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Unformatted text preview: SOLUTIONS FOR SELECTED PRACTICE PROBLEMS Section 1.1 3. (b) The plane in question is the set of all points of the form (3 , − 6 , 7) + t ( (3 , − 6 , 7) − ( − 2 , , − 4) ) + s ( (3 , − 6 , 7) − (5 , − 9 , − 2) ) = (3 + 5 t − 2 s, − 6 − 6 t + 3 s, 7 + 11 t + 9 s ) , with t,s ∈ R . Section 1.3 31. (a) Suppose first that v ∈ W , and show that, in this case, v + W = W (in particular, v + W would be a subset of V ). On the one hand, for every w ∈ W , we have v + w ∈ W , hence v + W ⊂ W . On the other hand, any w ∈ W can be written as w = v + u , with u = w − v ∈ W . Thus, v + W ⊃ W . Now suppose v / ∈ W , and show that W ′ = v + W is not a subspace of V . To this end, it suffices to prove that v ∈ W ′ , but 2 v / ∈ W ′ (subspaces are closed under multiplication by a scalar). Note first that 0 ∈ W , hence v +0 ∈ W ′ . Suppose, for the sake of contradiction, that 2 v ∈ W ′ . That is, there exists w ∈ W for which v + w = 2 v . Solving for w , we obtain w = v , which is impossible. Section 1.4 5. (g) Yes . We have to show that there exist x 1 ,x 2 ,x 3 ∈ R such that parenleftbigg 1 2 − 3 4 parenrightbigg = x 1 parenleftbigg 1 − 1 0 parenrightbigg + x 2 parenleftbigg 0 1 0 1 parenrightbigg + x 3 parenleftbigg 1 1 0 0 parenrightbigg . In other words, we need to know if the system of 4 linear equations x 1 + x 3 = 1 x 2 + x 3 = 2 − x 1 = − 3 x 2 = 4 has a solution. It is straightforward that the triple x 1 = 3, x 2 = 4, x 3 = − 2 solves this system. 1 2 SOLUTIONS FOR SELECTED PRACTICE PROBLEMS No , parenleftbigg 1 0 0 1 parenrightbigg does not belong to span( S ). If it did, there would exist a,b,c ∈ R satisfying parenleftbigg 1 0 0 1 parenrightbigg = a parenleftbigg 1 − 1 0 parenrightbigg + b parenleftbigg 0 1 0 1 parenrightbigg + c parenleftbigg 1 1 0 0 parenrightbigg = parenleftbigg a + c b + c − a b parenrightbigg Thus, the triple ( a,b,c ) should be a solution for the following system of four equations with three unknowns: a + c = 1 b + c = 0 − a = 0 b = 1 Equations 3 and 4 lead us to conclude that a = 0 and b = 1. By Equation 1, c = 1. Plugging these values into Equation 3, we obtain 2 = 0, which is false. Section 1.5 13. (b) Assume the set { u,v,w } is linearly independent, and prove that the set { u + v,u + w,v + w } is linearly independent – that is a 1 ( u + v ) + a 2 ( u + w ) + a 3 ( v + w ) = 0 implies a 1 = a 2 = a 3 = 0. Note that a 1 ( u + v ) + a 2 ( u + w ) + a 3 ( v + w ) = ( a 1 + a 2 ) u + ( a 1 + a 3 ) v + ( a 2 + a 3 ) w, hence, by the linear independence of { u,v,w } , a 1 ( u + v ) + a 2 ( u + w ) + a 3 ( v + w ) = 0 iff a 1 + a 2 = 0, a 1 + a 3 = 0, and a 2 + a 3 = 0. Solving this system of linear equations, we obtain a 2 = a 3 = − a 1 . Furthermore, 0 = a 2 + a 3 = − 2 a 1 , hence a 1 = a 2 = a 3 = 0....
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This note was uploaded on 02/02/2012 for the course MATH 115A 262398211 taught by Professor Fuckhead during the Spring '10 term at UCLA.
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