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Unformatted text preview: Math 136 Assignment 1 Solutions 1. Compute the following linear combinations. (a) 1 2 2 6 + 1 3 4 3 Solution: 1 2 2 6 + 1 3 4 3 = 1 3 + 4 / 3 1 = 7 / 3 4 (b) 2 3 3 1 2 1 / 4 1 / 3 Solution: 2 3 3 1 2 1 / 4 1 / 3 = 2 2 / 3 1 / 2 2 / 3 = 3 / 2 (c) √ 2 √ 2 √ 3 + 3 1 √ 6 Solution: √ 2 √ 2 √ 3 + 3 1 √ 6 = 2 √ 6 + 3 3 √ 6 = 5 4 √ 6 2. Describe geometrically the following sets and write a simplified vector equation for each. (a) span 1 1 , 2 2 , Solution: By Theorem 2, we have span 1 1 , 2 2 , = span 1 1 , 2 2 = span 1 1 Hence, it spans the line in R 2 with simplified vector equation ~x = t 1 1 , t ∈ R . (b) span 2 3 1 , 1 3 Solution: Since neither 2 3 1 nor 1 3 is a scalar multiple of the other the set B = 2 3 1 , 1 3 is linearly independent. Hence, the set spanned B is the plane in R 3 with vector equation ~x = s 2 3 1 + t 1 3 , s,t ∈...
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This note was uploaded on 02/22/2012 for the course MATH cs136 taught by Professor Mark during the Winter '12 term at Waterloo.
 Winter '12
 Mark

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