Chapter_1(Ans)

# Chapter_1(Ans) - 1.1 SOLUTIONS Notes The key exercises are...

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1 1.1 SOLUTIONS Notes : The key exercises are 7 (or 11 or 12), 19–22, and 25. For brevity, the symbols R1, R2,…, stand for row 1 (or equation 1), row 2 (or equation 2), and so on. Additional notes are at the end of the section. 1 . 12 57 27 5 xx += −− = 157 275   −−−  Replace R2 by R2 + (2)R1 and obtain: 2 39 x = 039 Scale R2 by 1/3: 2 3 x = 013 Replace R1 by R1 + (–5)R2: 1 2 8 3 x x =− = 10 8 01 3 The solution is ( x 1 , x 2 ) = (–8, 3), or simply (–8, 3). 2 . 24 4 1 1 24 4 571 1 Scale R1 by 1/2 and obtain: 22 1 1 12 2 1 Replace R2 by R2 + (–5)R1: 2 32 1 x −= 122 03 2 1 Scale R2 by –1/3: 2 7 x 01 7 Replace R1 by R1 + (–2)R2: 1 2 12 7 x x = 10 1 2 The solution is ( x 1 , x 2 ) = (12, –7), or simply (12, –7).

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2 CHAPTER 1 Linear Equations in Linear Algebra 3 . The point of intersection satisfies the system of two linear equations: 12 57 22 xx += −= 157 122 −− Replace R2 by R2 + (–1)R1 and obtain: 2 79 x 079 Scale R2 by –1/7: 2 9/7 x = 15 7 019 / 7 Replace R1 by R1 + (–5)R2: 1 2 4/7 x x = = 104 / 7 / 7 The point of intersection is ( x 1 , x 2 ) = (4/7, 9/7). 4 . The point of intersection satisfies the system of two linear equations: 51 37 5 15 1 5    Replace R2 by R2 + (–3)R1 and obtain: 2 82 x = 1 08 2 Scale R2 by 1/8: 2 1/4 x = 1 01 1 / 4 Replace R1 by R1 + (5)R2: 1 2 9/4 1/4 x x = = 109 / 4 011 / 4 The point of intersection is ( x 1 , x 2 ) = (9/4, 1/4). 5 . The system is already in “triangular” form. The fourth equation is x 4 = –5, and the other equations do not contain the variable x 4 . The next two steps should be to use the variable x 3 in the third equation to eliminate that variable from the first two equations. In matrix notation, that means to replace R2 by its sum with 3 times R3, and then replace R1 by its sum with –5 times R3. 6 . One more step will put the system in triangular form. Replace R4 by its sum with –3 times R3, which produces 16401 02704 00123 0005 1 5 . After that, the next step is to scale the fourth row by –1/5. 7 . Ordinarily, the next step would be to interchange R3 and R4, to put a 1 in the third row and third column. But in this case, the third row of the augmented matrix corresponds to the equation 0 x 1 + 0 x 2 + 0 x 3 = 1, or simply, 0 = 1. A system containing this condition has no solution. Further row operations are unnecessary once an equation such as 0 = 1 is evident. The solution set is empty.
1.1 • Solutions 3 8 . The standard row operations are: 1 490 1 490 1 400 1000 01 7 0 ~ 7 0 ~ 0 0 ~ 0 1 0 0 00 2 0 1 0 1 0 0 0 1 0 −−−    The solution set contains one solution: (0, 0, 0).

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## This note was uploaded on 09/06/2010 for the course MATH MATH1111 taught by Professor Forgot during the Fall '08 term at HKU.

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Chapter_1(Ans) - 1.1 SOLUTIONS Notes The key exercises are...

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