copyright © 2007 pearson education, inc. slide 7-1
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Copyright © 2007 Pearson Education, Inc. Slide 7-1
Copyright © 2007 Pearson Education, Inc. Slide 7-2
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
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.
063572
Copyright © 2007 Pearson Education, Inc. Slide 7-4
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
Copyright © 2007 Pearson Education, Inc. Slide 7-5
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
Copyright © 2007 Pearson Education, Inc. Slide 7-6
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
Copyright © 2007 Pearson Education, Inc. Slide 7-7
7.4 Matrix Addition
Analytic Solution
Graphing Calculator Solution
61601
)3(98866)4(5
3864 98
65 (a)
Copyright © 2007 Pearson Education, Inc. Slide 7-8
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
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
000000
.000000
643125
643125)(
then,643125and643
125
AA
AA
Copyright © 2007 Pearson Education, Inc. Slide 7-10
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
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.
64
31
52
2
128
62
104
64
31
52
64
31
52
Copyright © 2007 Pearson Education, Inc. Slide 7-12
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
Copyright © 2007 Pearson Education, Inc. Slide 7-13
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
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.
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
.237
195
.237195
9918612495187122
91812
964572
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 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.
Copyright © 2007 Pearson Education, Inc. Slide 7-17
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.
7.4 Matrix Multiplication
.233246
and405243
BA
Copyright © 2007 Pearson Education, Inc. Slide 7-18
7.4 Matrix Multiplication
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
Copyright © 2007 Pearson Education, Inc. Slide 7-19
Example Use 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
233246
405243 BA
Copyright © 2007 Pearson Education, Inc. Slide 7-20
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
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
B
22015020210
RanchColonial
Concrete Lumber Brick Shingles
C
2560
18020
ShinglesBrickLumberConcrete
Cost per Unit
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.
220150
20210
2020
2010
300
AB
Z
Y
X
Model
Model
Model
80400601200
60400401100
60600301500
Concrete Lumber Brick Shingles
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
2560
18020
804006012006040040110060600301500
)( CAB
Z
Y
X
Model
Model
Model
800,60
700,54
900,72
Cost
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 amountsmatrix. The total cost of the materials is $188,400.
7.4 Applying Matrix Algebra
20014001303800D
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