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Unformatted text preview: 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 . 1 2 1 2 5 7 2 7 5 x x x x + = =  1 5 7 2 7 5 ! " # $ % & Replace R2 by R2 + (2)R1 and obtain: 1 2 2 5 7 3 9 x x x + = = 1 5 7 3 9 ! " # $ % & Scale R2 by 1/3: 1 2 2 5 7 3 x x x + = = 1 5 7 1 3 ! " # $ % & Replace R1 by R1 + (–5)R2: 1 2 8 3 x x =  = 1 8 1 3 ! " # $ % & The solution is ( x 1 , x 2 ) = (–8, 3), or simply (–8, 3). 2 . 1 2 1 2 2 4 4 5 7 11 x x x x + =  + = 2 4 4 5 7 11 ! " # $ % & Scale R1 by 1/2 and obtain: 1 2 1 2 2 2 5 7 11 x x x x + =  + = 1 2 2 5 7 11 ! " # $ % & Replace R2 by R2 + (–5)R1: 1 2 2 2 2 3 21 x x x + =  = 1 2 2 3 21 ! " # $ % & Scale R2 by –1/3: 1 2 2 2 2 7 x x x + =  =  1 2 2 1 7 ! " # $ % & Replace R1 by R1 + (–2)R2: 1 2 12 7 x x = =  1 12 1 7 ! " # $ % & The solution is ( x 1 , x 2 ) = (12, –7), or simply (12, –7). 2 CHAPTER 1 • Linear Equations in Linear Algebra 3 . The point of intersection satisfies the system of two linear equations: 1 2 1 2 5 7 2 2 x x x x + = =  1 5 7 1 2 2 ! " # $ % & Replace R2 by R2 + (–1)R1 and obtain: 1 2 2 5 7 7 9 x x x + = =  1 5 7 7 9 ! " # $ % & Scale R2 by –1/7: 1 2 2 5 7 9/7 x x x + = = 1 5 7 1 9/7 ! " # $ % & Replace R1 by R1 + (–5)R2: 1 2 4/7 9/7 x x = = 1 4/7 1 9/7 ! " # $ % & 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: 1 2 1 2 5 1 3 7 5 x x x x = = 1 5 1 3 7 5 ! " # $ % & Replace R2 by R2 + (–3)R1 and obtain: 1 2 2 5 1 8 2 x x x = = 1 5 1 8 2 ! " # $ % & Scale R2 by 1/8: 1 2 2 5 1 1/4 x x x = = 1 5 1 1 1/4 ! " # $ % & Replace R1 by R1 + (5)R2: 1 2 9/4 1/4 x x = = 1 9/4 1 1/4 ! " # $ % & 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 1 6 4 1 2 7 4 1 2 3 5 15 ! " # $ # $ # $ # $ % & . 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....
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 Spring '03
 BUSS
 Linear Algebra, Addition, augmented matrix

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