HW2solutions_4

# HW2solutions_4 - MATH 115 SOLUTION SET 2 Â 1.4 1 a True b...

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Unformatted text preview: MATH 115 SOLUTION SET 2 Â§ 1.4: 1. a) True b) False (the span is { } ) c) True (if you think super hard about this, youâ€™ll see that it is the same as Theorem 1.5 in the book) d) False (you canâ€™t 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 âˆˆ 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 Ã— 2 ( F ) is of the form a b c d Â¶ , and this is indeed a linear combination of the given matrices: a b c d Â¶ = a 1 0 0 0 Â¶ + b 0 1 0 0...
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HW2solutions_4 - MATH 115 SOLUTION SET 2 Â 1.4 1 a True b...

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