Linear algebra with applications 5th edition otto bretscher solutions pdf

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Odd Solutions For Otto Bretscher Linear Algebra With Applications, 5th Edition

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Odd Solutions for Otto Bretscher Linear Algebra with Applications, 5th Edition

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Odd solutions for exercises in Linear Algrebra Otto Bretscher 5th edition

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Linear Algebra with Applications 5th Edition Bretscher Solutions Manual

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Linear Algebra with Applications 5th edition by Otto Bretscher Solution manual
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Linear Algebra with Applications 5th edition by Otto Bretscher Solution manual
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1. 1. Linear Algebra with Applications 5th edition b y Otto Bretscher Solution manual Full download link: https://bit.ly/3fz62W0
2. 2. Section 1.1 Chapter 1 Section 1.1 1.1.1 x + 2y = 1 2x + 3y = 1 −2 × 1st equation → x + 2y = 1 −y = −1 ÷(−1) → x + 2y = 1 y = 1 −2 × 2nd equation → x = −1 y = 1 , so that (x, y) = (−1, 1). 1.1.2 4x + 3y = 2 7x + 5y = 3 ÷4 → x + 3 4 y = 1 2 7x + 5y = 3 −7 × 1st equation → x + 3 4 y = 1 2 −1 4 y = −1 2 ×(−4) → x + 3 4 y = 1 2 y = 2 −3 4 × 2nd equation → x = −1 y = 2 , so that (x, y) = (−1, 2). 1.1.3 2x + 4y = 3 3x + 6y = 2 ÷2 → x + 2y = 3 2 3x + 6y = 2 −3 × 1st equation → x + 2y = 3 2 0 = −5 2 . So there is no solution. 1.1.4 2x + 4y = 2 3x + 6y = 3 ÷2 → x + 2y = 1 3x + 6y = 3 −3 × 1st equation → x + 2y = 1 0 = 0 This system has infinitely many solutions: if we choose y = t, an arbitrary real number, then the equation x + 2y = 1 gives us x = 1 − 2y = 1 − 2t. Therefore the general solution is (x, y) = (1 − 2t, t), where t is an arbitrary real number. 1.1.5 2x + 3y = 0 4x + 5y = 0 ÷2 → x + 3 2 y = 0 4x + 5y = 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). 1.1.6   x + 2y + 3z = 8 x + 3y + 3z = 10 x + 2y + 4z = 9   −I −I →   x + 2y + 3z = 8 y = 2 z = 1   −2(II) →   x + 3z = 4 y = 2 z = 1   −3(III) →   x = 1 y = 2 z = 1  , so that (x, y, z) = (1, 2, 1). 1.1.7   x + 2y + 3z = 1 x + 3y + 4z = 3 x + 4y + 5z = 4   −I −I →   x + 2y + 3z = 1 y + z = 2 2y + 2z = 3   −2(II) −2(II) →   x + z = −3 y + z = 2 0 = −1   This system has no solution. 1 Copyright c 2013 Pearson Education, Inc.
3. 3. Chapter 1 1.1.8   x + 2y + 3z = 0 4x + 5y + 6z = 0 7x + 8y + 10z = 0   −4(I) −7(I) →   x + 2y + 3z = 0 −3y − 6z = 0 −6y − 11z = 0   ÷(−3) →   x + 2y + 3z = 0 y + 2z = 0 −6y − 11z = 0   −2(II) +6(II) →   x − z = 0 y + 2z = 0 z = 0   +III −2(III) →   x = 0 y = 0 z = 0  , so that (x, y, z) = (0, 0, 0). 1.1.9   x + 2y + 3z = 1 3x + 2y + z = 1 7x + 2y − 3z = 1   −3(I) −7(I) →   x + 2y + 3z = 1 −4y − 8z = −2 −12y − 24z = −6   ÷(−4) →   x + 2y + 3z = 1 y + 2z = 1 2 −12y − 24z = −6   −2(II) +12(II) →   x − z = 0 y + 2z = 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 − 2z = 1 2 − 2t. Therefore, the general solution is (x, y, z) = t, 1 2 − 2t, t , where t is an arbitrary real number. 1.1.10   x + 2y + 3z = 1 2x + 4y + 7z = 2 3x + 7y + 11z = 8   −2(I) −3(I) →   x + 2y + 3z = 1 z = 0 y + 2z = 5   Swap : II ↔ III →   x + 2y + 3z = 1 y + 2z = 5 z = 0   −2(II) →   x − z = −9 y + 2z = 5 z = 0   +III −2(III) →   x = −9 y = 5 z = 0  , so that (x, y, z) = (−9, 5, 0). 1.1.11 x − 2y = 2 3x + 5y = 17 −3(I) → x − 2y = 2 11y = 11 ÷11 → x − 2y = 2 y = 1 +2(II) → x = 4 y = 1 , so that (x, y) = (4, 1). See Figure 1.1. Figure 1.1: for Problem 1.1.11. 1.1.12 x − 2y = 3 2x − 4y = 6 −2(I) → x − 2y = 3 0 = 0 2 Copyright c 2013 Pearson Education, Inc.
4. 4. Section 1.1 This system has infinitely many solutions: If we choose y = t, an arbitrary real number, then the equation x − 2y = 3 gives us x = 3 + 2y = 3 + 2t. Therefore the general solution is (x, y) = (3 + 2t, t), where t is an arbitrary real number. (See Figure 1.2.) Figure 1.2: for Problem 1.1.12. 1.1.13 x − 2y = 3 2x − 4y = 8 −2(I) → x − 2y = 3 0 = 2 , which has no solutions. (See Figure 1.3.) Figure 1.3: for Problem 1.1.13. 1.1.14 The system reduces to   x + 5z = 0 y − z = 0 0 = 1  , so that there is no solution; no point in space belongs to all three planes. Compare with Figure 2b. 1.1.15 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. 1.1.16 The system reduces to   x + 5z = 0 y − z = 0 0 = 0   , so the solutions are of the form (x, y, z) = (−5t, t, t), where t is an arbitrary number. The three planes intersect in a line; compare with Figure 2a. 1.1.17 x + 2y = a 3x + 5y = b −3(I) → x + 2y = a −y = −3a + b ÷(−1) → x + 2y = a y = 3a − b −2(II) 3 Copyright c 2013 Pearson Education, Inc.
5. 5. Chapter 1 x = −5a + 2b y = 3a − b , so that (x, y) = (−5a + 2b, 3a − b). 1.1.18   x + 2y + 3z = a x + 3y + 8z = b x + 2y + 2z = c   −I −I →   x + 2y + 3z = a y + 5z = −a + b −z = −a + c   −2(II) →   x − 7z = 3a − 2b y + 5z = −a + b −z = −a + c   ÷(−1) →   x − 7z = 3a − 2b y + 5z = −a + b z = a − c   +7(III) −5(III) →   x = 10a − 2b − 7c y = −6a + b + 5c z = a − c  , so that (x, y, z) = (10a − 2b − 7c, −6a + b + 5c, a − c). 1.1.19 The system reduces to   x + z = 1 y − 2z = −3 0 = k − 7  . a. The system has solutions if k − 7 = 0, or k = 7. b. If k = 7 then the system has infinitely many solutions. c. If k = 7 then we can choose z = t freely and obtain the solutions (x, y, z) = (1 − t, −3 + 2t, t). 1.1.20 The system reduces to   x − 3z = 1 y + 2z = 1 (k2 − 4)z = k − 2   This system has a unique solution if k2 − 4 6= 0, that is, if k 6= ±2. If k = 2, then the last equation is 0 = 0, and there will be infinitely many solutions. If k = −2, then the last equation is 0 = −4, and there will be no solutions. 1.1.21 Let x, y, and z represent the three numbers we seek. We set up a system of equations and solve systematically (although there are short cuts):
6. 6. x +y = 24 x +z = 28 y+ z = 30
7. 7. −(I) →
8. 8. x +y = 24 −y +z = 4 y+ z = 30
9. 9. ÷(−1) →
10. 10. x +y = 24 y −z = −4 y+ z = 30
11. 11. −(II) −(II) →
12. 12. x +z = 28 y −z = −4 2z = 34
13. 13. ÷2 →
14. 14. x +z = 28 y −z = −4 z = 17
15. 15. −(III) +(III) →
16. 16. x = 11 y = 13 z = 17
17. 17. We see that x = 11, y = 13, and z = 17. 4 Copyright c 2013 Pearson Education, Inc.
18. 18. Section 1.1 1.1.22 Let x = the number of male children and y = the number of female children. Then the statement “Emile has twice as many sisters as brothers” translates into y = 2(x − 1) and “Gertrude has as many brothers as sisters” translates into x = y − 1. Solving the system −2x + y = −2 x − y = −1 gives x = 3 and y = 4. There are seven children in this family. 1.1.23 a Note that the demand D1 for product 1 increases with the increase of price P2; likewise the demand D2 for product 2 increases with the increase of price P1. This indicates that the two products are competing; some people will switch if one of the products gets more expensive. b Setting D1 = S1 and D2 = S2 we obtain the system 70 − 2P1 + P2 = −14 + 3P1 105 + P1 − P2 = −7 + 2P2 , or −5P1 + P2 = −84 P1 − 3P2 = 112 , which yields the unique solution P1 = 26 and P2 = 46. 1.1.24 The total demand for the product of Industry A is 1000 (the consumer demand) plus 0.1b (the demand from Industry B). The output a must meet this demand: a = 1000 + 0.1b. Setting up a similar equation for Industry B we obtain the system a = 1000 + 0.1b b = 780 + 0.2a or a − 0.1b = 1000 −0.2a + b = 780 , which yields the unique solution a = 1100 and b = 1000. 1.1.25 The total demand for the products of Industry A is 310 (the consumer demand) plus 0.3b (the demand from Industry B). The output a must meet this demand: a = 310 + 0.3b. Setting up a similar equation for Industry B we obtain the system a = 310 + 0.3b b = 100 + 0.5a or a − 0.3b = 310 −0.5a + b = 100 , which yields the solution a = 400 and b = 300. 1.1.26 Since x(t) = a sin(t) + b cos(t) we can compute dx dt = a cos(t) − b sin(t) and d2 x dt2 = −a sin(t) − b cos(t). Substituting these expressions into the equation d2 x dt2 − dx dt − x = cos(t) and simplifying gives (b − 2a) sin(t) + (−a − 2b) cos(t) = cos(t). Comparing the coefficients of sin(t) and cos(t) on both sides of the equation then yields the system −2a + b = 0 −a − 2b = 1 , so that a = −1 5 and b = −2 5 . See Figure 1.4. Figure 1.4: for Problem 1.1.26. 5 Copyright c 2013 Pearson Education, Inc.
19. 19. Chapter 1 1.1.27 a Substituting λ = 5 yields the system 7x − y = 5x −6x + 8y = 5y or 2x − y = 0 −6x + 3y = 0 or 2x − 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 . c Proceedings as in part (a), we find only the solution (0, 0). 1.1.28 Use the distance from Stein to Schaffhausen (and from Stein to Constance) as the unit of length. Let the speed of the boat and the speed of the river flow be vb and vs, respectively. Using the formula speed = distance time and measuring time in hours we get: vb − vs = 1 vb + vs = 3 2 We find vb = 5 4 and vs = 1 4 . The time it takes to travel from Stein to Constance is time = distance speed = 1 vb = 4 5 hours, or 48 minutes. 1.1.29 Let v be the speed of the boat relative to the water, and s be the speed of the stream; then the speed of the boat relative to the land is v +s downstream and v −s upstream. Using the fact that (distance) = (speed)(time), we obtain the system 8 = (v + s)1 3 8 = (v − s)2 3 # ← downstream ← upstream The solution is v = 18 and s = 6. 1.1.30 The thermal equilibrium condition requires that T1 = T2+200+0+0 4 , T2 = T1+T3+200+0 4 , and T3 = T2+400+0+0 4 . We can rewrite this system as   −4T1 + T2 = −200 T1 − 4T2 + T3 = −200 T2 − 4T3 = −400   The solution is (T1, T2, T3) = (75, 100, 125). 1.1.31 To assure that the graph goes through the point (1, −1), we substitute t = 1 and f(t) = −1 into the equation f(t) = a + bt + ct2 to give −1 = a + b + c. Proceeding likewise for the two other points, we obtain the system   a + b + c = −1 a + 2b + 4c = 3 a + 3b + 9c = 13  . The solution is a = 1, b = −5, and c = 3, and the polynomial is f(t) = 1 − 5t + 3t2 . (See Figure 1.5.) 6 Copyright c 2013 Pearson Education, Inc.
