slide 7- 1. chapter 7 systems and matrices 7.1 solving systems of two equations

Slide 7- 1

Chapter 7

Systems and Matrices

7.1

Solving Systems of Two Equations

Slide 7- 4

Quick Review

3

2

1. Solve for in terms of . 2 3 6

Solve the equation algebraically.

2. 9

3. 5 6

4. Write the equation of the line that contains the point

22

3

0,

(1,1) and

is perpendicular to

3

6,1

y y xx x y

x x

x x

the line 2 3 6.

5. Write an equation equivalent to 5 with coefficient of

equal to 2.

31 ( 1)

2

2 2 10

y xx

x

x

y

y

y x

Slide 7- 5

What you’ll learn about

The Method of Substitution Solving Systems Graphically The Method of Elimination Applications

… and whyMany applications in business and science can be modeled using systems of equations.

Slide 7- 6

Solution of a System

A solution of a system of two equations in two

variables is an ordered pair of real numbers that

is a solution of each equation.

Slide 7- 7

Example Using the Substitution Method

Solve the system using the substitution method.

2 10

6 4 1

x y

x y

Slide 7- 8

Example Using the Substitution Method

Solve the system using the substitution method.

2 10

6 4 1

x y

x y

Solve the first equation for .

2 10

2 10

Substitute the expression for into the second equation:

6 4(2 10) 1

41

1441 29

2 10 2 1014 7

41 29The solution is the ordered pair , .

14 7

y

x y

y x

y

x x

x

y x

Slide 7- 9

Example Solving a Nonlinear System Algebraically

2

Solve the system algebraically.

6

8

y x x

y x

2

2

Substitute the values of from the first equation into

the second equation:

8 6

0 2

0, 2.

If 0, then 0. If 2, then 16.

The system of equations has two solutions: (0,0) and (2,16).

y

x x x

x x

x x

x y x y

Slide 7- 10

Example Using the Elimination Method

Solve the system using the elimination method.

5 3 21

3 2 5

x y

x y

Multiply the first equation by 2 and the second equation by 3 to obtain:

10 6 42

9 6 15

Add the two equations to eliminate the variable .

19 57

3

Substitue 3 into either of the two original equati

x y

x y

y

x

x

x

ons:

5(3) 3 21

2

The solution of the original system is (3, 2).

y

y

Slide 7- 11

Example Finding No Solution

Solve the system:

3 2 5

6 4 10

x y

x y

Multiply the first equation by 2.

6 4 10

6 4 10

Add the equations:

0 20

The last equation is true for no values of and .

The equation has no solution.

x y

x y

x y

Slide 7- 12

Example Finding Infinitely Many Solutions

Solve the system.

3 6 10

9 18 30

x y

x y

Multiply the first equation by 3.

9 18 30

9 18 30

Add the two equations.

0=0

The last equation is true for all values of and .

The system has infinitely many solutions.

x y

x y

x y

7.2

Matrix Algebra

Slide 7- 14

Quick Review

The points (a) (1, 3) and (b) ( , ) are reflected across the given line.

Find the coordinates of the r

(a) (1,3) (b) ( ,

eflected points.

1. The -axis

2. The line

3.

)

(a) ( 3,

Th

1) (b) ( , )

e

x yx

x x

x

y y

y

line

Expand the expression,

4.

(a) ( 3, 1) (b) ( , )

sin cos sin cossin( )

5. cos( cos cos sin s n) i

y x

x y y x

x y x

y

y

x

x y

x y

Slide 7- 15

What you’ll learn about

Matrices Matrix Addition and Subtraction Matrix Multiplication Identity and Inverse Matrices Determinant of a Square Matrix Applications

… and whyMatrix algebra provides a powerful technique to manipulate large data sets and solve the related problems that are modeled by the matrices.

Slide 7- 16

Matrix

11 12 1

21 22 2

1 2

Let and be positive integers. An (read " by matrix")

is a rectangular array of rows and columns of real numbers.

