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College Algebra Fifth Edition James Stewart Lothar Redlin Saleem Watson.

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Presentation on theme: "College Algebra Fifth Edition James Stewart Lothar Redlin Saleem Watson."— Presentation transcript:

1 College Algebra Fifth Edition James Stewart Lothar Redlin Saleem Watson

2 Matrices and Determinants 7

3 Inverses of Matrices and Matrix Equations 7.3

4 Introduction In the preceding section, we saw that, when the dimensions are appropriate, matrices can be added, subtracted, and multiplied. Here, we investigate division of matrices. With this operation, we can solve equations that involve matrices.

5 The Inverse of a Matrix

6 Identity Matrices First, we define identity matrices. These play the same role for matrix multiplication as the number 1 does for ordinary multiplication of numbers. That is, 1 · a = a · 1 = a for all numbers a.

7 Identity Matrices In the following definition, the term main diagonal refers to the entries of a square matrix whose row and column numbers are the same. These entries stretch diagonally down the matrix—from top left to bottom right.

8 Identity Matrix—Definition The identity matrix I n is the n x n matrix for which: Each main diagonal entry is a 1. All other entries are 0.

9 Identity Matrices Thus, the 2 x 2, 3 x 3, and 4 x 4 identity matrices are:

10 Identity Matrices Identity matrices behave like the number 1 in the sense that A · I n = A and I n · B = B whenever these products are defined.

11 E.g. 1—Identity Matrices The following matrix products show how: Multiplying a matrix by an identity matrix of the appropriate dimension leaves the matrix unchanged.

12 E.g. 1—Identity Matrices

13 Inverse of a Matrix If A and B are n x n matrices, and if AB = BA = I n, we say that B is the inverse of A, and we write B = A –1. The concept of the inverse of a matrix is analogous to that of the reciprocal of a real number.

14 Inverse of a Matrix—Definition Let A be a square n x n matrix. If there exists an n x n matrix A –1 with the property that AA –1 = A –1 A = I n then we say that A –1 is the inverse of A.

15 E.g. 2—Verifying that a Matrix Is an Inverse Verify that B is the inverse of A, where: We perform the matrix multiplications to show that AB = I and BA = I.

16 E.g. 2—Verifying that a Matrix Is an Inverse

17

18 Finding the Inverse of a 2 x 2 Matrix

19 The following rule provides a simple way for finding the inverse of a 2 x 2 matrix, when it exists. For larger matrices, there’s a more general procedure for finding inverses—which we consider later in this section.

20 Inverse of a 2 x 2 Matrix If then If ad – bc = 0, then A has no inverse.

21 E.g. 3—Finding the Inverse of a 2 x 2 Matrix Let A be the matrix Find A –1 and verify that AA –1 = A –1 A = I 2

22 E.g. 3—Finding the Inverse of a 2 x 2 Matrix Using the rule for the inverse of a 2 x 2 matrix, we get: To verify that this is indeed the inverse of A, we calculate AA –1 and A –1 A.

23 E.g. 3—Finding the Inverse of a 2 x 2 Matrix

24 Determinant of a Matrix The quantity ad – bc that appears in the rule for calculating the inverse of a 2 x 2 matrix is called the determinant of the matrix. If the determinant is 0, the matrix does not have an inverse (since we cannot divide by 0).

25 Finding the Inverse of an n x n Matrix

26 For 3 x 3 and larger square matrices, the following technique provides the most efficient way to calculate their inverses.

27 Finding the Inverse of an n x n Matrix If A is an n x n matrix, we first construct the n x 2n matrix that has the entries of A on the left and of the identity matrix I n on the right:

28 Finding the Inverse of an n x n Matrix We then use the elementary row operations on this new large matrix to change the left side into the identity matrix. This means that we are changing the large matrix to reduced row-echelon form.

29 Finding the Inverse of an n x n Matrix The right side is transformed automatically into A –1. We omit the proof of this fact.

30 E.g. 4—Finding the Inverse of a 3 x 3 Matrix Let A be the matrix (a)Find A –1. (b)Verify that AA –1 = A –1 A = I 3.

31 E.g. 4—Inverse of a 3 x 3 Matrix We begin with the 3 x 6 matrix whose left half is A and whose right half is the identity matrix. We then transform the left half of this new matrix into the identity matrix—by performing the following sequence of elementary row operations on the entire new matrix. Example (a)

32 E.g. 4—Inverse of a 3 x 3 Matrix Example (a)

33 E.g. 4—Inverse of a 3 x 3 Matrix Example (a)

34 E.g. 4—Inverse of a 3 x 3 Matrix We have now transformed the left half of this matrix into an identity matrix. This means we’ve put the entire matrix in reduced row-echelon form. Example (a)

35 E.g. 4—Inverse of a 3 x 3 Matrix Note that, to do this in as systematic a fashion as possible, we first changed the elements below the main diagonal to zeros—just as we would if we were using Gaussian elimination. Example (a)

36 E.g. 4—Inverse of a 3 x 3 Matrix Then, we changed each main diagonal element to a 1 by multiplying by the appropriate constant(s). Example (a)

37 E.g. 4—Inverse of a 3 x 3 Matrix Finally, we completed the process by changing the remaining entries on the left side to zeros. Example (a)

38 E.g. 4—Inverse of a 3 x 3 Matrix The right half is now A –1. Example (a)

39 E.g. 4—Inverse of a 3 x 3 Matrix We calculate AA –1 and A –1 A, and verify that both products give the identity matrix I 3. Example (b)

40 Using Graphing Calculators Graphing calculators are also able to calculate matrix inverses. On the TI-82 and TI-83 calculators, matrices are stored in memory using names such as [A], [B], [C],.... To find the inverse of [A], we key in: [A] x –1 ENTER

41 Using Graphing Calculators For the matrix of Example 4, this results in the output shown. We have also used the Frac command to display the output in fraction form rather than in decimal form.

