1. Welcome to Week 10 MA1310 College Math 2

2. Systems of Equations A set of two equations containing two variables... A set of three equations containing three variables... Are you seeing a pattern?

3. 2x + 5y = 38 x + 7y = 37 Lots of equations come in groups! Systems of Equations

4. If they give you values for “x” and “y” and ask if they are true values, plug them in to both equations – both equations must be true for the values Systems of Equations

5. Are x = 4 and y = 6 true values for the system: 2x + 5y = 38 x + 7y = 37 Plug in x = 4 and y = 6 Systems of Equations IN-CLASS PROBLEM

6. Plug in x = 4 and y = 6 2(4) + 5(6) =? 38 (4) + 7(6) =? 37 Are the values x = 4 and y = 6 a true solution to this system? Systems of Equations IN-CLASS PROBLEM

7. Systems of Equations To solve a system, you must have at least as many equations as you have variables

8. Suppose we wanted to find “x” and “y” for the equations: 2x + 5y = 38 x + 7y = 37 Systems of Equations IN-CLASS PROBLEM

9. Strategy: try to get one of the variables written in terms of the other variable: 2x + 5y = 38 x + 7y = 37 Look for a “loner” Systems of Equations IN-CLASS PROBLEM

10. x + 7y = 37 Use algebra to solve for “x”: x = 37 – 7y Now, plug in this “x” into the other equation Systems of Equations IN-CLASS PROBLEM

11. 2x + 5y = 38 x = 37 – 7y To get: 2(37-7y) + 5y = 38 Now, solve for “y”! Systems of Equations IN-CLASS PROBLEM

12. 2(37-7y) + 5y = 38 74 – 14y + 5y = 38 -9y = 38 – 74 = -36 y = -36/-9 y = 4 Then we plug in this value for “y” and solve for “x” Systems of Equations IN-CLASS PROBLEM

13. We already have: x = 37 – 7y y = 4 So just plug in y = 4: x = 37 – 7(4) x = 37 – 28 x = 9 Systems of Equations IN-CLASS PROBLEM

14. So, the true solution to the system: 2x + 5y = 38 x + 7y = 37 is x = 9, y = 4 or (x,y) = (9,4) Systems of Equations IN-CLASS PROBLEM

15. To find the true solution to the system: 2x + 5y = 38 3x + 7y = 50 Solve one of the equations for either “x” or “y” Systems of Equations IN-CLASS PROBLEM

16. 2x + 5y = 38 2x = 38 – 5y x = 38 – 5y 2 Then plug in this “x” into the other equation Systems of Equations IN-CLASS PROBLEM

17. 3x + 7y = 50 x = 38 – 5y 2 Systems of Equations IN-CLASS PROBLEM

18. 3(38/2 - 5y/2) + 7y = 50 57 – 7.5y + 7y = 50 Solve for “y”: -7.5y + 7y = 50 – 57 -0.5y = -7 y = -7/-0.5 y = 14 Systems of Equations IN-CLASS PROBLEM

19. You already have: x = 38 – 5y 2 and y = 14 so:x = 38 – 5(14) = 38-70 2 2 x = -32/2 = -16 Systems of Equations IN-CLASS PROBLEM

20. So the true solution to the system: 2x + 5y = 38 3x + 7y = 50 is x = -16, y = 14 or (x,y) = (-16,14) Systems of Equations IN-CLASS PROBLEM

21. Systems of Equations For more equations – Same stuff, just more of it!

22. Solve the following system of equations: x + z = 3 x + 2y – z = 1 2x – y + z = 3 Systems of Equations IN-CLASS PROBLEM

23. x + z = 3 is the easiest – let’s start there: z = 3 – x Plug that in! Systems of Equations IN-CLASS PROBLEM

24. z = 3 – x x + 2y – z = 1 2x – y + z = 3 become: x + 2y – (3 – x) = 1 2x – y + (3 – x) = 3 Now it’s just two equations! Systems of Equations IN-CLASS PROBLEM

