2. Needs Work Need to add –HW Quizzes

3. Chapter 13 Matrices and Determinants

4. 13.1 Matrices and Systems of Equations

5. row rows A matrix is a rectangular array of numbers. We subscript entries to tell their location in the array Matrices are identified by their size.

7. A matrix that has the same number of rows as columns is called a square matrix. main diagonal

8. Coefficient matrix If you have a system of equations and just pick off the coefficients and put them in a matrix it is called a coefficient matrix.

9. If you take the coefficient matrix and then add a last column with the constants, it is called the augmented matrix. Often the constants are separated with a line. Augmented matrix

10. We are going to work with our augmented matrix to get it in a form that will tell us the solutions to the system of equations. The three things above are the only things we can do to the matrix but we can do them together (i.e. we can multiply a row by something and add it to another row). Operations that can be performed without altering the solution set of a linear system 1. Interchange any two rows 2. Multiply every element in a row by a nonzero constant 3. Add elements of one row to corresponding elements of another row

11. After we get the matrix to look like our goal, we put the variables back in and use back substitution to get the solutions. We use elementary row operations to make the matrix look like the one below. The # signs just mean there can be any number here---we don’t care what.

12. Suse row operations to obtain row echelon form: The augmented matrix Work on this column first. Get the 1 and then use it as a “tool” to get zeros below it with row operations. We already have the 1 where we need it. We’ll take row 1 and multiply it by -3 and add to row 2 to get a 0. The notation for this step is -3r 1 + r 2 we write it by the row we replace in the matrix (see next screen).

13. -3r 1 + r 2 -3r 1 -3 -6 -3 -3 + r 2 3 5 1 3 0 -1 -2 0 Now we’ll use -2 times row 1 added to row 3 to get a 0 there. -2r 1 + r 3 -2r 1 -2 -4 -2 -2 + r 3 2 6 7 1 0 2 5 -1 Now our first column is like our goal.

14. - 1r 2 -2r 2 0 -2 -4 0 + r 3 0 2 5 -1 0 0 1 -1 -2r 2 + r 3 We’ll use row 2 with the 1 as a tool to get a 0 below it by multiplying it by -2 and adding to row 3 Now we’ll move to the second column and do row operations to get it to look like our goal. We need a 1 in the second row second column so we’ll multiply row 2 by a -1 the second column is like we need it now

15. Now we’ll move to the third column and we see for our goal we just need a 1 in the third row third column and we have it so we’ve achieved the goal and it’s time for back substitution. We put the variables and = signs back in. Substitute -1 in for z in second equation to find y x column y columnz column equal signs Substitute -1 in for z and 2 for y in first equation to find x. Solution is: (-2, 2, -1)

16. This is the only (x, y, z) that make ALL THREE equations true. Let’s check it. These are all true. Geometrically this means we have three planes that intersect at a point, a unique solution.

17. This method requires no back substitution. When you put the variables back in, you have the solutions. To obtain reduced row echelon form, you continue to do more row operations to obtain the goal below.

18. Let’s try this method on the problem we just did. We take the matrix we ended up with when doing row echelon form: Let’s get the 0 we need in the second column by using the second row as a tool. -2r 2 +r 1 Now we’ll use row 3 as a tool to work on the third column to get zeros above the 1. -2r 3 +r 2 3r3+r13r3+r1 Notice when we put the variables and = signs back in we have the solution

19. Let’s try another one: The augmented matrix: If we subtract the second row from the first we’ll get the 1 we need for the first column. r 1 -r 2 We’ll now use row 1 as our tool to get 0’s below it. -2r 1 +r 2 -7r 1 +r 3 We have the first column like our goal. On the next screen we’ll work on the next column.

20. If we multiply the second row by a -1/5 we’ll get the one we need in the second column. We’ll now use row 2 as our tool to get 0’s below it. -1/5r 2 10r 2 +r 3 Wait! If you put variables and = signs back in the bottom equation is 0 = -19 a false statement! INCONSISTENT - NO SOLUTION

21. One more: r1-r3r1-r3 -2r 1 +r 2 -4r 1 +r 3 1/3r 2 -9r 2 +r 3 Oops---last row ended up all zeros. Put variables and = signs back in and get 0 = 0 which is true. This is the dependent case. We’ll figure out solutions on next slide.

