Let v1 = (1,-1, 0), v2 = (1, 2, 2), w = (-1,-8,-6) be vectors from the vector sp
ID: 3099065 • Letter: L
Question
Let v1 = (1,-1, 0), v2 = (1, 2, 2), w = (-1,-8,-6) be vectors from the vector space R3 with the standard vector addition and scalar multiplication.(a) Does w belong to {v1,v2}?
(b) How many vectors are in the set {v1,v2}?
(c) How many vectors are in the set span {v1, v2}?
(d) Does w belong to the subspace spanned by the set {v1, v2}?
(e) Does u = (0, 1, 1) belong to span {v1v2}?
Explanation / Answer
a) no, w is not v1 or v2. b) two, v1 and v2. c) The space span{v1, v2} is all linear combinations of the vectors v1 and v2. There are infinitely many such vectors. For every pair of scalars a and b av1 + bv2 is in this space. thus a LOT. the set is infinite. d) let's check. to see, we'll set w = av1 + bv2, and see if we can find a suitable a and b. av1 = (a,-a,0), and bv2 = (b,2b,2b) so av1 + bv2 = (a + b,2b - a,2b) if this equals w, we have 3 equations: a + b = -1 2b - a = -8 2b = -6. the last one tells us b = -3. that makes the first two: a - 3 = -1 -6 - a = -8 a - 3 = -1 --> a = 2 -6 - a = -8 --> a = 2, so we can write: w = (-1,-8,-6) = 2(1,-1,0) - 3(1,2,2) = (2,-2,0) - (3,6,6) = (-1,-8,-6) w is in the span of {v1,v2} e) same deal, we set u = (a + b,2b - a,2b) and try to solve for a and b: a + b = 0 2b - a = 1 2b = 1 from equation 3, we have immediately b = 1/2 from equation 1, a = -1/2. 2b - a = 1 + 1/2 = 3/2. that's a problem. apparently, we can find no such a and b. so u does not belong to the span of {v1,v2}.
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