25. x + 2y – 3 + x = 1 2x – y + 3 – x = 3 or 2x + 2y = 4 x – y = 0 Use x – y = 0  x = y Systems of Equations IN-CLASS PROBLEM

26. x = y 2x + 2y = 4 2x + 2x = 4 4x = 4 x = 1 Systems of Equations IN-CLASS PROBLEM

27. So, now we have: x = 1 x = y z = 3 – x So, (x,y,z) = (1,1,2) Systems of Equations IN-CLASS PROBLEM

28. Uses for Systems of Equations Lots of business applications use systems of equations

29. The income formula is the total amount generated by selling the product: I(x) = (price per unit sold)  x x is the quantity sold Uses for Systems of Equations

30. The cost formula is the total cost of producing the product: C(x) = fixed cost + (cost per unit produced)  x x is the quantity produced Uses for Systems of Equations

31. The profit formula is the difference between the cost and the revenue formulas: P(x) = I(x) – C(x) x is the quantity of the product Uses for Systems of Equations

32. The x-solution for the two equations Cost and Income is the true solution of the system of the two equations (the Cost and Income equations) Uses for Systems of Equations

33. The solution to the system is called the “break-even point” Uses for Systems of Equations

34. Video Revenue = 25x – 0.01x2 Cost = 1600 + 5x + 10x

35. Revenue = 25x – 0.01x2 Cost = 1600 + 5x + 10x Systems of Equations IN-CLASS PROBLEM

36. Example: A widget-maker produces widgets Each widget sells for $300 They sell 20,000 widgets each year What is their annual income? Systems of Equations IN-CLASS PROBLEM

37. I(x) = (price per unit sold)  x I = $300 (20,000) = $6,000,000 Systems of Equations IN-CLASS PROBLEM

38. They have a fixed cost (rent, utilities, salaries, etc.) of $950,000 per year Each widget costs $10 to make They make 20,000 widgets each year What is their annual cost? Systems of Equations IN-CLASS PROBLEM

39. C(x) = fixed cost + (cost per unit produced)  x C = $950,000 + $10(20,000) = $1,150,000 Systems of Equations IN-CLASS PROBLEM

40. What is their annual profit? Systems of Equations IN-CLASS PROBLEM

41. P(x) = I(x) – C(x) P = $6,000,000 - $1,150,000 = $4,850,000 Systems of Equations IN-CLASS PROBLEM

42. Let’s buy Widget stock!!! Systems of Equations IN-CLASS PROBLEM

43. Break-even point: How many widgets you need to make for the cost to equal the income (profit of $0) Systems of Equations IN-CLASS PROBLEM

44. Break-even point: C(x) = I(x) fixed cost + (cost per unit)  x = (price per unit sold)  x Systems of Equations IN-CLASS PROBLEM

45. Break-even point: 950,000 + 10(x) = 300(x) Algebra magic… 950,000 = 300x - 10x 950,000 = 290x 950,000/290 = x So x = 3275.86 units Systems of Equations IN-CLASS PROBLEM

46. They will “break even” when they make and sell 3276 units Since they make and sell a lot more, they are making a huge profit! Systems of Equations IN-CLASS PROBLEM

47. If you graph the income and cost curves, the break-even point is where the two curves cross Uses for Systems of Equations

48. Actually this is true for all systems of equations – the true solution (x,y) is the point where the graph of the two curves cross Uses for Systems of Equations

49. What is the break-even point? Uses for Systems of Equations IN-CLASS PROBLEM

50. Questions?

51. Matrices Matrix (plural matrices) - a rectangular table of elements (or entries), numbers or abstract quantities Arranged in rows and columns

52. Matrices Matrices in everyday life

53. Matrices The term "matrix" for arrangements of numbers was introduced in 1850 by James Joseph Sylvester

54. Matrices Horizontal lines in a matrix are called rows Vertical lines are called columns

55. Matrices A matrix with m rows and n columns is called an m-by-n matrix (written m × n) m and n are called its dimensions

