Linear Independence I've come across a question in Linear Algebra that I can't quite figure out. I've tried a multitude of things that either don't work or aren't sufficient enough to convince me I understand linear independence well enough. 
I know a set of vectors, S, in vector space V are linearly independent if their linear combination, that is,
$\lambda_1 \mathbf{v}_1 + ... + \lambda_n \mathbf{v}_n = \mathbf{0}$
means all scalars are equal to each other and 0,
$\lambda_1 = ... = \lambda_n = 0.$
I can also show a set of vectors S is linearly independent if I'm given a set of vectors with numerical values - by creating a matrix and reducing it to row echelon form. However, my understanding isn't great enough that I can expand on this and answer questions such as the following: 
Assume the vectors u, v and w are linearly independent elements of a vector space V. 
For each of the following sets decide whether it is linearly independent.
A. {u + v + w,  v - 2w,  2u + 3w}
B. {u + 2w, v + 2w, 2w} 
C. {x, y, z}
where, 
x = u + 2v - w, 
y = 2x + u + 2v - w,
z = 3x - 2y. 
If anyone can explain to me the connection between this type of question and the definition of linear independence by answering A or providing a guideline of how to answer A then hopefully I can tackle B and C and any related questions. Thanks. 
 A: Hint: In both cases, write the coordinates of the vectors in the set with respect to the original set as rows in a matrix. Row reduce this matrix. 
A: You basically need to write the definition of linear independence. Suppose that 
$$
\alpha(u+v+w)+\beta(v-2w)+\gamma(2u+3w)=0. 
$$
We can rewrite this as 
$$
(\alpha+2\gamma)\,u+(\alpha+\beta)\,v+(\alpha-2\beta+3\gamma)\,w=0.
$$
Since $u,v,w$ are linearly independent, we get the equalities 
$$
\alpha+2\gamma=0,\ \ \alpha+\beta=0,\ \ \alpha-2\beta+3\gamma=0. 
$$
Now you can analyze this system. If the only solution is $\alpha=\beta=\gamma=0$, you'll know that the three vectors in A are linearly independent. If you produce a nonzero solution, you'll know that they are linearly dependent. 
A: You just need to use the definition. For the first case you can write:
$$\lambda_1(u+v+w)+\lambda_2(v-2w)+\lambda_3(2u+3w)=0$$
now rewrite like:
$$(\lambda_1+2\lambda_3)u+(\lambda_1+\lambda_2)v+(\lambda_1-2\lambda_2+3\lambda_3)w=0$$
Now use that $u,v,w$ are independent, solve the system and find $\lambda_1,\lambda_2,\lambda_3$.
Can you finish?
A: for A: we have
$$\alpha(u+v+w)+\beta(v-2w)+\gamma(2u+3w)=0$$
this is equivalant to
$$u(\alpha+2\gamma)+v(\alpha+\beta)+w(\alpha-2\beta+3\gamma)=0$$
can you finish?
for B: we get
$$u\alpha=0,v\beta=0,w(2\alpha+2\beta+2\gamma)=0$$