20. 20. Section 1.1 Figure 1.5: for Problem 1.1.31. 1.1.32 Proceeding as in the previous exercise, we obtain the system   a + b + c = p a + 2b + 4c = q a + 3b + 9c = r
21. 21. . The unique solution is   a = 3p − 3q + r b = −2.5p + 4q − 1.5r c = 0.5p − q + 0.5r  . Only one polynomial of degree 2 goes through the three given points, namely, f(t) = 3p − 3q + r + (−2.5p + 4q − 1.5r)t + (0.5p − q + 0.5r)t2 . 1.1.33 f(t) is of the form at2 + bt + c. So f(1) = a(12 ) + b(1) + c = 3, and f(2) = a(22 ) + b(2) + c = 6. Also, f′ (t) = 2at + b, meaning that f′ (1) = 2a + b = 1. So we have a system of equations:    a + b + c = 3 4a + 2b + c = 6 2a + b = 1    which reduces to    a = 2 b = −3 c = 4   . Thus, f(t) = 2t2 − 3t + 4 is the only solution. 1.1.34 f(t) is of the form at2 + bt + c. So, f(1) = a(12 ) + b(1) + c = 1 and f(2) = 4a + 2b + c = 0. Also, R 2 1 f(t)dt = R 2 1 (at2 + bt + c)dt = a 3 t3 + b 2 t2 + ct|2 1 = 8 3 a + 2b + 2c − (a 3 + b 2 + c) = 7 3 a + 3 2 b + c = −1. So we have a system of equations:   a + b + c = 1 4a + 2b + c = 0 7 3 a + 3 2 b + c = −1   which reduces to   a = 9 b = −28 c = 20  . 7 Copyright c 2013 Pearson Education, Inc.
22. 22. Chapter 1 Thus, f(t) = 9t2 − 28t + 20 is the only solution. 1.1.35 f(t) is of the form at2 + bt + c. f(1) = a + b + c = 1, f(3) = 9a + 3b + c = 3, and f′ (t) = 2at + b, so f′ (2) = 4a + b = 1. Now we set up our system to be    a + b + c = 1 9a + 3b + c = 3 4a + b = 1   . This reduces to    a − c 3 = 0 b + 4 3 c = 1 0 = 0   . We write everything in terms of a, revealing c = 3a and b = 1 − 4a. So, f(t) = at2 + (1 − 4a)t + 3a for an arbitrary a. 1.1.36 f(t) = at2 + bt + c, so f(1) = a + b + c = 1, f(3) = 9a + 3b + c = 3. Also, f′ (2) = 3, so 2(2)a + b = 4a + b = 3. Thus, our system is   a + b + c = 1 9a + 3b + c = 3 4a + b = 3  . When we reduce this, however, our last equation becomes 0 = 2, meaning that this system is inconsistent. 1.1.37 f(t) = ae3t + be2t , so f(0) = a + b = 1 and f′ (t) = 3ae3t + 2be2t , so f′ (0) = 3a + 2b = 4. Thus we obtain the system a + b = 1 3a + 2b = 4 , which reveals a = 2 b = −1 . So f(t) = 2e3t − e2t . 1.1.38 f(t) = a cos(2t) + b sin(2t) and 3f(t) + 2f′ (t) + f′′ (t) = 17 cos(2t). f′ (t) = 2b cos(2t) − 2a sin(2t) and f′′ (t) = −4b sin(2t) − 4a cos(2t). So, 17 cos(2t) = 3(a cos(2t) + b sin(2t)) + 2(2b cos(2t) − 2a sin(2t)) + (−4b sin(2t) − 4a cos(2t)) = (−4a + 4b + 3a) cos(2t) + (−4b − 4a + 3b) sin(2t) = (−a + 4b) cos(2t) + (−4a − b) sin(2t). So, our system is: −a + 4b = 17 −4a − b = 0 . This reduces to: a = −1 b = 4 . So our function is f(t) = − cos(2t) + 4 sin(2t). 8 Copyright c 2013 Pearson Education, Inc.
23. 23. Section 1.1 1.1.39 Plugging the three points (x, y) into the equation a + bx + cy + x2 + y2 = 0, leads to a system of linear equations for the three unknowns (a, b, c). a + 5b + 5c + 25 + 25 = 0 a + 4b + 6c + 16 + 36 = 0 a + 6b + 2c + 36 + 4 = 0 . The solution is a = −20, b = −2, c = −4. −20 − 2x − 4y + x2 + y2 = 0 is a circle of radius 5 centered at (1, 2). 1.1.40 Plug the three points into the equation ax2 + bxy + cy2 = 1. We obtain a system of linear equations a + 2b + 4c = 1 4a + 4b + 4c = 1 9a + 3b + c = 1 . The solution is a = 3/20, b = −9/40, c = 13/40. This is the ellipse (3/20)x2 − (9/40)xy + (13/40)y2 = 1. 1.1.41 The given system reduces to    x − z = −5a+2b 3 y + 2z = 4a−b 3 0 = a − 2b + c   . This system has solutions (in fact infinitely many) if a − 2b + c = 0. The points (a, b, c) with this property form a plane through the origin. 1.1.42 a x1 = −3 x2 = 14 + 3x1 = 14 + 3(−3) = 5 x3 = 9 − x1 − 2x2 = 9 + 3 − 10 = 2 x4 = 33 + x1 − 8x2 + 5x3 − x4 = 33 − 3 − 40 + 10 = 0, so that (x1, x2, x3, x4) = (−3, 5, 2, 0). b x4 = 0 x3 = 2 − 2x4 = 2 x2 = 5 − 3x3 − 7x4 = 5 − 6 = −1 x1 = −3 − 2x2 + x3 − 4x4 = −3 + 2 + 2 = 1, so that (x1, x2, x3, x4) = (1, −1, 2, 0). 1.1.43 a The two lines intersect unless t = 2 (in which case both lines have slope −1). To draw a rough sketch of x(t), note that limt→∞ x(t) = limt→−∞ x(t) = −1 the line x + t 2 y = t becomes almost horizontal and limt→2− x(t) = ∞, limt→2+ x(t) = −∞. 9 Copyright c 2013 Pearson Education, Inc.
24. 24. Chapter 1 Figure 1.6: for Problem 1.1.43a. Figure 1.7: for Problem 1.1.43a. Also note that x(t) is positive if t is between 0 and 2, and negative otherwise. Apply similar reasoning to y(t). (See Figures 1.6 and 1.7.) b x(t) = −t t−2 , and y(t) = 2t−2 t−2 . 1.1.44 We can think of the line through the points (1, 1, 1) and (3, 5, 0) as the intersection of any two planes through these two points; each of these planes will be defined by an equation of the form ax + by + cz = d. It is required that 1a + 1b + 1c = d and 3a + 5b + 0c = d. Now the system a +b +c −d = 0 3a +5b −d = 0 reduces to a +5 2 c −2d = 0 b −3 2 c +d = 0 . We can choose arbitrary real numbers for c and d; then a = −5 2 c + 2d and b = 3 2 c − d. For example, if we choose c = 2 and d = 0, then a = −5 and b = 3, leading to the equation −5x + 3y + 2z = 0. If we choose c = 0 and d = 1, then a = 2 and b = −1, giving the equation 2x − y = 1. 10 Copyright c 2013 Pearson Education, Inc.
25. 25. Section 1.1 We have found one possible answer: −5x +3y +2z = 0 2x −y = 1 . 1.1.45 To eliminate the arbitrary constant t, we can solve the last equation for t to give t = z − 2, and substitute z − 2 for t in the first two equations, obtaining x = 6 + 5(z − 2) y = 4 + 3(z − 2) or x − 5z = −4 y − 3z = −2 . This system does the job. 1.1.46 Let b = Boris’ money, m = Marina’s money, and c = cost of a chocolate bar. We are told that b + 1 2 m = c 1 2 b + m = 2c , with solution (b, m) = (0, 2c). Boris has no money. 1.1.47 Let us start by reducing the system:   x + 2y + 3z = 39 x + 3y + 2z = 34 3x + 2y + z = 26   −I −3(I) →   x + 2y + 3z = 39 y − z = −5 −4y − 8z = −91   Note that the last two equations are exactly those we get when we substitute x = 39 − 2y − 3z: either way, we end up with the system y − z = −5 −4y − 8z = −91 . 1.1.48 a We set up two equations here, with our variables: x1 = servings of rice, x2 = servings of yogurt. So our system is: 3x1 +12x2 = 60 30x1 +20x2 = 300 . Solving this system reveals that x1 = 8, x2 = 3. b Again, we set up our equations: 3x1 +12x2 = P 30x1 +20x2 = C , and reduce them to find that x1 = − P 15 + C 25 , while x2 = P 10 − C 100 . 1.1.49 Let x1 = number of one-dollar bills, x2 = the number of five-dollar bills, and x3 = the number of ten-dollar bills. Then our system looks like: x1 + x2 + x3 = 32 x1 + 5x2 + 10x3 = 100 , which reduces to give us solutions that fit: x1 = 15 + 5 4 x3, x2 = 17 − 9 4 x3, where x3 can be chosen freely. Now let’s keep in mind that x1, x2, and x3 must be positive integers and see what conditions this imposes on the variable x3. We see that since x1 and x2 must be integers, x3 must be a multiple of 4. Furthermore, x3 must be positive, and x2 = 17− 9 4 x3 must be positive as well, meaning that x3 68 9 . These constraints leave us with only one possibility, x3 = 4, and we can compute the corresponding values x1 = 15+ 5 4 x3 = 20 and x2 = 17− 9 4 x3 = 8. Thus, we have 20 one-dollar bills, 8 five-dollar bills, and 4 ten-dollar bills. 1.1.50 Let x1, x2, x3 be the number of 20 cent, 50 cent, and 2 Euro coins, respectively. Then we need solutions to the system: x1 +x2 +x3 = 1000 .2x1 +.5x2 +2x3 = 1000 11 Copyright c 2013 Pearson Education, Inc.
26. 26. Chapter 1 this system reduces to: x1 −5x3 = −5000 3 x2 +6x3 = 8000 3 . Our solutions are then of the form   x1 x2 x3   =   5x3 − 5000 3 −6x3 + 8000 3 x3  . Unfortunately for the meter maids, there are no integer solutions to this problem. If x3 is an integer, then neither x1 nor x2 will be an integer, and no one will ever claim the Ferrari. Section 1.2 1.2.1   1 1 −2 . . . 5 2 3 4 . . . 2   −2(I) →   1 1 −2 . . . 5 0 1 8 . . . −8   −II →   1 0 −10 . . . 13 0 1 8 . . . −8   x − 10z = 13 y + 8z = −8 −→ x = 13 + 10z y = −8 − 8z   x y z   =   13 + 10t −8 − 8t t  , where t is an arbitrary real number. 1.2.2   3 4 −1 . . . 8 6 8 −2 . . . 3   ÷3 →   1 4 3 −1 3 . . . 8 3 6 8 −2 . . . 3   −6(I) →   1 4 3 −1 3 . . . 8 3 0 0 0 . . . −13   This system has no solutions, since the last row represents the equation 0 = −13. 1.2.3 x = 4 − 2y − 3z y and z are free variables; let y = s and z = t.   x y z   =   4 − 2s − 3t s t  , where s and t are arbitrary real numbers. 1.2.4     1 1 . . . 1 2 −1 . . . 5 3 4 . . . 2     −2(I) −3(I) →     1 1 . . . 1 0 −3 . . . 3 0 1 . . . −1     ÷(−3) →     1 1 . . . 1 0 1 . . . −1 0 1 . . . −1     −II −II →     1 0 . . . 2 0 1 . . . −1 0 0 . . . 0    , so that x = 2 y = −1 . 12 Copyright c 2013 Pearson Education, Inc.
