copyright © 2013, 2009, 2005 pearson education, inc. 1 5 systems and matrices copyright © 2013,...

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 1

5Systems and Matrices

Copyright © 2013, 2009, 2005 Pearson Education, Inc.


5.7 Properties of Matrices•Basic Definitions•Adding Matrices•Special Matrices•Subtracting Matrices•Multiplying Matrices•Applying Matrix Algebra


Basic Definitions

It is customary to use capital letters to name matrices and to use subscript notation to name elements of a matrix, as in the following matrix A.

11 12 13 1

21 22 23 2

31 32 33 3

1 2 3

n

n

n

m m m mn

a a a a

a a a a

A a a a a

a a a a


Basic Definitions

Certain matrices have special names. Ann x n matrix is a square matrix 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.


Basic Definitions

Two matrices are equal if they are the same size and if corresponding elements,position by position, are equal. Using this definition, the matrices

2 1

3 5

and1 2

5 3

are not equal (even though they contain the same elements and have the same dimension), since the corresponding elements differ.


Example 1 FINDING VALUES TO MAKE TWO MATRICES EQUAL

Find the values of the variables for which each statement is true, if possible.

Solution

(a)2 1

1 0

x y

p q

From the definition of equality, the only way that the statement can be true is if 2 = x, 1 = y, p = – 1 and q = 0.


Example 1 FINDING VALUES TO MAKE TWO MATRICES EQUAL

Find the values of the variables for which each statement is true, if possible.

Solution

(b)1

4

0

x

y

This statement can never be true since the two matrices have different dimensions. (One is 2 1 and the other is 3 1. )


Addition of Matrices

To add two matrices of the same dimension, add corresponding elements. Only matrices of the same size can be added.


Example 2 ADDING MATRICES

Find each sum, if possible.

Solution

(a) 5 6 4 6

8 9 8 3

5 6 6

8 9 8 3

4

( ) 6 6

8 8 9 (

4

3)

5

6

1 0

16




Solution

(b)

2 6

5 3

8 12

2 6 4

5 3 8

8 12 20




Solution

(c)5 8

6 2

A + B, if A = and B = 3 9 1

4 2 5

The matrices5 8

6 2A

and

3 9 1

4 2 5B

have different dimension so A and B cannot be added. The sum A + B does not exist.


Special Matrices

A matrix containing only zero elements is called a zero matrix. A zero matrix can be written with any dimension.

1 3 zero matrix 2 3 zero matrix

0 0 0O0 0 0

0 0 0

O


Special Matrices

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

( ) 0 and 0.a a a a


Special Matrices

Given matrix A, there is a matrix – A such that A + (– A) = . The matrix – A has as elements the additive inverses of the elements of A. (Remember, each element of A is a real number and therefore has an additive inverse.) For example, if

5 2 1 5 2 1, then .

3 4 6 3 4 6A A


Special Matrices

5 2 1 5 2 1( )

3 4 6 3 4 6

0 0 0

0 0 0

A A

O

CHECK Test that A + (– A) equals the zero matrix, O.

Matrix – A is called the additive inverse, or negative, of matrix A. Every matrix has an additive inverse.


Subtraction of Matrices

If A and B are two matrices of the same dimension, then

A – B = A + (– B).


Example 3 SUBTRACTING MATRICES

(a)5 6 3 2

2 4 5 8

Find each difference, if possible.

Solution

6 2 ( ) 6 2

2 4 5 8 2 5 4 ( 8)

5 3 5 3

4

3 12

2



(b) 8 6 4 3 5 8


Solution

8 6 4 3 5 8 5 1 4



(c)


Solution

The matrices2 5

0 1A

and

3

5B

2 5

0 1

A – B, if A = and B = 3

5

have different dimensions and cannot be subtracted, so the difference A – B does not exist.


Example 4 MULTIPLYING MATRICES BY SCALARS

Find each product.

Solution

(a)2 3

50 4

2 3 (2) ( 3)5 55

50 4 (0 5) (4)

Multiply each element of the matrix by the scalar 5.

10 15

0 20


Example 4 MULTIPLYING MATRICES BY SCALARS

Find each product.

Solution

(b)20 363

4 12 16

3 34 4

(20) (36)20 3634 12 16

(13 34 4

2) ( 16)

Multiply each element of the matrix by the scalar 3

.4

15 27

9 12


Properties of Scalar Multiplication

If A and B are matrices of the same dimension, and let c and d be scalars. Then these properties hold.

( ) ( ) ( )

( ) ( ) ( )

c d A cA dA cA d cd A

c A B cA cB cd A c dA


Matrix Multiplication

We have seen how to multiply a real number (scalar) and a matrix. To find the product of two matrices, such as

6 43 4 2

and 2 3 ,5 0 4

3 2

A B

first locate row 1 of A and column 1 of B, which are shown shaded below. 6 4

3 4 2and 2 3

5 0 43 2

A B



Multiply corresponding elements, and find the sum of the products.

6 2( ) ( )3 4 2 3( ) 32 This result is the element for row 1, column 1 of the product matrix. Now use row 1 of A and column 2 of B to determine the element in row 1,column 2 of the product matrix.

6 43 4 2

2 3 ( ) ( ) ( ) 45 0 4

3 2

24 4 23 3


6 43 4 2

2 3 ( ) ( ) ( ) 125 0 4

3

4 3 20

2

5 4


Next, use row 2 of A and column 1 of B. This will give the row 2, column 1 entry of the product matrix.

Finally, use row 2 of A and column 2 of B to find the entry for row 2, column 2 of the product matrix.

6 43 4 2

2 3 ( ) ( ) ( ) 186 2 35

40 4

3 2

5 0



6 43 4 2 32 4

2 35 0 4 18 12

3 2

The product matrix can now be written.


Matrix MultiplicationTo find 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. (Note the shaded areas in the matrices below.) The sum of these products will give the row i, column j element of AB.



The number of columns of an m n matrix A is the same as the number of rows of an n p matrix B (i.e., both n). The element cij of the product matrix C = AB is found as follows:

1 1 2 2ij i j i j in njc a b a b a b

Matrix AB will be an m p matrix.


Example 5 DECIDING WHETHER TWO MATRICES CAN BE MULTIPLIED

Suppose A is a 3 2 matrix, while B is a 2 4 matrix.

Solution

(a) 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. (Both are 2.)

Matrix A Matrix B

3 2 2 4

must match

dimension of AB 3 4




Solution

(b) If AB can be calculated, what is its dimension?

As indicated in the diagram below, the product AB is a 3 x 4 matrix.

Matrix A Matrix B

3 2 2 4

must match

dimension of AB 3 4




Solution

(c)

Matrix B Matrix A

2 4 3 2

different

Can BA be calculated?

The diagram below shows that BA cannot be calculated.




Solution

(d) If BA can be calculated, what is its dimension?

The product BA cannot be calculated, because B has 4 columns and A has only 3 rows.


Example 6 MULTIPLYING MATRICES

Let1 3 1 0 1 2

and .7 2 3 1 4 1

A B

Find each product, if possible.

Solution

(a) AB

First decide whether AB can be found. Since A is 2 2 and B is 2 4, the product can be found and will be a 2 4 matrix.


Example 6

1 0 1 2

3 1 42

1

7

3

1AB

(1) ( )3 (0) ( )1 ( 1) ( )4 (2) ( )( 1)

(1) (3) (0) (1)7 2 7 2 7( 1) (4) (2)

1 3 1 3 1

( 1)2

3 1 3

2 7

Use the definition of matrix multiplication

8 3 13 5

13 2 1 12

Perform the operations.

MULTIPLYING MATRICES


Example 6

Let1 3 1 0 1 2

and .7 2 3 1 4 1

A B

Find each product, if possible.

Solution

(b) BA

Since B is a 2 4 matrix, and A is a 2 2 matrix, the number of columns of B (here 4) does not equal the number of rows of A (here 2). Therefore, the product BA cannot be found.

MULTIPLYING MATRICES


Example 7 MULTIPLYING SQUARE MATRICES IN DIFFERENT ORDERS

Let1 3 2 7

and .2 5 0 2

A B

Find each product.

Solution

(a) AB

1 3 2 7

2 5 0 2AB

1( 2) 3(0) 1(7) 3(2) 2 13

2( 2) 5(0) 2(7) 5(2) 4 4


Example 7

Let1 3 2 7

and .2 5 0 2

A B

Find each product.

Solution

(b) BA

2 7 1 3

0 2 2 5BA

( 2)1 7( 2) 2(3) 7(5) 16 29

0(1) 2( 2) 0(3) 2(5) 4 10

Note that AB ≠ BA.

MULTIPLYING SQUARE MATRICES IN DIFFERENT ORDERS


In general, if A and B are matrices, then AB ≠ BA . Matrix multiplication is not commutative.


Properties of Matrix MultiplicationIf A, B, and C are matrices such that all the following products and sums exist, then these properties hold.( ) ( ), ( ) , ( )AB C A BC A B C AB AC B C A BA CA


Example 8 USING MATRIX MULTIPLICATION TO MODEL PLANS FOR A SUBDIVISION

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 to build for a new 100-home subdivision. The amounts for each of the main materials used depend on the style of the house. These amounts are shown in matrix Q, while matrix R gives the cost in dollars for each kind of material. Concrete is measured here in cubic yards, lumber in 1000 board feet, brick in 1000s, and shingles in 100 square feet.



0 30

10 20

20 20

P

Colonial Ranch

Model A

Model B

Model C



10 2 0 2

50 1 20 2Q

Colonial

Ranch

Concrete Lumber Brick Shingles



Cost per Unit

Concrete

Lumber

Brick

Shingles

20

180

60

25

R



Solution

(a) What is the total cost of materials for all houses of each model?

To find the materials cost for each model, first find matrix PQ, which will show the total amount of each material needed for all houses of each model.

0 3010 2 0 2

10 2050 1 20 2

20 20

PQ



0 3010 2 0 2

10 2050 1 20 2

20 20

PQ

1500 30 600 60

1100 40 400 60

1200 60 400 80

Concrete Lumber Brick Shingles

Model A

Model B

Model C



201500 30 600 60

180( ) 1100 40 400 60

601200 60 400 80

25

PQ R

Multiplying PQ and the cost matrix R gives the total cost of materials for each model.

72,900

54,700

60,800

Cost

Model A

Model B

Model C



Solution

(b) How much of each of the four kinds of material must be ordered?

To find how much of each kind of material to order, refer to the columns of matrix PQ. The sums of the elements of the columns will give a matrix whose elements represent the total amounts of all materials needed for the subdivision. Call this matrix T, and write it as a row matrix.

3800 130 1400 200T



Solution

(c)

The total cost of all the materials is given by the product of matrix T, the total amounts matrix, and matrix R, the cost matrix. To multiply these matrices and get a 1 1 matrix, representing the total cost, requires multiplying a 1 4 matrix and a 4 1 matrix. This is why in part (b) a row matrix was written rather than a column matrix.

What is the total cost of the materials?



20

1803800 130 1400 200 188,400 .

60

25

T

The total cost of materials is $188,400. This total may also be found by summing the elements of the column matrix (PQ)R.

The total materials cost is given by TR, so

copyright © 2013, 2009, 2005 pearson education, inc. 1 5 systems and matrices copyright © 2013,...

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