chapter 1 linear equations and graphs - inter college mathematics 12e 4 learning objectives for...

Chapter 4

Systems of

Linear Equations;

Matrices

Section 1 – Review: Sys of Linear Eg in Two Var

Section 2 – Sys of Linear Eq and Aug Matr

Section 3 – Gauss-Jordan Elimination

Section 4 – Matrices: Basic Operations

Section 5 – Inverse of a Square Matrix

Section 6 – Matrix Eq and Sys of Linear Eq

Section 7 – Leontief Input-Output Analysis

2 Barnett/Ziegler/Byleen College Mathematics 12e

Introduction Chapter 4

Systems of linear equations can be used to solve resource

allocation problems in business and economics.

Such systems can involve many equations in many

variables.

• After reviewing methods for solving two linear equations

in two variables, we use matrices and matrix operations

to develop procedures that are suitable for solving linear

systems of any size.

• We also discuss the Wassily Leontief application of

matrices to economic planning of industrialized

countries.

Chapter 4

Systems of

Linear Equations;

Matrices

Section 1

Review: Systems of

Linear Equations in

Two Variables


Learning Objectives for Section 4.1

The student will be able to solve systems of linear equations in

two variables by graphing

The student will be able to solve these systems using

substitution

The student will be able to solve these systems using

elimination by addition

The student will be able to solve applications of linear

systems.

Review: Systems of Linear

Equations in Two Variables


Table of Content

Systems of Linear Equations in Two

Variables

Graphing

Substitution

Elimination by Addition

Applications


Terms

solution

solution set

consistent / inconsistent system

dependent / independent system

substitution

elimination by addition

equilibrium price / quantity


Opening Example

A restaurant serves two types of fish dinners- small for $5.99

and large for $8.99. One day, there were 134 total orders of

fish, and the total receipts for these 134 orders was $1,024.66.

How many small fish dinners and how many large fish

dinners were ordered?

x + y = 134

5.99x + 8.99y = 1,024.66

x fish dinner-small

y fish dinner-large


Opening Example

x fish dinner-small

y fish dinner-large

x + y = 134

5.99x + 8.99y = 1,024.66

Now we have a system of two linear equations in two

variables. It is easy to find ordered pairs (x, y) that satisfies

one or the other of these equations. To solve this system we

have to find an ordered pair of real number that satisfies both

equation at the same time. In general, we have the following

definition:


Systems of Two Equations

in Two Variables

We are given the linear system

ax + by = h

cx + dy = k a, b, c, d, h, k real constants

A solution is an ordered pair (x0, y0) that will satisfy each

equation (make a true equation when substituted into that

equation). The solution set is the set of all ordered pairs

that satisfy both equations. In this section, we wish to find

the solution set of a system of linear equations. To solve a

system is to find its solution set.


Opening Example

We will consider three methods of solving such systems:

• Graphing

• Susbtitution and

• Elimination by addition


Solve by Graphing

One method to find the solution of a system of linear

equations is to graph each equation on a coordinate plane

and to determine the point of intersection (if it exists). The

drawback of this method is that it is not very accurate in

most cases, but does give a general location of the point of

intersection. Lets take a look at an example:

Solve the following system by graphing:

3x + 5y = –9

x+ 4y = –10


Solve by Graphing

Solution

3x + 5y = –9

x+ 4y = –10

First line (intercept method)

If x = 0, y= –9/5

If y = 0 , x = –3

Plot points and draw line (2, –3)

Second line:

Intercepts are (0, –5/2) , ( –10,0)

From the graph we estimate that the point of intersection is (2,–3).

Check: 3(2) + 5(–3)= –9 and 2 + 4(–3)= –10. Both check.


Another Example

Now, you try one:

Solve the system by

graphing:

2x + 3 = y

x + 2y = –4


Another Example

Solution

Now, you try one:

Solve the system by

graphing:

2x + 3 = y

x + 2y = –4

First line (intercept method)

If x = 0, y= 3

If y = 0 , x = –3/2

Plot points and draw line

The solution is (–2,–1).

