copyright © 2007 pearson education, inc. slide 7-1


Chapter 7: Matrices and Systems of Equations and Inequalities

7.1 Systems of Equations

7.2 Solution of Linear Systems in Three Variables

7.3 Solution of Linear Systems by Row Transformations

7.4 Matrix Properties and Operations

7.5 Determinants and Cramer’s Rule

7.6 Solution of Linear Systems by Matrix Inverses

7.7 Systems of Inequalities and Linear Programming

7.8 Partial Fractions


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.

063572


7.4 Classifying Matrices by Dimension

Example Find 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.

154356

242321121

52561


7.4 Determining Equality of Matrices

Example

Solution Two 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.

thefind ,01

and 12 If

yx

BqpA

.such that and ,,, of values BAqpyx


7.4 Matrix Addition

Example Find 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.

3864 98

65 (a)

524

193 and 2685 if , (b) BABA


7.4 Matrix Addition

Analytic Solution

Graphing Calculator Solution

61601

)3(98866)4(5

3864 98

65 (a)


7.4 Matrix Addition

Analytic Solution

Graphing Calculator Solution The calculatorreturns a dimension mismatch error.

exist.not does sum the therefore,dimensionsdifferent

have 524193 and 26

85 matrices The (b)

BA


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

000000

.000000

643125

643125)(

then,643125and643

125

AA

AA


7.4 Matrix Subtraction

Example Find the difference of

Solution

If A and B are matrices with the same dimension, then A – B = A + (– B).

.8523

4265

12342

8523

4265

8523

4265


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.

64

31

52

2

128

62

104

64

31

52

64

31

52


7.4 Matrix Multiplication by a Scalar

Example Perform 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.

.40325

2001510

)4(5)0(5)3(5)2(5

40325


7.4 Matrix Multiplication

Example Suppose 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.

964572

or



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.


• 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


.237

195

.237195

9918612495187122

91812

964572


• The product AB can be found only if the number of columns of A is the same as the number of rows of B.


The product AB of an m × n matrix A and an n × k matrix B isfound 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.


Example Find 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.


.233246

and405243

BA



1218432

233246

405243

12)2(4)3(0)4)(5(326243

18)3(4)2(0)6(5234243

4)2(2)3(4)4)(3(326

405

32)3(2)2(4)6)(3(234

405

234

405

326

405

234243

326243


Example Use the graphing calculator to find the product BA of the two matrices from the previous problem.

Graphing Calculator Solution Notice AB BA.


233246

405243 BA


Example A 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

ZYX

ModelModel

Model A

20202010300

Colonial Ranch


(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.


B

22015020210

RanchColonial

Concrete Lumber Brick Shingles

C

2560

18020

ShinglesBrickLumberConcrete

Cost per Unit



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.

220150

20210

2020

2010

300

AB

Z

Y

X

Model

Model

Model

80400601200

60400401100

60600301500

Concrete Lumber Brick Shingles


Multiplying the total amount of materials matrix AB and the cost matrix C gives the total cost of materials.


2560

18020

804006012006040040110060600301500

)( CAB

Z

Y

X

Model

Model

Model

800,60

700,54

900,72

Cost


(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 amountsmatrix. The total cost of the materials is $188,400.


20014001303800D

copyright © 2007 pearson education, inc. slide 7-1

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