56. Matrices The dimensions of a matrix are always given with the number of rows first, then the number of columns: r x c

57. Matrices Matrices are usually given capital letter names like “A”

58. Matrices To designate a specific element in a matrix, you use small letters with subscripts r and c: a 37 means the element in matrix A in row 3 column 7

59. Matrices In Math class, matrices are shown by listing the elements inside square braces: [ ]

60. Matrices ┌ ┐ │ a 11 a 12 a 13 │ A = │ a 21 a 22 a 23 │ │ a 31 a 32 a 33 │ └ ┘

61. Matrices The Matrix Game

62. Questions?

63. Matrices A matrix can be used as a shorthand way of writing a system of equations

64. Matrices Called an augmented matrix: Each row is an equation Each column is the coefficient of a variable or a constant A vertical bar separates coefficients on the left side from constants on the right

65. Matrices For example: 3x + 7y – 4z = 10 2x + 5y + z = 9 x + 6z = 15

66. Matrices 3x + 7y – 4z = 10 2x + 5y + z = 9 x + 6z = 15 becomes: x y z k ┌ ┐ eq 1 │ 3 7 -4 │ 10│ eq 2 │ 2 5 1 │ 9│ eq 3 │ 1 0 6 │ 15│ └ ┘

67. Matrices It is traditional to use “k” to designate the constant rather than “c” (Don’t ask me why!)

68. What would be the matrix for the system of equations: 46x – 51y = 709 30x + 88y = -420 Matrices IN-CLASS PROBLEM

69. 46x – 51y = 709 30x + 88y = -420 x y k ┌ ┐ eq 1 │ 46 -51 │ 709│ eq 2 │ 30 88 │-420│ └ ┘ Matrices IN-CLASS PROBLEM

70. What would be the matrix for the system of equations: x = 7y + 8 y = x – 3 Careful! Matrices IN-CLASS PROBLEM

71. x = 7y + 8 x – 7y = 8 y = x – 3 -x + y = -3 x y k ┌ ┐ eq 1 │ 1 -7 │ 8 │ eq 2 │ -1 1 │-3 │ └ ┘ Matrices IN-CLASS PROBLEM

72. You can also recover the original equations from a matrix What would be the equations for the matrix: x y k ┌ ┐ eq 1 │ 10 4 │ 92│ eq 2 │-43 0 │ 87│ └ ┘ Matrices IN-CLASS PROBLEM

73. Matrices x y k ┌ ┐ eq 1 │ 10 4 │ 92│ eq 2 │-43 0 │ 87│ └ ┘ The equations would be: 10x + 4y = 92 -43x = 87

74. Matrices The goal of a system of equations is to solve the system for values of the variables that work in all of the equations in the system

75. Matrices Same thing with a matrix!

76. How would you solve: x y z k ┌ ┐ eq 1 │ 1 0 0 │ 10│ eq 2 │ 0 1 0 │ 9│ eq 3 │ 0 0 1 │ 15│ └ ┘ Matrices IN-CLASS PROBLEM

77. How would you solve: x y z k ┌ ┐ eq 1 │ 1 0 0 │ 10│ eq 2 │ 0 1 0 │ 9│ eq 3 │ 0 0 1 │ 15│ └ ┘ x = 10, y = 9, z = 15 Matrices IN-CLASS PROBLEM

78. Matrices This pattern is easy to solve: upper triangle ┌ ┐ eq 1 │ 1 0 0 │ 10│ eq 2 │ 0 1 0 │ 9│ eq 3 │ 0 0 1 │ 15│ └ ┘ lower triangle

79. What if you had: x y z k ┌ ┐ eq 1 │ 6 3 2 │ 19│ eq 2 │ 0 5 4 │ 15│ eq 3 │ 0 0 1 │ 5│ └ ┘ Matrices IN-CLASS PROBLEM

80. This is called a “lower triangular matrix” x y z k ┌ ┐ eq 1 │ 6 3 2 │ 19│ eq 2 │ 0 5 4 │ 15│ eq 3 │ 0 0 1 │ 5│ └ ┘ Matrices IN-CLASS PROBLEM

