Two sets of vectors that could be used to span a set in R^3 are:
Set 1: { (1, 0, 0), (0, 1, 0), (0, 0, 1) }
Set 2: { (1, 1, 0), (0, 1, 1), (1, 0, 1) }
Now, let's show how the vectors (-1, 2, 0) and (3, 4, 0) can be written as linear combinations of the vectors in each set:
For Set 1:
-1 = -1*(1, 0, 0) + 0*(0, 1, 0) + 0*(0, 0, 1)
2 = 0*(1, 0, 0) + 2*(0, 1, 0) + 0*(0, 0, 1)
So, (-1, 2, 0) can be written as a linear combination of the vectors in Set 1.
3 = 3*(1, 0, 0) + 0*(0, 1, 0) + 0*(0, 0, 1)
4 = 0*(1, 0, 0) + 4*(0, 1, 0) + 0*(0, 0, 1)
So, (3, 4, 0) can also be written as a linear combination of the vectors in Set 1.
For Set 2:
-1 = -1*(1, 1, 0) + 0*(0, 1, 1) + 0*(1, 0, 1)
2 = 0*(1, 1, 0) + 2*(0, 1, 1) + 0*(1, 0, 1)
So, (-1, 2, 0) can be written as a linear combination of the vectors in Set 2.
3 = 3*(1, 1, 0) + 0*(0, 1, 1) + 0*(1, 0, 1)
4 = 0*(1, 1, 0) + 4*(0, 1, 1) + 0*(1, 0, 1)
So, (3, 4, 0) can also be written as a linear combination of the vectors in Set 2.
Therefore, both sets can be used to span the set in R^3, and the vectors (-1, 2, 0) and (3, 4, 0) can be expressed as linear combinations of the vectors in each of these sets.