1. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-1 Systems of Linear Equations 5.1 Linear Systems ▪ Substitution Method ▪ Elimination Method ▪ Special Systems ▪ Applying Systems of Equations ▪ Solving Linear Systems with Three Unknowns (Variables) ▪ Using Systems of Equations to Model Data 5.1 5.3 5.5 5.6 5.7

2. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-2 Solve the system by substitution method. 5.1 Example 1 Solving a System by Substitution (page 495) Substitute 1 + 2y for x in L1: 4x – 3(2) = 14 4x – 6 = 14 4x = 20 x = 5 Solution: (5, 2) Check the solution Eq1: 4(5) – 3(2) = 20 – 6 = 14 Eq 2: (5) – 2(2) = 5 – 4 = 1 x = 1 + 2y

3. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-3 Solve the system algebraically. 5.1 Example 2 Solving a System by Elimination (page 496) 13x = 52 x = 4 2(4) + 3y = -1 8 + 3y = -1 3y = -9 y = -3 Solution: (4, -3) Check the solution: 2(4) + 3(-3) = 8 – 9 = -1 3(4) – 2(-3) = 12 + 6 = 18 2x + 3y = -1 3x – 2y = 18 (2) (3) 4x + 6y = -2 9x – 6y = 54

4. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-4 5.1 Example 3 Solving an Inconsistent System (page 496) Solve the system. The system is inconsistent. Solution set: ø The graphs of the equations are parallel and never intersect. 14x + 6y = –10 –14x – 6y = –10 0 = –20 (2)

5. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-5 5.1 Example 4 Solving a System with Infinitely Many Solutions (page 497) Solve the system. The solution set with an arbitrary y is {( 1 / 3 y + 8 / 3, y)}. The graphs of the two equations coincide. 3x – y = 8 Solve for x 3x = y + 8 x = 1 / 3 y + 8 / 3 (3) 0 = 0 Always true Infinite Solutions Find the solution with arbitrary y

6. 5-6 5.1 Example 5 Using a Linear System to Solve an Application (page 498) In the 2006 Winter Olympics in Torino, Italy, the two countries winning the most medals were Germany and the United States. The total number of medals won by these two countries was 54, with Germany winning 4 more medals than the United States. How many medals were won by each country? (Source: www.infoplease.com) Let U = the # of medals won by the U.S. Let G = the # of medals won by Germany. Then, we have the system: U + G = 54 -U + G = 4 The U.S. won 25 medals and Germany won 29 medals. U + G = 54 G – U = 4 2G = 58 G = 29 U = 25

7. 5-7 5.1 Ex 6 Solving a System of Three Equations with Three Variables (pg 500) Solve the system Planning to eliminate z 5x + 5y = 5 3y = 9 y = 3 Add lines 1 & 2 and 2 & 3 5x + 5(3) = 5 5x + 15 = 5 5x = -10 x = -2 3(-2) + 4(3) – 2z = 14 -6 + 12 – 2z = 14 6 – 2z = 14 –2z = 8 z = -4 Solution: (-2, 3, -4) (-2) 3x + 4y – 2z = 14 2x + y + 2z = -9 -2x + 2y – 2z = 18

8. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-8 5.1 Example 6 Solving a System of Three Equations with Three Variables (cont.) Solution set:{(–2, 3, –4)} Verify that (–2, 3, –4) satisfies all three equations in the original system.

9. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-9 5.1 Example 8 Find equation of the parabola (page 502) Find the equation of the parabola that passes through the points (–5, 7), (–1, –2), and (3, 5). y = ax 2 + bx + c x 2 a + xb + c = y (–5, 7) (–1, –2) (3, 5) 25a 1a 9a – 5b – b + 3b + c = 7 = –2 = 5 25a – 5b + c = 7 –a + b – c = 2 9a + 3b + c = 5 24a – 4b = 9 8a + 4b = 7 32a = 16 a = ½ 24(½) – 4b = 9 12 – 4b = 9 – 4b = -3 b = ¾ (½) – (¾) + c = -2 -¼ + c = -2 c = - 7 / 4 y = 0.5x 2 + 0.75x – 1.75

10. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-10 5.1 add on: If infinite solutions, arbitrary z 4x – 3y + z = 0 x + y = 0 -16x + 12y – 4z = 0 Solve. Answer: (,, z) –1/7z–1/7z 1/7z1/7z x = - 1 / 7 z y = 1 / 7 z Plan to eliminate z (4) 16x – 12y + 4z = 0 x + y = 0 -16x + 12y – 4z = 0 4x – 3y + z = 0 3x + 3y = 0 Add L1 & L2 solve for x by eliminating y Add L1 & L2 solve for y by eliminating x 7x + z = 0 4x – 3y + z = 0 -4x – 4y = 0 – 7y + z = 0 Infinite solutions so go back to original equations

11. 5.1 Summary Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-11 Systems of Linear Equations Substitution Method Elimination Method Special Systems: no solution and infinite solution Applying Systems of Equations Solving Systems with Three Unknowns (Variables) Using Systems of Equations to Model Data Solve with an arbitrary X ( X, ___________, __________ ) solve for Z eliminate Y solve for Y eliminate Z Solve with an arbitrary Z ( ___________, __________, Z ) solve for X eliminate Y solve for Y eliminate X

12. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-12 Determinant Solution of Linear Systems 5.3 Determinants ▪ Cofactors ▪ Evaluating n × n Determinants ▪ Cramer’s Rule

13. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-13 5.3 Example 1 Evaluating a 2 × 2 Determinant (page 523) You are being asked to find the determinant of matrix B Det = -20 – (–16) = = -20 + 16 = -4 minus

14. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-14 5.3 Example 3 Evaluating a 3 × 3 Determinant (page 526) Evaluate, by using diagonals. minus -48 + 0 + 10 – (-20) – (18) – (0) Det = -36

15. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-15 5.3 Example 3 Evaluating a 3 × 3 Determinant (page 526) Evaluate, by using expansion by minors. Start with row 1: keep it, change it, keep it Det = 2 + 3 + 5 = 2(-24 – -10) + 3(-6 – 0) + 5(2 – 0) = 2(-14) + 3(-6) + 5(2) = -28 – 18 + 10 = -36

16. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-16 5.3 Ex 4 Applying Cramer’s Rule to a 2 × 2 System (page 528) Use Cramer’s rule to solve the system. By Cramer’s rule, Find D first, since if D = 0, Cramer’s rule does not apply. If D ≠ 0, find D x and D y. D = -8 – 3 D x = -4 – 15 D y = 20 – 2 = -11 = -19 = 18

17. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-17 5.3 Example 5 Applying Cramer’s Rule to a 3 × 3 System (page 529) Use Cramer’s rule to solve the system. Rewrite each equation in the form ax + by + cz = k. By Cramer’s rule, Find D first, since if D = 0, Cramer’s rule does not apply.

18. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-18 5.3 Example 5 Applying Cramer’s Rule to a 3 × 3 System (cont.) D = 2(12+1) + 4(8+1) + 1(4-6) = 60D x = 19(12+1)+4(-30+11)+1(-15-66) = 90 D y = 2(-30+11) –19(8+1)+1(44+15) = -150D z = 2(66+15) +4(44+15)+19(4–6) = 360

19. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-19 5.3 Example 6 Showing That Cramer’s Rule Does Not Apply (page 530) Show that Cramer’s rule does not apply to the system. We need to show that D = 0. Thus, Cramer’s rule does not apply to the system. = 1(12 – 12) + 3(36 – 36) + 2(-9 + 9) = 0

20. Additional – Area of a Triangle Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-20 Find the area of a triangle with the given vertices: (-2,3), (4,5), and (-10,-3) = 10

21. 5.3 Summary Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-21 Determinant Solution of Linear Systems Determinants Evaluating n × n Determinants Cramer’s Rule

22. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-22 Nonlinear Systems of Equations 5.5 Solving Nonlinear Systems with Real Solutions ▪ Solving Nonlinear Systems with Nonreal Complex Solutions ▪ Applying Nonlinear Systems

23. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-23 Solve the system. 5.5 Example 1 Solving a Nonlinear System by Substitution (page 543) Solution set: {(–2, 2), (1, 5)} x 2 + x – 2 = 0 (x + 2)(x – 1) = 0 Add lines 1 & 2 x 2 + x = 2 x = -2 x = 1 Substitute x = -2 in line 2 (-2) – y = -4 -y = -2 y = 2 Substitute x =1 in line 2 (1) – y = -4 -y = -5 y = 5

24. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-24 Solve the system. 5.5 Example 2 Solving a Nonlinear System by Elimination (page 544) Multiply line 1 by (–4) Substitute x = 0 in line (1), then solve for y. Solution set: {(0, –3),(0, 3)} (0) 2 + y 2 = 9 y 2 = 9 y = ±3 –4x 2 – 4y 2 = –36 9x 2 + 4y 2 = 36 5x 2 = 0 x 2 = 0 x = 0

25. Solve the system. 5.5 Example 3 Solving a Nonlinear System with Nonreal Complex Numbers in its Solutions (page 547) Solution set: Mult L1 by (-2) -2x 2 – 2y 2 = -12 3x 2 + 2y 2 = 8 x 2 = –4 x = ± √-4 x = ±2i Substitute x = 2i which means Substitute x 2 = -4 into L1 -4 + y 2 = 6 y 2 = 10 y = ±√10 Substitute x = -2i which means Substitute x 2 = -4 into L1 -4 + y 2 = 6 y 2 = 10 y = ±√10

26. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-26 The length of the hypotenuse of a right triangle is 41 cm. One of the legs is 31 cm longer than the other. Find the lengths of the two legs of the triangle. 5.5 Example 4 Using a Nonlinear System to Find the Dimensions of a Box (page 547) X X+31 41 X 2 + (X + 31) 2 = 41 2 X 2 + X 2 + 62X + 961 = 1681 2X 2 + 62X – 720 = 0 X 2 + 31X – 360 = 0 (X + 40) (X – 9) = 0 X = -40 or X = 9 The lengths of the legs are 9 cm and 40 cm.

27. Additional Problem Solve the system of equations:x + y = 21 xy = 80 Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-27 From the 1 st line y = -x + 21 sub this into the 2 nd equation x(-x + 21) = 80 -x 2 + 21x = 80 0 = x 2 – 21x + 80 0 = (x – 5) (x – 16) x = 5 or x = 16 Solution: (5,16), (16,5)

28. Additional Problem Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-28 Solve the system of equations:x 2 + y 2 = 13 xy = 6 Since solving for x or y in the 1 st equation would lead to a really ugly equation with the square root of a binomial we will solve for y in the 2 nd equation y = 6 / x Sub y = 6 / x into 1 st equation x 2 + ( 6 / x ) 2 = 13 x 2 = 4 or x 2 = 9 x = ±2 or x = ±3 When x = 2, y = 3 When x = -2, y = -3 When x = 3, y = 2 When x = -3, y = -2 Answers: {(2,3),(-2,-3),(3,2),(-3,-2)}

29. Add on to notes Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-29 Solve: –x 2 + xy + 2y 2 = 26 x 2 – xy + y 2 = 13 3y 2 = 39 y 2 = 13

30. 5.5 add on Find two numbers whose ratio is 9 to 2 and whose product is 162. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-30 Let 9x and 2x represent the numbers 18x 2 = 162 x 2 = 9 x = ±3 Solutions: 27 and 6 or –27 and –6

31. 5.5 add on A box with an open top has a square base and four sides of equal height. The volume of the box is 360 ft 3. If the surface area is 276 ft 2, find the dimensions of the box. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-31 x x y V = 360 x 2 y = 360 A = 276 x 2 + 4xy = 276 y = 360/x 2 x 2 + 4x(360/x 2 ) = 276 x 2 + 1440/x = 276 x 3 + 1440 = 276x x 3 – 276x + 1440 = 0 Continued on next slide

32. Cont from previous slide Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-32 x 3 – 276x + 1440 = 0 V = 360 x 2 y = 360 A = 276 x 2 + 4xy = 276 Use synthetic division to find a zero 61 0 -276 1440 1 6 6 36 -240 -1440 0 Then use Quad Formula with a = 1, b = 6, and c = -240 Keep the positive answer x x y Disc = 996 x = (-6 ± √(996))/2 x ≈12.780 Solutions: x = 6ft and y = 10ft x = 12.780ft and y = 2.204ft

