# MAT 102 ```MATH
130
MATH 130
Finite Mathematics
Mathematics
Finite
41
2-Math 130
Course Guide
Self-paced study. Anytime. Anywhere!
Math 130
Finite Mathematics
University of Idaho
3 Semester-Hour Credits
Prepared by:
Judith Terrio, MS
Instructor of Mathematics
University of Idaho
RV: 2/2012
2-Math 130
Copyright Independent Study in Idaho/Idaho State Board of Education
2-Math 130
Welcome! ........................................................................................................................................................ 1
Policies and Procedures .................................................................................................................................. 1
Course Description .......................................................................................................................................... 1
Course Materials ............................................................................................................................................. 1
Course Delivery ............................................................................................................................................... 2
Course Introduction ........................................................................................................................................ 2
Course Objectives ........................................................................................................................................... 2
Lessons ............................................................................................................................................................ 3
Exams .............................................................................................................................................................. 4
About the Course Developer .......................................................................................................................... 5
Disability Support Services .............................................................................................................................. 5
Assignment Submission Log ............................................................................................................................ 6
Lesson 1:
Lesson 2:
Lesson 3:
Linear Functions [Chapter 1, sections 1, 2, and 3] ...................................................................... 7
Linear Systems [Chapter 2, sections 1 and 2] .............................................................................. 10
Matrices [Chapter 2, sections 3, 4, 5] .......................................................................................... 16
Exam 1 Information: Covers Lessons 1-3 ...............................................................................................19
Lesson 4:
Lesson 5:
Lesson 6:
Matrix Equations [Chapter 2, section 6] ...................................................................................... 21
Linear Programming: Geometric Approach [Chapter 3, sections 1, 2, 3, and 4]......................... 26
Linear Programming: The Simplex Method [Chapter 4, sections 1, 2, and 3] ............................. 33
Exam 2 Information: Covers Lessons 4-6 ...............................................................................................40
Lesson 7:
Lesson 8:
Lesson 9:
Lesson 10:
Sets and Counting I [Chapter 6, sections 1, 2, and 3] .................................................................. 42
Sets and Counting II [Chapter 6, sections 4, 5, and 6] ................................................................. 46
Probability I [Chapter 7, sections 1, 2, and 3].............................................................................. 50
Probability II [Chapter 7, sections 4, 5, and 6]............................................................................. 53
Exam 3 Information: Covers Lessons 7-10 .............................................................................................56
Lesson 11: Self-study: Markov Chains [Chapter 7, section 7] ....................................................................... 58
Final Exam Information: Covers Lessons 1-11 ........................................................................................60
Lesson Submission Form ......................................................................................................................73
2-Math 130
Math 130: Finite Mathematics
3 Semester-Hour Credits: U-Idaho
Welcome!
Whether you are a new or returning student, welcome to the Independent Study in Idaho (ISI) program.
Below, you will find information pertinent to your course including the course description, course
materials, course objectives, as well as information about lessons, exams, and grading.
Policies and Procedures
Important!
As you read this section, you will see the following icon:
Use this icon to direct yourself to essential ISI information. Students are responsible for following ISI’s
policies. Refer to ISI’s website at www.uidaho.edu/isi, select About ISI, Policies for the most current
policies, procedures and course information. The Registration Confirmation Letter sent to you upon
registration provides your course instructor’s contact information and lesson guidelines. If you have any
Course Description
Systems of linear equations and inequalities, matrices, linear programming, and probability.
Prerequisite: Sufficient score on SAT, ACT, or COMPASS Math Test; or Math 108 with a C or better.
Required test scores can be found here:
www.uidaho.edu/registrar/registration/placement/math.htm.
U-Idaho students: May be used as core credit in J-3-c.
10 graded lessons, 1 self-study lesson, 4 proctored exams
Course Materials
Required Course Materials
Rolf, Howard L. Finite Mathematics, Enhanced Edition (with Enhanced WebAssign with eBook for One
Term Math and Science Printed Access Card). 7th ed. Belmont, CA: Thomson Brooks/Cole, 2010.
ISBN-10: 0-538-49732-7. ISBN-13: 978-0-538-49732-9.
Recommended Course Materials
Rolf, Howard L. Student Solutions Manual for Rolf’s Finite Mathematics. 7th ed. Belmont, CA: Thomson
Brooks/Cole, 2007. ISBN-10: 0-495-11843-5. ISBN-13: 978-0495118435.
If you are unsure if your background is sufficient, or if it has been a long time since you have done any
formal mathematics, I suggest you look at the Appendices, starting on page 777 in Finite Mathematics,
Enhanced Edition (with Enhanced WebAssign with eBook for One Term Math and Science Printed Access
Card). The answers to the odd numbered problems are in the back of the text. Work some of these from
each of the four sections. Also, the material in Chapter 1 “Functions and Lines”, sections 1 and 2 and
Chapter 2 “Linear Systems”, section 1, should mostly be review from algebra. If it seems unfamiliar to
you, then you should consider taking the algebra course, Math 108 [Intermediate Algebra], before
proceeding with this course.
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Independent Study in Idaho course materials are available for purchase at the VandalStore (University of
Idaho bookstore). Your Registration Confirmation Letter contains the VandalStore’s contact information.
Independent Study in Idaho courses are updated and revised periodically. Ordering course materials
from the VandalStore at the time of registration allows you to purchase the correct edition(s) of
textbooks, course guides, and supplemental materials. Contact the VandalStore directly for questions
regarding course materials that you have ordered.
If purchasing textbooks from another source, refer to the ISBN(s) for the textbook(s) listed for this
course to ensure that you obtain the correct edition(s).
Course Delivery
This course is available online. An electronic course guide is accessible through BbLearn at no additional
cost. Refer to your Registration Confirmation Letter for instructions on how to access BbLearn.
Course Introduction
The goal of the first two-thirds of this course is to teach you to solve a particular type of application,
called linear programming. These problems occur in numerous settings in the “real” world, as you will
see from the variety of applications we will do. We will be studying two methods for solving this type of
problem: a graphical method and a matrix method. The first two chapters lay the groundwork for linear
programming, and then the graphical approach is presented in Chapter 3 “Linear Programming,” and the
matrix method in Chapter 4 “Linear Programming: The Simplex Method.”
The last one-third of this course is devoted to studying probability. Probability lays the groundwork for
statistical methods, which most students now study, regardless of their majors. Chapters 6 “Sets and
Counting” and 7 “Probability” will discuss counting and probabilities.
Please note that the textbook is set up in the same manner as most of the math textbooks you have
used in the past. The notation “Section 2.3” means Chapter 2, Section 3. The exercise sets are at the end
of each section and there is a set of review exercises at the end of every chapter. Also, the answers to
the odd numbered problems are all in the back to the textbook beginning on page 807.
Course Objectives
Upon successful completion of this course, the student will be able to:
 Solve systems of linear equations by the Elimination Method.
 Solve systems of linear equations by the Gauss-Jordan Method.
 Solve systems of linear equations using Matrix Inversion.
 Solve applications of linear equations.
 Solve linear programming problems by graphing.
 Solve linear programming problems by the Simplex Method.
 Use sets and Venn diagrams to solve counting problems.
 Use permutations and combinations to solve counting problems.
 Find probabilities for equally likely events and compound events.
 Use conditional probabilities and independence to solve probability problems.
 Use Bayes’ Rule to determine probabilities.
 Solve Markov Chain problems.
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Lessons
Each lesson includes the following components:
• overview
• self-study practice problems
Reading mathematics takes practice, time, and effort. You should spend a lot of time reading the text.
Read very slowly, thinking about each sentence. Read with a pencil in hand so you can write in the
margins and work through the given examples. Often you will need to fill in a step or two. Authors
sometimes leave out a few steps of arithmetic or algebra in order to keep the number of pages down to
a manageable size. If you have any questions on the reading, you can always contact the instructor.
I hope that you will find the textbook to be very readable. I chose it because students seem to like it and
find the author to be highly understandable. There are two features that I particularly like that I would
like you to look at right now. In the margins of some pages, the author sometimes writes little
“cautions” (see p. 20) or “notes” (see p. 23). These are bits of information that I would have included in
the overview, but don’t need to because this author is very thorough. So, pay attention to these tidbits
of information. The second really nice feature, especially for an independent learner, is the direct link
that the author provides between examples and exercises. Please look at the end of example 12 on page
23 in the textbook. Note where it says, in blue, “Now you are ready to Work Exercise 92, pg. 30.” Now
turn to that exercise and notice how it refers you to the example. I hope you will find this to be a very
helpful feature. And, while we are looking at the textbook, notice the very large number of examples
that are in each section. Quite frankly, I love this textbook for a course like this. I hope you appreciate it
and find that it eases your learning.
In the overview section of each lesson, I point out material that is especially important, any topics in the
reading that you may skip, some tips of my own and, sometimes, I give other examples that I think you
The self-study practice problems are intended to help develop your understanding of the basic concepts
and to increase your competence with the algebraic and arithmetic manipulations needed to solve
problems.
There is an old saying: “Math is not a spectator sport.” I can read and understand the rules of basketball
or directions on how to drive a car. However, unless I train my muscles and practice, I would not be a
very competent basketball team member, nor would you want to ride in a car with me driving!
Practice is the key to succeeding in a math class. It is a lot of work; that is true. However, the rewards
come when each problem gets a little bit easier and then each lesson gets a little bit easier as you
exercise your math brain muscles. There are a lot of false starts in mathematics. That’s okay. The key is
to be aware of this and to not get frustrated. Take a break, go for a walk and clear your head. This is
what we all have to do; it is the nature of math problem solving and indeed, of all problem solving in life.
These practice problems are the same ones that I assign to my on-campus students. The answers to the
self-study practice problems are in the back of the textbook and in the Student’s Solution Manual for
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Rolf’s Finite Mathematics. Do not turn these problems in, but do not neglect them. Feel free to contact
the grader if you need help on any of these.
Graded assignments take the place of in-class quizzes. They give both of us a way to access which
concepts you need more practice on before taking an exam. Equally important is that you learn to write
mathematics correctly. These assignments need to be done neatly and completely. You are going to be
graded on the thoroughness of your work as well as its accuracy.
Study Hints:
 Complete all the self-study problems before doing the graded problems.
• Set a schedule allowing for course completion one month prior to your personal deadline. An
Assignment Submission Log is provided for this purpose.
• Web pages and URL links in the World Wide Web are continuously changing. Contact your instructor
if you find a broken webpage or URL.
 Keep a copy of every lesson submitted. Submit the entire lesson at one time (partial or incomplete
Refer to your Registration Confirmation Letter for further details on your instructor’s lesson
guidelines and requirements. Also refer to the ISI Policies and Procedures for essential ISI
policies on submitting lessons to your instructor.
Exams
•
•