27. 27. Section 1.2 1.2.5       0 0 1 1 . . . 0 0 1 1 0 . . . 0 1 1 0 0 . . . 0 1 0 0 1 . . . 0       swap : I ↔ III →       1 1 0 0 . . . 0 0 1 1 0 . . . 0 0 0 1 1 . . . 0 1 0 0 1 . . . 0       −I →       1 1 0 0 . . . 0 0 1 1 0 . . . 0 0 0 1 1 . . . 0 0 −1 0 1 . . . 0       −II +II →        1 0 −1 0 . . . 0 0 1 1 0 . . . 0 0 0 1 1 . . . 0 0 0 1 1 . . . 0        +III −III −III →        1 0 0 1 . . . 0 0 1 0 −1 . . . 0 0 0 1 1 . . . 0 0 0 0 0 . . . 0          x1 + x4 = 0 x2 − x4 = 0 x3 + x4 = 0   −→   x1 = −x4 x2 = x4 x3 = −x4      x1 x2 x3 x4    =    −t t −t t   , where t is an arbitrary real number. 1.2.6 The system is in rref already.   x1 = 3 + 7x2 − x5 x3 = 2 + 2x5 x4 = 1 − x5   Let x2 = t and x5 = r.      x1 x2 x3 x4 x5      =      3 + 7t − r t 2 + 2r 1 − r r      1.2.7        1 2 0 2 3 . . . 0 0 0 1 3 2 . . . 0 0 0 1 4 −1 . . . 0 0 0 0 0 1 . . . 0        −II →        1 2 0 2 3 . . . 0 0 0 1 3 2 . . . 0 0 0 0 1 −3 . . . 0 0 0 0 0 1 . . . 0        −2(III) −3(III) →        1 2 0 0 9 . . . 0 0 0 1 0 11 . . . 0 0 0 0 1 −3 . . . 0 0 0 0 0 1 . . . 0        −9(IV ) −11(IV ) +3(IV ) →       1 2 0 0 0 . . . 0 0 0 1 0 0 . . . 0 0 0 0 1 0 . . . 0 0 0 0 0 1 . . . 0          x1 + 2x2 = 0 x3 = 0 x4 = 0 x5 = 0    −→    x1 = −2x2 x3 = 0 x4 = 0 x5 = 0    13 Copyright c 2013 Pearson Education, Inc.
28. 28. Chapter 1 Let x2 = t.      x1 x2 x3 x4 x5      =      −2t t 0 0 0      , where t is an arbitrary real number. 1.2.8 0 1 0 2 3 . . . 0 0 0 0 4 8 . . . 0 # ÷4 → 0 1 0 2 3 . . . 0 0 0 0 1 2 . . . 0 # −2(II)   0 1 0 0 −1 . . . 0 0 0 0 1 2 . . . 0   x2 − x5 = 0 x4 + 2x5 = 0 −→ x2 = x5 x4 = −2x5 Let x1 = r, x3 = s, x5 = t.      x1 x2 x3 x4 x5      =      r t s −2t t      , where r, t and s are arbitrary real numbers. 1.2.9     0 0 0 1 2 −1 . . . 2 1 2 0 0 1 −1 . . . 0 1 2 2 0 −1 1 . . . 2     swap : I ↔ II →     1 2 0 0 1 −1 . . . 0 0 0 0 1 2 −1 . . . 2 1 2 2 0 −2 1 . . . 2     −I →     1 2 0 0 1 −1 . . . 0 0 0 0 1 2 −1 . . . 2 0 0 2 0 −2 2 . . . 2     swap : II ↔ III →     1 2 0 0 1 −1 . . . 0 0 0 2 0 −2 2 . . . 2 0 0 0 1 2 −1 . . . 2     ÷2 →     1 2 0 0 1 −1 . . . 0 0 0 1 0 −1 1 . . . 1 0 0 0 1 2 −1 . . . 2       x1 + 2x2 + x5 − x6 = 0 x3 − x5 + x6 = 1 x4 + 2x5 − x6 = 2   −→   x1 = −2x2 − x5 + x6 x3 = 1 + x5 − x6 x4 = 2 − 2x5 + x6   Let x2 = r, x5 = s, and x6 = t.        x1 x2 x3 x4 x5 x6        =        −2r − s + t r 1 + s − t 2 − 2s + t s t        , where r, s and t are arbitrary real numbers. 1.2.10 The system reduces to   x1 + x4 = 1 x2 − 3x4 = 2 x3 + 2x4 = −3   −→   x1 = 1 − x4 x2 = 2 + 3x4 x3 = −3 − 2x4   14 Copyright c 2013 Pearson Education, Inc.
29. 29. Section 1.2 Let x4 = t.    x1 x2 x3 x4    =    1 − t 2 + 3t −3 − 2t t   , where t is an arbitrary real number. 1.2.11 The system reduces to   x1 + 2x3 = 0 x2 − 3x3 = 4 x4 = −2   −→   x1 = −2x3 x2 = 4 + 3x3 x4 = −2  . Let x3 = t.    x1 x2 x3 x4    =    −2t 4 + 3t t −2    1.2.12 The system reduces to    x1 + 3.5x5 + x6 = 0 x2 + x5 = 0 x3 − 5 3 x6 = 0 x4 + 3x5 + x6 = 0    −→    x1 = −3.5x5 − x6 x2 = −x5 x3 = 5 3 x6 x4 = −3x5 − x6   . Let x5 = r and x6 = t.        x1 x2 x3 x4 x5 x6        =        −3.5r − t −r 5 3 t −3r − t r t        1.2.13 The system reduces to   x − z = 0 y + 2z = 0 0 = 1  . There are no solutions. 1.2.14 The system reduces to x + 2y = −2 z = 2 −→ x = −2 − 2y z = 2 . Let y = t.   x y z   =   −2 − 2t t 2   15 Copyright c 2013 Pearson Education, Inc.
30. 30. Chapter 1 1.2.15 The system reduces to   x = 4 y = 2 z = 1  . 1.2.16 The system reduces to x1 + 2x2 + 3x3 +5x5 = 6 x4 +2x5 = 7 −→ x1 = 6 − 2x2 − 3x3 − 5x5 x4 = 7 − 2x5 . Let x2 = r, x3 = s, and x5 = t.      x1 x2 x3 x4 x5      =      6 − 2r − 3s − 5t r s 7 − 2t t      1.2.17 The system reduces to            x1 = −8221 4340 x2 = 8591 8680 x3 = 4695 434 x4 = −459 434 x5 = 699 434            . 1.2.18 a No, since the third column contains two leading ones. b Yes c No, since the third row contains a leading one, but the second row does not. d Yes 1.2.19    0 0 0 0    and    1 0 0 0    1.2.20 a = 1 (by property a. on page 16), c = 0 (by property b. on page 16), and e = 0 (by property c. on page 16). For the values of b and d we have two possibilities: b = 0, d = 1 or b arbitrary, d = 0. 1.2.21 e = 0 (by property c. on page 16), c = 1 (c must be a leading 1), d = 0 (by property b. on page 16), b = 0 (by property b. on page 16), and 16 Copyright c 2013 Pearson Education, Inc.
31. 31. Section 1.2 a is arbitrary. 1.2.22 Four, namely 0 0 0 0 , 1 k 0 0 , 0 1 0 0 , 1 0 0 1 (k is an arbitrary constant.) 1.2.23 Four, namely   0 0 0 0 0 0   ,   1 k 0 0 0 0   ,   0 1 0 0 0 0   ,   1 0 0 1 0 0   (k is an arbitrary constant.) 1.2.24 Seven, namely 0 0 0 0 0 0 , 1 a b 0 0 0 , 0 1 c 0 0 0 , 0 0 1 0 0 0 , 1 0 d 0 1 e , 1 f 0 0 0 1 , 0 1 0 0 0 1 . Here, a, b, . . . , f are arbitrary constants. 1.2.25 The conditions a, b, and c for the reduced row-echelon form correspond to the properties P1, P2, and P3 given on Page 13. The Gauss-Jordan algorithm, summarized on Page 15, guarantees that those properties are satisfied. 1.2.26 Yes; each elementary row operation is reversible, that is, it can be “undone.” For example, the operation of row swapping can be undone by swapping the same rows again. The operation of dividing a row by a scalar can be reversed by multiplying the same row by the same scalar. 1.2.27 Yes; if A is transformed into B by a sequence of elementary row operations, then we can recover A from B by applying the inverse operations in the reversed order. 1.2.28 Yes, by Exercise 27, since rref(A) is obtained from A by a sequence of elementary row operations. 1.2.29 No; whatever elementary row operations you apply to   1 2 3 4 5 6 7 8 9  , you cannot make the last column equal to zero. 1.2.30 Suppose (c1, c2, . . . , cn) is a solution of the system   a11x1 + a12x2 + · · · + a1nxn = b1 a21x1 + a22x2 + · · · + a2nxn = b2 . . . . . . . . .  . To keep the notation simple, suppose we add k times the first equation to the second; then the second equation of the new system will be (a21 + ka11)x1 + · · · + (a2n + ka1n)xn = b2 + kb1. We have to verify that (c1, c2, . . . , cn) is a solution of this new equation. Indeed, (a21 + ka11)c1 + · · · + (a2n + ka1n)cn = a21c1 + · · · + a2ncn + k(a11c1 + · · · + a1ncn) = b2 + kb1. We have shown that any solution of the “old” system is also a solution of the “new.” To see that, conversely, any solution of the new system is also a solution of the old system, note that elementary row operations are reversible (compare with Exercise 26); we can obtain the old system by subtracting k times the first equation from the second equation of the new system. 1.2.31 Since the number of oxygen atoms remains constant, we must have 2a + b = 2c + 3d. 17 Copyright c 2013 Pearson Education, Inc.
32. 32. Chapter 1 Considering hydrogen and nitrogen as well, we obtain the system   2a + b = 2c + 3d 2b = c + d a = c + d   or   2a + b − 2c − 3d = 0 2b − c − d = 0 a − c − d = 0  , which reduces to   a − 2d = 0 b − d = 0 c − d = 0  . The solutions are    a b c d    =    2t t t t   . To get the smallest positive integers, we set t = 1: 2NO2 + H2O −→ HNO2 + HNO3 1.2.32 Plugging the points into f(t), we obtain the system    a = 1 a + b + c + d = 0 a − b + c − d = 0 a + 2b + 4c + 8d = −15    with unique solution a = 1, b = 2, c = −1, and d = −2, so that f(t) = 1 + 2t − t2 − 2t3 . (See Figure 1.8.) Figure 1.8: for Problem 1.2.32. 1.2.33 Let f(t) = a + bt + ct2 + dt3 + et4 . Substituting the points in, we get      a + b + c + d + e = 1 a + 2b + 4c + 8d + 16e = −1 a + 3b + 9c + 27d + 81e = −59 a − b + c − d + e = 5 a − 2b + 4c − 8d + 16e = −29      This system has the unique solution a = 1, b = −5, c = 4, d = 3, and e = −2, so that f(t) = 1−5t+4t2 +3t3 −2t4 . (See Figure 1.9.) 1.2.34 The requirement f′ i (ai) = f′ i+1(ai) and f′′ i (ai) = f′′ i+1(ai) ensure that at each junction two different cubics fit “into” one another in a “smooth” way, since they must have the same slope and be equally curved. The requirement that f′ 1(a0) = f′ n(an) = 0 ensures that the track is horizontal at the beginning and at the end. How 18 Copyright c 2013 Pearson Education, Inc.
33. 33. Section 1.2 Figure 1.9: for Problem 1.2.33. many unknowns are there? There are n pieces to be fit, and each one is a cubic of the form f(t) = p+qt+rt2 +st3 , with p, q, r, and s to be determined; therefore, there are 4n unknowns. How many equations are there? fi(ai) = bi for i = 1, 2, . . . , n gives n equations fi(ai−1) = bi−1 for i = 1, 2, . . . , n gives n equations f′ i (ai) = f′ i+1(ai) for i = 1, 2, . . . , n − 1 gives n − 1 equations f′′ i (ai) = f′′ i+1(ai) for i = 1, 2, . . . , n − 1 gives n − 1 equations f′ 1(a0) = 0, f′ n(an) = 0 gives 2 equations Altogether, we have 4n equations; convince yourself that all these equations are linear. 1.2.35 Let f(t) = a + bt + ct2 + dt3 , so that f′ (t) = b + 2ct + 3dt2 . Substituting the given points into f(t) and f′ (t) we obtain the system    a + b + c + d = 1 a + 2b + 4c + 8d = 5 b + 2c + 3d = 2 b + 4c + 12d = 9    This system has the unique solution a = −5, b = 13, c = −10, and d = 3, so that f(t) = −5 + 13t − 10t2 + 3t3 . (See Figure 1.10.) Figure 1.10: for Problem 1.2.35. 1.2.36 We want all vectors   x y z   in R3 such that   x y z   ·   1 3 −1   = x + 3y − z = 0. The endpoints of these vectors form a plane. 19 Copyright c 2013 Pearson Education, Inc.