33. 5.5 Summary Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-33 Nonlinear Systems of Equations Solving Nonlinear Systems with Real Solutions Solving Nonlinear Systems with Nonreal Complex Solutions

34. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-34 Systems of Inequalities and Linear Programming 5.6 Solving Linear Inequalities ▪ Solving Systems of Inequalities ▪ Linear Programming

35. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. Graph 3x – 2y ≥ 6. 5.6 Example 1 Graphing a Linear Inequality (page 555) The boundary of the graph is the straight line 3x – 2y = 6, which can be graphed using the x-intercept 2 and the y-intercept –3. The boundary is included in the graph, so draw a solid line. Don’t forget to shade the true side of the line. Or solve the equation for y. 5-35 3x – 2y > 6 -2y > -3x + 6 y < 3 / 2 x – 3

36. 5-36 Graph the system 5.6 Example 2(a) Graphing Systems of Inequalities (page 556) The graph of inequality (1) is a dashed parabola with vertex (0, 4) and x-intercepts (–2, 0) and (2,0) and shade. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. The graph of inequality (2) is a dashed line with x-int of –1 and y-int of 1 and shade.

37. 5-37 Graph the system 5.6 Example 2(b) Graphing Systems of Inequalities (page 556), so the graph consists of the points between and on the lines y = 1 and y = –1. The graph of x ≥ 0 includes the points on or to the right of the y-axis. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. The graph of V graph, 2xsteepness, shifted up 1, dotted, and shade.

38. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-38 5.6 Example 2(b) Graphing Systems of Inequalities (page 556) Since there are no points in common, the solution set is ø. x > 0

39. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-39 5.6 Example 3 Finding a Maximum Profit Model (page 557) A company makes two products – MP3 players and DVD players. Each MP3 player gives a profit of $50, and each DVD player gives a profit of $20. The company must manufacture at least 40, but no more than 60, MP3 players per day. The number of DVD players cannot exceed 75 per day, and the number of MP3 players cannot exceed the number of DVD players. How many of each should the company manufacture to obtain the maximum profit? Let x = the number of MP3 players produced daily Let y = the number of DVD players produced daily #MP3 #DVD The total daily profit is 50x + 20y. Find the maximum profit The maximum profit of $4500 will be reached when 60 MP3 players and 75 DVD players are produced.

40. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-40 5.6 Example 4 Finding a Minimum Cost Model (page 559) Robin takes vitamin pills each day. She wants at least 16 units of Vitamin A, at least 5 units of Vitamin B 1, and at least 20 units of Vitamin C daily. She can choose between red pills, costing 20¢ each, that contain 8 units of A, 1 of B 1, and 2 of C, or blue pills, costing 10¢ each, that contain 2 units of A, 1 of B 1, and 7 of C. How many of each pill should she take each day to minimize her cost and yet fulfill her daily requirements? Let x = the number of red pills to buy Let y = the number of blue pills to buy The total daily cost is 20x + 10y. Find the minimum cost The minimum cost of 60¢ will be obtained when she takes 1 red pill and 4 blue pills per day. #Red # Blue

41. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-41 5.6 Add on An object function and a system of linear inequalities representing constraints are given. Graph the system of inequalities representing the constraints. Find the value of the objective function at each corner of the graphed region. Use these values to determine the maximum value of the objective function and the values of x and y for which the maximum occurs. Objective Function z = 8x + 5y Constraints: 0 < x < 10 0 < y < 5 3x + 2y > 6 (0,5)(10,5) (10,0) (2,0) (0,3) 8(0) + 5(5) = 25 8(10) + 5(5) = 105 8(10) + 5(0) = 80 8(2) + 5(0) = 16 8(0) + 5(3) = 15 Maximum value is 105 @ (10,5)