Second line:

Intercepts are (0, –2) , ( –4,0)

From the graph we estimate

that the point of intersection is

(-2,–1).


Example 2

Solve the following systems by graphing:

(A) x – 2y = 2

x + y = 5

(B) x + 2y = -4

2x + 4y = 8

(C) 2x + 4y = 8

x + 2y = 4


Example 2 - Solution

(4,1)

(B) Lines are parallel

(each has slope -1/2) –

no solutions.

(C) Lines coincide –

infinite number of

solutions.

(A) Interception at one point –

exactly one solution.

x – 2y = 2

x + y = 5

x + 2y = -4

2x + 4y = 8

2x + 4y = 8

x + 2y = 4


Basic Terms

A consistent linear system is one that has one or more

solutions.

• If a consistent system has exactly one solution (unique

solution), it is said to be independent. An independent

system will occur when the two lines have different slopes.

• If a consistent system has more than one solution, it is said

to be dependent. A dependent system will occur when the

two lines have the same slope and the same y intercept. In

other words, the graphs of the lines will coincide. There

will be an infinite number of points of intersection. Two

systems of equations are equivalent if they have the same

solution set.


Basic Terms

An inconsistent linear system is one that has no solutions.

This will occur when two lines have the same slope but

different y intercepts. In this case, the lines will be parallel

and will never intersect.

Referring to the three systems in the Example 2, the system in

part (A) is consistent and independent with the unique solution

x = 4, y = 1. The system in part (B) is inconsistent. And the

system in part (C) is consistent and dependent with an

infinitive number of solutions.


Example 2 - Solution

(4,1)

(B) Lines are parallel

(each has slope -1/2) –

no solutions.

(C) Lines coincide –

infinite number of

solutions.

(A) Interception at one point –

exactly one solution.

x – 2y = 2

x + y = 5

x + 2y = -4

2x + 4y = 8

2x + 4y = 8

x + 2y = 4

consistent and independent

inconsistent

consistent and dependent


Basic Terms

Example: Determine if the system is consistent,

independent, dependent or inconsistent:

2x – 5y = 6

–4x + 10y = –1


Solution of Example

Solve each equation for y to obtain the slope intercept form of the

equation:

Since each equation has the same slope but different y intercepts,

they will not intersect. This is an inconsistent system.

2𝑥 − 5𝑦 = 6

−5𝑦 = −2𝑥 + 6

5𝑦 = 2𝑥 − 6

𝑦 =2

5𝑥 −

6

5

−4𝑥 + 10𝑦 = −1

10𝑦 = 4𝑥 − 1

𝑦 =4

10𝑥 −

1

10

𝑦 =2

5𝑥 −

1

10


Basic Terms

By graphing a system of two linear equations in two variables,

we gain useful information about the solution set of the

system. In general, any two lines in a coordinate plane must

intersect in exactly one point, be parallel, or coincide (have

identical graphs). The example 2 illustrate the only three

possible types of solutions for systems of two linear equations

in two variables. This is summarized as follows…


Possible Solutions to a Linear

System

A linear system

ax + by = h

cx + dy = k

must have

A. Exactly one solution Consistent and independent

or

B. No solution Inconsistent

or

C. Infinitely many solutions Consistent and dependent

There are no other possibilities.

is one that has no solutions. This will occur when two lines

have the same slope but different y intercepts. In this case,

the lines will be parallel and will never intersect.

Example: Determine if the system is consistent,

independent, dependent or inconsistent:

2x – 5y = 6

–4x + 10y = –1


Method of Substitution

Although the method of graphing is intuitive, it is not very

accurate in most cases, unless done by calculator. There is

another method that is 100% accurate. It is called the

method of substitution. This method is an algebraic one.