You must wait for grades and comments on lessons prior to taking subsequent exams.
In this course guide, on the Exam Information pages, there are test outlines containing textbook
exercises that are similar to what you will be tested on. You should do all of those problems to
prepare for the test. You may find that you need to do them more than once. Keep track of the
problems you had to look up in the textbook, in this guide, or on your past homework before you
could solve the problems. Do those over again until you can do every problem without any
references.
Refer to Grading for specific information on lesson/exam points and percentages.
Proctor Selection/Scheduling Exams
All exams require a proctor.
Refer to the ISI Policies and Procedures for guidelines on how to choose a proctor and
schedule exams. Complete the Proctor Information Form and send it to the ISI office at least
two weeks prior to scheduling your first exam.
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2-Math 130
The course grade will be based upon the following considerations:
Assignments
(10 @ 10 points each)
Unit Exams
(3 @ 100 points each)
Comprehensive Final Exam
Total
100 points
300 points
100 points
500 points
A letter grade will be assigned at the end of the course based on the standard 10 percent scale.
A = 500 – 450 points
B = 449 – 400 points
C = 399 – 350 points
D = 349 – 300 points
F = 299 points or lower
Self-study Lessons
Lesson 11 is the only lesson for which there is no graded assignment. As with the self-study exercises
listed in the other lessons, the answers to lesson 11’s problems are in the back of the textbook.
The final course grade is issued after all lessons and exams have been graded.
Refer to the ISI Policies and Procedures for information about confidentiality of student
grades, course completion, time considerations, and requesting a transcript.
Judith Terrio got her math ability from her father and her love of people from her mother. She grew up
and went to college on the east coast. Judi obtained her secondary teaching certificate and taught a few
years in Colorado before moving to beautiful Idaho. In north Idaho, Judi taught high school for many
years before earning a master’s degree in mathematics and subsequently becoming an instructor at the
University of Idaho. She thinks she is one of the most fortunate people alive because her passion and
her job are one in the same. She loves teaching! She is also a cat whisperer. 
You will receive specific course requirements and instructor contact information in your
Registration Confirmation Letter, which you will receive upon registration. Please copy the ISI
office at [email protected] when corresponding and submitting lessons by email to your
instructor.
Disability Support Services
Refer to the ISI Policies and Procedures for information on Disability Support Services (DSS).
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Assignment Submission Log
Send the completed Proctor Information Form to the ISI office at least two weeks prior to taking your first exam.
Lesson
Projected Date for
Completion
Date
Submitted
1
2
3
It is time to make arrangements with your proctor to take Exam 1.
Exam 1
4
5
6
It is time to make arrangements with your proctor to take Exam 2.
Exam 2
7
8
9
10
It is time to make arrangements with your proctor to take Exam 3.
Exam 3
It is time to make arrangements with your proctor to take the Final Exam.
Exam 4
6
Cumulative
Point Totals
2-Math 130
Lesson 1
Linear Functions
Preface “To the Student,” pp. xi–xii
Chapter 1, Functions and Lines, Sections 1.1, 1.2, and 1.3
Overview
As mentioned in the introduction, the goal of the first two-thirds of this course is to teach you to solve a
particular type of application, called linear programming. One of the very nice things about these
problems is that the variables are only to the first power and are only added or subtracted to one
another (i.e., no square roots, no xy or x/y in the equations). Sounds great, doesn’t it? This means that
we will be concerned mostly with lines. In actuality, applications may involve many, many variables, but
fortunately we have technology to lend a helping hand in solving those more complex problems. The
technological methods used are just an extension of the methods used with lines or with three or four
variables. So, we start with the study of functions, particularly linear functions, as the foundation for
this course.
Section 1.1 Functions
Please pay special attention to how variables are defined when using mathematics to describe an
application. For instance, let’s say I’m going to the store to buy bananas. To define a variable by only
saying, “let x = bananas,” is not a sufficient definition. Is x the number of individual bananas, the number
of bunches, or the number of pounds of bananas? You must include a quantifier (like “number of” or
“pounds of”) when defining a variable.
Section 1.2 Lines
This section provides you with an opportunity to review and refresh your algebra skills concerning lines.
Here is a list of formulas that you will need.
 Slope of a line between two given points (x1, y1) and (x2, y2) is m 
y y2  y1
, provided x2 ≠x1

x x2  x1
Note that this is said “delta y over delta x” or “change in y over change in x.” Δ is the capital Greek
letter “delta” and “change” means the difference in two numbers (like the change you get when you
make a cash purchase.)





Slope-Intercept Form
Point-Slope Form
Standard Form
Horizontal Line
Vertical Line
y = mx + b, where m is the slope and the point (0, b) is the y-intercept
y – y1 = m(x –x1), where m is the slope and (x1, y1) is a known point
Ax + By = C, where A is a whole number greater than or equal to 0
y = b [m = 0, y-intercept is (0, b), there is no x-intercept]
x = a [slope is undefined, there is no y-intercept, x-intercept is (a, 0)]
*There are basically two methods for sketching a line:
(1) Using a point on the line and its slope (see textbook examples 4 [pp. 17-8], 9 [p. 21], and 10 [pp.
21-2])
(2) Using the line’s x- and y- intercepts
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The second method is the one that we will use most often, so it is important for you to understand how
to find these two points.
Example: Given 2x  3 y  12.
A point on the x-axis has a y -coordinate of zero,
A point on the y -axis has a x-coordinate of zero,
so find the x-intercept by setting y  0.
so find the y -intercept by setting x  0.
2 x  3(0)  12
2 x  12
x6
2(0)  3 y  12
- 3 y  12
y  4
To sketch the line, connect the two points (6, 0) and (0, -4) with a line which continues in both directions.
*When given two points, there are two ways to find the equation of a line:
(1) Using the slope-intercept form (which is the method you are probably the most familiar with)
(2) Using the point-slope form
The second method is the one that we will use most often. It is actually less work and less prone to
errors.
Example: Find an equation of the line passing through  3, 1 and  2, 9  .
y 1  9 10