34. 34. Chapter 1 These vectors are of the form   x y z   =   −3r + t r t  , where r and t are arbitrary real numbers. 1.2.37 We need to solve the system   x1 + x2 + x3 + x4 = 0 x1 + 2x2 + 3x3 + 4x4 = 0 x1 + 9x2 + 9x3 + 7x4 = 0  , which reduces to   x1 + 0.25x4 = 0 x2 − 1.5x4 = 0 x3 + 2.25x4 = 0  . The solutions are of the form    x1 x2 x3 x4    =    −0.25t 1.5t −2.25t t   , where t is an arbitrary real number. 1.2.38 Writing the equation ~ b = x1~ v1 + x2~ v2 + x3~ v3 in terms of its components, we obtain the system    x1 + 2x2 + 4x3 = −8 4x1 + 5x2 + 6x3 = −1 7x1 + 8x2 + 9x3 = 9 5x1 + 3x2 + x3 = 15    The system has the unique solution x1 = 2, x2 = 3, and x3 = −4. 1.2.39 Compare with the solution of Exercise 1.1.25. The diagram tells us that   x1 = 0.2x2 + 0.3x3 + 320 x2 = 0.1x1 + 0.4x3 + 90 x3 = 0.2x1 + 0.5x2 + 150   or   x1 − 0.2x2 − 0.3x3 = 320 −0.1x1 + x2 − 0.4x3 = 90 −0.2x1 − 0.5x2 + x3 = 150  . This system has the unique solution x1 = 500, x2 = 300, and x3 = 400. 1.2.40 a ~ v1 =   0 0.1 0.2   , ~ v2 =   0.2 0 0.5   , ~ v3 =   0.3 0.4 0   , ~ b =   320 90 150   b Recall that xj is the output of industry Ij, and the ith component aij of ~ vj is the demand of Industry Ij on industry Ij for each dollar of output of industry Ij. Therefore, the product xjaij (that is, the ith component of xj~ vj), represents the total demand of industry Ij on Industry Ii (in dollars). c x1~ v1 + · · · + xn~ vn +~ b is the vector whose ith component represents the total demand on industry Ii (consumer demand and interindustry demand combined). d The ith component of the equation x1~ v1 + · · · + xn~ vn +~ b = ~ x expresses the requirement that the output xi of industry Ii equal the total demand on that industry. 1.2.41 a These components are zero because neither manufacturing not the energy sector directly require agricultural products. 20 Copyright c 2013 Pearson Education, Inc.
35. 35. Section 1.2 b We have to solve the system x1~ v1 + x2~ v2 + x3~ v3 +~ b = ~ x or   0.707x1 = 13.2 −0.014x1 + 0.793x2 − 0.017x3 = 17.6 −0.044x1 + 0.01x2 + 0.784x3 = 1.8   The unique solution is approximately x1 = 18.67, x2 = 22.60, and x3 = 3.63. 1.2.42 We want to find m1, m2, m3 such that m1 + m2 + m3 = 1 and 1 1 m1 1 2 + m2 2 3 + m3 4 1 = 2 2 , that is, we have to solve the system   m1 + m2 + m3 = 1 m1 + 2m2 + 4m3 = 2 2m1 + 3m2 + m3 = 2  . The unique solution is m1 = 1 2 , m2 = 1 4 , and m3 = 1 4 . We will put 1 2 kg at the point 1 2 and 1 4 kg at each of the two other vertices. 1.2.43 We know that m1~ v1 + m2~ v2 = m1 ~ w1 + m2 ~ w2 or m1(~ v1 − ~ w1) + m2(~ v2 − ~ w2) = ~ 0 or   −3m1 + 2m2 = 0 −6m1 + 4m2 = 0 −3m1 + 2m2 = 0  . We can conclude that m1 = 2 3 m2. 1.2.44 Let x1, x2, x3, and x4 be the traffic volume at the four locations indicated in Figure 1.11. Figure 1.11: for Problem 1.2.44. We are told that the number of cars coming into each intersection is the same as the number of cars coming out: 21 Copyright c 2013 Pearson Education, Inc.
36. 36. Chapter 1    x1 + 300 = 320 + x2 x2 + 300 = 400 + x3 x3 + x4 + 100 = 250 150 + 120 = x1 + x4    or    x1 − x2 = 20 x2 − x3 = 100 x3 + x4 = 150 x1 + x4 = 270    The solutions are of the form    x1 x2 x3 x4    =    270 − t 250 − t 150 − t t   . Since the xi must be positive integers (or zero), t must be an integer with 0 ≤ t ≤ 150. The lowest possible values are x1 = 120, x2 = 100, x3 = 0, and x4 = 0, while the highest possible values are x1 = 270, x2 = 250, x3 = 150, and x4 = 150. 1.2.45 Plugging the data into the function S(t) we obtain the system     a + b cos 2π·47 365 + c sin 2π·47 365 = 11.5 a + b cos 2π·74 365 + c sin 2π·74 365 = 12 a + b cos 2π·273 365 + c sin 2π·273 365 = 12     The unique solution is approximately a = 12.17, b = −1.15, and c = 0.18, so that S(t) = 12.17 − 1.15 cos 2πt 365 + 0.18 sin 2πt 365 . The longest day is about 13.3 hours. (See Figure 1.12.) Figure 1.12: for Problem 1.2.45. 1.2.46 Kyle first must solve the following system: x1 +x2 +x3 = 24 3x1 +2x2 +1 2 x3 = 24 . This system reduces to x1 −1.5x3 = −24 x2 +2.5x3 = 48 . Thus, our solutions will be of the form   x1 x2 x3   =   1.5x3 − 24 −2.5x3 + 48 x3  . Since all of our values must be non-negative integers (and x3 must be even), we find the following solutions for   lilies roses daisies  :   0 8 16   and   3 3 18  . Since Olivia loves lilies, Kyle spends his 24 dollars on 3 lilies, 3 roses and 18 daisies. 22 Copyright c 2013 Pearson Education, Inc.
37. 37. Section 1.2 1.2.47 a When k 6= 1 and k 6= 2, we can see that this will continue to reduce to a consistent system with a unique solution. b When k = 1, our bottom row reveals the inconsistency 0 = 2. c When k = 2, the second row and third row both represent the equation z = 2, meaning that the third row will be replaced with the equation 0 = 0 during further reduction. This reveals that we will have an infinite number of solutions. 1.2.48 a We reduce our matrix in the following steps:     0 1 2k . . . 0 1 2 6 . . . 2 k 0 2 . . . 1     swap : I ↔ II →     1 2 6 . . . 2 0 1 2k . . . 0 k 0 2 . . . 1     −k(I) →     1 2 6 . . . 2 0 1 2k . . . 0 0 −2k 2 − 6k . . . 1 − 2k     −2(II) +2k(II) →     1 0 6 − 4k . . . 2 0 1 2k . . . 0 0 0 2 − 6k + 4k2 . . . 1 − 2k     →     1 0 6 − 4k . . . 2 0 1 2k . . . 0 0 0 2(2k − 1)(k − 1) . . . −(2k − 1)     . We see that there will be a unique solution when the 2(2k − 1)(k − 1) term is not equal to zero, when 2k − 1 6= 0 and k − 1 6= 0, or k 6= 1 2 and k 6= 1. b We will have no solutions when the term 2(2k − 1)(k − 1) is equal to zero, but the term −(2k − 1) is not. This occurs only when k = 1. c We will have infinitely many solutions when the last row represents the equation 0 = 0. This occurs when 2k − 1 = 0, or k = 1 2 . 1.2.49 a So −1 2 x1 + x2 − 1 2 x3 = 0 and −1 2 x2 + x3 − 1 2 x4 = 0. After reduction of the system, we find that our solutions are all of the form    x1 x2 x3 x4    = s    −2 −1 0 1    + t    3 2 1 0   . b Yes, from our solution in part (a), if we plug in 1 for x1 and 13 for x4, we obtain 3t − 2s = 1 and s = 13, which leads to t = 9, and x2 = 5, x3 = 9. So we have the solution: x1 = 1, x2 = 5, x3 = 9 and x4 = 13, which is an arithmetic progression. 1.2.50 It is required that xk = 1 2 (xk−1 + xk+1), or 2xk = xk−1 + xk+1, or xk − xk−1 = xk+1 − xk. This means that the difference of any two consecutive terms must be the same; we are looking at the finite arithmetic sequences. Thus the solutions are of the form (x1, x2, x3, . . . , xn) = (t, t + r, t + 2r, . . . , t + (n − 1)r), where t and r are arbitrary constants. 23 Copyright c 2013 Pearson Education, Inc.
38. 38. Chapter 1 1.2.51 We begin by solving the system. Our augmented matrix begins as:     2 1 0 . . . C 0 3 1 . . . C 1 0 4 . . . C     and is reduced to      1 0 0 . . . 9 25 C 0 1 0 . . . 7 25 C 0 0 1 . . . 4 25 C      . In order for x, y and z to be integers, C must be a multiple of 25. We want the smallest positive choice, so C = 25. 1.2.52 f(t) = a+bt+ct2 +dt3 and we learn that f(0) = a = 3, f(1) = a+b+c+d = 2, f(2) = a+2b+4c+8d = 0. Also, Z 2 0 f(t)dt = at + 1 2 bt2 + 1 3 ct3 + 1 4 dt4 |2 0 = 2a + 2b + 8 3 c + 4d = 4. Now, we set up our matrix,       1 0 0 0 . . . 3 1 1 1 1 . . . 2 1 2 4 8 . . . 0 2 2 8 3 4 . . . 4       . However, when we reduce this, the last line becomes 0 = 1, meaning that the system is inconsistent. In introductory calculus you may have seen the approximation formula: Z b a f(t)dt ≈ b − a 6 (f(a) + 4f( a + b 2 ) + f(b)), the simplest form of Simpson’s Rule. For polynomials f(t) of degree ≤ 3, Simpson’s Rule gives the exact value of the integral. Thus, for the f(t) in our problem, Z 2 0 f(t)dt = 2 6 (f(0) + 4f(1) + f(2)) = 1 3 (3 + 8 + 0) = 11 3 . Thus it is impossible to find such a cubic with Z 2 0 f(t)dt = 4, as required. 1.2.53 The system of linear equations is c1 = 0 c1 + c2 + c4 = 0 c1 + 2c2 + 4c4 = 0 c1 + c3 + c6 = 0 c1 + 2c3 + 4c6 = 0 It has the solution (c1, c2, c3, c4, c5, c6) = (0, 0, 0, 0, c5, 0). This is the conic xy = 0. 24 Copyright c 2013 Pearson Education, Inc.
39. 39. Section 1.2 -1 -1 3 2 1 3 2 1 x y 1.2.54 The system of linear equations is c1 = 0 c1 + 2c2 + 4c4 = 0 c1 + 2c3 + 4c6 = 0 c1 + 2c2 + 2c3 + 4c4 + 4c5 + 4c6 = 0 c1 + c2 + 3c3 + c4 + 3c5 + 9c6 = 0 It has the solution (c1, c2, c3, c4, c5, c6) = (0, −6c6, −2c6, 3c6, 0, c6). This is the conic −6x − 2y + 3x2 + y2 = 0 or 3(x − 1)2 + (y − 1)2 = 4, the ellipse centered at (1, 1) with semiaxis 2/ √ 3 and 2. -1 -1 3 2 1 3 2 1 x y 1.2.55 The system of linear equations is c1 = 0 c1 + c2 + c4 = 0 c1 + 2c2 + 4c4 = 0 c1 + 3c2 + 9c4 = 0 c1 + c2 + c3 + c4 + c5 + c6 = 0 This system has the solution (c1, c2, c3, c4, c5, c6) = (0, 0, −c5 −c6, 0, c5, c6). Setting c5 = a and c6 = b, we get the family of conics a(xy − y) + b(y2 − y) = ay(x − 1) + by(y − 1) = 0 where a 6= 0 or b 6= 0. Each such conic is the union of the x axis with some line through the point (1, 1). Two sample solutions are shown in the accompanying figures. 25 Copyright c 2013 Pearson Education, Inc.