81. Not quite as easy! What value do you know? x y z k ┌ ┐ eq 1 │ 6 3 2 │ 19│ eq 2 │ 0 5 4 │ 15│ eq 3 │ 0 0 1 │ 5│ └ ┘ Matrices IN-CLASS PROBLEM

82. x y z k ┌ ┐ eq 1 │ 6 3 2 │ 19│ eq 2 │ 0 5 4 │ 15│ eq 3 │ 0 0 1 │ 5│ z = 5 ! └ ┘ Matrices IN-CLASS PROBLEM

83. We’ll use the same techniques you used in solving systems of equations to solve for y and x x y z k ┌ ┐ eq 1 │ 6 3 2 │ 19│ eq 2 │ 0 5 4 │ 15│ eq 3 │ 0 0 1 │ 5│ └ ┘ Matrices IN-CLASS PROBLEM

84. Matrices Remember - each row in an augmented matrix represents an equation So…anything that you can do to an equation can be done to the row of a matrix!

85. Matrices Scalar multiplication: 3x + 7y = 15 2(3x + 7y) = 2(15) You can multiply a row of an augmented matrix by a constant

86. Matrices Scalar multiplication uses a special notation: 2R 1 This means multiply all of the elements in row 1 by the constant value 2

87. Matrices Just like you can add two polynomial equations, you can also add two rows of an augmented matrix: 3x + 7y = 15 5x - 2y = 40 8x + 5y = 55

88. Matrices You can multiply by a constant and add rows of a matrix at the same time

89. Matrices 3R1 + R2 means multiply the elements in row 1 by three, add it to the elements in row 2, and replace the elements in ROW 2 with the result

90. So… back to our problem: x y z k ┌ ┐ eq 1 │ 6 3 2 │ 19│ eq 2 │ 0 5 4 │ 15│ eq 3 │ 0 0 1 │ 5│ └ ┘ Matrices IN-CLASS PROBLEM

91. If you multiply row 3 by -4 and add it to row 2, you can solve for y easily! x y z k ┌ ┐ eq 1 │ 6 3 2 │ 19│ eq 2 │ 0 5 4 │ 15│ eq 3 │ 0 0 1 │ 5│ └ ┘ Matrices IN-CLASS PROBLEM

92. x y z k ┌ ┐ eq 1 │ 6 3 2 │ 19│ eq 2 │ 0 5 4 │ 15│ eq3  -2 0 0 -4 -20 new eq2 0 5 0 -5 eq 3 │ 0 0 1 │ 5│ └ ┘ Matrices IN-CLASS PROBLEM

93. x y z k ┌ ┐ eq 1 │ 6 3 2 │ 19│ eq 2 │ 0 5 0 │ -5│ eq 3 │ 0 0 1 │ 5│ └ ┘ Now what? Matrices IN-CLASS PROBLEM

94. x y z k ┌ ┐ eq 1 │ 6 3 2 │ 19│ eq 2 │ 0 5 0 │ -5│ eq 3 │ 0 0 1 │ 5│ └ ┘ Divide row 2 by 5! Matrices IN-CLASS PROBLEM

95. Matrices IN-CLASS PROBLEM x y z k ┌ ┐ eq 1 │ 6 3 2 │ 19│ eq 2 │ 0 1 0 │ -1│ eq 3 │ 0 0 1 │ 5│ └ ┘ Now what values do you know?