 2
x 3   2
5
Then substitute the slope and either point into the form y  y1  m( x  x1 )
First find the line's slope: m 
y  1  2( x  3)
y  1  2 x  6
y  2 x  5
y  9  2( x   2)
y  9  2 x  4
y  2 x  5
Section 1.3 Linear Models
This section deals with the use of math to model real-world situations. The major focus in this section is
on the cost, revenue, and profit functions.
If x = number of items sold, then:
The cost function C(x) = ax + b; where C(x) is the total cost for producing x items,
a = unit cost, the cost for producing each item, which includes such
things as material and labor costs; and
b = fixed costs, which don’t depend on the number of items made, such
as factory rent, heating and electricity, office staff salaries and
insurance, etc.
The revenue function R(x) = px; where R(x) is the gross amount of money received for selling x
items and,
p = unit selling price
The profit function P(x) = R(x) – C(x); the amount of money received from selling x items minus
the cost of producing those x items.
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The break-even point is the number of units needed to be sold so that P(x) = 0 or R(x) = C(x).
Example: Textbook exercise #40, p. 52.
There are three values given in this problem:
Fixed costs of \$28,000, unit costs of \$48/each, and selling price of \$62/each.
So, if we let x = the number of TV stands made, then the cost function is C(x) = 48x + 28000 and
the revenue function is R(x) = 62x.
To find the break-even point, equate the revenue function and the cost function:
R ( x)  C ( x)
62 x  48 x  28000
14 x  28000
x  2000
So, 2000 TV stands must be sold to break even.
Self-Study Practice Problems
These are not to be handed in. They are for you to check your understanding of the material; to exercise
your math brain. Make sure you attempt each problem without looking in the solutions manual. As I
mentioned above, it is crucial for you to practice, to do math, and not just read about it. The answers to
the self-study practice problems are in the back of the textbook and in the Student Solution Manual for
Rolf’s Finite Mathematics.
Section 1.1 (p. 6):
Section 1.2 (p. 28):
Section 1.3 (p. 50)
#1, 9, 13–23 odd
#7–43 every other odd, 45–83 odd, 87, 89, 93, 95, 99, 103, 109
#1–29 odd, 37, 43, 51
Before beginning the first written assignment, refer to your Registration Confirmation Letter
the ISI office at [email protected]
Show all your work in a neat and organized manner, or you will receive no credit.
Good luck and have some fun! Treat yourself when you finish the lesson. Your labors deserve a reward.
(16 problems)
Section 1.1 (p. 6):
Section 1.2 (p. 28):
Section 1.3 (p. 50):
#10, 18
#14, 16, 18, 54, 56, 64, 66, 92, 96
#14, 16, 18, 30, 48
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Lesson 2
Linear Systems
Chapter 2, Linear Systems, Sections 2.1 and 2.2
Overview
In this chapter we will investigate systems of lines and systems of planes. First, we will look at geometric
and algebraic methods for finding the relationship between lines in a plane or planes in space. Then we
will look at a method, which may be new to you, that uses something called an augmented matrix.
Section 2.1 Algebraic Elimination
In this section, we will review, from algebra, the solving of systems of two lines in the same plane. Also,
we will look at some applications for systems of lines.
To solve a system of equations means to find any points in common. If I have two lines in the same
plane, then there are three possible arrangements.
The lines intersect in one point
and have one unique solution
The lines are parallel and
have no solutions.
The equations are the same
line and have an infinite
number of solutions.
The solution is a point (x1, x2).
There is no solution.
The solution is the set of all
points on the line (k, mk+b).
*There are two basic ways to solve a system of two linear equations:
(1)
by substitution, or
(2)
by elimination
You most likely used both of these methods in algebra. In this class, the elimination method is the
method we will use most often.
3x  4 y  5
Example: Given 
2 x  3 y  4
We need to get the coefficients in front of either x or y to be the same, except opposite signs. That way
we can add the two equations together and one of the variables drops out. Since the signs in front of
the y’s are already opposite, I will multiply the first equation by three and the second equation by four.
In the next box, notice how I indicate what I am doing above and below the arrow:
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3 x  4 y  5 3 R1 9 x  12 y  15 add
1

 17 x  0  1 
x 

4 R2  
17
2x  3y  4
8 x  12 y  16
To get the corresponding y -coordinate,
substitute the x-value into either of the given equations.
multiply by 17
3( 171 )  4 y  5 

 3  28 y  85 
 28 y  88 
y
88 22

28
7
The solution to this system is the point ( 171 , 722 ).
The textbook, Section 2.1, examples 7(pp. 72-3) and 8 (pp. 73-4) show you what happens when you use
the elimination method on a system of parallel equations or a system of two identical lines.
Demand, Supply, and Equilibrium
In Section 1.3, when we considered the revenue function R(x) = px (p. 41), we assumed that the unit
price was a constant amount, independent of the number of items sold. Actually, price can depend on
the amount that consumers want or the price can depend on the amount that is available.
When x = number of items in demand
price usually rises with demand.
When x = number of items produced,
price usually lessens as more units are
P = ax + b is called the demand equation
P = cx + d is called the supply equation
x
x
The equilibrium point is the point (x, p) which satisfies both of these equations.
Example: Textbook exercise #46, p. 78. Let x be the number of CD players.
For the demand equation, we are given two points (10, 130) and (25, 100). Finding the equation of the
line, using the slope and one of the points (see Lesson 1, Section 1.2 Lines) we find the demand equation
to be
p = -2x + 150.
Similarly, for the supply equation, we are given two points (40, 130) and (10, 100). Finding the equation
of the line, using the slope and one of the points we find the supply equation to be p = x + 90. In this
case it is easiest to solve for the equilibrium point using substitution
-2x + 150 = x + 90
-3x = -60
x = 20 and substituting 20 into either the demand or the supply equations gives p = 110
So, the equilibrium quantity is 20 CD players at a price of \$110 each.
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Section 2.2 Augmented Matrices and the Gauss-Jordan Method
The textbook starts this section by algebraically solving systems of three equations with three variables.
I am not going to require you to solve these in this way. The method we are going to use starts on p. 86.
It is known as the Gauss-Jordan Method (after two mathematicians who developed it) and uses a
Suppose we are trying to solve the following system of equations to find the point or points, if any, that
all three equations have in common.
x – 2y + z = 7
3x + y – z = 2
2x + 3y + 2z = 7
1 2 1 7 
1 0 0 a 

 followingthena set of rules 

 0 1 0 b 
change into
 3 1 1 2  
 2 3 2 7 
0 0 1 c 
The first step is to
write the system as
an augmented matrix.
Once you have all 1’s in the diagonals and zeros everywhere else to the left of the vertical line, you now
have a new system of equations:
x + 0y + 0 z = a
0x + y + 0 z = b
0x + 0y + z = c
x=a
y = b The solution point is (a, b, c).
z=c
or, simply
So, the heart of the Gauss-Jordan Method is how to change the first augmented matrix into diagonal
form.
The following are the three legal rules and the notation we use to show what has been done between
any two matrices:
R1 ↔ R 2
c R 1 → R1
c R 1 + d R2 → R1
Swap any two rows, in this case, rows 1 and 2.
Replace any row with a nonzero multiple of itself.
Replace any row with the sum of a multiple of itself and a multiple of another row.
So, how does one start? There is an order to follow or else you might find yourself undoing some of the
zeros that you got in one step while performing another step. There are three possible cases for the first
crucial step.
1. If the first row, first column is not a 1, but there is a 1 in the first column, then use the first rule
to swap two rows and get the one in the a11 position.
2. If there are no 1’s in the first column use a nonzero a11 (swapping rows if necessary) with rule
1
two to multiply row one by a .
11
3. However, if this results in messy fractions, you can work with any nonzero number in the first
position. Getting ones makes the other steps easier, but it is not a crucial step. Concentrate on
getting the zeros. Get all the zeros in the first column and then the zeros in the second column
and finally the last column.
For example, solve the following system using the Gauss-Jordan method:
12
2-Math 130
0 2
2 y  3z  7
3 7




set-up
  3 6 12 3 
 3 x  6 y  12 z  3 
5 x  2 y  2 z  7
5 2
2 7 

Notice that we do not have any 1’s in the first
column. However, if we multiply the second row by
⅓ and swap the first and second rows, we will have a
1 in the top left corner and no fractions in that row.
0 2
0 2 3 7 
1
3 7 1
2 4 1

