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Unformatted text preview: MATH 115 SOLUTION SET 2 1.4: 1. a) True b) False (the span is { } ) c) True (if you think super hard about this, youll see that it is the same as Theorem 1.5 in the book) d) False (you cant multiply by 0) e) True f) False (for instance, the system x + y = 2 and x + y = 4 makes a system of linear equations with no solution). 2. a) { r (1 , 1 , , 0) + s ( 3 , , 2 , 1) + (5 , , 4 , 0)  r,s R } 6. There are many ways to do this. One way is to show that the standard basis vectors e 1 , e 2 and e 3 all lie in the span of the three given vectors. It would then follow that the span of { e 1 ,e 2 ,e 3 } , which is F 3 itself, is also contained in the span of the given vectors. Indeed, if v 1 = (1 , 1 , 0), v 2 = (1 , , 1) and v 3 = (0 , 1 , 1) we have e 1 = 1 2 ( v 1 + v 2 v 3 ) e 2 = 1 2 ( v 1 v 2 + v 3 ) e 3 = 1 2 ( v 1 + v 2 + v 3 ) . (Strictly speaking, this only works when F has characteristic not equal to two...) 8. An arbitrary matrix in M 2 2 ( F ) is of the form a b c d , and this is indeed a linear combination of the given matrices: a b c d = a 1 0 0 0 + b 0 1 0 0...
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This note was uploaded on 02/01/2010 for the course MATH math115a taught by Professor Shalom during the Spring '10 term at UCLA.
 Spring '10
 SHALOM
 Math, Linear Algebra, Algebra

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