96. So now we know 2 values: x y z k ┌ ┐ eq 1 │ 6 3 2 │ 19│ eq 2 │ 0 1 0 │ -1│ y = -1 eq 3 │ 0 0 1 │ 5│ z = 5 └ ┘ Matrices IN-CLASS PROBLEM

97. Use the same strategy to solve for x! x y z k ┌ ┐ eq 1 │ 6 3 2 │ 19│ eq 2 │ 0 1 0 │ -1│ eq 3 │ 0 0 1 │ 5│ └ ┘ Matrices IN-CLASS PROBLEM

98. -3R2 + R1: x y z k ┌ ┐ eq 1 │ 6 3 2 │ 19│ +0 -3 +0 3 6 0 2 22 eq 2 │ 0 1 0 │ -1│ eq 3 │ 0 0 1 │ 5│ └ ┘ Matrices IN-CLASS PROBLEM

99. Now what? x y z k ┌ ┐ eq 1 │ 6 0 2 │ 22│ eq 2 │ 0 1 0 │ -1│ eq 3 │ 0 0 1 │ 5│ └ ┘ Matrices IN-CLASS PROBLEM

100. -2R3 + R1: x y z k ┌ ┐ eq 1 │ 6 0 2 │ 22│ +0 +0 -2 -10 6 0 0 12 eq 2 │ 0 1 0 │ -1│ eq 3 │ 0 0 1 │ 5│ └ ┘ Matrices IN-CLASS PROBLEM

101. x y z k ┌ ┐ eq 1 │ 6 0 0 │ 12│ eq 2 │ 0 1 0 │ -1│ eq 3 │ 0 0 1 │ 5│ └ ┘ What now? Matrices IN-CLASS PROBLEM

102. x y z k ┌ ┐ eq 1 │ 6 0 0 │ 12│ eq 2 │ 0 1 0 │ -1│ eq 3 │ 0 0 1 │ 5│ └ ┘ Divide row 1 by 6! Matrices IN-CLASS PROBLEM

103. x y z k ┌ ┐ eq 1 │ 1 0 0 │ 2│ eq 2 │ 0 1 0 │ -1│ eq 3 │ 0 0 1 │ 5│ └ ┘ Now what values do we know? Matrices IN-CLASS PROBLEM

104. Now what do we know? x y z k ┌ ┐ eq 1 │ 1 0 0 │ 2│ x = 2 eq 2 │ 0 1 0 │ -1│ y = -1 eq 3 │ 0 0 1 │ 5│ z = 5 └ ┘ Matrices IN-CLASS PROBLEM

105. Matrices So… if you have an upper and lower triangular matrix, the solution is obvious: upper triangle ┌ ┐ eq 1 │ 1 0 0 │ 10│ eq 2 │ 0 1 0 │ 9│ eq 3 │ 0 0 1 │ 15│ └ ┘ lower triangle

106. Matrices And… if you have only one zero triangle, it’s more work: ┌ ┐ eq 1 │ 1 1 2 │ 19│ eq 2 │ 0 1 0 │ 3│ eq 3 │ 0 0 1 │ 5│ └ ┘ lower triangular matrix

107. Matrices The upper and lower triangular form is called the “Gauss- Jordan” solution, the lower triangular form is called the “Gauss” solution

108. Carl Friedrich Gauss Wilhelm Jordan Matrices

109. Wilhelm Jordan Camille Jordan

110. Matrices Computers and calculators manipulate the matrix to achieve the Gauss-Jordan solution

111. Matrices Computers and calculators do these processes using a strategy to reduce the matrix to the Gauss-Jordan “Easy-To- Solve” form

112. Excel Solver Demo Suppose you had the system of equations: 7u + 6v + 15w – 44x + 18y + 11z = 70 9u + 25v + 10w + 3x – 14y + 8z = 161 13u + 8v + 2w + 6x + 62y + 15z = 671 –2u + v – 5w + 57x + 9y – 3z = 354 5u – 17v + 18w + 9x – 7y + 5z = 82 4u + 3v + 16w + 14x + 2y + 7z = 258 Solve for u, v, w, x, y, z

113. Questions?

114. Liberation! Be sure to turn in your assignments from last week to me before you leave Don’t forget your homework due next week! Have a great rest of the week!