42. 5.6 add on Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-42 Cabinet X costs $100, requires 10 sq ft of floor space, and holds 30 cu ft of files. Cabinet Y costs $140, requires 4 sq ft of floor space, and holds 12 cu ft. To get maximum storage, how many of each should be purchased with a budget limit of $1400 and floor space of 90 sq ft? X > 0, and Y > 0 Want Max Storage Storage=30X + 12Y Money: 100X + 140Y < 1400 (14,0) (0,10) Floor Space: 10X + 4Y < 90 (9,0) (0,22.5) Find the pt. of intersection of green and purple eq If there are two values that are maxima, the office manager will purchase at least one of each cabinet (7,5) Value @ (0,10) = 120 Value @ (7,5) = 270 Value @ (9,0) = 270 Purchase 7 of Cabinet X and 5 of Cabinet Y

43. 5.6 Summary Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-43 Systems of Inequalities and Linear Programming Solving Linear Inequalities Solving Systems of Inequalities Linear Programming

44. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-44 Properties of Matrices 5.7 Basic Definitions ▪ Adding Matrices ▪ Special Matrices ▪ Subtracting Matrices ▪ Multiplying Matrices ▪ Applying Matrix Algebra

45. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-45 5.7 Example 1 Finding Values to Make Two Matrices Equal (page 565) Find the values of the variables for which each statement is true, if possible. (a) (b) a = –3, b = 9, c = –5, and d = 0. a + 3 = 0 b – 4 = 5 -5 = c 5 = d + 5 Not possible, since the matrices do not have the same dimensions

46. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-46 5.7 Example 2 Adding/Subtracting Matrices (page 566) Find each sum, if possible. (a) (b)

47. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-47 5.7 Example 3 Subtracting Matrices (cont.) A and B cannot be subtracted because A is a 2 × 3 matrix, while B is a 2 × 2 matrix. (c)A – B if and

48. Add on to notes Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-48 Solve for each variable x + 3 + 2x = 9 y – 2 + 5 = 10 5 + z = 4 2w – 4 + 4 = –12 3x = 6 x = 2 y = 7 z = –1 2w = –12 w = –6

49. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-49 5.7 Example 4 Multiplying Matrices by Scalars (page 568) Find each product. (a) (b)

50. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-50 5.7 Example 5 Deciding Whether Two Matrices Can Be Multiplied (page 570) Suppose C is a 2 × 5 matrix and D is a 4 × 2 matrix. (a, b)Can the product CD be calculated? If so, what size is it? No, CD cannot be calculated. (c, d)Can the product DC be calculated? If so, what size is it? Yes, DC can be calculated. The result is a 4 × 5 matrix.

51. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-51 5.7 Example 6 Multiplying Matrices (page 571) Find each product, if possible. (a)AB (b)BA AB cannot be calculated. Matrix B Matrix A 2 x 2 2 x 3 BA can be calculated and would yield a 2 x 3 matrix

52. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-52 5.7 Example 7 Multiplying Square Matrices (page 571) Find each product. (a)CD (b)DC

53. 5.7 Summary Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-53 Properties of Matrices Basic Definitions Adding Matrices Special Matrices Subtracting Matrices Multiplying Matrices Applying Matrix Algebra

54. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-54 Matrix Solution of Linear Systems 5.2 The Gauss-Jordan Method ▪ Special Systems

55. 5.2 Add on Use the given row transformation to change each matrix as indicated Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-55 4 times row 1 added to row 2 –7times row 1 added to row 3

56. 5.2 add on Write the augmented matrix for the system and give its dimensions. Do NOT solve. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-56 3x + 5y = –13 2x + 3y = –9 2 x 3 Write the system of equations associated with the augmented matrix. Do NOT solve. 3x + 2y + z = 1 2y + 4z = 22 –x – 2y + 3y = 15 +3z

57. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-57 Solve the system. 5.2 Example 1 Using the Gauss-Jordan Method (page 512) The system has the augmented matrix

58. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-58 5.2 Example 1 Using the Gauss-Jordan Method (page 512) 1 st goal: element 2,1 = 0 3 -2 R 1 + R 2 2 nd goal: element 1,2 = 0 -17 9 R 2 + R 1 3 rd goal: 1’s in the main diagonal R 1 ∕-102 and R 2 ∕153 Solution: (-4,5)

59. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-59 Solve the system. 5.2 Example 2 Using the Gauss-Jordan Method (page 514) Solution set: {(3, –1, –2)} We will use elimination instead of Gauss-Jordan Method (4) x + y – 4z = 10 8x – 12y + 4z = 28 12x – 4y – 4z = 48 Add L1+L2 : L2+L3 9x – 11y = 38 20x – 16y = 76 (-20) (9) –180x + 220y = –760 180x – 144y = 684 76y = –76 y = –1 9x – 11(-1) = 38 9x + 11 = 38 9x = 27 x = 3 (3) + (-1) – 4z = 10 2 – 4z = 10 –4z = 8 z = –2

60. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-60 Use the Gauss-Jordan method to solve the system. 5.2 Example 3 Solving an Inconsistent System (page 515) Solution set: ø We will use elimination instead of Gauss-Jordan Method (3) 6x – 9y = 21 –6x + 9y = 0 0 = 21 Never true

61. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-61 Use the Gauss-Jordan method to solve the system. Write the solution set with z arbitrary. 5.2 Ex 4 Solving a System with Infinitely Many Solutions (page 516) Solution set: {(–3z + 18, 5z – 31, z)} Eliminate y (-2) -4x – 2y – 2z = -10 3x + 2y – z = -8 -x = 3z – 18 x = -3z + 18 -x – 3z = -18 Eliminate x (-3) (2) -6x – 3y – 3z = -15 6x + 4y – 2z = -16 y = 5z – 31 y – 5z = -31 We will use elimination instead of Gauss-Jordan Method

62. 5.2 additional example Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-62 Write the system of linear equations represented by the augmented matrix. Use x, y, z, and if necessary, w for the variables. Then find the solution. x + y + 3z = 6 y + 7z = 5 z = 9 y + 7(9) = 5 y + 63 = 5 y = -58 x + (-58) + 3(9) = 6 x – 58 + 27 = 6 x = 37 Solution set: {(37, –58, 9)}

63. 5.2 add on A chemist has two prepared acid solutions, one of which is 2% acid by volume, the other 7% acid. How many cubic centimeters of each should the chemist mix together to obtain 40cm 3 of a 3.2% acid solution? Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-63 X + Y = 40 2%X + 7%Y = 3.2%(40) (100) X + Y = 40 2X + 7Y = 128 Then solve by either substitution or elimination

64. 5.2 add on In a special diet for a hospital patient, the total amount per meal of food groups A, B, and C must equal 400 g. The diet should include one-third as much of group A as of group B. The sum of the amounts of group A and group C should equal twice the amount of group B. How many grams of each food group should be included? Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-64 A + B + C = 400 3A = B A + C = 2B A B A + B + C = 400 3A – B = 0 A – 2B + C = 0 Then solve by either substitution or elimination

65. 5.2 Summary Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-65 Matrix Solution of Linear Systems The Gauss-Jordan Method Special Systems

66. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-66 Matrix Inverses 5.8 Identity Matrices ▪ Multiplicative Inverses ▪ Solving Systems Using Inverse Matrices

67. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-67 5.8 Example 1 Verifying the Identity Property of I 3 (page 580) The 3 ×3 identity matrix is

68. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-68 5.8 Example 2 Finding the Inverse of a 2 x 2 Matrix (page 583) Find A -1 Det [A] = 3(5) – (2)(-1) = 17

69. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-69 5.8 Example 2 Finding the Inverse of a 3 × 3 Matrix (page 583) 1 in first row, first column

70. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-70 5.8 Example 2 Finding the Inverse of a 3 × 3 Matrix (page 583) 0 in second row, first column 0 in third row, second column 1 in second row, second column; 1 in third row, third column

71. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-71 5.8 Example 2 Finding the Inverse of a 3 × 3 Matrix (page 583) 0 in second row, third column 0 in first row, second column

72. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-72 5.8 Example 2 Finding the Inverse of a 3 × 3 Matrix (page 583) Graphing calculator solution The graphing calculator screens support the algebraic solution. Use the calculator to find the inverse of 3x3 matrices

73. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-73 5.8 Example 4(a) Solving Systems of Equations Using Matrix Inverses (page 585) Use the inverse of the coefficient matrix to solve the system. Write the system in matrix form. Solution set: {(–3, 7)}

74. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-74 5.8 Example 4(b) Solving Systems of Equations Using Matrix Inverses (page 585) Use the inverse of the coefficient matrix to solve the system. Write the system in matrix form. For these questions you will use elimination method as shown in previous sections …

75. 5.8 Summary Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-75 Matrix Inverses Identity Matrices Multiplicative Inverses Solving Systems Using Inverse Matrices Use the calculator to find the inverse of a 3x3 matrix Use elimination method to solve a system of equations with 3 variables instead of using the inverse matrix method

76. View Example: 5.1B #9,10 Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-76 Solve the system of equations Answer: (43,8) 30 20 42x – 6 + 5y + 50 = 1890 42x + 5y = 1846 5x + 5 – 4x + 4y = 80 x + 4y = 75 Now solve: 42x + 5y = 1846 x + 4y = 75 6 5 5-4 Original problem

77. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-77 5.7 Example 8 Using Matrix Multiplication to Model Plans for a Subdivision (page 572) A contractor builds three kinds of houses, models A, B, and C, with a choice of two styles, colonial or ranch. Matrix M shows the number of each kind of house the contractor is planning to build for a new 150-home subdivision. The amounts for each of the main materials used depend on the style of the house. These amounts are shown in matrix Q, while matrix R gives the cost in dollars for each kind of material. Concrete is measured here in cubic yards, lumber in 1000 board feet, brick in 1000s, and shingles in 100 square feet.

78. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-78 5.7 Example 8 Using Matrix Multiplication to Model Plans for a Subdivision (cont.)

79. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-79 5.7 Example 8 Using Matrix Multiplication to Model Plans for a Subdivision (cont.) (a)What is the total cost of materials for all houses of each model? To calculate the total cost of material for all houses of each model, first find MQ, which will show the total amount of each material needed for all houses of each model.

80. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-80 5.7 Example 8 Using Matrix Multiplication to Model Plans for a Subdivision (cont.)

81. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-81 5.7 Example 8 Using Matrix Multiplication to Model Plans for a Subdivision (cont.) Multiplying MQ and the cost matrix R gives the total cost of material for each model. The cost of materials for model A is $79,000. For model B, the cost is $69,900. For model C, the cost is $124,600.

82. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-82 5.7 Example 8 Using Matrix Multiplication to Model Plans for a Subdivision (cont.) (b)How much of each of the four kinds of material must be ordered? To find how much of each kind of material to order, total each column of matrix MQ. Write this as a row matrix, 5500 units of concrete, 200 units of lumber, 2000 units of brick and 300 units of shingles must be ordered.

83. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-83 5.7 Example 8 Using Matrix Multiplication to Model Plans for a Subdivision (cont.) (c)What is the total cost of the materials? To find the total cost of the materials, find TR. The total cost of the materials is $273,500.

84. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-84 5.8 Example 1 Verifying the Identity Property of I 3 (cont.) Graphing calculator solution The identity matrix for n = 3. The graphing calculator screens support the algebraic solution.

85. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-85 5.8 Example 3 Identifying a Matrix With No Inverse (page 584) Write the augmented matrix : Since there is no way to convert the third element in the third row to a 1, does not exist.

86. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-86 5.8 Example 4(b) Solving Systems of Equations Using Matrix Inverses (cont.) Find : Write the augmented matrix. 1 in first row, first column 0 in second row, first column

87. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-87 5.8 Example 4(b) Solving Systems of Equations Using Matrix Inverses (cont.) 0 in second row, third column 1 in second row, second column 0 in first row, second column; 0 in third row, second column

88. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-88 5.8 Example 4(b) Solving Systems of Equations Using Matrix Inverses (cont.) 1 in third row, third column

89. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-89 5.8 Example 4(b) Solving Systems of Equations Using Matrix Inverses (cont.) Now find : Solution set: {(–3, 0, 5)}

90. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-90 5.8 Example 4(b) Solving Systems of Equations Using Matrix Inverses (cont.) Graphing calculator solution The graphing calculator screens support the algebraic solution.