We also use the shorth

n

n

m m mn

m n m n

m n

a a a

a a a

a a a

matrix

m × n

and notation for this matrix.ij

a

Slide 7- 17

Matrix Vocabulary

Each element, or entry, aij, of the matrix uses

double subscript notation. The row subscript is

the first subscript i, and the column subscript is

j. The element aij is the ith row and the jth

column. In general, the order of an m × n

matrix is m×n.

Slide 7- 18

Example Determining the Order of a Matrix

What is the order of the following matrix?

1 4 5

3 5 6

The matrix has 2 rows and 3 columns so it has order 2 3.

Slide 7- 19

Matrix Addition and Matrix Subtraction

Let and be matrices of order .

1. The is the matrix .

2. The is the matrix .

ij ij

ij ij

ij ij

A a B b m n

m n A B a b

m n A B a b

sum +

difference

A B

A - B

Suppose that

5 1 3 2

3 7 and 12 4

0 8 6 9

A B

A B

5 3 1 2

3 12 7 4

0 6 8 9

( )

( )

8 3

9 3

6 17

Adding Matrices

Slide 7- 21

Example Matrix Addition

1 2 3 2 3 4

4 5 6 5 6 7

3 5 7

9 11 13

Slide 7- 22

Example Using Scalar Multiplication

1 2 3

34 5 6

3 6 9

12 15 18

Slide 7- 23

The Zero Matrix

The matrix 0 [0] consisting entirely of zeros is the .m n zero matrix

Slide 7- 24

Additive Inverse

Let be any matrix.

The matrix consisting of the additive inverses

of the entries of is the because

0.

ij

ij

A a m n

m n B a

A

A B

additive inverse of A

Slide 7- 25

Matrix Multiplication

1 1 2 2

Let be any matrix and be any matrix.

The product is the matrix where

+ ... .

ij ij

ij

ij i j i j ir rj

A a m r B b r n

AB c m n

c a b a b a b

Let A denote an m by r matrix and let B denote an r by n matrix. The product AB is defined as the m by n matrix whose entry in row i, column j is the product of the ith row of A and the jth column of B.

Multiplying Matrices

Note: If we multiply a matrix by a constant, this is equivalent to multiplying each term in the matrix by the constant.

Slide 7- 27

Example Matrix Multiplication

Find the product if possible.

1 01 2 3

and 2 1 0 1 1

0 1

AB

A B

Slide 7- 28

Example Matrix Multiplication

11

12

The number of columns of is 3 and the number of rows of is 3,

so the product is defined. The product is a 2 2 matrix where

1

1 2 3 2 1 1 2 2 3 0 5,

0

0

1 2 3 1 1 0 2

1

ij

A B

AB c

c

c

21

22

1 3 1 1,

1

0 1 1 2 0 1 1 2 1 0 2,

0

0

0 1 1 1 0 0 1 1 1 1 2.

1

5 1Thus .

2 2

c

c

AB

7 11 4 6

2 0 and 3 8 5

3 5

A B

7 1 1 3 7 4 1 8 7 6 1 5

2 1 0 3 2 4 0 8 2 6 0 5

3 1 5 3 3 4 5 8 3 6 5 5

10 20 37

2 8 12

18 28 7

Example: Find the product AB

An by square matrix where

are 1 and are 0 is called the identity

matrix

n n a i n

a i j

I

I I

ii

ij

n

,

,

1

1 0

0 1

1 0 0

0 1 0

0 0 12 3

The Identity Matrix

Slide 7- 31

Identity Matrix

The matrix with 1's on the main diagonal and 0's elsewhere

is the .

1 0 0 0

0 1 0 0

0 0 1 0

0

0 0 0 0 1

n

n

n n I

I

identity matrix of order

n n

Slide 7- 32

Inverse of a Square Matrix

-1

Let be an matrix. If there is a matrix such that

, then is the of .