42 E.g. 5—Matrix that Does Not Have an Inverse Find the inverse of the matrix.

43 E.g. 5—Matrix that Does Not Have an Inverse

44

45 At this point, we would like to change the 0 in the (3, 3) position of this matrix to a 1, without changing the zeros in the (3, 1) and (3, 2) positions.

46 E.g. 5—Matrix that Does Not Have an Inverse However, there is no way to accomplish this. No matter what multiple of rows 1 and/or 2 we add to row 3, we can’t change the third zero in row 3 without changing the first or second zero as well.

47 E.g. 5—Matrix that Does Not Have an Inverse Thus, we cannot change the left half to the identity matrix. So, the original matrix doesn’t have an inverse.

48 Matrix that Does Not Have an Inverse If we encounter a row of zeros on the left when trying to find an inverse—as in Example 5—then the original matrix does not have an inverse.

49 Singular Matrix If we try to calculate the inverse of the matrix from Example 5 on a TI-83 calculator, we get the error message shown here. A matrix that has no inverse is called singular.

50 Matrix Equations

51 We saw in Example 6 in Section 7.2 that a system of linear equations can be written as a single matrix equation.

52 Matrix Equations For example, the system is equivalent to the matrix equation

53 Coefficient Matrix If we let this matrix equation can be written as: AX = B The matrix A is called the coefficient matrix.

54 Matrix Equations We solve this matrix equation by multiplying each side by the inverse of A—provided the inverse exists. AX = B A –1 (AX) = A –1 B (A –1 A)X = A –1 B Associative Property I 3 X = A –1 B Property of Inverses X = A –1 B

55 Matrix Equations In Example 4, we showed that:

56 Matrix Equations So, from X = A –1 B, we have: Thus, x = –11, y = –23, z = 7 is the solution of the original system.

57 Matrix Equations We have proved that the matrix equation AX = B can be solved by the following method.

58 Solving a Matrix Equation Let: A be a square n x n matrix that has an inverse A –1. X be a variable matrix, with n rows. B be a known matrix, with n rows. Then, the solution of the matrix equation AX = B is given by: X = A –1 B

59 E.g. 6—Solving a System Using a Matrix Inverse (a) Write the system of equations as a matrix equation. (b) Solve the system by solving the matrix equation.

60 E.g. 6—Using a Matrix Inverse We write the system as a matrix equation of the form AX = B: Example (a)

61 E.g. 6—Using a Matrix Inverse Using the rule for finding the inverse of a 2 x 2 matrix, we get: Example (b)

62 E.g. 6—Using a Matrix Inverse Multiplying each side of the matrix equation by the inverse matrix, we get: Thus, x = 30 and y = 9. Example (b)

63 Modeling with Matrix Equations

64 Applications Suppose we need to solve several systems of equations with the same coefficient matrix. Then, converting the systems to matrix equations provides an efficient way to obtain the solutions. We only need to find the inverse of the coefficient matrix once.

65 Applications This procedure is particularly convenient if we use a graphing calculator to perform the matrix operations—as in the next example.

66 E.g. 7—Modeling Nutritional Requirements A pet-store owner feeds his hamsters and gerbils different mixtures of three types of rodent food: KayDee Food, Pet Pellets, Rodent Chow He wishes to feed his animals the correct amount of each brand to satisfy their daily requirements for protein, fat, and carbohydrates exactly.

67 E.g. 7—Modeling Nutritional Requirements Suppose that each day: Hamsters require 340 mg of protein 280 mg of fat 440 mg of carbohydrates Gerbils need 480 mg of protein 360 mg of fat 680 mg of carbohydrates

68 E.g. 7—Modeling Nutritional Requirements The amount of each nutrient (in mg) in one gram of each brand is given here. How many grams of each food should the storekeeper feed his hamsters and gerbils daily to satisfy their nutrient requirements?

69 E.g. 7—Modeling Nutritional Requirements We let x 1, x 2, and x 3 be the respective amounts (in grams) of KayDee Food, Pet Pellets, and Rodent Chow that the hamsters should eat. We let y 1, y 2, and y 3 be the corresponding amounts for the gerbils.

70 E.g. 7—Modeling Nutritional Requirements Hamster eqn. Gerbil eqn. Then, we want to solve the matrix equations

71 E.g. 7—Modeling Nutritional Requirements Let:

72 E.g. 7—Modeling Nutritional Requirements Then, we can write these matrix equations as: AX = B Hamster eqn. AY = C Gerbil eqn. We want to solve for X and Y. So, we multiply both sides of each equation by A –1, the inverse of the coefficient matrix.

73 E.g. 7—Modeling Nutritional Requirements We could find A –1 by hand. However, it is more convenient to use a graphing calculator.

74 E.g. 7—Modeling Nutritional Requirements From the calculator displays, we see:

75 E.g. 7—Modeling Nutritional Requirements Daily, each hamster should be fed: 10 g of KayDee Food. 3 g of Pet Pellets. 12 g of Rodent Chow. Daily, each gerbil should be fed: 8 g of KayDee Food. 4 g of Pet Pellets. 20 g of Rodent Chow.


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