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Unformatted text preview: MATH321 – HOMEWORK SOLUTIONS HOMEWORK #2 Section 1.5: Problems 1, 2(d)(e), 3, 4, 8 Section 1.6: Problems 1, 2(a)(b), 3(a), 4, 5, 9, 13, 14, 17 Krzysztof Galicki Problem 1.5.1 (See Answers to Selected Exercises ). Problem 1.5.2 (a) Linearly dependent. The second matrix A 2 is obtained from the first one A 1 by multiplying with 2. Hence, 2 A 1 + A 2 = O . (d) Linearly independent. We write a ( x 3 x ) + b (2 x 2 + 4) + c ( 2 x 3 + 3 x 2 + 2 x + 6) = 0 . ( a 2 c ) x 3 + (2 b + 3 a ) x 2 + ( a + 2 c ) x + 4 b + 6 c = 0 . This is easily seen to have only trivial solutions. a = 2 c but 2 b = 3 a = 3 c which makes a = c . Put together this implies that a = b = c = 0. Problem 1.5.3 The sum of the first three matrices equals to the sum of the last two. Hence, the set is linearly dependent. Problem 1.5.4 We have a 1 e 1 + ··· a n e n = a 1 (1 , ,..., 0) + ··· a n (0 ,..., , 1) = ( a 1 ,a 2 ,...,a n ) = (0 ,..., 0) which can happen if and only if a 1 = a 2 = ··· = a n = 0. Hence, the set { e 1 ,...,e n } is linearly independent in F n (for any field F )....
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This note was uploaded on 11/30/2009 for the course MATH 115A taught by Professor Liu during the Winter '07 term at UCLA.
 Winter '07
 Liu
 Math, Linear Algebra, Algebra

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