40. 40. Chapter 1 -1 -1 3 2 1 3 2 x y 1 -1 -1 3 2 1 3 2 x y 1 1.2.56 The system of linear equations is c1 = 0 c1 + c2 + c3 + c4 + c5 + c6 = 0 c1 + 2c2 + 2c3 + 4c4 + 4c5 + 4c6 = 0 c1 + 3c2 + 3c3 + 9c4 + 9c5 + 9c6 = 0 c1 + c2 + c4 = 0 It has the solution (c1, c2, c3, c4, c5, c6) = (0, c5 + c6, −c5 − c6, −c5 − c6, c5, c6). Setting c5 = a and c6 = b, we get the family of conics a(x − y + xy − x2 ) + b(x − y − x2 + y2 ) = a(x − y)(1 − x) + b(x − y)(1 − x − y) = 0 where a 6= 0 or b 6= 0. Each such conic is the union of the line y = x with some line through the point (1, 0). Two sample solutions are shown in the accompanying figures. -1 -1 3 2 1 3 2 1 x y -1 -1 3 2 1 3 2 1 x y 1.2.57 The system of linear equations is c1 = 0 c1 + c2 + c4 = 0 c1 + c3 + c6 = 0 c1 + c2 + c3 + c4 + c5 + c6 = 0 It has the solution (c1, c2, c3, c4, c5, c6) = (0, −c4, −c6, c4, 0, c6). Setting c4 = a and c6 = b, we get the family of conics a(x2 − x) + b(y2 − y) = ax(x − 1) + by(y − 1) = 0, where a 6= 0 or b 6= 0. Two sample solutions are shown in the accompanying figures. 26 Copyright c 2013 Pearson Education, Inc.
41. 41. Section 1.2 1 x y 1 1 x y 1 1.2.58 The system of linear equations is c1 = 0 c1 + c2 + c4 = 0 c1 + c3 + c6 = 0 c1 + c2 − c3 + c4 − c5 + c6 = 0 It has the solution (c1, c2, c3, c4, c5, c6) = (0, −c4, −c6, c4, 2c6, c6). Setting c4 = a and c6 = b, we get the family of conics a(x2 − x) + b(y2 + 2xy − y) = ax(x − 1) + by(y + 2x − 1) = 0, where a 6= 0 or b 6= 0. Two sample solutions are shown in the accompanying figures. -1 -2 2 1 -1 3 1 x y 2 -1 -2 2 1 -1 3 1 x y 2 1.2.59 The system of linear equations is c1 + 5c2 + 25c4 = 0 c1 + c2 + 2c3 + c4 + 2c5 + 4c6 = 0 c1 + 2c2 + c3 + 4c4 + 2c5 + c6 = 0 c1 + 8c2 + c3 + 64c4 + 8c5 + c6 = 0 c1 + 2c2 + 9c3 + 4c4 + 18c5 + 81c6 = 0 It has the solution (c1, c2, c3, c4, c5, c6) = (25c6, −10c6, −10c6, c6, 0, c6). This is the conic 25−10x−10y+x2 +y2 = 0, a circle of radius 5 centered at (5, 5). -4 -4 12 8 4 12 8 4 x y (5, 5) 27 Copyright c 2013 Pearson Education, Inc.
42. 42. Chapter 1 1.2.60 The system of linear equations is c1 + c2 + c4 = 0 c1 + 2c2 + 4c4 = 0 c1 + 2c2 + 2c3 + 4c4 + 4c5 + 4c6 = 0 c1 + 5c2 + 2c3 + 25c4 + 10c5 + 4c6 = 0 c1 + 5c2 + 6c3 + 25c4 + 30c5 + 36c6 = 0 It has the solution (c1, c2, c3, c4, c5, c6) = (2c6, −3c6, 2c6, c6, −2c6, c6). This is the conic 2−3x+2y+x2 −2xy+y2 = 0. -2 -2 6 4 2 6 4 2 x y 1.2.61 The system of linear equations is c1 = 0 c1 + c2 + c4 = 0 c1 + 2c2 + 4c4 = 0 c1 + c3 + c6 = 0 c1 + 2c3 + 4c6 = 0 c1 + c2 + c3 + c4 + c5 + c6 = 0 It has the only solution (c1, c2, c3, c4, c5, c6) = (0, 0, 0, 0, 0, 0). There is no conic which goes through these points. Alternatively, note that the only conic through the first five points is xy = 0, according to Exercise 51. But that conic fails to run through the point (1, 1), so that there is no conic through all six points. -1 -1 3 2 1 3 2 1 x y 1.2.62 The system of linear equations is c1 = 0 28 Copyright c 2013 Pearson Education, Inc.
43. 43. Section 1.2 c1 + 2c2 + 4c4 = 0 c1 + 2c3 + 4c6 = 0 c1 + 2c2 + 2c3 + 4c4 + 4c5 + 4c6 = 0 c1 + c2 + 3c3 + c4 + 3c5 + 9c6 = 0 c1 + 4c2 + c3 + 16c4 + 4c5 + c6 = 0 It has the solution (c1, c2, c3, c4, c5, c6) = (0, 0, 0, 0, 0, 0). There is no conic which goes through these points. -1 3 4 2 4 3 2 x y 1 -1 1 1.2.63 Let x1 be the cost of the environmental statistics book, x2 be the cost of the set theory text and x3 be the cost of the educational psychology book. Then, from the problem, we deduce the augmented matrix     1 1 0 . . . 178 2 1 1 . . . 319 0 1 1 . . . 147    . We can reduce this matrix to     1 0 0 . . . 86 0 1 0 . . . 92 0 0 1 . . . 55    , revealing that x1 = 86, x2 = 92 and x3 = 55. Thus, the environmental statistics book costs $ 86, the set theory book costs $ 92 and the educational psychology book is only priced at $ 55. 1.2.64 Let our vectors   x1 x2 x3   represent the numbers of the books   grammar Werther LinearAlg.   . Then we can set up the matrix     1 1 0 . . . 64 1 0 1 . . . 98 0 1 1 . . . 76    . This system yields one solution,   43 21 55   , meaning that the grammar book costs 43 Euro, the novel costs 21 Euro, and the linear algebra text costs 55 Euro. 1.2.65 The difficult part of this problem lies in setting up a system from which we can derive our matrix. We will define x1 to be the number of “liberal” students at the beginning of the class, and x2 to be the number of “conservative” students at the beginning. Thus, since there are 260 students in total, x1 + x2 = 260. We need one more equation involving x1 and x2 in order to set up a useful system. Since we know that the number of conservative students at the end of the semester is equal to the number of liberal student initially, we obtain the equation 3 10 x1 + 6 10 x2 = x1, or − 7 10 x1 + 6 10 x2 = 0. We then use 1 1 . . . 260 − 7 10 6 10 . . . 0 # to obtain 1 0 . . . 120 0 1 . . . 140 # . 29 Copyright c 2013 Pearson Education, Inc.
44. 44. Chapter 1 Thus, there are initially 120 liberal students, and 140 conservative students. Since the number of liberal students initially is the same as the number of conservative students in the end, the class ends with 120 conservative students and 140 liberal students. 1.2.66 Let x1 and x2 be the initial number of students in Sections A and B, respectively. Then, since there are 55 students total, x1 + x2 = 55. Also, interpreting the change of students from the perspective of Section B, we gain 0.2x1, lose 0.3x2, and in the process, lose 4 students. Thus, 0.2x1 − 0.3x2 = −4.0. Our matrix becomes 1 1 . . . 55 .2 −.3 . . . −4 # , which reduces to 1 0 . . . 25 0 1 . . . 30 # . This reveals that there are initially 25 students in Section A and 30 students in Section B. 1.2.67 We are told that five cows and two sheep cost ten liang, and two cows and five sheep cost eight liang of silver. So, we let C be the cost of a cow, and S be the cost of a sheep. From this we derive 5C +2S = 10 2C +5S = 8 . This reduces to C = 34 21 S = 20 21 which gives the prices: 34 21 liang silver for a cow, and 20 21 liang silver for a sheep. 1.2.68 Letting x1, x2, and x3 be the prize, in coins, of cows, sheep and pigs, respectively, we can represent the system in a matrix:     2 5 −13 . . . 1000 3 −9 3 . . . 0 −5 6 8 . . . −600     . We reduce this matrix to     1 0 0 . . . 1200 0 1 0 . . . 500 0 0 1 . . . 300     . The prize of a cow, a sheep, and a pig is 1200, 500 and 300 coins, respectively. 1.2.69 The second measurement in the problem tells us that 4 sparrows and 1 swallow weigh as much as 1 sparrow and 5 swallows. We will immediately interpret this as 3 sparrows weighing the same as 4 swallows. The other measurement we use is that all the birds together weigh 16 liang. Setting x1 to be the weight of a sparrow, and x2 to be the weight of a swallow, we find the augmented matrix   3 −4 . . . 0 5 6 . . . 16   representing these two equations. We reduce this to   1 0 . . . 32 19 0 1 . . . 24 19  , meaning that each sparrow weighs 32 19 liang, and each swallow weighs 24 19 liang. 1.2.70 This problem gives us three different combinations of horses that can pull exactly 40 dan up a hill. We condense the statements to fit our needs, saying that, One military horse and one ordinary horse can pull 40 dan, two ordinary and one weak horse can pull 40 dan and one military and three weak horses can also pull 40 dan. With this information, we set up our matrix:     1 1 0 . . . 40 0 2 1 . . . 40 1 0 3 . . . 40     , which reduces to      1 0 0 . . . 40 7 0 1 0 . . . 120 7 0 0 1 . . . 40 7      . Thus, the military horses can pull 40 7 dan, the ordinary horses can pull 120 7 dan and the weak horses can pull 40 7 dan each. 1.2.71 Here, let W be the depth of the well. 30 Copyright c 2013 Pearson Education, Inc.
45. 45. Section 1.2 Then our system becomes      2A +B −W = 0 3B +C −W = 0 4C +D −W = 0 5D +E −W = 0 A +6E −W = 0      . We transform this system into an augmented matrix, then perform a prolonged reduction to reveal            1 0 0 0 0 −265 721 . . . 0 0 1 0 0 0 −191 721 . . . 0 0 0 1 0 0 −148 721 . . . 0 0 0 0 1 0 −129 721 . . . 0 0 0 0 0 1 − 76 721 . . . 0            . Thus, A = 265 721 W, B = 191 721 W, C = 148 721 W, D = 129 721 W and E = 76 721 W. If we choose 721 to be the depth of the well, then A = 265, B = 191, C = 148, D = 129 and E = 76. 1.2.72 We let x1, x2 and x3 be the numbers of roosters, hens and chicks respectively. Then, since we buy a total of a hundred birds, and spend a hundred coins on them, we find the equations x1 + x2 + x3 = 100 and 5x1 + 3x2 + 1 3 x3 = 100. We fit these into our matrix, 1 1 1 . . . 100 5 3 1 3 . . . 100 # , which reduces to   1 0 −4 3 . . . −100 0 1 7 3 . . . 200   . So, x1 − 4 3 x3 = −100, and x2 + 7 3 x3 = 200. Now, we can write our solution vectors in terms of x3:   x1 x2 x3   =   4 3 x3 − 100 −7 3 x3 + 200 x3  . Since all of our values must be non-negative, x1 must be greater than or equal to zero, or 4 3 x3 − 100 ≥ 0, which means that x3 ≥ 75. Also, x3 must be greater than or equal to zero, meaning that −7 3 x3 + 200 ≥ 0 or x3 ≤ 600 7 . Since x3 must be an integer, this forces x3 ≤ 85. Thus, we are looking for solutions where 75 ≤ x3 ≤ 85. We notice, however, that x1 and x2 are only integers when x3 is a multiple of 3. Thus, the possible values for x3 are 75, 78, 81 and 84. Now the possible solutions for   roosters hens chicks   are   0 25 75   ,   4 18 78   ,   8 11 81   , and   12 4 84  . 1.2.73 We let x1, x2, x3 and x4 be the numbers of pigeons, sarasabirds, swans and peacocks respectively. We first determine the cost of each bird. Each pigeon costs 3 5 panas, each sarasabird costs 5 7 panas, the swans cost 7 9 panas apiece and each peacock costs 3 panas. We use these numbers to set up our system, but we must remember to make sure we are buying the proper amount of each to qualify for these deals when we find our solutions (for example, the number of sarasabirds we buy must be a multiple of 7). 31 Copyright c 2013 Pearson Education, Inc.