We write .

ij

n

A a n n B

AB BA I B A

B A

inverse

Slide 7- 33

Inverse of a 2 × 2 Matrix

1

1If 0, then .

a b d bad bc

c d c aad bc

Slide 7- 34

Determinant of a Square Matrix

Let be a matrix of order ( 2). The determinant

of , denoted by det or | | , is the sum of the entries in any row

or any column multiplied by their respective cofactors. For

example, expa

ijA a n n n

A A A

1 1 2 2

nding by the ith row gives

det | | ... .i i i i in in

A A a A a A a A

Slide 7- 35

Inverses of n × n Matrices

An n × n matrix A has an inverse if and only if

det A ≠ 0.

Slide 7- 36

Example Finding Inverse Matrices

1 3Find the inverse matrix if possible.

2 5A

1

Since det 1 5 2 3 1 0, must have an inverse.

5 31 1Use the formula

2 11

5 3 .

2 1

A ad bc A

d bA

c aad bc

Slide 7- 37

Properties of MatricesLet A, B, and C be matrices whose orders are such that the following sums, differences, and products are defined.1. Community propertyAddition: A + B = B + AMultiplication: Does not hold in general2. Associative propertyAddition: (A + B) + C = A + (B + C)Multiplication: (AB)C = A(BC)3. Identity propertyAddition: A + 0 = AMultiplication: A·In = In·A = A4. Inverse propertyAddition: A + (-A) = 0Multiplication: AA-1 = A-1A = In |A|≠05. Distributive propertyMultiplication over addition: A(B + C) = AB + AC (A + B)C = AC + BCMultiplication over subtraction: A(B - C) = AB - AC (A - B)C = AC - BC

Slide 7- 38

Matrices and Transformations

P(x,y)

P’(x’,y’)

sin

cos

y

x

)sin('

)cos('

y

x

cossincossin'

sinsincoscos'

y

x

Slide 7- 39

Matrices and Transformations

Rotation through an angle

The rotation through an angle maps each point P(x,y) in the rectangular coordinate planeto the point P’(x’,y’). where

cossin'

sincos'

yxy

yxx

Slide 7- 40

Matrices and Transformations

cossin'

sincos'

yxy

yxx

or

y

x

y

x

cossin

sincos

'

'

Slide 7- 41

Matrices and Transformations

Find the rotation matrix about theorigin whose angle is /3.

3cos

3sin

3sin

3cos

2

1

2

3

2

3

2

1

Slide 7- 42

Matrices and Transformations

Where does the point (4,-2) move?

2

1

2

3

2

3

2

1

2

4

132

32

7.3

Multivariate Linear Systems and Row Operations

Slide 7- 44

Quick Review

3

1. Find the amount of pure acid in 45L of a 58%

acid solution.

2. Find the amount of water in 30 L of a 28%

acid solution.

3. Is the point (0, 1) on the graph of the function

26.1 L

21.6 L

( ) 4 1f x x x

?

4. Solve for in terms of the other variables:

2

2 15. Find the inverse of th

yes

2

1e matrix .

/2 1/ 6

0 1 30 3 /

x z

x

x wz w

Slide 7- 45

What you’ll learn about

Triangular Forms for Linear Systems Gaussian Elimination Elementary Row Operations and Row Echelon Form Reduced Row Echelon Form Solving Systems with Inverse Matrices Applications

… and whyMany applications in business and science are modeled by systems of linear equations in three or more variables.

Slide 7- 46

Equivalent Systems of Linear Equations

The following operations produce an equivalent

system of linear equations.

1. Interchange any two equations of the system.

2. Multiply (or divide) one of the equations by any nonzero real number.

3. Add a multiple of one equation to any other equation in the system.

Slide 7- 47

Row Echelon Form of a Matrix

A matrix is in row echelon form if the following

conditions are satisfied.

1. Rows consisting entirely of 0’s (if there are any) occur at the bottom of the matrix.

2. The first entry in any row with nonzero entries is 1.

3. The column subscript of the leading 1 entries increases as the row subscript increases.

Slide 7- 48

Elementary Row Operations on a Matrix

A combination of the following operations will

transform a matrix to row echelon form.