It works well when the coefficients of x or y are either 1 or

–1. For example, let’s solve the previous system

2x + 3 = y

x + 2y = -4

using the method of substitution.


Method of Substitution

(continued)

The steps for this method are as follows:

1) Solve one of the equations for either x or y.

2) Substitute that result into the other equation to obtain an

equation in a single variable (either x or y).

3) Solve the equation for that variable.

4) Substitute this value into any convenient equation to

obtain the value of the remaining variable.


Example

(continued)

2x + 3 = y

x + 2y = –4

x + 2(2x + 3) = –4

x + 4x + 6 = –4

5x + 6 = –4

5x = –10

x = –2

Substitute y from first equation into second equation

Solve the resulting equation

After we find x = –2, then from the first equation, we have 2(–2) + 3 = y or y = –1. Our solution is (–2, –1)


Another Example

Solve the system using

substitution:

3x – 2y = 7

y = 2x – 3


Another Example

Solution

Solve the system using

substitution:

3x – 2y = 7

y = 2x – 3

Solution:

3 2 7

2 3

3 2(2 3) 7

3 4 6 7

1

1

2 1 3 5

x y

y x

x x

x x

x

x

y y



The method of substitution is not preferable if none of the

coefficients of x and y are 1 or –1. For example, substitution is

not the preferred method for the system below:

2x – 7y = 3

–5x + 3y = 7

A better method is elimination by addition. The following

operations can be used to produce equivalent systems:

• Two equations can be interchanged.

• An equation can be multiplied by a non-zero constant.

• A constant multiple of one equation and is added to

another equation.



Example

For our system, we will seek to

eliminate the x variable. The

coefficients are 2 and –5. Our goal is

to obtain coefficients of x that are

additive inverses of each other.

We can accomplish this by

multiplying the first equation by 5,

and the second equation by 2.

Next, we can add the two equations

to eliminate the x-variable.

Solve for y

Substitute y value into original

equation and solve for x

Write solution as an ordered pair

{

2x 7 y 3

5x 3y 7

5(2x 7 y) 5(3)

2(5x 3y) 2(7)

10x 35y 15

10x 6y 14

0x 29y 29

y 1

2x 7(1) 3

2x 7 3

2x 4

x 2

(2,1)

{



Another Example

Solve 2x – 5y = 6

–4x + 10y = –1



Another Example

Solve 2x – 5y = 6

–4x + 10y = –1

Solution:

1. Eliminate x by multiplying

equation 1 by 2 .

2. Add two equations

3. Upon adding the equations,

both variables are eliminated

producing the false equation

0 = 11

2 5 6

4 10 1

4 10 12

4 10 1

0 11

x y

x y

x y

x y

4. Conclusion: If a false

equation arises, the system

is inconsistent and there is

no solution.



Additional Considerations

Let see what happens in the elimination process when a system

has either no solution or infinitely many solutions.




Consider the following system:

2x – 6y = -3

x + 3y = 2

Multiplying the second equation by -2 and adding, we obtain:

2x – 6y = -3

-2x – 6y = 2

0 = -7 not possible

We have obtained a contradiction. The initial assumption that the

system has solutions is false. The system has no solutions and

the graphs of the equations are parallel lines.




Now consider the system:

x – ½y = 4

-2x + y = -8

Multiplying the top equation by 2 and adding, we obtain:

2x – y = 8

-2x + y = -8

0 = 0

This result implies that the equations are equivalent; that is, their

graphs coincide and the system is dependent.




If we let x = k, where k is any real number, and solve either

equation for y, we obtain y = 2k – 8. So (k, 2k – 8) is a solution

to the system for any real number k.

The variable k is called a parameter and replacing k with a real

number produces a particular solution to the system.

For example, some particular solutions to the system are:

k = -1 k = -1 k = -1 k = -1

(-1, -10) (2, -4) (5, 2) (9.4, 10.8)


Application

A man walks at a rate of 3 miles per hour and jogs at a rate

of 5 miles per hour. He walks and jogs a total distance of

3.5 miles in 0.9 hours. How long does the man jog?


