chapter 5.7
DESCRIPTION
Chapter 5.7. Properties of Matrices. Basic Definitions. It is necessary to use capital letters to name matrices. Also, subscript notation is often used to name element of a matrix, as in the following Matrix A. With this notation, the first row, first column element is - PowerPoint PPT PresentationTRANSCRIPT
Chapter 5.7
Properties of Matrices
Basic Definitions
It is necessary to use capital letters to name matrices.
Also, subscript notation is often used to name element of a matrix, as in the following Matrix A.
With this notation, the first row, first column element is a11 (read “a-sub-one-one”);
The second row, third column element is a23
and in general the i th row, j th column element is aij
Certain matrices have special names:
because the number of rows is equal to the number of columns.
A matrix with just one row is a row matrix, and a matrix with just one column is a column matrix.
matrix square a ismatrix n nAn
Two matrices are equal if they are the same size and if the corresponding element, position by position are equal.
Using this definition, the matrices
are not equal (even though they contain the same elements and are the same size), since the corresponding elements differ.
35
21 and
53
12
Find the values of the variables for which each statement is true, if possible.
01
yx
qp
12
Find the values of the variables for which each statement is true, if possible.
0
4
1
y
x
Adding MatricesAddition of matrices is defined as follows.
It can be shown that matrix addition satisfies the commutative, associative, closure, identity, and inverse properties.
Find the sum if possible
38
64
98
65
Find the sum if possible
12
3
6
8
5
2
Find the sum if possible
B, A
26
85A if
524
193 B and
Special MatricesA matrix containing only zero elements is called a zero matrix. A zero matrix can be written with any size.
By the additive inverse property, each real number has an additive inverse: if a is a real number, then there is a real number –a such that
a + (-a) = 0 and –a + a = 0
Given matrix A, there is a matrix –A such that
A + -A = 0
The matrix –A has as elements the additive inverses of the elements of A.
For example, if
643
125A
643
125A -then
For example, if
643
125-A A
643
125
000
000
Matrix –A is called the additive inverse, or negative, of matrix A. Every matrix has an additive inverse
Subtracting MatricesThe real number b is subtracted from the real number a, written a – b, by adding a and the additive inverse of b.That is,
a – b = a + (-b)
In practice, the difference of two matrices of the same size is found by subtracting corresponding elements.
Find the difference if possible
85
23
42
65
Find the difference if possible
468 853
Find the difference if possible
B, A
10
52A if
5
3 B and
Multiplying MatricesIn work with matrices, a real number is called a scalar to distinguish it from a matix. The product of a scalar K and a matrix X is the matrix kX, each of whose elements is k times the corresponding element of X.
Find the product
40
325
)4(5)0(5
)3(5)2(5
200
1510
Find the product
1612
3620
4
3
129
2715
We have seen how to multiply a real number (scalar) and a matrix. To find the product of two matrices, such as
405
243A
23
32
46-
B and
first locate row 1 of A
405
243A
23
32
46-
B and
first locate row 1 of Aand column 1 of B,
405
243A
23
32
46-
B and
Multiply corresponding elements, and find the sum of the products.
405
243A
23
32
46-
B and
(-3)(-6) + (4) (2) + (2)(3) 18+ 8 + 6 = 32
The result is the first element for row 1, column 1 in the product matrix.
405
243A
23
32
46-
B and
(-3)(-6) + (4) (2) + (2)(3) 18+ 8 + 6 = 32
Now use row 1 of A and column 2 of B to determine the element in row 1 and column 2 of the product matrix.
405
243A
23
32
46-
B and
Now use row 1 of A and column 2 of B to determine the element in row 1 and column 2 of the product matrix.
405
243A
23
32
46-
B and
(-3)(4) + (4) (3) + (2)(-2) -12 + 12 + -4 = -4
Now use row 2 of A and column 1 of B to determine the element in row 2 and column 1 of the product matrix.
405
243A
23
32
46-
B and
Now use row 2 of A and column 1 of B to determine the element in row 2 and column 1 of the product matrix.
405
243A
23
32
46-
B and
(5)(-6) + (0) (2) + (4)(3) -30 + 0 + 12 = -18
Now use row 2 of A and column 2 of B to determine the element in row 2 and column 2 of the product matrix.
405
243A
23
32
46-
B and
Now use row 2 of A and column 2 of B to determine the element in row 2 and column 2 of the product matrix.