1. Interchange any two rows.

2. Multiply all elements of a row by a nonzero real number.

3. Add a multiple of one row to any other row.

Slide 7- 49

Example Finding a Row Echelon Form

Solve the system:

2 3 1

5 3 10

3 6 5

x y z

x y z

x y z

Slide 7- 50

Example Finding a Row Echelon Form

21 1 2

Apply elementary row operations to find a row echelon form of the augmented matrix.

2 3 1 1 1 5 3 10 1 5 3 10

1 5 3 10 2 3 1 1 2 0 13 5 21 3

3 1 6 5 3 1 6 5 3 1 6 5

R R R R

���������������������������� 1 3

2 2 3 3

1 5 3 10 1 5 3 101 5 3 10

1 5 21 5 21 130 13 5 21 0 1 14 0 1

13 13 13 13 13 310 14 3 25

0 14 3 25 31 310 0

13 13

1 5 3 10

5 210 1

13 130 0 1 1

R

R R R R

��������������

������������������������������������������

Convert the matrix to equations and solve by substitution.

1; 5 /13 21/13 so 2; 10 3 10 so 3.

The solution is 3, 2,1 .

z y y x x

Slide 7- 51

Reduced Row Echelon Form

If we continue to apply elementary row

operations to a row echelon form of a matrix, we

can obtain a matrix in which every column that

has a leading 1 has 0’s elsewhere. This is the

reduced echelon form.

Slide 7- 52

Example Solving a System Using Inverse Matrices

Solve the system

2 3 0

2 2 10

x y

x y

Slide 7- 53

Example Solving a System Using Inverse Matrices

Solve the system

2 3 0

2 2 10

x y

x y

-1

Write the system as a matrix equation.

2 3 0Let , , and .

2 2 10

2 3 2 3Then so that

2 2 2 2

, where is the coefficient matrix of the system.

xA X B

y

x x yA X

y x y

AX B A

A

-1

exists since det 0. Use grapher to find

15. The solution of the system is (15,10).

10

A

X A B

Slide 7- 54

Multivariate Linear Systems and Row Operations

Page 602

Slide 7- 55

Multivariate Linear Systems and Row Operations

Page 602

Slide 7- 56

Multivariate Linear Systems and Row Operations

7.4

Partial Fractions

Slide 7- 58

Quick Review

Perform the indicated operations and write your answer as a

single reduced fraction.

1 21.

1 2

3 22.

1 2

3. Divide ( ) by ( ) to obtain as quotient ( ) an

3

1 2

8

1

d remainder

2

(

x

xx x

x x

f x d x

x

x

x x

q x r

3

2

2

).

Write a summary statement in fraction form: ( ) ( ) / ( ).

( ) 1, ( ) 2

4. Write the polynomials as a product of linear and irreducible quadra

2 3/( 2)

tic

factors with real coefficient

x

q x r x d

x x x

x

f x x x d x x

3 2

2

2

2

s. 2 2

5. Assume that ( ) ( ). What can you conclude about , , , and ?

( ) 2

( ) 3 2 3

2 1

3, 2, 5

x x x

f x g x A B C D

f x Ax Bx C

g x x x

x x

A B C

Slide 7- 59

What you’ll learn about

Partial Fraction Decomposition Denominators with Linear Factors Denominators with Irreducible Quadratic Factors Applications

… and whyPartial fraction decompositions are used in calculus in integration and can be used to guide the sketch of the graph of a rational function.

Slide 7- 60

Partial Fraction Decomposition of f(x)/d(x)

1. Degree of degree of : Use the division algorithm to divide by

( ) ( )to obtain the quotient and remainder and write ( ) .

( ) ( )

2. Factor ( ) into a product of factors of the form (

f d f d

f x r xq r q x

d x d x

d x

2 2

1 2

2 1 2

) or

( ) , where is irreducible.

3. For each factor ( ) : The partial fraction decomposition of ( ) / ( )

must include the sum ... , where , ,...,

are r

u

v

u

u

u u

mx n

ax bx c ax bx c

mx n r x d x

AA AA A A

mx n mx n mx n

2

1 1 2 2

22 2 2

1 2 1 2

eal numbers.