46. 46. Chapter 1 Our matrix then is 1 1 1 1 . . . 100 3 5 5 7 7 9 3 . . . 100 # which reduces to   1 0 −5 9 −20 . . . −250 0 1 14 9 21 . . . 350  . Thus, x1 = 5 9 x3 + 20x4 − 250 and x2 = −14 9 x3 − 21x4 + 350. Then our solutions are of the form    5 9 x3 + 20x4 − 250 −14 9 x3 − 21x4 + 350 x3 x4   . We determine the possible solutions by choosing combinations of x3 and x4 of the correct multiples (9 for x3, 3 for x4) that give us non-negative integer solutions for x1 and x2. Thus it is required that x1 = 5 9 x3 + 20x4 − 250 ≥ 0 and x2 = −14 9 x3 − 21x4 + 350 ≥ 0. Solving for x3 we find that 225 − 27 2 x4 ≥ x3 ≥ 450 − 36x4. To find all the solutions, we can begin by letting x4 = 0, and finding all corresponding values of x3. Then we can increase x4 in increments of 3, and find the corresponding x3 values in each case, until we are through. For x4 = 0 we have the inequality 225 ≥ x3 ≥ 450, so that there aren’t any solutions for x3. Likewise, there are no feasible x3 values for x4 = 3, 6 and 9, since 450 − 36x4 exceeds 100 in these cases. In the case of x4 = 12 our inequality becomes 63 ≥ x3 ≥ 18, so that x3 could be 18, 27, 36, 45, 54 or 63. In the next case, x4 = 15, we have 45 2 ≥ x3 ≥ −90, so that the non-negative solutions are 0, 9 and 18. If x4 is 18 or more, then the term 225 − 27 2 x4 becomes negative, so that there are only negative solutions for x3. (Recall that it is required that 225 − 27 2 x4 ≥ x3.) We have found nine solutions. If we compute the corresponding values of x1 = 5 9 x3+20x4−250 and x2 = −14 9 x3−21x4+350, we end up with the following vectors for:    number of pigeons number of sarasabirds number of swans number of peacocks    to be:    0 70 18 12   ,    5 56 27 12   ,    10 42 36 12   ,    15 28 45 12   ,    20 14 54 12    ,    25 0 63 12   ,    50 35 0 15   ,    55 21 9 15   ,    60 7 18 15   . 1.2.74 We follow the outline of Exercise 72 to find the matrix 1 1 1 . . . 100 4 1 5 1 . . . 100 # , which reduces to   1 0 4 19 . . . 400 19 0 1 15 19 . . . 1500 19   . Thus, our solutions are of the form   400−4x3 19 1500−15x3 19 x3  . We find that our solutions are bound by 0 ≤ x3 ≤ 100. However, since both 400−4x3 19 = 4100−x3 19 and 1500−15x3 19 = 15100−x3 19 must be non-negative integers, the quantity 100−x3 19 must be a non-negative integer, k, so that x3 = 100 − 19k. The condition x3 ≥ 0 now leaves us with the possibilities k = 0, 1, 2, 3, 4, 5. 32 Copyright c 2013 Pearson Education, Inc.
47. 47. Section 1.2 Thus, we find our solutions for   ducks sparrows roosters   :   0 0 100   .   4 15 81   ,   8 30 62   ,   12 45 43   ,   16 60 24   and   20 75 5  . 1.2.75 We let x1 be the number of sheep, x2 be the number of goats, and x3 be the number of hogs. We can then use the two equations 1 2 x1 + 4 3 x2 + 7 2 x3 = 100 and x1 +x2 +x3 = 100 to generate the following augmented matrix:   1 2 4 3 7 2 . . . 100 1 1 1 . . . 100   then reduce it to   1 0 −13 5 . . . 40 0 1 18 5 . . . 60  . With this, we see that our solutions will be of the form   40 + 13 5 s 60 − 18 5 s s  . Now all three components of this vector must be non-negative integers, meaning that s must be a non-negative multiple of 5 (that is, s = 0, 5, 10, . . .) such that 60 − 18 5 s ≥ 0, or, s ≤ 50 3 . This leaves the possible solutions x3 = s = 0, 5, 10 and 15, and we can compute the corresponding values of x1 = 40 + 13 5 s and x2 = 60 − 18 5 s in each case. So we find the following solutions:   40 60 0  ,   53 42 5  ,   66 24 10   and   79 6 15  . 1.2.76 This problem is similar in nature to Exercise 72, and we will follow that example, revealing the ma- trix: 1 1 1 . . . 100 3 2 1 2 . . . 100 # . We reduce this to   1 0 −3 2 . . . −100 0 1 5 2 . . . 200   , which yields solutions of the form   3 2 x3 − 100 −5 2 x3 + 200 x3  . Since all the values must be positive (there are at least one man, one woman and one child), we see that 66 x3 80, and x3 must be even. From this, we use x3 to find our solutions:   2 30 68   ,   5 25 70   ,   8 20 72   ,   11 15 74   ,   14 10 76   and   17 5 78   . 1.2.77 Rather than setting up a huge system, here we will reason this out logically. Since there are 30 barrels, each son will get 10 of them. If we use the content of a full barrel as our unit for wine, we see that each brother will get 15 3 = 5 barrel-fulls of wine. Thus, the ten barrels received by each son will, on average, be half full, meaning that for every full barrel a son receives, he also receives an empty one. Now let x1, x2, and x3 be the numbers of half-full barrels received by each of the three sons. The first son, receiving x1 half-full barrels will also gain 10 − x1 other barrels, half of which must be full and half of which must be empty, each equal to the quantity 10−x1 2 . Thus, x1 must be even. The same works for x2 and x3. Since x1 + x2 + x3 = 10, we have boiled down our problem to simply finding lists of three non-negative even numbers that add up to 10. We find our solutions by inspection:   10 0 0   ,   8 2 0   ,   8 0 2   ,   6 4 0   ,   6 2 2   ,   6 0 4   ,   4 6 0   ,   4 4 2   ,   4 2 4   ,   4 0 6   , 33 Copyright c 2013 Pearson Education, Inc.
48. 48. Chapter 1   2 8 0   ,   2 6 2   ,   2 4 4   ,   2 2 6   ,   2 0 8   ,   0 10 0   ,   0 8 2   ,   0 6 4   ,   0 4 6   ,   0 2 8   and   0 0 10  . As we stated before, the number of full and empty barrels is dependent on the number of half-full barrels. Thus, each solution here translates into exactly one solution for the overall problem. Here we list those solutions, for   first son second son third son  , using triples of the form (full barrels, half-full barrels, empty barrels) as our entries:   (0, 10, 0) (5, 0, 5) (5, 0, 5)   ,   (1, 8, 1) (4, 2, 4) (5, 0, 5)   ,   (1, 8, 1) (5, 0, 5) (4, 2, 4)   ,   (2, 6, 2) (3, 4, 3) (5, 0, 5)   ,   (2, 6, 2) (4, 2, 4) (4, 2, 4)   ,   (2, 6, 2) (5, 0, 5) (3, 4, 3)   ,   (3, 4, 3) (2, 6, 2) (5, 0, 5)   ,   (3, 4, 3) (3, 4, 3) (4, 2, 4)   ,   (3, 4, 3) (4, 2, 4) (3, 4, 3)   ,   (3, 4, 3) (5, 0, 5) (2, 6, 2)   ,   (4, 2, 4) (1, 8, 1) (5, 0, 5)   ,   (4, 2, 4) (2, 6, 2) (4, 2, 4)   ,   (4, 2, 4) (3, 4, 3) (3, 4, 3)   ,   (4, 2, 4) (4, 2, 4) (2, 6, 2)   ,   (4, 2, 4) (5, 0, 5) (1, 8, 1)   ,   (5, 0, 5) (0, 10, 0) (5, 0, 5)   ,   (5, 0, 5) (1, 8, 1) (4, 2, 4)   ,   (5, 0, 5) (2, 6, 2) (3, 4, 3)   ,   (5, 0, 5) (3, 4, 3) (2, 6, 2)   ,   (5, 0, 5) (4, 2, 4) (1, 8, 1)   and   (5, 0, 5) (5, 0, 5) (0, 10, 0)  . 1.2.78 We let x1 be the amount of gold in the crown, x2 be the amount of bronze, x3 be the amount of tin and x4 be the amount of iron. Then, for example, since the first requirement in the problem is: “Let the gold and bronze together form two-thirds,” we will interpret this as x1 +x2 = 2 3 (60). We do this for all three requirements, and use the fact that all combined will be the total weight of the crown as our fourth. So we find the matrix         1 1 0 0 . . . 2 3 (60) 1 0 1 0 . . . 3 4 (60) 1 0 0 1 . . . 3 5 (60) 1 1 1 1 . . . 60         , which has the solution    gold bronze tin iron    =    30.5 9.5 14.5 5.5    . 1.2.79 Let xi be the number of coins the ith merchant has. We interpret the statement of the first merchant, “If I keep the purse, I shall have twice as much money as the two of you together” as x1 + 60 = 2(x2 + x3), or −x1 +2x2 +2x3 = 60. We interpret the other statements in a similar fashion, translating this into the augmented matrix,     −1 2 2 . . . 60 3 −1 3 . . . 60 5 5 −1 . . . 60    . The reduced row echelon form of this matrix is     1 0 0 . . . 4 0 1 0 . . . 12 0 0 1 . . . 20    . Thus we deduce that the first merchant has 4 coins, the second has 12, and the third is the richest, with 20 coins. 34 Copyright c 2013 Pearson Education, Inc.
49. 49. Section 1.3 1.2.80 For each of the three statements, we set up an equation of the form (initial amount of grass) + (grass growth) = (grass consumed by cows), or (#offields)x + (#offields)(#ofdays)y = (#ofcows)(#ofdays)z. For the first statement, this produces the equation x + 2y = 6z, or x + 2y − 6z = 0. Similarly, we obtain the equations 4x + 16y − 28z = 0 and 2x + 10y − 15z = 0 for the other two statements. From this information, we write the matrix     1 2 −6 . . . 0 4 16 −28 . . . 0 2 10 −15 . . . 0     , which reduces to     1 0 −5 . . . 0 0 1 −1 2 . . . 0 0 0 0 . . . 0     . Thus our solutions are of the form   x y z   =   5t 1 2 t t   , where t is an arbitrary positive real number. Section 1.3 1.3.1 a No solution, since the last row indicates 0 = 1. b The unique solution is x = 5, y = 6. c Infinitely many solutions; the first variable can be chosen freely. 1.3.2 The rank is 3 since each row contains a leading one. 1.3.3 This matrix has rank 1 since its rref is   1 1 1 0 0 0 0 0 0  . 1.3.4 This matrix has rank 2 since its rref is   1 0 −1 0 1 2 0 0 0   1.3.5 a x 1 3 + y 2 1 = 7 11 b The solution of the system in part (a) is x = 3, y = 2. (See Figure 1.13.) 1.3.6 No solution, since any linear combination x~ v1 + y~ v2 of ~ v1 and ~ v2 will be parallel to ~ v1 and ~ v2. 1.3.7 A unique solution, since there is only one parallelogram with sides along ~ v1 and ~ v2 and one vertex at the tip of ~ v3. 1.3.8 Infinitely many solution. There are at least two obvious solutions. Write ~ v4 as a linear combination of ~ v1 and ~ v2 alone or as a linear combination of ~ v3 and ~ v2 alone. Therefore, this linear system has infinitely many solutions, by Theorem 1.3.1. 35 Copyright c 2013 Pearson Education, Inc.
50. 50. Chapter 1 Figure 1.13: for Problem 1.3.5. 1.3.9   1 2 3 4 5 6 7 8 9     x y z   =   1 4 9   1.3.10   1 2 3   ·   1 −2 1   = 1 · 1 + 2 · (−2) + 3 · 1 = 0 1.3.11 Undefined since the two vectors do not have the same number of components. 1.3.12 [1 2 3 4] ·    5 6 7 8    = 1 · 5 + 2 · 6 + 3 · 7 + 4 · 8 = 70 1.3.13 1 2 3 4 7 11 = 7 1 3 + 11 2 4 = 29 65 or 1 2 3 4 7 11 = 1 · 7 + 2 · 11 3 · 7 + 4 · 11 = 29 65 1.3.14 1 2 3 2 3 4   −1 2 1   = −1 1 2 + 2 2 3 + 1 3 4 = 6 8 or 1 2 3 2 3 4   −1 2 1   = 1 · (−1) + 2 · 2 + 3 · 1 2 · (−1) + 3 · 2 + 4 · 1 = 6 8 1.3.15 [1 2 3 4]    5 6 7 8    = 5 · 1 + 6 · 2 + 7 · 3 + 4 · 8 = 70 either way. 1.3.16 0 1 3 2 2 −3 = 0 · 2 + 1 · (−3) 3 · 2 + 2 · (−3) = −3 0 36 Copyright c 2013 Pearson Education, Inc.
