Download Sec 7.4

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
Copyright © 2007 Pearson Education, Inc.
Slide 7-2
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.
2
7

5

 has dimension 2×3.
e.g. The matrix
 3  6 0
•
A square matrix has the same number of rows as
it does columns. The dimension of a square
matrix is n × n.
Copyright © 2007 Pearson Education, Inc.
Slide 7-3
7.4 Classifying Matrices by Dimension
Example Find the dimension of each matrix.
 6 5
(a) The matrix  3 4 is a 3 × 2 matrix.
5 1
 1 2 1
(b) The matrix  1  2 3 is a 3 × 3 square matrix.
2 4 2
(c) The matrix 1 6 5  2 5 is a 1 × 5 row
matrix.
Copyright © 2007 Pearson Education, Inc.
Slide 7-4
7.4 Determining Equality of Matrices
x y
2
1



and B 
, find the
Example If A 
 p q 
 1 0 
values of x, y, p, and q such that A  B.
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.
Copyright © 2007 Pearson Education, Inc.
Slide 7-5
7.4 Matrix Addition
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.
Example Find each sum.
5

6

4
6




(a)

8 9  8  3
5
8
3
9
1




(b) A  B, if A 
and B 
6 2
4 2 5
Copyright © 2007 Pearson Education, Inc.
Slide 7-6
7.4 Matrix Addition
Analytic Solution
5  (4)  6  6
5

6

4
6





(a)


9  8  3  8  8 9  (3)
8
  1 0
16 6
Graphing Calculator Solution
Copyright © 2007 Pearson Education, Inc.
Slide 7-7
7.4 Matrix Addition
Analytic Solution
5
8
3
9
1



 have
(b) The matrices A 
and B 
6 2
4 2 5
different dimensions , therefore the sum does not exist.
Graphing Calculator Solution The calculator
returns a dimension mismatch error.
Copyright © 2007 Pearson Education, Inc.
Slide 7-8
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
0 0 0 
0 0 0
is the 2 × 3 zero matrix.
The elements of matrix –A are the additive inverses of
the elements of matrix A. For example, if

5
2

1
5

2
1



 , then
A
and  A 
 3 4  6
 3  4 6
A  ( A)   5 2  1   5  2 1  0 0 0.
 3 4  6  3  4 6 0 0 0
Copyright © 2007 Pearson Education, Inc.
Slide 7-9
7.4 Matrix Subtraction
If A and B are matrices with the same dimension,
then A – B = A + (– B).
Example Find the difference of
 5 6   3 2.
 2 4  5  8
Solution
  5 6    3 2    5 6   3  2
 2 4  5  8  2 4  5 8
  2 4 
  3 12
Copyright © 2007 Pearson Education, Inc.
Slide 7-10
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.
2
1

4
5 2
3   1
 
6 4
5 4
3  2
 
6  8
10
2
6  2  1


12
4
5
3

6
• 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.
Copyright © 2007 Pearson Education, Inc.
Slide 7-11
7.4 Matrix Multiplication by a Scalar
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.
Example
Perform the multiplication
52  3.
0 4
Solution
5(2) 5(3) 10  15
2

3



5


0 4 5(0) 5(4)  0 20
Copyright © 2007 Pearson Education, Inc.
Slide 7-12
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.
or
Copyright © 2007 Pearson Education, Inc.
2 7 5
4 6 9
Slide 7-13
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 7-14
7.4 Matrix Multiplication
• 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
195 
 237 . The


product of matrices can be written as
12 2  12
2
7
5

 18  
4 6 9 9 4  12
 
Copyright © 2007 Pearson Education, Inc.
 7  18
 6  18
 5  9   195 .
 9  9  237
Slide 7-15
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.
• The product AB can be found only if the number of
columns of A is the same as the number of rows of
B.
Copyright © 2007 Pearson Education, Inc.
Slide 7-16
7.4 Matrix Multiplication
Example Find the product AB of the two matrices
A   3 4 2 and
 5 0 4
  6 4
B   2 3.
 3  2
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.
Copyright © 2007 Pearson Education, Inc.
Slide 7-17
7.4 Matrix Multiplication
  6 4
 3 4 2  2 3
 5 0 4 3  2


(3)( 6)  4(2)  2(3)  32
  6 4
  3 4 2  2 3
 5 0 4 3  2


(3)( 4)  4(3)  2(2)  4
 6 4
 3 4 2  2 3
 5 0 4 3  2


5(6)  0(2)  4(3)  18
 6 4
  3 4 2  2 3
 5 0 4 3  2


(5)( 4)  0(3)  4(2)  12
 6 4
 3 4 2  2 3   32  4
 5 0 4 3  2  18 12


Copyright © 2007 Pearson Education, Inc.
Slide 7-18
7.4 Matrix Multiplication
Example Use the graphing calculator to find the
product BA of the two matrices from the previous
problem.
A    3 4 2
 5 0 4
  6 4
B   2 3
 3  2
Graphing Calculator Solution
Copyright © 2007 Pearson Education, Inc.
Notice AB  BA.
Slide 7-19
7.4 Applying Matrix Algebra
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.
Colonial
Model X  0
Model Y 10
Model Z 20
Copyright © 2007 Pearson Education, Inc.
Ranch
30
20  A
20
Slide 7-20
7.4 Applying Matrix Algebra
Concrete Lumber
Colonial
Ranch
10
50
2
1
Brick
0
20
Shingles
2  B
2
Cost per Unit
Concrete
Lumber
Brick
Shingles
 20
180  C
 60
 25
(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.
Copyright © 2007 Pearson Education, Inc.
Slide 7-21
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.
 0 30 10 2 0 2

AB  10 20 
20 20 50 1 20 2


Concrete
1500
 1100

1200
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Lumber
Brick
30
40
60
600
400
400
Shingles
60
60

80 
Model X
Model Y
Model Z
Slide 7-22
7.4 Applying Matrix Algebra
Multiplying the total amount of materials matrix AB
and the cost matrix C gives the total cost of materials.
20

1500
30
600
60

 180
( AB)C  1100 40 400 60
60

1200
60
400
80

  25
 
Cost
72,900 Model X
 54,700 Model Y


60,800 Model Z
Copyright © 2007 Pearson Education, Inc.
Slide 7-23
7.4 Applying Matrix Algebra
(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.
D  3800 130 1400 200
(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.
Copyright © 2007 Pearson Education, Inc.
Slide 7-24