 3 R2  R2 
 R1  R2 

 1
2 4 1 
 0 2 3 7 
 3 6 12 3 
5 2
5 2
5 2
2 7 
2 7 
2 7 
If you can do so correctly, you may perform two
operations at once, but no more than two. Here
is the same matrix reduced by doing two steps
in one.
0
1
2
3 7 1
2 4 1

 3 R2  R2 

2
3 7
R1  R2  0
3 6 12 3 
5 2
5 2
2 7 
2 7 
R
R
1
2
or one could do 
 and get the same result.
1
R  R1
3 1
The next step is to get all zeros in the first
column under the one. Since the second row
already has a zero, we next get a zero where the
five is by using the 1 in row one.
Notice that all the new values in row three are
now divisible by 2 (or -2). Sometimes it will make
your work easier to reduce a row so you have
smaller numbers to work with. You can do this in
a separate step or in the same step if you notice it
at the time.
1 2 4 1
1
2
5 R1  R3  R3 


2
0 2 3 7   0
5 2 2 7 
0 12
4 1

3 7
22 2 
1 2 4 1
1 2 4 1
5 R1  R3  R3 



3 7
1
0 2 3 7  
0 2
 R3  R3
2
5 2 2 7 
0 6 11 1 
For this problem, we now have a choice as to how to proceed depending on your competence with
1
fractions. After getting column one to look like  0 , the next step is to get a 1 in the second column,

 0
second row. We would do this by using rule two: &frac12; R2 → R2. For this matrix, we would now have to work
with fractions in the second row. Halves aren’t too messy to work with, but we can do something else to
avoid the fractions at this stage and that is to use rule three. Please recall the third step, which I wrote
immediately prior to this example. Concentrate on the zeros; we can get the 1’s at the end. So, to avoid
fractions, proceed as follows, similar to the elimination method for a 2 x 2 system.
*Notice that when getting the zeros in column two, we multiply the second row to get the number
needed for our zeros. This element, the second row and second column is called the pivot element.
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2-Math 130
1 2 4 1
1 0 7 8  1
1 0 7 8 

 1R2  R1  R1 
 2 R3  R3 

3 7  
3 7   0 2
3 7
3 R2  R3  R3
0 2
0 2
0 6 11 1 
0 0 20 20 
0 0
1 1 
1 0 7 8
1 0 0 1 1
1 0 0 1

 7 R3  R1 R1 
 2 R2  R2 

 0 1 0 2 
3 R3  R2  R2
0 2 3 7  
0 2 0 4  
0 0 1 1 
0 0 1 1 
0 0 1 1 
substituting the values into all three of
the original equations to see if they are
satisfied.
Now we get the
zeros in the third
column by pivoting
on the 1 in the third
row, third column.
The solution is now
determined to be x = -1,
y = 2, and z = 1.
2 y  3z  7
2(2)  3(1)  7


x 1, y  2, z 1
 3(1)  6(2)  12(1)  3
 3x  6 y  12 z  3 
5 x  2 y  2 z  7
5(1)  2(2)  2(1)  7

Section 2.2 Applications
You may find it helpful for some of the applications to use a table to organize the given data.
Example: Textbook exercise #54, pp. 100
Let A = the price (\$) of a medium T-shirt
Let B = the price (\$) of a large T-shirt
Let C = the price (\$) of an extra-large T-shirt
Organizing the given information into a table with the variables horizontally across the top makes
translation into the system of equations straightforward.
Alpha Beta
Beta Gamma
Pi Beta
A B C
3 8 13
16 9 0
6 11 14
Total Cost (\$)
357.50
326.00
453.00
14
3 A  8 B  13C  357.50
16 A  9 B
 326.00
6 A  11B  14C  453.00
2-Math 130
Example: Textbook exercise #62, p. 101
Let x = number of minutes jogging
Let y = number of minutes playing handball
Let z = number of minutes cycling
Sometimes using a table doesn’t to capture all of given information. In this exercise, only the
information about calories fits into table format. The other information needs to be translated into an
equation as you learned in algebra class.
x
y z
Total
Calories
13 11 7
640
The total number of minutes is one hour, so
And minutes jogging = minutes cycling
So, x = z or
13 x  11 y  7 z  640 (calories)
x  y  z  60 (minutes)
x
 z 0
Self-Study Practice Problems
These are not to be handed in. They are for you to check your understanding of the material; to exercise
your math brain. Make sure you attempt each problem without looking in the solutions manual. It is
crucial for you to practice, to do math, and not just read about it.
Section 2.1 (p. 77):
Section 2.2 (p. 98):
#17 – 33 odd, 41 – 53 odd, 59
#11, 21, 23, 27, 29, 37, 39, 41, 45, 53, 55, 57, 59, 63
Show all your work in a neat and organized manner or you will receive no credit.
(10 problems)
Section 2.1 (p. 77):
Section 2.2 (p. 98):
#20, 24, 48, 54
#40, 42, 46, 60, 64, 70 [set up, but do not solve #64]
15
2-Math 130
Lesson 3
Matrices
Chapter 2, Linear Systems, Sections 2.3, 2.4, and 2.5
Overview
Section 2.3 Gauss-Jordan Method for General Systems of Equations
Recall that back in the beginning of Lesson 2 (in textbook Section 2.1 Algebraic Elimination and this
course guide) we found that there are three possibilities for solutions of a 2 x 2 system of equations. In
this lesson, we will extend that idea to systems of equations and variables larger than 2 x 2. We want to
investigate what happens when using the Gauss-Jordan Method in these cases.
The textbook has 11 examples that are done out in complete detail, so I will not fill in all the details
below. You might try for yourself to see if you can.
If you haven’t done so yet, please read pages 108 through 110 before proceeding.
It’s not always going to be possible, in these cases, to reduce the augmented matrix into the diagonal
form.
1 0 0 a 


0 1 0 b  This form is actually a particular case of what is called the reduced echelon form.
0 0 1 c 


The basic structure for the reduced echelon form is to put a nonzero number in the first row, first
column, just as before and then get every other number in the column to be zeros. Then locate the first
nonzero number in the second row and make all the others in that row a zero. Note that the first
nonzero number in the second row may not be in the second column. That is okay. Then proceed in this
manner until you get to the bottom. If along the way, any row turns into all zeros, then move it to the
bottom.
Take for instance, the following matrix and its reduced form.
1

2
 1
2
3
5
1
3 9
 using the set of rules 
1 4  
 0
change into
0
4 2 
0
1
0
0 5

1 2 
0 3 
16
Notice that we can go no further since all
the numbers in the bottom row that we
would work with are zeros. Let’s consider
what this bottom means. Using the
variables x, y, and z, the bottom row
represents: 0x + 0y + 0z = 3 or 0 = 3. This
statement is obviously false. Therefore,
we conclude that the original system has
no solution.
2-Math 130
Here’s a second type that you will encounter:
 1 4
2 3 using the set of rules 1 4 0 7 


 

change into
 2 8 1 9
0 0 1 5 
Once again we can go no further; the first
row, first column is a one and under it is a
zero. The first nonzero in the second row
has a zero above it.
Let’s consider what these two rows are telling us. The first row gives us x + 4y = -7 and the second row tells
us that z = 5. We have no way to determine a unique solution for x and y, so this is an example where there
are an infinite number of solutions. Let’s consider how we shall represent these solutions. Solving the first
equation for x gives us x = -7 – 4y and since we know no information about y, we say that y can be any
number. This is where the infinite number of solutions comes from. As a system, we now have:
x = 7 – 4y
y = any number
z=5
Instead of writing out “any number” we use another letter called a parameter. The
letter k is usually used if there is only one parameter. If there are two, then it is
customary to use j and k. So, we would write the solution as:
x = 7 – 4k
y=k
z=5
And in ordered triplet form, we would write: (-7 – 4k, k, 5). Note that we can now
find as many particular solutions as we need (recall there are an infinite number)
by choosing values for k. There’s no trick to this. I’m just picking whatever numbers
I want for values of k.
1
4
If k = 0 then
If k = 4 then
If k = -1 then
If k =
then
x = -7 – 4(0) = -7
x = -7 – 4(4) = -23
x = -7 – 4(-1) = -3
x = -7 – 4(
y=0
y=4
y = -1
y=
z=5
z=5
z=5
z=5
1
4
) = -6
1
4
So we have identified four of the infinite number of solutions: (-7, 0, 5), (-23, 4, 5), (-3, -1, 5), and (-6,
and we could find however many we need in this manner.
1
4
, 5)
Section 2.4 Matrix Operations
help.
Section 2.5 Multiplication of Matrices
After Exam 1, in textbook section 2.6 and Lesson 4, you will learn a final method for solving a system of
simultaneous equations. In this section you will learn how to combine matrices by multiplying them
together; a skill that will be needed in the next section.
The first type of multiplication is for multiplying a row matrix times a column matrix. Recall that the
dimensions for a row matrix are of the form 1 x c and the dimensions for a column matrix are of the
form r x 1. In order to multiply these matrices “c” must equal “r”. This type of multiplication of a row
matrix times a column matrix is called a dot product and results in a number, not a matrix.
17
2-Math 130
Definition: The dot product of a row matrix times a column matrix is found as follows:
 c1 
The number found by computing:
c 
2