Application

A man walks at a rate of 3 miles per hour and jogs at a rate

of 5 miles per hour. He walks and jogs a total distance of

3.5 miles in 0.9 hours. How long does the man jog?

Solution: Let x represent the amount of time spent walking

and y represent the amount of time spent jogging. Since the

total time spent walking and jogging is 0.9 hours, we have

the equation x + y = 0.9.

We are given the total distance traveled as 3.5 miles. Since

Distance = Rate x Time, distance walking = 3x and

distance jogging = 5y. Then total distance is 3x + 5y= 3.5.


Application

(continued)

We can solve the system using substitution.

1. Solve the first equation for y

2. Substitute this expression into the second equation.

3. Solve second equation for x

4. Find the y value by substituting this x value back into the first equation.

5. Answer the question: Time spent jogging is 0.4 hours.

Solution:

0.9

3 5 3.5

3 5( ) 3.5

3

0.9

0.9

4.5 5 3.5

2 1

0.5

0.5 0.9

0.4

x y

y

x y

x

x x

x

x

x

x

y

y


Supply and Demand

The quantity of a product that people are willing to buy during

some period of time depends on its price. Generally, the higher

the price, the less the demand; the lower the price, the greater

the demand.

Similarly, the quantity of a product that a supplier is willing to

sell during some period of time also depends on the price.

Generally, a supplier will be willing to supply more of a

product at higher prices and less of a product at lower prices.

The simplest supply and demand model is a linear model

where the graphs of a demand equation and a supply equation

are straight lines.


Supply and Demand

(continued)

In supply and demand problems we are usually interested in

finding the price at which supply will equal demand. This is

called the equilibrium price, and the quantity sold at that

price is called the equilibrium quantity.

If we graph the the supply equation and the demand equation

on the same axis, the point where the two lines intersect is

called the equilibrium point. Its horizontal coordinate is the

value of the equilibrium quantity, and its vertical coordinate is

the value of the equilibrium price.


Supply and Demand

Example

Example: Suppose that the supply equation for long-life

light bulbs is given by

p = 1.04 q – 7.03,

and that the demand equation for the bulbs is

p = –0.81q + 7.5

where q is in thousands of cases. Find the equilibrium price

and quantity, and graph the two equations in the same

coordinate system.


Supply and Demand

(Example continued)

If we graph the two equations on a graphing calculator or

manually, or solve by substitution, and find the intersection

point, we see the graph below.

Thus the equilibrium point is (7.854, 1.14), the

equilibrium price is $1.14 per bulb, and the equilibrium

quantity is 7,854 cases.

Supply Curve

Demand Curve

1.04 q – 7.03 = –0.81q + 7.5

1.85 q = 14.53

q = 7.854

p = 1.04 q – 7.03

p = 1.04 (7.854) – 7.03 = 1.14


Now, Solve the Opening Example

A restaurant serves two types of fish dinners- small for

$5.99 each and large for $8.99. One day, there were 134

total orders of fish, and the total receipts for these 134

orders was $1024.66. How many small dinners and how

many large dinners were ordered?

x fish dinner-small

y fish dinner-large

x + y = 134

5.99x + 8.99y = 1,024.66


Solution

Multiply the first equation by -8.99 and add to the second equation:

(-8.99) (x + y) = 134(-8.99)

-8.99x - 8.99y = -1,204.66

5.99x + 8.99y = -1,024.66

-3x + 0 = -180.00

x = 60

x + y = 134

60 + y = 134

y = 134 – 60 = 74

Answer: 60 small orders and 74 large orders

Chapter 4

Systems of

Linear Equations;

Matrices

Section 1

Review: Systems of

Linear Equations in

Two Variables

END Last Update: Abril 9/2013

chapter 1 linear equations and graphs - inter college mathematics 12e 4 learning objectives for...

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