22. x y z Let’s go one step further and get a 0 above the 1 in the second column 3r 2 +r 1 Infinitely many solutions where z is any real number No restriction on z put variables back in solve for x & y

23. Infinitely many solutions where z is any real number What this means is that you can choose any real number for z and put it in to get the x and y that go with it and these will solve the equation. You will get as many solutions as there are values of z to put in (infinitely many). Let’s try z = 1. Then y = 2 and x = 3 works in all 3 Let’s try z = 0. Then y = 1 and x = 2 The solution can be written: (z + 2, z + 1, z)

26. HW #13.1 Pg 572 1-4, 6-7, 9-11, 15

27. HW Quiz 13.1 Friday, September 11, 2015 Row 1, 3, 5 1.4 2.6 3.10 4.15 Row 2, 4 1.2 2.4 3.6 4.10

28. Chapter 13 Matrices and Determinants Section 13.2 Addition and Subtraction of Matrices To Add and Subtract matrices To find the additive inverse of a matrix

29. To compare matrices they must have the same dimensions and have the same entries in the same positions

31. You can only add or subtract matrices when they have exactly the same dimensions

32. 4. The operation is not possible

33. The additive inverse of a matrix can be obtained by replacing each element by its additive inverse. Finding the Additive Inverse of a Matrix Find the additive inverse of the matrix

34. Find the Additive Inverse of each Matrix

35. Subtracting by finding the Additive Inverse Subtract by finding the Additive Inverse

36. Exercises for Example 4 Subtract by finding the Additive Inverse

37. HW # 13.2 Pg 575 1-32

38. Chapter 13 Matrices and Determinants Section 13.3 Cramer’s Rule

39. Objective: Evaluate a 2 x 2 Determinant

41. A B Your Turn Hidden

42. Objective: Solve a system of 2 equations and 2 variables using Cramer’s Rule

44. Objective: Evaluate a 3 x 3 Determinant

46. Your Turn Hidden C

47. Objective: Solve a system of 3 equations and 3 variables using Cramer’s Rule

49. E D Your Turn Hidden

50. HW #13.3 Pg 580 1-33 odd, 34-42

51. Chapter 13 Matrices and Determinants Section 13.4 Multiplying Matrices

52. 4 X 3  3 X 5  4 X 5 A  B  AB M ULTIPLYING T WO M ATRICES 4 rows 3 columns 3 rows 5 columns

53. 4 X 3  3 X 5  4 X 5 M ULTIPLYING T WO M ATRICES 4 rows 5 columns 4 rows 5 columns A  B  AB

54. If A is a 4 X 3 matrix and B is a 3 X 5 matrix, then the product AB is a 4 X 5 matrix. M ULTIPLYING T WO M ATRICES

55. m X n  n X p  m X p A  B  AB M ULTIPLYING T WO M ATRICES m rows n columns n rows p columns

56. m X n  n X p  m X p M ULTIPLYING T WO M ATRICES m rows p columns m rows p columns A  B  AB

57. If A is an m X n matrix and B is an n X p matrix, then the product AB is an m X p matrix. M ULTIPLYING T WO M ATRICES

58. Finding the Product of Two Matrices – 23 1– 4 60 – 13– 24– 13– 24 Find AB ifA =and B = Use a similar procedure to write the other entries of the product. Because A is a 3 X 2 matrix and B is a 2 X 2 matrix, the product AB is defined and is a 3 X 2 matrix. To write the entry in the first row and first column of AB, multiply corresponding entries in the first row of A and the first column of B. Then add. S OLUTION

59. (– 2)(– 1) + (3)(– 2)(– 2)(3) + (3)(4) (1)(– 1) + (– 4)(– 2)(1)(3) + (– 4)(4) (6)(– 1) + (0)(– 2)(6)(3) + (0)(4) 3 X 2  2 X 2  3 X 2 A  B  AB – 23 1– 4 60 – 13– 24– 13– 24 Finding the Product of Two Matrices

60. 3 X 2  2 X 2  3 X 2 A  B  AB Finding the Product of Two Matrices (– 2)(– 1) + (3)(– 2)(– 2)(3) + (3)(4) (1)(– 1) + (– 4)(– 2)(1)(3) + (– 4)(4) (6)(– 1) + (0)(– 2)(6)(3) + (0)(4) – 23 1– 4 60 – 13– 24– 13– 24

61. 3 X 2  2 X 2  3 X 2 A  B  AB Finding the Product of Two Matrices (– 2)(– 1) + (3)(– 2)(– 2)(3) + (3)(4) (1)(– 1) + (– 4)(– 2)(1)(3) + (– 4)(4) (6)(– 1) + (0)(– 2)(6)(3) + (0)(4) – 23 1– 4 60 – 13– 24– 13– 24

