Unformatted text preview: R 3 (see 1d). 1b) No, 4 vectors in R 3 must be LD. 1c) No, the set is LD since v 1 + v 4v 2 = 0 (or explain with a determinant). 1d) Combining the vectors into a matrix A , we see det A =1 6 = 0, so the columns form a basis of R 3 . 1e) It is the matrix A used in 1d). Do NOT compute its inverse! 2a) This was done in class (and probably also appears in my help pages). A good proof should contain the key equation below (or the equivalent): c 1 v 1 + c 2 v 2 + ... + c j v j + 0 v j + 1 + ... + 0 v n = 2b) This is essentially the ﬁrst half of Thm 3.3.2 (page 151). The main goal is to prove ‘uniquely’, but you should also mention that v is a LC of the other vectors (since they are a basis, they span V ... ). 1...
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 Spring '09
 JULIANEDWARDS
 Linear Algebra, Vectors, Vector Space, basis, V1, Prof. S. Hudson

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