4. For each factor ( ) : The partial fraction decomposition of ( ) / ( )

must include the sum ... ,

where , ,..., and , ,..., a

v

v v

v

v v

ax bx c r x d x

B x CB x C B x C

ax bx c ax bx c ax bx c

B B B C C C

re real numbers.

The partial fraction decomposition of the original rational function is the sum of

( ) and the fractions in parts 3 and 4.q x

Slide 7- 61

Example Decomposing a Fraction with Distinct Linear Factors

3 3

Find the partial fraction decomposition of .1 2

x

x x

Slide 7- 62

Example Decomposing a Fraction with Distinct Linear Factors

3 3

Find the partial fraction decomposition of .1 2

x

x x

1 2

1 2

1 2 1 2

1 2

1 2

3 3

1 2 1 2

3 3 ( 2) ( 1) multiply both sides by ( 1)( - 2)

3 3 ( ) ( 2 )

Compare coefficients on the left and right side of the equation to find

3

2 3

Solv

A Ax

x x x x

x A x A x x x

x A A x A A

A A

A A

1 2e the system of equations to find 2 and 1.

3 3 2 1Thus .

1 2 1 2

A A

x

x x x x

Slide 7- 63

Example Decomposing a Fraction with an Irreducible Quadratic Factor

2

2

3 1Find the partial fraction decomposition of .

2 1

x x

x x

Slide 7- 64

Example Decomposing a Fraction with an Irreducible Quadratic Factor

2

2

3 1Find the partial fraction decomposition of .

2 1

x x

x x

2

2 2

2 2

2 2

3 1

2 1 1 2

3 1 2 1

3 1 ( ) ( ) 2

Compare coefficients to find the system of equations:

1

3

2 1

Use any method to solve the system and find 1,

x x A Bx C

x x x x

x x A x Bx C x

x x A B x B C x A C

A B

B C

A C

A B

2

2 2

0, and 3.

3 1 1 3Thus, .

2 1 1 2

C

x x

x x x x

7.5

Systems of Inequalities in Two Variables

Slide 7- 66

Quick Review Solutions

(0,6) and (8,0)

(0,30) and

Find the - and -intercepts of the line.

1. 3 4 24

2. 1 20 30

Find the point of intersection of the two lines.

3. 3 and 2 5

(2

0,0)

(8/3

4. 1 a

,1/3)

x y

x y

x y

x y x y

x y

nd 3 1

5. 7 3 10 an

(

d

0,1)

(1.3,1 ) 0.3

y x

x y x y

Slide 7- 67

What you’ll learn about

Graph of an Inequality Systems of Inequalities Linear Programming

… and whyLinear programming is used in business and industry to maximize profits, minimize costs, and to help management make decisions.

Slide 7- 68

Steps for Drawing the Graph of an Inequality in Two Variables

1. Draw the graph of the equation obtained by replacing the inequality sign by an equal sign. Use a dashed line if the inequality is < or>. Use a solid line if the inequality is ≤ or ≥.

2. Check a point in each of the two regions of the plane determined by the graph of the equation. If the point satisfies the inequality, then shade the region containing the point.

Slide 7- 69

Example Graphing a Linear Inequality

Draw the graph of 2 4. State the boundary of the region.y x

Slide 7- 70

Example Graphing a Linear Inequality

Draw the graph of 2 4. State the boundary of the region.y x

Because of " ," the graph of 2 4 is part of the graph of the inequality.

The point (0,0) satisfies the inequality because 0 2(0) 4.

Thus the graph of 2 4 consists of all of the points on or below

y x

y x

the line

2 4.y x

Slide 7- 71

Example Solving a System of Inequalities Graphically

2Solve the system 2 3 4 and .x y y x

Slide 7- 72

Example Solving a System of Inequalities Graphically

2Solve the system 2 3 4 and .x y y x

Graph both inequalities and find their intersection.

