{(9x 2y, 3x – 4y, 6x + 4y) I x,y arbitrary numbers}
{(x,y,z) I 3x + 6y + 2z + 9}
{(x,y,z) I 7x + 5y = 0, 8x – 9z = 0}
{(x,y,z) I 7x – 5y + 8z = 0}
{(x, x + 7, x – 5) I x arbitrary number}
{(x,y,z) I x < y < z}
If you can give an explanation too because I don’t quite get this
3 Answers

I will number these six sets from i) to vi).
i) (9x 2y, 3x – 4y, 6x + 4y) = x(9, 3, 6) + y(2, 4, 4), and x and y are arbitrary.
So this set is the span of the vectors (9, 3, 6) and (2, 4, 4). The span of any collection of vectors is always a subspace, so this set is a subspace.
ii) This problem doesn’t make sense, since 3x + 6y + 2z + 9 has no equals sign and therefore is not a condition. If this problem were modified correctly, I would suspect that the set is not a subspace because of the nonzero constant term 9.
iii) and iv) are solution sets of systems of linear equations with zeros for all the righthand constants and therefore must be subspaces, since the solution set of any system of linear equations with zeros for all the righthand constants is always a subspace.
v) This set is not a subspace, since it is not closed under scalar multiplication.
For example, (0, 7, 5) is in the set, but 2(0, 7, 5) = (0, 14, 10) is not in the set.
vi) This set is not a subspace, since it is not closed under scalar multiplcation.
For example, (2, 3, 4) is in the set, but (1)(2, 3, 4) = (2, 3, 4) is not in the set.
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RE:
Which of the following sets are subspaces of R3?
{(9x 2y, 3x – 4y, 6x + 4y) I x,y arbitrary numbers}
{(x,y,z) I 3x + 6y + 2z + 9}
{(x,y,z) I 7x + 5y = 0, 8x – 9z = 0}
{(x,y,z) I 7x – 5y + 8z = 0}
{(x, x + 7, x – 5) I x arbitrary number}
{(x,y,z) I x < y < z}
If you can give an explanation too because I don’t quite get this

sets subspaces r3