hw2solutions - Homework 2 Solutions Josh Hernandez 1.4...

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Homework 2 Solutions Josh Hernandez October 27, 2009 1.4 - Linear Combinations and Systems of Linear Equations 2. Solve the following systems of linear equations. b. 2 x 1 - 7 x 2 + 4 x 3 = 10 x 1 - 2 x 2 + x 3 = 3 2 x 1 - x 2 - 2 x 3 = 6 Solution: 1. Scaling down from first pivot: 1 · ( 2 x 1 - 7 x 2 + 4 x 3 = 10) -2 · ( x 1 - 2 x 2 + x 3 = 3) -1 · ( 2 x 1 - x 2 - 2 x 3 = 6) 2. Summing down from first row: 2 x 1 - 7 x 2 + 4 x 3 = 10 -3 x 2 + 2 x 3 = 4 -6 x 2 + 6 x 3 = 4 3. Scaling down from second pivot: 2 x 1 - 7 x 2 + 4 x 3 = 10 2( -3 x 2 + 2 x 3 = 4) -1( -6 x 2 + 6 x 3 = 4) 4. Summing down from second row: 2 x 1 - 7 x 2 + 4 x 3 = 10 -3 x 2 + 2 x 3 = 4 -2 x 3 = 4 5. Scaling up from second pivot: -3 · ( 2 x 1 - 7 x 2 + 4 x 3 = 10) 7 · ( -3 x 2 + 2 x 3 = 4) -2 x 3 = 4 6. Summing up from second pivot: -6 x 1 + 2 x 3 = -2 -3 x 2 + 2 x 3 = 4 -2 x 3 = 4 7. Summing up from third pivot: -6 x 1 = 2 -3 x 2 = 8 -2 x 3 = 4 8. Solution: x 1 = -1 / 3 x 2 = -8 / 3 x 3 = -2 d. x 1 + 2 x 2 + 2 x 3 = 2 x 1 + 8 x 3 + 5 x 4 = -6 x 1 + x 2 + 5 x 3 + 5 x 4 = 3 1
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Solution: 1. Scaling down from first pivot: x 1 + 2 x 2 + 2 x 3 = 2 -1 · ( x 1 + 8 x 3 + 5 x 4 = -6) -1 · ( x 1 + x 2 + 5 x 3 + 5 x 4 = 3) 2. Summing down from first row: x 1 + 2 x 2 + 2 x 3 = 2 2 x 2 + -6 x 3 - 5 x 4 = 8 x 2 + -3 x 3 - 5 x 4 = -1 3. Scaling down from second pivot: x 1 + 2 x 2 + 2 x 3 = 2 2 x 2 + -6 x 3 - 5 x 4 = 8 -2 · ( x 2 + -3 x 3 - 5 x 4 = -1) 4. Summing down from second row: x 1 + 2 x 2 + 2 x 3 = 2 2 x 2 + -6 x 3 - 5 x 4 = 8 5 x 4 = 10 5. Scaling up from second pivot: -1 · ( x 1 + 2 x 2 + 2 x 3 = 2) 2 x 2 + -6 x 3 - 5 x 4 = 8 5 x 4 = 10 6. Summing up from second pivot: - x 1 + -8 x 3 - 5 x 4 = 6 2 x 2 + -6 x 3 - 5 x 4 = 8 5 x 4 = 10 7. Summing up from third pivot: - x 1 + -8 x 3 = 16 2 x 2 + -6 x 3 = 18 5 x 4 = 10 8. Solution: x 1 = -16 + - x 3 x 2 = 9 + 3 x 3 x 3 = x 3 x 4 = 2 This solution set is the line through the point (-16 , 9 , 0 , 2), in the direction (-1 , 3 , 1 , 0). 3. For each of the following lists of vectors in R 3 , determine whether the first vector can be expressed as a linear combination of the other two. b. (1 , 2 , -3), (-3 , 2 , 1), (2 , -1 , -1). 2
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Solution: We can find such a linear combination i.e. a, b R such that a (1 , 2 , -3) + b (-3 , 2 , 1) = (2 , -1 , -1) if we can find a solution to the linear system a + -3 b = 2 2 a + 2 b = -1 -3 a + b = -1 1. Scaling down from first pivot: 6 · ( a + -3 b = 2) -3 · ( 2 a + 2 b = -1) 2 · ( -3 a + b = -1) 2. Summing down from first row: a + -3 b = 2 -24 b = 15 -16 b = 10 The bottom two rows are equivalent. 3. Scaling up from the second pivot: 8 · ( a + -3 b = 2) 3 · ( 8 b = -5) 3. Summing up from the second row: a = 1 8 b = -5 4. Solution: a = 1 b = -5 / 8 So these vectors are indeed linearly dependent. d. (2 , -1 , 0), (1 , 2 , -3), (1 , -3 , 2). Solution: Suppose there existed such a linear combination, a (2 , -1 , 0) + b (1 , 2 , -3) = (1 , -3 , 2) Then there must be a solution to the linear system 2 a + b = 1 - a + 2 b = -3 -3 b = 2 So b = -2 / 3. Plugging this into the two first equations, 2 a + -2 / 3 = 1 - a + 2(-2 / 3) = -3 Solving for a gives conflicting results: a = 5 / 6, and 5 / 3. This set of equations has no solution, so the three vectors are linearly independent. 5h. Determine whether the given vector is in the span of S : v = 1 0 0 1 , S = v 1 = 1 0 -1 0 , v 2 = 0 1 0 1 , v 3 = 1 1 0 0 3
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Solution: Suppose there existed a, b, c R such that av 1 + bv 2 + cv 2 = v . A quick check of the set reveals that only v 1 has a nonzero element in the bottom-left position. Since v has a zero in that position, then a = 0. Likewise, considering the bottom-right element, we see that b = 1.
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