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Unformatted text preview: R 4 . Solution B: Another way to solve the problem is to do it directly. Let b = b 1 b 2 b 3 b 4 be a vector in R 4 . b is in the span of { v 1 , v 2 , v 3 } if the system 1 1 b 11 b 21 b 3 11 b 4 has a solution. We can row reduce the matrix. 1 1 b 11 b 21 b 3 11 b 4 III + I→ 1 1 b 11 b 2 1 b 1 + b 3 11 b 4 IV + II→ 1 1 b 11 b 2 1 b 1 + b 31 b 2 + b 4 IV + III→ 1 1 b 11 0 b 2 1 b 1 + b 3 b 1 + b 2 + b 3 + b 4 Therefore the span is the set of vectors where b 1 + b 2 + b 3 + b 4 = 0 Since this is not all of R 4 , the vectors do not span....
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 Fall '09
 Math, Vector Space, ........., British B class submarine, B type proanthocyanidin

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