Linear Algebra by otto brestscher chapter 01

# Linear Algebra with Applications (3rd Edition)

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SSM: Linear Algebra Section 1.1 Chapter 1 1.1 1. x + 2 y = 1 2 x + 3 y = 1 2 × 1st equation x + 2 y = 1 y = 1 ÷ ( 1) x + 2 y = 1 y = 1 2 × 2nd equation x = 1 y = 1 , so that ( x, y ) = ( 1 , 1). 3. 2 x + 4 y = 3 3 x + 6 y = 2 ÷ 2 x + 2 y = 3 2 3 x + 6 y = 2 3 × 1st equation x + 2 y = 3 2 0 = 5 2 So there is no solution. 5. 2 x + 3 y = 0 4 x + 5 y = 0 ÷ 2 x + 3 2 y = 0 4 x + 5 y = 0 4 × 1st equation x + 3 2 y = 0 y = 0 ÷ ( 1) x + 3 2 y = 0 y = 0 3 2 × 2nd equation x = 0 y = 0 , so that ( x, y ) = (0 , 0). 7. x + 2 y + 3 z = 1 x + 3 y + 4 z = 3 x + 4 y + 5 z = 4 I I x + 2 y + 3 z = 1 y + z = 2 2 y + 2 z = 3 2( II ) 2( II ) x + z = 3 y + z = 2 0 = 1 This system has no solution. 9. x + 2 y + 3 z = 1 3 x + 2 y + z = 1 7 x + 2 y 3 z = 1 3( I ) 7( I ) x + 2 y + 3 z = 1 4 y 8 z = 2 12 y 24 z = 6 ÷ ( 4) x + 2 y + 3 z = 1 y + 2 z = 1 2 12 y 24 z = 6 2( II ) +12( II ) x z = 0 y + 2 z = 1 2 0 = 0 This system has infinitely many solutions: if we choose z = t , an arbitrary real number, then we get x = z = t and y = 1 2 2 z = 1 2 2 t . Therefore, the general solution is ( x, y, z ) = ( t, 1 2 2 t, t ) , where t is an arbitrary real number. 11. x 2 y = 2 3 x + 5 y = 17 3( I ) x 2 y = 2 11 y = 11 ÷ 11 x 2 y = 2 y = 1 +2( II ) x = 4 y = 1 , so that ( x, y ) = (4 , 1). See Figure 1.1. 1

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Chapter 1 SSM: Linear Algebra Figure 1.1: for Problem 1.1.11 . Figure 1.2: for Problem 1.1.13 . 13. x 2 y = 3 2 x 4 y = 8 2( I ) x 2 y = 3 0 = 2 , which has no solutions. (See Figure 1.2.) 15. The system reduces to x = 0 y = 0 z = 0 so the unique solution is ( x, y, z ) = (0 , 0 , 0). The three planes intersect at the origin. 17. x + 2 y = a 3 x + 5 y = b 3( I ) x + 2 y = a y = 3 a + b ÷ ( 1) x + 2 y = a y = 3 a b 2( II ) x = 5 a + 2 b y = 3 a b , so that ( x, y ) = ( 5 a + 2 b, 3 a b ). 19. a. Note that the demand D 1 for product 1 increases with the increase of price P 2 ; likewise the demand D 2 for product 2 increases with the increase of price P 1 . This indicates that the two products are competing; some people will switch if one of the products gets more expensive. 2
SSM: Linear Algebra Section 1.1 b. Setting D 1 = S 1 and D 2 = S 2 we obtain the system 70 2 P 1 + P 2 = 14 + 3 P 1 105 + P 1 P 2 = 7 + 2 P 2 , or 5 P 1 + P 2 = 84 P 1 3 P 2 = 112 , which yields the unique solution P 1 = 26 and P 2 = 46. 21. The total demand for the products of Industry A is 310 (the consumer demand) plus 0 . 3 b (the demand from Industry B). The output a must meet this demand: a = 310 + 0 . 3 b . Setting up a similar equation for Industry B we obtain the system a = 310 + 0 . 3 b b = 100 + 0 . 5 a or a 0 . 3 b = 310 0 . 5 a + b = 100 , which yields the solution a = 400 and b = 300. 23. a. Substituting λ = 5 yields the system 7 x y = 5 x 6 x + 8 y = 5 y or 2 x y = 0 6 x + 3 y = 0 or 2 x y = 0 0 = 0 . There are infinitely many solutions, of the form ( x, y ) = ( t 2 , t ) , where t is an arbitrary real number. b. Proceeding as in part (a), we find ( x, y ) = ( 1 3 t, t ) .

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