62. 3 X 2  2 X 2  3 X 2 A  B  AB Finding the Product of Two Matrices – 46 7– 13 – 618  (– 2)(– 1) + (3)(– 2)(– 2)(3) + (3)(4) (1)(– 1) + (– 4)(– 2)(1)(3) + (– 4)(4) (6)(– 1) + (0)(– 2)(6)(3) + (0)(4)

63. 3 x 2 2 x 2 AB will be 3 x 2

64. 2 x 2 AB will be 2 x 2 2 x 2 BA will be 2 x 2

67. Properties of Matrix Arithmetic For any matrices A, B, C of dimensions appropriate for them to be added or multiplied. –Commutative Property of Addition A + B = B + A –Associative Property A + (B + C) = (A + B) + C A(BC) = (AB)C

68. Properties of Matrix Arithmetic For any matrices A, B, C of dimensions appropriate for them to be added or multiplied. –Additive Identity There exists a unique matrix O such that A + 0 = 0 + A = A –Additive Inverse There Exists a unique matrix –A such that A + (-A) = -A + A = 0

69. Properties of Matrix Arithmetic For any matrices A, B, C of dimensions appropriate for them to be added or multiplied. –Distributive Property A(B + C) = AB + AC

70. Properties of Matrix Arithmetic For any real numbers k and m  k(A + B) = kA + kB  (k + m)A = kA + mA  (km)A = k(mA)

72. HW #13.4 Pg 587-588 1-39 Odd, 40-46

73. HW Quiz 13.4 Friday, September 11, 2015 Row 1, 3, 5 Write the answer to 1. 9 2. 15 3. 17 4. 39 Row 2, 4, 6 Write the answer to 1. 9 2. 15 3. 17 4. 39

75. Chapter 13 Matrices and Determinants Section 13.5 Inverses of Matrices To write a matrix equation equivalent to a system of matrices To determine when two matrices are multiplicative inverses and find the multiplicative inverse of a 2 x 2 matrix

77. Two n x n matrices are inverses of each other if their product (in both orders) is the n x n identity matrix. AA -1 = A -1 A = 1

78. To determine if two matrices A and B are inverses of each other you need to make sure AB = BA = I Determine if A and B are multiplicative inverses of each other No Determine if C and D are multiplicative inverses of each other Yes

79. Tell whether the matrices are multiplicative inverses of each other.

81. Computing an Inverse Matrix 2 x 2

83. Let’s Prove it!

84. Use the shortcut

85. Find the inverse of the given matrix

86. HW #13.5 Pg 591-592 1-19, 21-25 Odd

87. 13.6 Inverses and Systems

88. Writing Linear Systems as Matrix Equations Consider the system Let A = Let X = Let B = Write the equation AX = B using the above matrices Coefficient Matrix Matrix of Variables Matrix of Constants

89. For a linear system of equations written as a matrix equation the matrix A is the coefficient matrix of the system, X is the matrix of variables, and B is the matrix of constants.

90. Write the system of linear equations as a matrix equation

91. SOLUTION OF A LINEAR SYSTEM Let AX = B represent a system of linear equations. If the determinant of A is nonzero, then the linear system has exactly one solution, which is X = A -1 B.

94. Solve system of linear equations.

95. HW 13.6 Pg 596-597 1-19 Odd

96. 13-7 Using Matrices

97. Application Two stores sell the exact same brand and style of a dresser, a night stand, and a bookcase. Matrix A gives the retail prices (in dollars) for the items. Matrix B gives the number of each item sold at each store in one month. Calculate AB and interpret the entries of AB

105. Your Turn

106. HW #13.7 Pg 600-601 1-7

107. Chapter 13 Test Review Part 1

108. Helpful Hints What does it mean when the determinant of a matrix is 0 –In terms of a system of Linear Equations –In Terms of a Matrix Rules of Multiplication/Addition/Equality Scalar Multiplication Shortcut for finding the inverse of a 2 x 2 matrix

109. Solve

110. Evaluate the determinant of the matrix.

111. Use Cramer’s rule to solve the system of equations.

112. Use Augmented Matrices to solve the system of equations.

113. Use the matrix equation AX = B to solve the system.

114. When does a matrix fail to have an inverse? Study all the challenge problems in the book Know how to multiply and add matrices

115. Chapter 13 Test Review Part 2

116. This test will have only 1 part

117. HW #R-13 Pg 609 1-14