Answer: a) The set W contains the zero vector, since (0,0,0) = (3s,-2s,s) for s = 0.
The set W is closed under addition and scalar multiplication:
If (3s1, -2s1, s1) and (3s2, -2s2, s2) are in W, then their sum, (3(s1 + s2), -2(s1 + s2), (s1 + s2)), is also in W.
If (3s, -2s, s) is in W and c is any scalar, then (3cs, -2cs, cs) is also in W.
The set W has a basis of {(3, -2, 1)}. To see this, we can write any vector in W as a scalar multiple of (3, -2, 1). For example, (3s, -2s, s) = s(3, -2, 1). The dimension of the set W is 1.
b) The set W contains the zero vector, since (t-2,0,6-3t) = (t-2,0,6-3t) for t = 2.
The set W is closed under addition and scalar multiplication:
If (t1 - 2, 0, 6 - 3t1) and (t2 - 2, 0, 6 - 3t2) are in W, then their sum, ((t1 + t2) - 2, 0, 6 - 3(t1 + t2)), is also in W.
If (t - 2, 0, 6 - 3t) is in W and c is any scalar, then (ct - 2c, 0, 6c - 3ct) is also in W.
The set W has a basis of {(1, 0, -3)}. To see this, we can write any vector in W as a scalar multiple of (1, 0, -3). For example, (t - 2, 0, 6 - 3t) = t(1, 0, -3) - 2(1, 0, -3). The dimension of the set W is 1.
c) The set H contains the zero vector, since (2b, a, 3b) = (0,0,0) for b = 0 and a = 0.
The set H is closed under addition and scalar multiplication:
If (2b1, a1, 3b1) and (2b2, a2, 3b2) are in H, then their sum, (2(b1 + b2), a1 + a2, 3(b1 + b2)), is also in H.
If (2b, a, 3b) is in H and c is any scalar, then (2cb, ca, 3cb) is also in H.
The set H has a basis of {(2,1,3)}. To see this, we can write any vector in H as a scalar multiple of (2,1,3). For example, (2b, a, 3b) = b(2,1,3) + a(0,1,0). The dimension of the set H is 2.
d) The set H contains the zero vector, since (−b + 2c, b, 4c) = (0,0,0) for b = 2c.
Explanation: