Post on 04-Jan-2016
Slide 9.2- 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
OBJECTIVES
Matrix Algebra
Learn the definition of equality of matrices.Learn the definition of matrix addition and scalar multiplication.Learn the definition of matrix multiplication.Learn an application of matrix multiplication to computer graphics.
SECTION 9.2
1
2
3
4
Slide 9.2- 3 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EQUALITY OF MATRICES
Two matrices A = [aij] and B = [bij] are said to be equal, written A = B, if
1. A and B have the same order m n (that is, A and B have the same number m of rows and the same number n of columns.)
2. aij = bij for all i and j. (The (i, j)th entry of A is equal to the corresponding (i, j)th entry of B.)
Slide 9.2- 4 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
MATRIX ADDITION
A B aij bij ,
If A =[aij] and B = [bij] are two m n matrices, their sum A + B is the matrix defined by the m n matrix defined by
for all i and j.
Slide 9.2- 5 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
SCALAR MULTIPLICATION
cA caij .
If A = [aij] be an m n matrices, and let c be a real number. Then the scalar product of A and c is denoted by cA and is defined by
Slide 9.2- 6 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
MATRIX SUBTRACTION
A B A 1 B.
If A and B are two m n matrices, their difference is defined by
Subtraction A – B is performed by subtracting the corresponding entries of B from A.
Slide 9.2- 7 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
MATRIX ADDITION AND SCALAR MULTIPLICATION PROPERTIES
Let A, B, and C be m n matrices and c and d be scalars.
1. A + B = B + A2. A + (B + C) = (A + B) + C3. A + 0 = 0 + A = A4. A + (–A) = (–A) + A = 05. (cd)A = c(dA)6. 1A = A7. c(A + B) = cA + cB8. (c + d)A = cA + dA
Slide 9.2- 8 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 4 Solving a Matrix Equation
Solve the matrix equation 3A + 2X = 4B for X,
A 2 0
4 6
and B
1 3
5 2
.
Solution
3A 2X 4B
2X 4B 3A
X 1
24B 3A
where
Substitute for A and B.
Slide 9.2- 9 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 4 Solving a Matrix Equation
Solution continued
1
24
1 3
5 2
3
2 0
4 6
1
2
4 12
20 8
6 0
12 18
1
2
2 12
8 10
1 6
4 5
You should check that the matrix X satisfies the given matrix equation.
Slide 9.2- 10 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
RULE FOR DEFINING THE PRODUCT ABIn order to define the product AB of two matrices A and B, the number of columns of A must be equal to the number of rows of B. If A is an m p matrix and B is a p n matrix, then the product AB is an m n matrix.
Slide 9.2- 11 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
PRODUCT OF 1 x n AND n x 1 MATRICES
Suppose
is a 1 n row matrix and
is a n 1 column matrix. We define the product
A a1 a2 a3 L an
B
b1
b2
M
bn
AB a1b1 a2b2 ... anbn .
Slide 9.2- 12 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
MATRIX MULTIPLICATION
Let A = [aij] be an m p matrix and B = [bij] be a p n matrix. Then the product AB is the m n matrix C = [cij], where the entry cij of C is obtained by multiplying the ith row (matrix) of A by the jth column (matrix) of B. The definition of the product AB says that
cij ai1b1 j ai2b2 j ... aipbpj .
Slide 9.2- 13 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 8 Finding the Product of Two Matrices
Find the products AB and BA, where
A 1 2
1 3
and B
3 2 1
1 2 3
.
Solution
Since A is of order 2 2 and the order of B is 2 3, the product AB is defined and has order 2 3.If AB = C = [cij], each entry cij of C is obtained by multiplying the ith row of A by the jth column of B.
Slide 9.2- 14 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 8 Finding the Product of Two Matrices
1 2
* *
3 * *
1 * *
1 3 2 1 * *
* * *
Solution continued
C11 is obtained by multiplying the first row of A by the first column of B.
Thus AB 1 2
1 3
3 2 1
1 2 3
AB 1 3 2 1 1 2 2 2 1 1 2 3 1 3 3 1 1 2 3 2 1 1 3 3
Slide 9.2- 15 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 8 Finding the Product of Two Matrices
Solution continued
The product BA is not defined, because B is of order 2 3 and A is of order 2 2; that is, the number of columns of B is not the same as the number of rows of A.
AB 5 2 7
0 8 8
Slide 9.2- 16 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
PROPERTIES OF MATRIX MULTIPLICATION
Let A, B, and C be matrices and let c be a scalar. Assume that each product and sum is defined. Then
1. (AB)C = A(BC)
2. (i) A(B + C) = AB + AC
(ii) (A + B)C = AC + BC
3. c(AB) = (cA)B = A(cB)
Slide 9.2- 17 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 10 Transforming a Letter
The capital letter L in the figure is determined by six points (or vertices) P1 – P6 . The coordinates of the six points can be stored in a data matrix D, together with instructions stating that these vertices are connected by lines.
Slide 9.2- 18 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 10 Transforming a Letter
P1 P2 P3 P4 P5 P6
0 4 4 1 1 0
0 0 1 1 6 6
D
Given
x-coordinatey-coordinate
Vertex
compute AD. Graph theA 1 0.25
0 1
,
figure corresponding to the matrix AD by connecting the images of the vertices with the
Slide 9.2- 19 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 10 Transforming a Letter
Solution
The columns of the product matrix AD represent the images of the vertices of the letter L.
appropriate lines. This produces the complete transformed image of the original figure.
AD 1 0.25
0 1
0 4 4 1 1 0
0 0 1 1 6 6
10 0.250 14 0.250 14 0.25100 10 04 10 04 11
…
Slide 9.2- 20 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 10 Transforming a Letter
Solution continued
The transformed vertices are plotted on the next slide, along with connecting line segments that correspond to those in the original figure.
11 0.251 11 0.256 10 0.2560111 0116 00 16
…
P1 P2 P3 P4 P5 P6
AD 0 4 4.25 1.25 2.5 1.5
0 0 1 1 6 6
Slide 9.2- 21 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 10 Transforming a Letter
Solution continued
The figure is a result of the transformation represented by the product matrix AD.