this point we don't allow zero to be in any linearly independent set because it isn't a direction at all. Linear in this context means linear combination: a linear combination is one made of vector addition and scalar multiplication.
Now to your question. Let
$$v_1:=\left[\begin{array}{c}1\\2\end{array}\right]=\mathbf{i}+2\mathbf{j},\,\,v_2:=\left[\begin{array}{c}2\\1\end{array}\right]=2\mathbf{i}+\mathbf{j},\,\,v_3:=\left[\begin{array}{c}3\\2\end{array}\right]=3\mathbf{i}+2\mathbf{j}.$$
The set is not linearly independent because $v_1$, for example, can be written in terms of $v_2$ and $v_3$. To do this just solve
$$v_1=c_2 v_2+c_3 v_3,$$
to get
$$v_1=-4v_2+3v_3.$$
Now the question is where does the definition of linear independence that you have come from?
Let us use another more natural definition for now:

A set $\{v_1,v_2,\dots,v_n\}$ is linearly independent if and only if none of the $v_i$ can be written as a linear combination of the other 'directions' $v_i$ ($i\neq j$)

Suppose that your set $\{v_1,v_2,\dots,v_n\}$ is linearly dependent. That is the set of 'directions' isn't independent and one direction $v_j$ can be written in terms of the others:
$$v_j=c_1v_1+c_2v_2+\cdots+c_nv_n,$$
but with $c_j=0$ but not all of the other $c_i=0$. Then we can rewrite this as
$$c_1v_1+c_2v_2+\cdots-v_j+\cdots+c_nv_n=0.$$
So that when there are non-zero scalars $c_i$ such that a sum like this is zero then the set is linearly dependent. 
Now suppose that the set is linearly independent and that
$$c_1v_1+c_2v_2+\cdots+c_nv_n=0$$
and at least one of the $c_i$ are non-zero. If there is only one, say $c_j$, then $c_jv_j=0\Rightarrow v_j=0$ but this can't happen. Therefore there must be at least two. Suppose that $c_j$ is one of them. Rewriting we get
$$v_j=-\frac{1}{c_j}\left(c_1v_1+\dots+c_nv_n\right),$$
where the sum on the right contains at least one non-zero term but no $v_j$ term. This is a contradiction to the claim that the set is linearly independent and so all of the $c_i$ must be zero.
So we have our usual definition:

A set $\{v_1,v_2,\dots,v_n\}$ is linearly independent if and only if $$c_1v_1+c_2v_2+\cdots+c_nv_n=0$$ implies that all of the $c_i=0$.

You could learn two things from this. 


*

*A lot of things in finite-dimensional linear algebra generalise from $\mathbb{R}^2$, so if you are ever stuck on a concept try and work it out with $\mathbf{i}$ and $\mathbf{j}$.

*You can think in equivalent definitions even if the more usual definitions might be more useful for proving propositions and theorems.
For example, consider the two definitions below:

Definition 1: A basis of a vector space is a linearly independent spanning set.
Definition 2: A basis of a vector space is a set of vectors $\{v_1,\dots,v_n\}$ such that every vector $v$ can be represented as a linear combination of the $v_i$ uniquely.

Maybe prove in 1, think in 2?