rn  
 r1  c1  r2  c2  r3  c3 
 rn  cn
 r1 r2
 
 
cn 
 4
For example,  2 3 5  7   (2)(4)  (3)(7)  (5)(6)  8   21  30  17
 
 6 
To multiply two matrices together the number of columns of the first matrix must equal the number of
rows of the second matrix, similar to the dot product, except the result is going to a matrix.
(matrix A)(matrix B) = matrix C
with sizes
(r x n) (n x c) = (r x c)
One helpful trick to keep track of your place is to use the index finger on your left hand to run under the
rows in the first matrix while using the index finger of your right hand to follow down the columns in the
second matrix. Try this on example 2, page 144 in the textbook. With enough practice you can do a lot of
the arithmetic in your head. In this textbook example, using my index fingers as described above, I
would think to myself (or talk out loud, if that helps), “7 (1 times 7) + -6 (3 times -2) + -6 (2 times -3) is 5,” which goes in the first row and first column.
Using the first row again, but now with the second column, repeat the process. Then move on to the
second row in the first matrix dotted with each column in the second matrix.
Two final notes:
1. Notice, from example 5 (p. 146), that unlike multiplication with numbers, multiplication of
matrices is order dependent, i.e. AB does not necessarily equal BA.
2. The identity matrix, discussed on pages 148 – 150 in the textbook, is an important concept that
Self-Study Practice Problems
These are not to be handed in. They are for you to check your understanding of the material; to exercise
your math brain. Make sure you attempt each problem without looking in the solutions manual. It is
crucial for you to practice, to do math, and not just read about it.
Section 2.3 (p. 120):
Section 2.4 (p. 135):
Section 2.5 (p. 151):
# 31, 33, 37, 43, 47, 49, 57, 67, 69
#1, 5, 9, 11, 15, 19, 23 – 33 odd, 37, 43, 45, 47, 51, 55
#1 – 15 odd, 21 – 31 odd, 37, 41, 43, 47, 51, 53, 55, 57
Show all your work in a neat and organized manner or you will receive no credit.
(12 problems)
Section 2.3 P. 120):
Section 2.5 (p. 151):
#2, 4, 6, 8, 32, 34, 60
#14, 16, 28, 38, 54
18
2-Math 130
Exam 1 Information
Make arrangements with your proctor to schedule Exam 1.
Prior to taking this exam:
• You must submit lessons 1, 2, and 3 to your instructor before taking this exam.
• Please do not take this exam until you have received graded lessons 1, 2, and 3 back from your
instructor.
• Do not submit any subsequent lessons until you have taken this exam.
 Below, there is a test outline containing textbook exercises that are similar to what will be on the
test. You should do all of these problems to prepare for the test. You may find that you need to do
them more than once. Keep track of the problems you had to look up in the textbook, in this course
guide, or on your past homework before you could solve the problems. Do those over again until
you can do every problem without any references.
Exam components:
• This exam covers lessons 1, 2, and 3.
• This is a closed-book, closed-notes exam.
• Calculators are NOT allowed.
 The format and questions on this exam are similar to the textbook problems.
• This exam is worth 100 points.
• Time limit: 75 minutes.
Items to take to the exam:
• Photo identification
• V number
• Pencil and eraser
• Graded exams will not be returned to you. However, arrangements can be made to view graded
Exam 1 Topic Outline (Refer to your lessons for page numbers)
1. Find the slope of a line, given 2 points on the line. [1.2 #15, 17]
2. Find the x- and y- intercepts of a line. [1.2 #63, 65]
3. Find an equation of a line. [1.2 #23, 31, 39, 51, 53, 55]
4. Linear Models [1.1 #1, 9, 13–23 odd; 1.2 #93, 95, 101, 103, 109]
5. Cost, revenue, and profit functions [1.3 #1–15 odd, 25, 27, 29, 37]
6. Elimination Method for 2 x 2 systems. [2.1 #17, 29, 31]
7. Gauss-Jordan Method for 3 x 3 systems. [2.2 #39, 41, 45; 2.3 #31, 35, 37]
8. Gauss-Jordan Method for general systems. [2.3 #33, 43, 47, 49, 57]
9. Systems’ Applications: Define variables and set up a system of equations.
[2.1 #43, 45, 51, 59; 2.2 #53, 55, 57, 59, 63; 2.3 #67, 69]
10. Matrices [2.4 #11, 15, 19, 23, 27, 29, 31, 33, 37, 43, 45, 47, 55]
11. Matrix multiplication [2.5 #1, 5, 9–31 odd, 37, 41, 51, 53, 55]
Extra Practice Problems are on the next page.
19
2-Math 130
Extra Practice
The following problems do not necessarily cover all the topics. Make sure you can do the exam outline
problems above. It is also a good idea to make sure you can do the examples that were presented in this
course guide: pp. 61–63 #7, 13, 15c, e, f, 25, 27, 29, 31, 36, 39, 41, 43, 47, 51
pp. 202–204 #3, 7, 9, 11, 15–25 odd, 31, 33, 35, 37, 39, 41
20
2-Math 130
Lesson 4
Matrix Equations
Chapter 2, Linear Systems, Section 2.6
Overview
In this section you will be learning another method for solving systems of equations. “What? Yet another
method?” you say. Yes, but this method, which uses matrices, has many advantages over the methods
you studied in the first three lessons. The most important of which is that this efficiently handles large
numbers of variables. In real science and business applications, you will find that you may be dealing
with dozens or even hundreds of variables. Using algebraic elimination or the Gauss-Jordan method can
be very unwieldy. The method of matrix equations, however, is easily adaptable for the use of a
computer. Consequently, this method is very important.
Section 2.6 Inverse matrices
When dealing multiplication of regular numbers the number 1 is called the identity because any number
multiplied by 1 is itself, i.e. a 1  a and 1 a  a . Also, for numbers with multiplication, the inverse of a
1
number is its reciprocal because any number times its reciprocal equals the identity, i.e. a   1
a
1
and  a  1 .
a
In the last section, you saw that for matrices with multiplication, the square matrix with diagonal
containing 1’s and 0’s in all the other entries is the identity matrix, i.e. A  I  A and I  A  A . Our
immediate goal is to find the inverse of a square matrix, if one exists. The general notation for the
inverse of a matrix A is to write A1 . Use caution here; this is not the algebraic notation meaning
reciprocal. If A is a matrix, A1  1 and actually 1 is meaningless.
A
A
So, given a matrix A, we are searching for a matrix A1 for which A  A1  I and A1  A  I .
First, I will show a quick method that works only for 2  2 matrices.
a b 
d b 
Let A  
. Find the determinant, D  ad  bc . Then A1  1 
. Notice that a and d

D  c a 
c d 
swapped places, while b and c stayed in their positions but became opposite signs.
21
2-Math 130
 3 2 
For example: Find the inverse of 
A
5 4 
(i) D  (3)(4)  (2)(5)  12  10  22
2 2
1
4




4
2


1
(ii) A1 
  22 22    11 11 


22  5 3  5 3   5 3 
 22 22   22 22 
(iii) You can easily check this by multiplying A  A1 to get the identity matrix, I.
1   6 10
3 6 
2


3 2   11 11   11 22 11 22  1 0 

(iv) 


 

 

5 4   5 3   20  20 10  12  0 1 
 22 22   22 22 22 22 
Unfortunately, this short-cut method only works for 2 2 matrices. To find the inverse of any size square
matrix, we use a process similar to the Gauss-Jordan method. Take matrix A and append the identity
matrix to it:  A I  . Use the same rules as the Gauss-Jordan method to turn matrix A into the identity
matrix. When you have accomplished that, matrix I will become the inverse matrix, A1 ,
i.e.  A I 
G-J Rules
 I A1 




