1. Copyright © 2007 Pearson Education, Inc. Slide 7-1

2. Copyright © 2007 Pearson Education, Inc. Slide 7-2 Chapter 7: Matrices and Systems of Equations and Inequalities 7.1Systems of Equations 7.2Solution of Linear Systems in Three Variables 7.3Solution of Linear Systems by Row Transformations 7.4Matrix Properties and Operations 7.5Determinants and Cramer’s Rule 7.6Solution of Linear Systems by Matrix Inverses 7.7Systems of Inequalities and Linear Programming 7.8Partial Fractions

3. Copyright © 2007 Pearson Education, Inc. Slide 7-3 7.4 Matrix Properties and Operations Matrices are classified by their dimensions: the number of rows by the number of columns. A matrix with m rows and n columns has dimension m × n. e.g. The matrix has dimension 2×3. A square matrix has the same number of rows as it does columns. The dimension of a square matrix is n × n.

4. Copyright © 2007 Pearson Education, Inc. Slide 7-4 7.4 Classifying Matrices by Dimension ExampleFind the dimension of each matrix. (a) The matrix is a 3 × 2 matrix. (b)The matrix is a 3 × 3 square matrix. (c)The matrix is a 1 × 5 row matrix.

5. Copyright © 2007 Pearson Education, Inc. Slide 7-5 7.4 Determining Equality of Matrices Example SolutionTwo matrices are equal if they have the same dimension and if corresponding elements, position by position, are equal. This is true in this case if 2 = x, 1 = y, p = –1, and q = 0.

6. Copyright © 2007 Pearson Education, Inc. Slide 7-6 7.4 Matrix Addition ExampleFind each sum. The sum of two m × n matrices A and B is the m × n matrix A + B in which each element is the sum of the corresponding elements of A and B.

7. Copyright © 2007 Pearson Education, Inc. Slide 7-7 7.4 Matrix Addition Analytic Solution Graphing Calculator Solution

8. Copyright © 2007 Pearson Education, Inc. Slide 7-8 7.4 Matrix Addition Analytic Solution Graphing Calculator Solution The calculator returns a dimension mismatch error.

9. Copyright © 2007 Pearson Education, Inc. Slide 7-9 7.4 The Zero Matrix A matrix with only zero elements is called a zero matrix. For example, [0 0 0] is the 1 × 3 zero matrix while is the 2 × 3 zero matrix. The elements of matrix –A are the additive inverses of the elements of matrix A. For example, if

10. Copyright © 2007 Pearson Education, Inc. Slide 7-10 7.4 Matrix Subtraction ExampleFind the difference of Solution If A and B are matrices with the same dimension, then A – B = A + (– B).

11. Copyright © 2007 Pearson Education, Inc. Slide 7-11 7.4 Matrix Multiplication by a Scalar If a matrix A is added to itself, each element is twice as large as the corresponding element of A. In the last expression, the 2 in front of the matrix is called a scalar. A scalar is a special name for a real number.

12. Copyright © 2007 Pearson Education, Inc. Slide 7-12 7.4 Matrix Multiplication by a Scalar ExamplePerform the multiplication Solution The product of a scalar k and a matrix A is the matrix kA, each of whose elements is k times the corresponding elements of A.

13. Copyright © 2007 Pearson Education, Inc. Slide 7-13 7.4 Matrix Multiplication ExampleSuppose you are the manager of a video store and receive the following order from two distributors: from Wholesale Enterprises, 2 videotapes, 7 DVDs, and 5 video games; from Discount Distributors, 4 videotapes, 6 DVDs, and 9 video games. We can organize the information in table format and convert it to a matrix. or

14. Copyright © 2007 Pearson Education, Inc. Slide 7-14 7.4 Matrix Multiplication Suppose each videotape costs the store $12, each DVD costs $18, and each video game costs $9. To find the total cost of the products from Wholesale Enterprises, we multiply as follows. The products from Wholesale Enterprises cost a total of $195.

15. Copyright © 2007 Pearson Education, Inc. Slide 7-15 The result is the sum of three products: 2($12) + 7($18) + 5($9) = $195. In the same way, using the second row of the matrix and the three costs gives the total from Discount Distributors: 4($12) + 6($18) + 9($9) = $237. The total costs from the distributors can be written as a column matrix. The product of matrices can be written as 7.4 Matrix Multiplication

16. Copyright © 2007 Pearson Education, Inc. Slide 7-16 The product AB can be found only if the number of columns of A is the same as the number of rows of B. 7.4 Matrix Multiplication The product AB of an m × n matrix A and an n × k matrix B is found as follows: To get the ith row, jth column element of AB, multiply each element in the ith row of A by the corresponding element in the jth column of B. The sum of these products will give the element of row i, column j of AB. The dimension of AB is m × k.

17. Copyright © 2007 Pearson Education, Inc. Slide 7-17 ExampleFind the product AB of the two matrices Analytic Solution A has dimension 2 × 3 and B has dimension 3 × 2, so they are compatible for multiplication. The product AB has dimension 2 × 2. 7.4 Matrix Multiplication

18. Copyright © 2007 Pearson Education, Inc. Slide 7-18 7.4 Matrix Multiplication

19. Copyright © 2007 Pearson Education, Inc. Slide 7-19 ExampleUse the graphing calculator to find the product BA of the two matrices from the previous problem. Graphing Calculator Solution Notice AB  BA. 7.4 Matrix Multiplication

20. Copyright © 2007 Pearson Education, Inc. Slide 7-20 ExampleA contractor builds three kinds of houses, models X, Y, and Z, with a choice of two styles, colonial or ranch. Matrix A below shows the number of each kind of house the contractor is planning to build for a new 100-home subdivision. The amounts are shown in matrix B, while matrix C gives the cost in dollars for each kind of material. Concrete is measured in cubic yards, lumber in 1000 board feet, brick in 1000s, and shingles in 100 square feet. 7.4 Applying Matrix Algebra Colonial Ranch

21. Copyright © 2007 Pearson Education, Inc. Slide 7-21 (a)What is the total cost of materials for all houses of each model? (b)How much of each of the four kinds of material must be ordered? (c)Use a graphing calculator to find the total cost of the materials. 7.4 Applying Matrix Algebra Concrete Lumber Brick Shingles Cost per Unit

22. Copyright © 2007 Pearson Education, Inc. Slide 7-22 7.4 Applying Matrix Algebra Solution (a)To find the materials cost for each model, first find AB, the total amount of each material needed for all the houses of each model. Concrete Lumber Brick Shingles

23. Copyright © 2007 Pearson Education, Inc. Slide 7-23 Multiplying the total amount of materials matrix AB and the cost matrix C gives the total cost of materials. 7.4 Applying Matrix Algebra Cost

24. Copyright © 2007 Pearson Education, Inc. Slide 7-24 (b)The totals of the columns of matrix AB will give a matrix whose elements represent the total amounts of each material needed for the subdivision. Call this matrix D, and write it as a row matrix. (c)The total cost of all materials is given by the product of matrix C, the cost matrix, and matrix D, the total amounts matrix. The total cost of the materials is $188,400. 7.4 Applying Matrix Algebra