Did your

**textbook** actually say "de vectors"? Bad editting!

The problem says "being a basis". Yes, every basis is independent- if these are not already a basis for whatever set they span (I presume that these span some subset of

), then they must be dependent.

In each of those equations, you can solve for one of the variables, say z, in terms of the other two:

z= -3x- 2y, z= -(1/3)x- (2/3)y, z= -(5/2)x+ (1/2)y, and z= (1/4)x+ (3/4)y. That is, the subspaces spanned by each one separately are

(x, y, -3x- 2y)= x(1, 0, -3)+ y(0, 1, -2)

(x, y, -(1/3)x- (2/3)y)= x(1, 0, -1/3)+ y(0, 1, -2/3)

(x, y, -(5/2)x+ (1/2)y)= x(1, 0, -5/2)+ y(0, 1, 1/2)

(x, y, (1/4)x+ (3/4)y= x(1, 0, 1/4)+ y(0, 1, 3/4).

Of course, if it is true that you can separate these into sets that each span the same subspace, then that suspace must have dimension 2- you can separate them into two sets of two vectors each. I would suggest just trying pairs: do (v1, v2) and (v3, v4) span the same subspace? Do (v1, v3) and (v2, v4)? What about (v1, v4) and (v2, v3)?