Slide 7- 73

Chapter Test

1 3 2 11. Given , . Find (a) (b) (c) 2 ,

4 0 4 3

and (d) 3 2 .

Find and , or state that a given product is not possible.

1 2 2 3 1

2. 3 1 , 2 1 0

4 3 1 2 3

3. 1 4

A B A B A B A

A B

AB BA

A B

A

5 3

,2 1

B

Slide 7- 74

Chapter Test

1 0 1

4. Find the inverse matrix if it has one. 2 1 1

1 1 1

5. Find the reduced row echelon form of the matrix

2 1 1 1

3 1 2 1

5 2 2 3

6. Use Gaussian elimination to solve the system of equations.

2

3

3 2 3 8

x z w

x y z

x y z w

Slide 7- 75

Chapter Test

2

7. Solve the system of equations by finding the reduced row echelon

form of the augmented matrix.

2 2 8

2 7 7 2 25

3 3 11

3 28. Find the partial fraction decomposition of .

3 49.Find th

x y z w

x y z w

x y z w

x

x x

e minimum and maximum, if they exist, of the objective

function , subject to the constraints.

Objective function: 7 6

Constraints: 7 7 100

2 5 50

0, 0

f

f x y

x y

x y

x y

Slide 7- 76

Chapter Test

10. A stockbroker sold a customer 200 shares of stock A, 400 shares

of stock B, 600 shares of stock C, and 250 shares of stock D. The

price per share of A, B, C, and D are $80, $120, $200, and $300,

respectively.

(a) Write a 1 4 matrix representing the number or share of each

stock the customer bought.

(b) Write a 1 4 matrix representing the price per share of each

stock.

(c) Write a matrix product t

N

P

hat gives the total cost of the stocks that

the customer bought.

Slide 7- 77

Chapter Test Solutions

1 3 2 11. Given , . Find (a) (b)

4 0 4 3

(c) 2 , and (d) 3 2 .

Find and , or state that a given produ

1 2 3 4

8 3 0 3

2 6 7 11

ct is no

8 0 4 6

t possible.

2.

A B A B A B

A A B

AB BA

A

151 2 2 3 1

3 1 , 2 1 0

4 3 1 2 3

5 33. 1 4 ,

2 1

4

not possible; 1 3

5 13

3 7 ; not possible

B

A B

Slide 7- 78

Chapter Test Solutions

0.4 0.2 0.2

0.2 0.4 0.6

0.6 0

1 0 1

4. Find the inverse matrix if it has one. 2 1 1

1 1 1

5. Find the reduced row echelon form of the matrix

2 1 1 1

3 1 2 1

5 2 2 3

.2 0.2

1 0 0 1

0 1 0 2

0 0 1 3

6. Use Gaussian elimination to solve the system of equations.

2

3

3 2 3 2, 18 , ,

x z w

x y z

x y z w z w w z w

Slide 7- 79

Chapter Test Solutions

7. Solve the system of equations by finding the reduced row echelon

form of the augmented matrix.

2 2 8

2 7 7 2 25

3 3 11

38. Find the partial fraction decomposition

2, 3,

f

,

o

x y z w

x y z w

x y z w

x

w z z w

2

2.

3 49.Find the minimum and maximum, if they exist, of the objective

function , subject

1 2

1 4

minimum is 106 at (10,6); no max to the constraints.

Objective function: 7 6

Constraints:

u

im m

x x

f

f

x x

x y

7 7 100

2 5 50

0, 0

x y

x y

x y

Slide 7- 80

Chapter Test Solutions

10. A stockbroker sold a customer 200 shares of stock A, 400 shares

of stock B, 600 shares of stock C, and 250 shares of stock D. The

price per share of A, B, C, and D are $80, $120, $200, and $300,

res

pectively.

(a) Write a 1 4 matrix representing the number or share of each

stock the customer bought.

(b) Write a 1 4 matrix representing the price per share of each

stock.

200 400 600 250

$80 $120 $200 $30

N

P

(c) Write a matrix product that gives the total cost of the stocks that

the customer bought.

0

$259, 0 00TNP