51. 51. Section 1.3 1.3.17 Undefined, since the matrix has three columns, but the vector has only two components. 1.3.18   1 2 3 4 5 6   1 2 = 1   1 3 5   + 2   2 4 6   =   5 11 17   1.3.19   1 1 −1 −5 1 1 1 −5 3     1 2 3   = 1   1 −5 1   + 2   1 1 −5   + 3   −1 1 3   =   0 0 0   1.3.20 a   9 8 7 6 6 6   b 9 −9 18 27 36 45 1.3.21    158 70 81 123    1.3.22 By Theorem 1.3.4, the rref is   1 0 0 0 1 0 0 0 1   1.3.23 All variables are leading, that is, there is a leading one in each column of the rref:    1 0 0 0 1 0 0 0 1 0 0 0   . 1.3.24 By Theorem 1.3.4, rref (A) =    1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1   . 1.3.25 In this case, rref(A) has a row of zeros, so that rank(A) 4; there will be a free variable. The system A~ x = ~ c could have infinitely many solutions (for example, when ~ c = ~ 0) or no solutions (for example, when ~ c = ~ b), but it cannot have a unique solution, by Theorem 1.3.4. 1.3.26 From Example 4c we know that rank(A) = 3, so that rref(A) =    1 0 0 0 1 0 0 0 1 0 0 0   . Since all variables are leading, the system A~ x = ~ c cannot have infinitely many solutions, but it could have a unique solution (for example, if ~ c = ~ b) or no solutions at all (compare with Example 4d). 37 Copyright c 2013 Pearson Education, Inc.
52. 52. Chapter 1 1.3.27 By Theorem 1.3.4, rref (A) =    1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1   . 1.3.28 There must be a leading one in each column: rref (A) =      1 0 0 0 1 0 0 0 1 0 0 0 0 0 0      . 1.3.29 A is of the form   a 0 0 0 b 0 0 0 c   and   a 0 0 0 b 0 0 0 c     5 3 −9   =   5a 3b −9c   =   2 0 1  . So a = 2 5 , b = 0 and c = −1 9 , and A =   2 5 0 0 0 0 0 0 0 −1 9   1.3.30 We must satisfy the equation   a b c d e f g h i     5 3 −9   =   2 0 1  . Thus, 5a + 3b − 9c = 2, 5d + 3e − 9f = 0, and 5g + 3h − 9i = 1. One way to force our matrix to have rank 1 is to make all the entries in the second and third columns zero, meaning that a = 5 2 , d = 0, and g = 1 5 . Thus, one possible matrix is   2 5 0 0 0 0 0 1 5 0 0   . 1.3.31 A is of the form   a b c 0 d e 0 0 f   and   a b c 0 d e 0 0 f     5 3 −9   =   5a + 3b − 9c 3d − 9e −9f   =   2 0 1  . Clearly, f must equal −1 9 . Then, since 3d = 9e, we can choose any non-zero value for the free variable e, and d will be 3e. So, if we choose 1 for e, then d = 3e = 3. Lastly, we must resolve 5a + 3b − 9c = 2. Here, b and c are the free variables, and a = 2−3b+9c 5 . If we let b = c = 1. Then, a = 2−3(1)+9(1) 5 = 8 5 . So, in our example, A =   8 5 1 1 0 3 1 0 0 −1 9   1.3.32 For this problem, we set up the same three equations as in Exercise 30. However, here, we must enforce that our matrix, A, contains no zero entries. One possible solution to this problem is the matrix   −2 −2 −2 3 1 2 −1 −1 −1   . 38 Copyright c 2013 Pearson Education, Inc.
53. 53. Section 1.3 1.3.33 The ith component of A~ x is [0 0 . . . 1 . . . 0]        x1 x2 . . . xi . . . xn        = xi. (The 1 is in the ith position.) Therefore, A~ x = ~ x. 1.3.34 a A~ e1 =   a d g   , A~ e2 =   b e h  , and A~ e3 =   c f k  . b B~ e1 = [~ v1 ~ v2 ~ v3]   1 0 0   = 1~ v1 + 0~ v2 + 0~ v3 = ~ v1. Likewise, B~ e2 = ~ v2 and B~ e3 = ~ v3. 1.3.35 Write A = [~ v1 ~ v2 . . . ~ vi . . . ~ vm], then A~ ei = [~ v1 ~ v2 . . . ~ vi . . . ~ vm]        0 0 . . . 1 . . . 0        = 0~ v1 + 0~ v2 + · · · + 1~ vi + · · · + 0~ vm = ~ vi = ith column of A. 1.3.36 By Exercise 35, the ith column of A is A~ ei, for i = 1, 2, 3. Therefore, A =   1 4 7 2 5 8 3 6 9  . 1.3.37 We have to solve the system x1 + 2x2 = 2 x3 = 1 or x1 = 2 − 2x2 x3 = 1 . Let x2 = t. Then the solutions are of the form   x1 x2 x3   =   2 − 2t t 1  , where t is an arbitrary real number. 1.3.38 We will illustrate our reasoning with an example. We generate the “random” 3 × 3 matrix A =   0.141 0.592 0.653 0.589 0.793 0.238 0.462 0.643 0.383  . Since the entries of this matrix are chosen from a large pool of numbers (in our case 1000, from 0.000 to 0.999), it is unlikely that any of the entries will be zero (and even less likely that the whole first column will consist of zeros). This means that we will usually be able to apply the Gauss-Jordan algorithm to turn the first column into   1 0 0  ; this is indeed possible in our example:   0.141 0.592 0.653 0.589 0.793 0.238 0.462 0.643 0.383   −→   1 4.199 4.631 0 −1.680 −2.490 0 −1.297 −1.757  . 39 Copyright c 2013 Pearson Education, Inc.
54. 54. Chapter 1 Again, it is unlikely that any entries in the second column of the new matrix will be zero. Therefore, we can turn the second column into   0 1 0  . Likewise, we will be able to clear up the third column, so that rref(A) =   1 0 0 0 1 0 0 0 1  . We summarize: As we apply Gauss-Jordan elimination to a random matrix A (of any size), it is unlikely that we will ever encounter a zero on the diagonal. Therefore, rref(A) is likely to have all ones along the diagonal. 1.3.39 We will usually get rref(A) =   1 0 0 a 0 1 0 b 0 0 1 c  , where a, b, and c are arbitrary. 1.3.40 We will usually have rref(A) =    1 0 0 0 1 0 0 0 1 0 0 0   . (Compare with the summary to Exercise 38.) 1.3.41 If A~ x = ~ b is a “random” system, then rref(A) will usually be   1 0 0 0 1 0 0 0 1  , so that we will have a unique solution. 1.3.42 If A~ x = ~ b is a “random” system of three equations with four unknowns, then rref(A) will usually be   1 0 0 a 0 1 0 b 0 0 1 c   (by Exercise 39), so that the system will have infinitely many solutions (x4 is a free variable). 1.3.43 If A~ x = ~ b is a “random” system of equations with three unknowns, then rref[A . . .~ b] will usually be       1 0 0 . . .0 0 1 0 . . .0 0 0 1 . . .0 0 0 0 . . .1       , so that the system is inconsistent. 1.3.44 Let E = rref(A), and note that all the entries in the last row of E must be zero, by the definition of rref. If ~ c is any vector in Rn whose last component isn’t zero, then the system E~ x = ~ c will be inconsistent. Now consider the elementary row operations that transform A into E, and apply the opposite operations, in reversed order, to the augmented matrix h E . . . ~ c i . You end up with an augmented matrix h A . . . ~ b i that represents an inconsistent system A~ x = ~ b, as required. 40 Copyright c 2013 Pearson Education, Inc.
55. 55. Section 1.3 1.3.45 Write A = [~ v1 ~ v2 . . . ~ vm] and ~ x =   x1 . . . xm  . Then A(k~ x) = [~ v1 . . . ~ vm]   kx1 . . . kxm   = kx1~ v1 + · · · + kxm~ vm and k(A~ x) = k(x1~ v1 + · · · + xm~ vm) = kx1~ v1 + · · · + kxm~ vm. The two results agree, as claimed. 1.3.46 Since a, d, and f are all nonzero, we can divide the first row by a, the second row by d, and the third row by f to obtain   1 b a c a 0 1 e d 0 0 1  . It follows that the rank of the matrix is 3. 1.3.47 a ~ x = ~ 0 is a solution. b This holds by part (a) and Theorem 1.3.3. c If ~ x1 and ~ x2 are solutions, then A~ x1 = ~ 0 and A~ x2 = ~ 0. Therefore, A(~ x1 + ~ x2) = A~ x1 + A~ x2 = ~ 0 + ~ 0 = ~ 0, so that ~ x1 + ~ x2 is a solution as well. Note that we have used Theorem 1.3.10a. d A(k~ x) = k(A~ x) = k~ 0 = ~ 0 We have used Theorem 1.3.10b. 1.3.48 The fact that ~ x1 is a solution of A~ x = ~ b means that A~ x1 = ~ b. a. A(~ x1 + ~ xh) = A~ x1 + A~ xh = ~ b +~ 0 = ~ b b. A(~ x2 − ~ x1) = A~ x2 − A~ x1 = ~ b −~ b = ~ 0 c. Parts (a) and (b) show that the solutions of A~ x = ~ b are exactly the vectors of the form ~ x1 + ~ xh, where ~ xh is a solution of A~ x = ~ 0; indeed if ~ x2 is a solution of A~ x = ~ b, we can write ~ x2 = ~ x1 + (~ x2 − ~ x1), and ~ x2 − ~ x1 will be a solution of A~ x = ~ 0, by part (b). Geometrically, the vectors of the form ~ x1 + ~ xh are those whose tips are on the line L in Figure 1.14; the line L runs through the tip of ~ x1 and is parallel to the given line consisting of the solutions of A~ x = ~ 0. Figure 1.14: for Problem 1.3.48c. 41 Copyright c 2013 Pearson Education, Inc.
56. 56. Chapter 1 1.3.49 a This system has either infinitely many solutions (if the right-most column of rref[A . . .b] does not contain a leading one), or no solutions (if the right-most column does contain a leading one). b This system has either a unique solution (if rank[A . . .~ b] = 3), or no solution (if rank[A . . .~ b] = 4). c The right-most column of rref[A . . .~ b] must contain a leading one, so that the system has no solutions. d This system has infinitely many solutions, since there is one free variable. 1.3.50 The right-most column of rref[A . . .~ b] must contain a leading one, so that the system has no solutions. 1.3.51 For B~ x to be defined, the number of columns of B, which is s, must equal the number of components of ~ x, which is p, so that we must have s = p. Then B~ x will be a vector in Rr ; for A(B~ x) to be defined we must have m = r. Summary: We must have s = p and m = r. 1.3.52 A(B~ x) = A 0 −1 1 0 x1 x2 = 1 0 1 2 −x2 x1 = −x2 2x1 − x2 = 0 −1 2 −1 x1 x2 , so that C = 0 −1 2 −1 . 1.3.53 Yes; write A = [~ v1 . . . ~ vm], B = [~ w1 . . . ~ wm], and ~ x =   x1 . . . xm  . Then (A + B)~ x = [~ v1 + ~ w1 . . . ~ vm + ~ wm]   x1 . . . xm   = x1(~ v1 + ~ w1) + · · · + xm(~ vm + ~ wm) and A~ x + B~ x = [~ v1 . . . ~ vm]   x1 . . . xm   + [~ w1 . . . ~ wm]   x1 . . . xm   = x1~ v1 + · · · + xm~ vm + x1 ~ w1 + · · · + xm ~ wm. The two results agree, as claimed. 1.3.54 The vectors of the form c1~ v1 + c2~ v2 form a plane through the origin containing ~ v1 and ~ v2; in Figure 1.15 we draw a typical vector in this plane. 1.3.55 We are looking for constants a and b such that a   1 2 3   + b   4 5 6   =   7 8 9  . The resulting system   a + 4b = 7 2a + 5b = 8 3a + 6b = 9   has the unique solution a = −1, b = 2, so that   7 8 9   is indeed a linear combination of the vector   1 2 3   and   4 5 6  . 42 Copyright c 2013 Pearson Education, Inc.
57. 57. Section 1.3 Figure 1.15: for Problem 1.3.54. 1.3.56 We can use technology to determine that the system      30 −1 38 56 62      = x1      1 7 1 9 4      + x2      5 6 3 2 8      + x3      9 2 3 5 2      + x4      −2 −5 4 7 9      is inconsistent; therefore, the vector      30 −1 38 56 62      fails to be a linear combination of the other four vectors. 1.3.57 Pick a vector on each line, say 2 1 on y = x 2 and 1 3 on y = 3x. Then write 7 11 as a linear combination of 2 1 and 1 3 : a 2 1 + b 1 3 = 7 11 . The unique solution is a = 2, b = 3, so that the desired representation is 7 11 = 4 2 + 3 9 . 4 2 is on the line y = x 2 ; 3 9 is on line y = 3x. 1.3.58 We want   3 b c   = k1   1 3 2   + k2   2 6 4   + k3   −1 −3 −2   , for some k1, k2 and k3. Note that we can rewrite this right-hand side as k1   1 3 2   + 2k2   1 3 2   − k3   1 3 2   = (k1 + 2k2 − k3)   1 3 2  . It follows that k1 + 2k2 − k3 = 3, so that b = 9 and c = 6. 43 Copyright c 2013 Pearson Education, Inc.