 3 2 
For example, using the matrix A from the short-cut example above A  

5 4 
3 2 1 0
3 2 1 0
 A I   

 5R1  3R2  R2 
0 22 5 3
5 4 0 1
R2  11R1  R1 
33 0 6

0 1 5
1


22
R2  R2
22
3
1
 R1
3
33R1

22 

1 0
0 1

6
33
5
22
3
33 
=  I A1 
3 
22 
So, the inverse of matrix A,
3 2
1
6



33 33
11 11 
A1  


 5 3   5 3 
 22 22   22 22 
The same method can be used for any size square matrix. Your textbook section 2.6, example 4 (pp. 1634), shows you the procedure for a 3  3 matrix.
One final note before we see how to solve equations using matrix inversion. Does every matrix have an
inverse? Let’s start with a simpler question, “Does every number have an inverse?” Can you think of a
number that doesn’t have a reciprocal? How about the number zero? So, not every number has an
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2-Math 130
inverse when dealing with multiplication. Now, go back and look at the short-cut method for finding the
inverse of a 2  2 matrix. What could go wrong? Do you see a possible problem? What if the
determinant, D, is zero? In that case, 1 would not exist and the given matrix would not have an
D
inverse. For  4 6  , D  (4)(3)  (6)(2)  0 so this matrix has no inverse. For larger matrices using the
 2 3
Gauss-Jordan method, if a matrix doesn’t have an inverse it will not be possible to turn the matrix into
the identity, similar to the difficulties back in textbook section 2.3 (beginning on p. 107).
Section 2.6 Matrix Equations
Before proceeding, would you please look back in your self-study assignments for lesson 3,
section 2.5 #53 and 55 and also graded problem #54. Notice how for each of these when a square
matrix containing numbers is multiplied by a column matrix containing variables the result is a system of
“equations” without the equals portion. So, our first step is to take a system of equations and rewrite it
in Matrix Form.
For instance,
3x  5 y  4 z  10
4 x  2 y  3z  12 Can be written as
x
 z  2
 3 5 4   x   10 
 4 2 3    y    12

   

 1 0 1   z   2 
This is the first step: to write a system of equations in matrix form A  X  B , where A is the square
coefficient matrix, X is the column variable matrix, and B is the column constants matrix.
Before discussing how to solve such a matrix, let’s look at how a mathematician solves a simple algebra
equation if one had to show all the steps.
This is the same process we will use to solve a matrix equation.
Given:
3
x9
4
4 3
4
 x  9
3 4
3
Multi Multiply both sides on the left by the reciprocal of
1x  6
A number times its inverse is the identity.
x6
The identity times x is still x.
3
.
4
Given: A  x  B
A1  A  x  A1  B
Multiply both sides on the left by the inverse matrix, A1
I  x  A1  B
A matrix times its inverse is the identity matrix, I.
x  A1  B
The identity matrix times x equals x.
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2-Math 130
At the beginning of this lesson we found, using two different methods, that
1
2
3 2 
 11
5 4    5


 22
1
11 
.
3
22 
We can use this information to solve the following system:
3x  2 y  12
matrix
3 2  x  12 
form


    
5 x  4 y  2
5 4   y   2
1
1
3 2 3 2  x  3 2 12 
 
 
   
  
5 4  5 4   y  5 4   2
times by
A1
2
x 
 I      11
 y   5
 22
1

11  12 

3   2 
 
22 
1
since A  A  I
 24 2   22 
 x   11  11   11   2
 
  
 y   60  6   66   3
 22 22   22 
Therefore x  2 and y  3 .
Check by substituting into the original system:
3  2   2  3  6  6  12
5  2   4  3  10  12  2
The same process can be used on any n  n system of equations. Your textbook section 2.6 example 7
shows you the procedure for solving a 3  3 system using a matrix equation.
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2-Math 130
Self-Study Practice Problems
These are not to be handed in. They are for you to check your understanding of the material; to exercise
your math brain. Make sure you attempt each problem without looking in the solutions manual. It is
crucial for you to practice, to do math, and not just read about it.
Section 2.6 (p. 170): #1, 3, 5, 9 – 33 odd, 37, 39, 41, 43, 47
For #3 and #5 you are to check that AB = I.
You may use the short-cut, shown above, for finding the inverse of a 2 x 2 matrix.
Show all your work in a neat and organized manner or you will receive no credit.
(7 problems)
Section 2.6 (p. 170):
#14, 28c, 30c, 32, 34, 40, 46
You may use the short-cut, shown above, for finding the inverse of a 2 x 2 matrix.
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2-Math 130
Lesson 5
Linear Programming: Geometric Approach
Chapter 3, Linear Programming, Sections 3.1, 3.2, 3.3, and 3.4
Overview
Congratulations! You are now ready to delve into the application type we’ve been heading toward called
linear programming (LP).
Linear programming is a method or a process for finding optimal values and solutions. In business, one is
concerned with maximizing profits and minimizing waste; what’s the optimal production schedule that
will accomplish those goals? You will see many other scenarios in the applications where linear
programming is applicable.
We will be looking at two methods for solving LP problems: in Chapter 3 Linear Programming, a
geometrical or graphical method, and in Chapter 4 Linear Programming: The Simplex Method, a
numerical method which has similarities to matrices. The geometric method has the advantage that you
can picture all the possible solutions and then choose the optimal one. Its major disadvantage is that it
only works with two unknowns since we are restricted by our graphing ability. The method in Chapter 4
Linear Programming: The Simplex Method is advantageous because it works with many unknowns and is
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2-Math 130
Section 3.1 Linear Inequalities in Two variables
this material. Here is one additional example for you.
Section 3.1 Exercise #22 p. 211
Let x = number of tables
y = number of desks
Then 2.5x  4 y  350 hours and x  0, y  0
(i) Sketch 2.5x  4 y  350 by finding the intercepts
x-int : Let y  0  2.5x  0  350  x  140  (140,0)
y-int: Let x  0  0  4 y  350  y  87.5  (0,87.5)
Therefore,
y = # of desks
87.5
10
x = # of tables
10
140
(ii) Determine whether the solution set is above or below this line by choosing any
point, not on the line, to see if it satisfies the inequality.
Choosing (10,10) we find:
2.5(10)  4(10)  25  40  65  350 , so (10, 10) is included in the solution and
we shade the triangular region below the line segment.
y
87.5
10
*
x
10
140
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Section 3.2 Systems of Linear Inequalities; Feasible sets
In order to solve LP problems, you are going to have to find the region where inequalities overlap and
also the corners of such regions. The region of common points is called the feasible region. Here are two
examples of feasible regions:
bounded feasible region
unbounded feasible region
y
y
A
A
B
C
D
x
B
4 corners: A, B, C, and D
2 corners: A and B
28
x
2-Math 130
For example: Sketch the feasible region and find the corners for:
x  2y  4
4 x  4 y  4
x  0, y  0
The basic procedure is to sketch each line using the intercepts. Then use a checkpoint
y-int
x-int
(0,0)
included?
l:
(0,2)
(4,0)
x  2y  4
0  0  4 , yes
m: 4 x  4 y  4
(0,1)
(-1,0)
0  0  4 , yes
x  0, y  0 feasible set in the first quadrant only
y
l
Corners:
A (0, 0) the origin
B (0, 1) the y-int of line m
C l m see below
D (4, 0) the x-int of line l
m
C
B
x
A
D
l m : Point C is the intersection of lines l and m:
x  2y  4
4 x  4 y  4
2R1
2x  4 y  8
4 x  4 y  4
6x  4
4 2
x 
6 3
Then using line l: 2  2 y  4
3
10
5
2y   y 
3
3
Therefore, point C is
( 23 , 53 ) .
As you will see in the next section, the importance of finding the corners of a feasible set is that
mathematicians have proven that the optimal solution will be at a corner point.
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2-Math 130
Section 3.3 Linear Programming
We now are ready to combine all the steps into the process for solving a linear programming problem by
graphing.
1. Define the variables.
2. Write the objective function, sometimes called the target. This is the function that we are
trying to optimize.
3. Write the constraint inequalities, sometimes called the restrictions.
4. Sketch the constraint inequalities to find the feasible region.
5. Find the corners of the feasible region.
6. Evaluate the objective function at each corner to find the optimal solution.
Since there are many steps to solving one of these problems, please concentrate on doing your work in
an organized manner.
A NASA nutritionist wants to use two different tubes of food to meet the requirements of
at least 42 units of protein, 20 units of carbohydrate, and 18 units of fat, while keeping
the weight of the food at a minimum. Each tube of food A supplies 4 units of protein, 2
units of carbohydrate, 2 units of fat, and weighs 3 pounds. Each tube of food B contains 3
units of protein, 6 units of carbohydrate, and 1 unit of fat and weighs 2 pounds. How
many tubes of each food should be given to the astronauts?
Organize the data:
A
protein 4
carbs
2
fat
2
weight 3
B
3
6
1
2
total
42
30
18
min
Define the variables:
Let x  number of tubes of food A
Let y  number of tubes of food B
Write the objective function (target):
Minimize W  3x  2 y
Write the constraint inequalities (restrictions):
y-int
x-int
(0,0)?
l1 : 4 x  3 y  42
(0,14) (10.5,0)
no
2
x

6
y

30
l2 :
(0,5)
(15,0)
no
l3 : 2x  y  18
(0,18)
(9,0)
no
x  0, y  0
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2-Math 130
Sketch the inequalities and the feasible set:
y
l3
18
A
l1
14
B
l2
5
C
D
x
9 10.5
15
Find the corners and evaluate the target:
W  3x  2 y
A: y-int of l3= (0,18)
0  36  36
B: l1 l3  (6, 6)
18  12  30
C: l1 l2  (9, 2)
27  4  31
D: x-int of l2 = (15,0)
45  0  45
[Use algebraic eliminations to find B and C]
Solution
To minimize the weight, she should supply 6 tubes of each food to each astronaut.
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2-Math 130
Section 3.4 applications
This section deals with setting up applications that have more than two variables. You will learn how to
solve these in Chapter 4 Linear Programming: The Simplex Method.
The three examples in the text illustrate the types of problems you are expected to be able to set up.
Here is an alternative way to organize the data in example 3 (pp. 253-5):
Calvert  80
x1
A  62
x2
x3
Hico  65
B  90
x4
If you find this diagram helpful, it will suffice for the “defining the variables” step, and you can set up the
inequalities directly from the diagram.
Self-Study Practice Problems
These are not to be handed in. They are for you to check your understanding of the material; to exercise
your math brain. Make sure you attempt each problem without looking in the solutions manual. It is
crucial for you to practice, to do math, and not just read about it.
Section 3.1 (p. 211):
Section 3.2 (p. 219):
Section 3.3 (p. 240):
Section 3.3 (p. 240):
Section 3.4 (p. 254):
#3, 7, 11, 13, 17, 19, 21, 27, 29, 31
#1, 3, 7, 9, 13, 15, 17, 25, 27
Set-up, but do not solve the L.P. problem for the following:
#1, 39, 43, 45, 47, 49, 53, 59
Solve each of the following: #3, 5, 7, 9, 17, 19, 23, 27
#1, 3, 5, 7, 9
Show all your work in a neat and organized manner or you will receive no credit.
(11 problems)
Section 3.2 (p. 219):
Section 3.3 (p. 240):
Section 3.4 (p. 254):
#14, 22, 26
#24, 26, 38, 40, 42
#4, 6, 10
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2-Math 130
Lesson 6
Linear Programming: Simplex Method
Chapter 4, Linear Programming: The Simplex Method, Sections 4.1, 4.2, and 4.3
Overview
The graphical method for solving LP problems that you learned in the previous lesson is nice because
you have a picture and can see what is happening during the process. On the other hand, it is very
limited in its usefulness since one can only solve problems with no more than two variables. In reality, LP
problems usually have dozens of variables to consider. Those problems would be solved using the
method in this lesson and a computer. Since you are just learning this method and doing the problems
by hand, we will limit the number of initial variables to no more than four, with most problems having
just three variables.
The Simplex Method has variations depending upon the objective and constraints in the problem. We
will only be looking at two of the many variations. The first is for a standard maximization problem and
the second is a method for taking a standard minimization problem and turning it into a standard
maximization problem. Some of the justifications and reasoning behind these methods are given in the
textbook and in the overview below, but most of the mathematics behind these methods is beyond the
scope of what we are studying.
Section 4.1 and 4.2 The Simplex Method
A Standard Maximization Problem has three components:
1. The objective function is to be maximized.
2. All (independent) variables must be non-negative (  0 )
3. Constraints must be of the form: expression  non-negative constant, except for constraints
which state that the variables themselves are non-negative (i.e. x  0 ).
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2-Math 130
Consider the following example:
Maximize the objective function: P  2 x1  3x2  x3
Subject to the constraints: x  x  x  7
1
2
3
3 x1  x2  2 x3  9
x1  0, x2  0, x3  0
This simplex method requires that we use equations, not inequalities, so we add slack variables to make
up the difference.
s1 takes up the slack ( s1  0)
x1  x2  x3  7 becomes x1  x2  x3  s1  7
3x1  x2  2x3  9 becomes 3x1  x2  2 x3  s2  9
s2 takes up the slack (s2  0)
This method also requires that the objective be in the equation form: expression + P = 0.
So, P  2 x1  3x2  x3 becomes 2 x1  3x2  x3  P  0 by subtracting the variables over to the left.
So, our standard maximization problem:
Maximize: P  2 x1  3x2  x3
x1  x2  x3  7
Subject
to:
3 x1  x2  2 x3  9
x1  0, x2  0, x3  0
Becomes:
x1  x2  x3  s1
3 x1  x2  2 x3
 2 x1  3 x2  x3
7
 s2
9
P 0
x1  0, x2  0, x3  0, s1  0, s2  0
The next step is to set up a matrix from the system of equations we developed. This matrix is called the
simplex tableau.
x1 x2 x3 s1 s2 P
x1  x2  x3  s1
7
Convert to matrix
3x1  x2  2 x3
 s2
 9 

initial simplex tableau
2 x1  3x2  x3
P 0
 1 1
1 1

1 2 0
 3
 2 3 1 0

0
1
0
0 7

0 9
1 0 
In a simplex tableau, the variables with only 1’s and 0’s in their columns are called the basic variables;
the other variables are called non-basic. In our set-up above, the variables s1 , s2 , and P are currently the
basic variables. Some or all of the variables will transition between basic and non-basic as we
manipulate the simplex tableau.
A feasible solution (though not optimal) is at hand, if all non-basic variables are set to zero.
If x1  0, x2  0, and x3  0, then x1  x2  x3  s1  7 becomes 0  0  0  s1  7.
Therefore s1  7 and, similarly, s2  9 and P  0.
The simplex method moves from a non-optimal solution to one at least as good, if not better. The
method cycles through solutions until an optimal solution (maximum P value) is found. Section 4.1 (p.
262) deals with setting up the initial simplex tableau, and then section 4.2 (p. 273) deals with solving the
tableau for the optimal solution.
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2-Math 130
A solution is optimal if no negative numbers are in the bottom row, left of the vertical line. Negative
numbers in the bottom row mean the maximum solution has not yet been found, so we must pivot.
To pivot:
1. Choose the column that has the most negative number in the bottom row. This is called the
pivot column. In the above example, it would be the x2 – column.
2. Form row ratios: For each row with a positive number in the pivot column, divide the right most
entry in that row by the number in the pivot column, forming a row ratio. The pivot row has the
smallest row ratio.
3. The pivot element is the intersection of the pivot column and the pivot row.
4. To pivot on the pivot element, use row operations to make a unit column. That is, make the
pivot element a one (1) and all other entries in the pivot column zeros (0).
5. To determine if this is now the optimal, maximum solution, check for negative numbers in the
bottom row, left of the line. If there are still negative numbers, repeat the process starting back
at step one.
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2-Math 130
Example for section 4.2:
Maximize P  80 x1  70 x2 subject to
6 x1  3 x2  96
x1  x2  18
2 x1  6 x2  72
x1  0, x2  0
Step 1: Add the slack variables to make a system of equations and rewrite the objective equation.
6 x1  3 x2  s1
 96
x1  x2
 s2
 18
2 x1  6 x2
 s3
 72
80 x1  70 x2
P  0
Note to remember:
x1, x2 , s1, s2 , s3  0, and we want P to be as large as possible.
Step 2: Set up the initial tableau.
x1
x2 s1 s2 s3 P
3
 6
 1
1

 2
6

 80 70
1
0
0
0
0
1
0
0
0
0
1
0
0 96 
0 18
0 72 

1 0 
In order to solve for the optimal solution recall:
 Pivot Column – The column with the most negative number in the bottom row. In the above
example, that would be the x1 – column.