405
243A
23
32
46-
B and
(5)(4) + (0) (3) + (4)(-2) 20 + 0 + -8 = 12
The product matrix can now be written.
405
243
matrix. 22 a ismatrix 23 a and
matrix 32 a ofproduct thehere see As
1218
432
23
32
46-
Can the product AB be calculated?
The following diagram shows that AB can be calculated, because the number of columns of A is equal to the number of rows of B.
23
AMatrix
matrix. 42 a is B while
matrix, 23 a isA Suppos
42
BMatrix
same size
If AB can be calculated, how big is it?
23
AMatrix
matrix. 42 a is B while
matrix, 23 a isA Suppos
42
BMatrix
same size
size of AB = 3 x 4
If BA can be calculated?
BA cannot be calculated?
23
AMatrix
matrix. 42 a is B while
matrix, 23 a isA Suppos
different size
42
BMatrix
Find the product
27
31
1413
2101
)3)(3()1)(1(
Find the product
27
31
1413
2101
8
Find the product
27
31
1413
2101
)1)(3()0)(1(8
Find the product
27
31
1413
2101
38
Find the product
27
31
1413
2101
38
Find the product
27
31
1413
2101
)4)(3()1)(1(38
Find the product
27
31
1413
2101
1338
Find the product
27
31
1413
2101
1338
Find the product
27
31
1413
2101
)1)(3()2)(1(1338
Find the product
27
31
1413
2101
51338
Find the product
27
31
1413
2101
)3)(2()1)(7(
51338
Find the product
27
31
1413
2101
-8 -3 -13 513
Find the product
27
31
1413
2101
)1)(2()0)(7(13
51338
Find the product
27
31
1413
2101
-8 -3 -13 513 2
Find the product
27
31
1413
2101
)4)(2()1)(7(213
51338
Find the product
27
31
1413
2101
-8 -3 -13 513 2 1
Find the product
27
31
1413
2101
)1)(2()2)(7(1213
51338
Find the product
27
31
1413
2101
-8 -3 -13 513 2 1 12
Find the product
27
31
1413
2101
Since the first matrix is a 2 x 4and the second matrix is a 2 x 2the product can not be found.
Find the product
20
72
52
31
Find the product
20
72
52
31
)0)(3()2)(1(
Find the product
20
72
52
31
2
Find the product
20
72
52
31
)2)(3()7)(1(2
Find the product
20
72
52
31
132
Find the product
20
72
52
31
)0)(5()2)(2(
132
Find the product
20
72
52
31
4
132
Find the product
20
72
52
31
)2)(5()7)(2(4
132
Find the product
20
72
52
31
44
132
Find the product
52
31
20
72
Find the product
)2)(7()1)(2(
52
31
20
72
Find the product
16
52
31
20
72
Find the product
)5)(7()3)(2(16
52
31
20
72
Find the product
2916
52
31
20
72
Find the product
)2)(2()1)(0(
2916
52
31
20
72
Find the product
4
2916
52
31
20
72
Find the product
)5)(2()3)(0(4
2916
52
31
20
72
Find the product
104
2916
52
31
20
72
The products are not commutative.
104
2916
52
31
20
72
A B B A
A contractor builds three kinds of houses, models A, B, and C, with a choice of two styles, colonial or ranch.
Matrix P shows the number of each kind of house the contractor is planning for a new 100-home subdivision.
The amounts for each of the main materials used depend on the style of the house.
20
20
30
20
10
0
C Model
B Model
A ModelColonial Ranch
The amounts are shown in the matrix.
220150
20210
Ranch
ColonialConcrete Lumber Brick Shingles
Concrete is measured here in cubic yards, lumber in 1000 board feet, brick in 1000s, and shingles in 100 square feet.
220150
20210
Ranch
ColonialConcrete Lumber Brick Shingles
25
60
100
20
Shingles
Brick
Lumber
Concrete
Cost per Unit
What is the total cost of materials for all houses of each model?
20
20
30
20
10
0
220150
20210
80
60
60
400
400
600
60
40
30
1200
1100
1500
C Model
B Model
A ModelConcrete Lumber Brick Shingles
What is the total cost of materials for all houses of each model?
80
60
60
400
400
600
60
40
30
1200
1100
1500
25
60
100
20
C Model
B Model
A Model
60800
54700
72900
Cost
How much of each of the four kinds of material must be ordered?
80
60
60
400
400
600
60
40
30
1200
1100
1500
20014001303800
What is the total cost of the materials?
25
60
80
20
20014001303800 188400