58. 58. Chapter 1 1.3.59    5 7 c d    = a    1 1 1 1    + b    1 2 3 4    =    a + b a + 2b a + 3b a + 4b   . So we have a small system: a +b = 5 a +2b = 7 , which we quickly solve to find a = 3 and b = 2. Then, c = a + 3b = 3 + 6 = 9 and d = a + 4b = 3 + 8 = 11. 1.3.60 We need    a b c d    = k1    0 0 3 0    + k2    1 0 4 0    + k3    2 0 5 6    =    k2 + 2k3 0 3k1 + 4k2 + 5k3 6k3   . From this we see that a, c and d can be any value, while b must equal zero. 1.3.61 We need to solve the system   1 c c2   = x   1 2 4   + y   1 3 9   with augmented matrix      1 1 . . . 1 2 3 . . . c 4 9 . . . c2      . The matrix reduces to      1 1 . . . 1 0 1 . . . c − 2 0 0 . . . c2 − 5c + 6      . This system is consistent if and only if c = 2 or c = 3. Thus the vector is a linear combination if c = 2 or c = 3. 1.3.62 We need to solve the system   1 c c2   = x   1 a a2   + y   1 b b2   with augmented matrix      1 1 . . . 1 a b . . . c a2 b2 . . . c2      . The matrix reduces to      1 1 . . . 1 0 b − a . . . c − a 0 0 . . . (c − a)(c − b)      . This system is consistent if and only if c = a or c = b. Thus the vector is a linear combination if c = a or c = b. 44 Copyright c 2013 Pearson Education, Inc.
59. 59. True or False 1.3.63 This is the line parallel to ~ w which goes through the end point of the vector ~ v. 1.3.64 This is the line segment connecting the head of the vector ~ v to the head of the vector ~ v + ~ w. 1.3.65 This is the full parallelogram spanned by the two vectors ~ v and ~ w. 1.3.66 Write b = 1 − a and a~ v + b~ w = a~ v + (1 − a)~ w = ~ w + a(~ v − ~ w) to see that this is the line segment connecting the head of the vector ~ v to the head of the vector ~ w. 1.3.67 This is the full triangle with its vertices at the origin and at the heads of the vectors ~ v and ~ w. 1.3.68 Writing ~ u · ~ v = ~ u · ~ w as ~ u · (~ v − ~ w) = 0, we see that this is the line perpendicular to the vector ~ v − ~ w. 1.3.69 We write out the augmented matrix:     0 1 1 . . . a 1 0 1 . . . b 1 1 0 . . . c     and reduce it to      1 0 0 . . . −a+b+c 2 0 1 0 . . . a−b+c 2 0 0 1 . . . a+b−c 2      . So x = −a+b+c 2 , y = a−b+c 2 and z = a+b−c 2 . 1.3.70 We find it useful to let s = x1 + x2 + · · · + xn. Adding up all n equations of the system, and realizing that the term xi is missing from the ith equation, we see that (n − 1)s = b1 + · · · + bn, or, s = b1+···+bn n−1 . Now the ith equation of the system can be written as s − xi = bi, so that xi = s − bi = b1+···+bn n−1 − bi. True or False Ch 1.TF.1 T, by Theorem 1.3.8 Ch 1.TF.2 T, by Definition 1.3.9 Ch 1.TF.3 T, by Definition. Ch 1.TF.4 F; Consider the equation x + y + z = 0, repeated four times. Ch 1.TF.5 F, by Example 3a of Section 1.3 Ch 1.TF.6 T, by Definition 1.3.7 Ch 1.TF.7 T, by Theorem 1.3.4 Ch 1.TF.8 F, by Theorem 1.3.1 Ch 1.TF.9 F, by Theorem 1.3.4 Ch 1.TF.10 F; As a counter-example, consider the zero matrix. 45 Copyright c 2013 Pearson Education, Inc.
60. 60. Chapter 1 Ch 1.TF.11 T; The last component of the left-hand side is zero for all vectors ~ x. Ch 1.TF.12 T; A = 3 0 4 0 , for example. Ch 1.TF.13 T; Find rref Ch 1.TF.14 T; Find rref Ch 1.TF.15 F; Consider the 4 × 3 matrix A that contains all zeroes, except for a 1 in the lower left corner. Ch 1.TF.16 F; Note that A 2 2 = 2A 1 1 for all 2 × 2 matrices A. Ch 1.TF.17 F; The rank is 1. Ch 1.TF.18 F; The product on the left-hand side has two components. Ch 1.TF.19 T; Let A =   −3 0 −5 0 −7 0  , for example. Ch 1.TF.20 T; We have   1 2 3   = 2   4 5 6   −   7 8 9  . Ch 1.TF.21 F; Let A =   1 0 0 1 1 1   and ~ b =   2 3 5  , for example. Ch 1.TF.22 T, by Exercise 1.3.44. Ch 1.TF.23 F; Find rref to see that the rank is always 2. Ch 1.TF.24 T; Note that ~ v = 1~ v + 0~ w. Ch 1.TF.25 F; Let ~ u = 1 0 ,~ v = 2 0 , ~ w = 0 1 , for example. Ch 1.TF.26 T; Note that ~ 0 = 0~ v + 0~ w Ch 1.TF.27 F; Let A =   0 1 0 0 0 1 0 0 0   and B =   1 0 0 0 1 0 0 0 0  , for example. We can apply elementary row operations to A all we want, we will always end up with a matrix that has all zeros in the first column. Ch 1.TF.28 T; If ~ u = a~ v + b~ w and ~ v = c~ p + d~ q + e~ r, then ~ u = ac~ p + ad~ q + ae~ r + b~ w. 46 Copyright c 2013 Pearson Education, Inc.
61. 61. True or False Ch 1.TF.29 F; The system x = 2, y = 3, x + y = 5 has a unique solution. Ch 1.TF.30 F; Let A = 0 1 0 0 , for example. Ch 1.TF.31 F; If A   1 2 3   = ~ 0, then ~ x =   1 2 3   is a solution to A . . .~ 0 . However, since rank(A) = 3, rref A . . .~ 0 =    1 0 0 0 1 0 0 0 1 0 0 0 . . . 0 0 0 0   , meaning that only ~ 0 is a solution to A~ x = ~ 0. Ch 1.TF.32 F; If ~ b = ~ 0, then having a row of zeroes in rref(A) does not force the system to be inconsistent. Ch 1.TF.33 T; By Example 4d of Section 1.3, the equation A~ x = ~ 0 has the unique solution ~ x = ~ 0. Now note that A(~ v − ~ w) = ~ 0, so that ~ v − ~ w = ~ 0 and ~ v = ~ w. Ch 1.TF.34 T; Note that rank(A) = 4, by Theorem 1.3.4 Ch 1.TF.35 F; Let ~ u = 2 0 , ~ v = 1 0 , ~ w = 0 1 , for example.. Ch 1.TF.36 T; We use rref to solve the system A~ x = ~ 0 and find ~ x =   −2t −3t t  , where t is an arbitrary constant. Letting t = 1, we find [~ u ~ v ~ w]   −2 −3 1   = −2~ u − 3~ v + ~ w = ~ 0, so that ~ w = 2~ u + 3~ v. Ch 1.TF.37 F; Let A = B = 1 0 0 1 , for example. Ch 1.TF.38 T; Matrices A and B can both be transformed into I =    1 0 . . . 0 0 1 . . . 0 . . . . . . . . . 0 0 0 0 1   . Running the elementary operations backwards, we can transform I into B. Thus we can first transform A into I and then I into B. Ch 1.TF.39 T; If ~ v = a~ u + b~ w, then A~ v = A(a~ u + b~ w) = A(a~ u) + A(b~ w) = aA~ v + bA~ w. Ch 1.TF.40 T; check that the three defining properties of a matrix in rref still hold. F; If ~ b = ~ 0, then having a row of zeroes in rref(A) does not force the system to be inconsistent. Ch 1.TF.41 T; A~ x = ~ b is inconsistent if and only if rank A . . .~ b = rank(A)+1, since there will be an extra leading one in the last column of the augmented matrix: (See Figure 1.16.) 47 Copyright c 2013 Pearson Education, Inc.
62. 62. Chapter 1 Figure 1.16: for Problem T/F 41. Ch 1.TF.42 T; The system A~ x = ~ b is consistent, by Example 4b, and there are, in fact, infinitely many solutions, by Example 4c. Note that A~ x = ~ b is a system of three equations with four unknowns. Ch 1.TF.43 T; Recall that we use rref A . . .~ 0 to solve the system A~ x = ~ 0. Now, rref A . . .~ 0 = rref(A) . . .~ 0 = rref(B) . . .~ 0 = rref B . . .~ 0 . Then, since rref(A) . . .~ 0 = rref(B) . . .~ 0 , they must have the same solutions. Ch 1.TF.44 F; Consider 1 2 0 0 . If we remove the first column, then the remaining matrix fails to be in rref. Ch 1.TF.45 T; First we list all possible matrices rref(M), where M is a 2 × 2 matrix, and show the corresponding solutions for M~ x = ~ 0: rref(M) solutions of M~ x = ~ 0 1 0 0 1 {~ 0} 1 a 0 0 −at t , for an arbitrary t 0 1 0 0 t 0 , for an arbitrary t 0 0 0 0 R2 Now, we see that if rref(A) 6= rref(B), then the systems A~ x = ~ 0 and B~ x = ~ 0 must have different solutions. Thus, it must be that if the two systems have the same solutions, then rref(A) = rref(B). Ch 1.TF.46 T . First note that the product of the diagonal entries is nonzero if (and only if) all three diagonal entries are nonzero. If all the diagonal entries are nonzero, then A =   a 0 0 b c 0 d e f   ÷a ÷c ÷f →   1 0 0 b′ 1 0 d′ e′ 1   →   1 0 0 0 1 0 0 0 1  , showing that rank A = 3. Conversely, if a = 0 or c = 0 or f = 0, then it is easy to verify that rref A will contain a row of zeros, so that rank A ≤ 2. For example, if a and c are nonzero but f = 0, then A =   a 0 0 b c 0 d e 0   ÷a ÷c →   1 0 0 b′ 1 0 d e 0   →   1 0 0 0 1 0 0 0 0  , with rank A = 2. 48 Copyright c 2013 Pearson Education, Inc.
63. 63. True or False Ch 1.TF.47 T. If a 6= 0, then a b c d ÷a → 1 b/a c d −c(I) → 1 b/a 0 (ad − bc)/a a/(ad − bc) → 1 b/a 0 1 → 1 0 0 1 , showing that rank a b c d = 2. If a = 0, then b and c are both nonzero, so that 0 b c d reduces to 1 0 0 1 as claimed. Ch 1.TF.48 T. If ~ w = a~ u + b~ v, then ~ u + ~ v + ~ w = (a + 1)~ u + (b + 1)~ v = (a − b)~ u + (b + 1)(~ u + ~ v) Ch 1.TF.49 T. If A~ v = ~ b and A~ w = ~ c, then A(~ v + ~ w) = ~ b + ~ c, showing that the system A~ x = ~ b + ~ c is consistent. Suppose A is an n × m matrix. Since A~ x = ~ b has a unique solution, rank A must be m (by Example 1.3.3c), implying that the system A~ x = ~ b + ~ c has a unique solution as well (by Example 1.3.4d). Ch 1.TF.50 F. Think about constructing a 0-1 matrix A of size 3 × 3 with rank A = 3 row by row. The rows must be chosen so that rref A will not contain a row of zeros, which implies that no two rows of A can be equal. For the first row we have 7 = 23 − 1 choices: anything except [ 0 0 0 ]. For the second row we have six choices left: anything except the row of zeros and the first row. For the third row we have at most five choices, since we cannot chose the row of zeros, the first row, or the second row. Thus, at most 7 × 6 × 5 = 210 of the 0-1-matrices of size 3 × 3 have rank 3, out of a total of 29 = 512 matrices. 49 Copyright c 2013 Pearson Education, Inc.