hwsol1 - Homework 1 Solutions Course Reader Exercises 0 4 2...

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Homework 1 Solutions Course Reader Exercises 1.7 (a) 2 1 1 + 2 - 4 0 - 4 2 - 2 = 0 - 5 3 (b) 2 1 1 + 2 - 3 - 4 2 = 2 1 1 + - 6 - 8 4 = - 4 - 7 5 (c) ± 4 6 ² - ± - 9 6 ² = ± 13 0 ² (d) ± 6 9 ² + ± - 6 4 ² = ± 0 13 ² (e) 2(1) + 1( - 2) + 1(0) = 0 (f) 2(5) + 1(0) + 1( - 2) = 8 (g) 2( - 3) + 3(2) = 0 (h) p (2 - ( - 3)) 2 + (3 - 2) 2 = 26 2.3 The first two vectors are nonzero and neither is a scalar multiple of the other, so they span a plane. To see if the third vector is a linear combination of the first two, consider the equation c 1 2 1 1 + c 2 5 8 0 = 1 6 - 2 which is equivalent to the system 2 c 1 + 5 c 2 = 1 c 1 + 8 c 2 = 6 c 1 = - 2 The solution is c 1 = - 2, c 2 = 1. Thus - 2 2 1 1 + 5 8 0 = 1 6 - 2 and therefore the set spans a plane. 2.4 We know the span of these vectors is not a point, because the vectors are not all zero; and it is not a line, because clearly the second vector is not a multiple of the first. So we know the span is either a plane, or all of R 3 . 1
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To distinguish between these possibilities, it is enough to determine if the given vectors are dependent or independent. We therefore attempt to find solutions to the system of equations c 1 2 1 1 + c 2 5 8 0 + c 3 2 6 - 2 = 0 0 0 We row reduce the corresponding matrix: 2 5 2 0 1 8 6 0 1 0 - 2 0 1 0 - 2 0 1 8 6 0 2 5 2 0 1 0 - 2 0 0 8 8 0 0 5 6 0 1 0 - 2 0
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This note was uploaded on 03/01/2008 for the course MATH 51 taught by Professor Staff during the Spring '07 term at Stanford.

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hwsol1 - Homework 1 Solutions Course Reader Exercises 0 4 2...

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