Pivot Element – First, it must be positive. Second, choose the element with the smallest nonnegative ratio with the last column. In the above example,
x1
ratio
96 / 6  16  smallest ratio,
96 
6
 1 ... 18
18 /1  18
so pivot on the 6


2
72 
72 / 2  36




You have found the maximum value when the bottom row contains no negative numbers.
Read through the textbook examples carefully, stopping to do each exercise as indicated. The simplex
method is a long process, so take the time to master each step as you go along.
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2-Math 130
Section 4.3 The Dual Method
In this section, you will learn how to change a standard minimization problem into a standard
maximization problem so that the simplex method can be used to solve it. A standard minimum problem
still assumes that all variables are non-negative, but all the constraints are of the form:
[expression  non-negative constant].
First, you need a new concept called the transpose of a matrix. To transpose a matrix, you swap its rows
and columns, so an n  m matrix becomes an m  n matrix.
For example:
1 
1 2 3  2
3
t
 3 2
 4 1   3 4 5

  2 1 8

 5 8 
t
We use this concept of transposition to change a standard minimum problem, called the primal system
into a standard maximum problem, called its dual system. In general, using matrix notation:
Primal system
Dual System
 At  Y  C t
A X  B
 transposed 
X  0    Y  0
t

min C  X 
 max B  Y
The dual system can be solved using the simplex method and the maximum value of the dual equals the
minimum value of the primal.
Example of transposing a primal system into a dual system:
Primal System
Dual System
2 x1  3x2  1 
2 y1  3 y2  1y3  3
3x1  4 x2  2 


 3 ineq transposed 2 ineq 3 y1  4 y2  1y3  5
1x1  1x2  4  and  and 
2 var
3 var  y1  0, y2  0, y3  0
x1  0, x2  0 


max  1y1  2 y2  4 y3
min  3x1  5 x2 

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2-Math 130
Complete example 1:
Step 1: Convert to the dual system.
Primal System
Dual System
3 y1  6 y2  18

3x1  5 x2  2 x3  4 

5 y1
 20
3i
 2i

6 x1
 8 x3  9  3v

2v
2 y1  8 y2  2
  
x1  0, x2  0, x3  0 

y1  0, y2  0

min C  18 x1  20 x2  2 x3 


max P  4 y1  9 y2
Step 2: Add slack variables x1 , x2 , and x3 .
3 y1  6 y2  x1
 18
 x2
5 y1
2 y1  8 y2
 20
 x3
4 y1  9 y2
 2
P 0
Step 3: Set up the simplex tableau.
y1
y2 x1 x2 x3 P
6 1 0 0
 3
 5
0 0 1 0

 2 8 0 0 1

 4 9 0 0 0
0 18
0 20 
0
2

1 0 
Step 4: Solve by pivoting as in section 4.2.
y1 y2 x1 x2 x3 P
0
1

0

0
1 16
0 0
0 43
0 32
Primal Solution
1
10
1
5
2
5
1
10
0
0
1
0
0
5
0
4 
0 34 

1 29 
Dual Solution
max  29
when y1  4
y2  5
min  29
1 , x 0
when x1  32 , x2  10
3
Notice that the solution to the original minimum problem is found simply by reading the values in the
bottom row under the x-columns.
38
2-Math 130
Complete example 2:
Primal
Dual
y1
3
x1  x2  x3  20
y1  y2  5
2i
3i
x2  2 x3  0
3v
2v
y1  2 y2  1


x1  0, x2  0, x3  0
y1  0, y2  0
min C  3 x1  5 x2  x3
max P  20 y1
You may skip step 2 and go directly to the tableau, if you wish.
y1 y2 x1 x2 x3 P
 1
 1

 1

 20
0
1
2
0
1
0
0
0
0
1
0
0
y1
3
5
pivoting


1

0 
Solution: min  20
0
0
1
0
0
0
0
1
y2 x1 x2 x3 P
0 2 1 0 1
0 1 0 1 1

1 2 0 0
1

0 40 0 0 20
0
2
0
4 
0
1

1 20
when x1  0, x2  0, x3  20
Self-Study Practice Problems
These are not to be handed in. They are for you to check your understanding of the material; to exercise
your math brain. Make sure you attempt each problem without looking in the solutions manual. It is
crucial for you to practice, to do math, and not just read about it.
Section 4.1 (p. 269):
#1 – 11 odd
Section 4.2 (p. 288):
#5 – 15 odd, 23, 29, 33, 35
Section 4.3 (p. 305):
#1 – 15 odd;
On #5 and #7, if you can do part (c) without first doing (a) &amp; (b) then that is acceptable.
Show all your work in a neat and organized manner or you will receive no credit.
(12 problems)
Section 4.1 (p. 269):
Section 4.2 (p. 288):
Section 4.3 (p. 305):
#8, 10
#6, 22, 26, 28
#6c, 8c, 10, 12, 14, 16
39
2-Math 130
Exam 2 Information
Make arrangements with your proctor to schedule Exam 2.
Prior to taking this exam:
• You must submit lessons 4, 5, and 6 to your instructor before taking this exam.
• Please do not take this exam until you have received graded lessons 4, 5, and 6 back from your
instructor.
• Do not submit any subsequent lessons until you have taken this exam.
 Below, there is a test outline containing textbook exercises that are similar to what will be on the
test. You should do all of these problems to prepare for the test. You may find that you need to do
them more than once. Keep track of the problems you had to look up in the textbook, in this course
guide, or on your past homework before you could solve the problems. Do those over again until
you can do every problem without any references.
Exam components:
• This exam covers lessons 4, 5, and 6.
• This is a closed-book, closed-notes exam.
• Calculators are NOT allowed.
 The format and questions on this exam are similar to the textbook problems.
• This exam is worth 100 points.
• Time limit: 75 minutes.
Items to take to the exam:
• Photo identification
• V number
• Pencil and eraser
• Graded exams will not be returned to you. However, arrangements can be made to view graded
Exam 2 Topic Outline (Refer to your lessons for page numbers)
1.
The “short-cut” method for finding the inverse of a 2 x 2 matrix. [2.6 #9, 11]
2.
The inverse of a square matrix. [2.6 #11, 13, 23, 25]
3.
Matrix form [2.6 #27c, 29c, 31, 33]
4.
Solving a system by matrix inversion. [2.6 #37, 41, 46]
5.
Systems of inequalities; feasible set; corners [3.2 # 25, 27]
6.
Geometric approach to linear programming [3.3 #21, 23, 25]
7.
Converting a system of inequalities, with its objective, into a system of equations. [4.1
#5, 7]
8.
Simplex tableau for a maximum problem [4.1 #9, 11]
9.
Simplex method for linear programming [4.2 #7, 9, 11, 13, 15, 23, 25, 33]
10.
The dual of a minimum problem [4.3#5c, 7c, 9, 11]
11.
Solving a minimum linear programming problem using its dual. [4.3 #13, 15]
40
2-Math 130
Extra Practice Problems are on the next page.
Extra Practice – The following problems do not necessarily cover all the topics. Make sure you can do
the exam outline problems above. It’s also a good idea to make sure you can do the examples that were
presented in this study guide.
p. 203 #27, 29
pp. 258 – 259 #3, 5, 9, 11, 13, 17, 21, 23
pp. 358 – 359 #1 – 13 odd, 19, 27, 29, 37, 41
41
2-Math 130
Lesson 7
Sets and Counting, Part I
Chapter 6, Sets and Counting, Sections 6.1, 6.2, and 6.3,
Overview
Congratulations! You have now completed two-thirds of the course and will be studying completely
different concepts. Chapters 6 and 7 deal with sets, counting, and probability. These chapters form a
foundation for future study on statistics, which nearly all degrees now require.
Before starting this new material, I invite you to re-read the paragraphs under the “self-study practice
problems” on page 3 of this course study guide. Remember that false starts and practicing many
problems are necessary to master new concepts in mathematics. Good Luck!
Section 6.1 Sets
A set is simply a group or a collection. This section covers the basic terms and notation when dealing
with the concept of a set. Read it carefully; take notes to ensure you understand the key concepts as
listed below:
 Set
 Element or member
 Empty or null set
Note: We use either empty braces { } or the symbol  to denote an empty set. Do not
combine the two.  is a set within a set and, therefore, it is not empty.
(Strange, I know.)
 Equal sets
 Explicit or roster vs. implicit or set-builder notation
e.g. a, e, i, o, u  x x is a vowel of the English alphabet
Note:
To write simply  x x is a vowel would not be a well-defined set since there are
vowels in other languages besides a, e, i, o, u.
 Universe or universal set
 Subset
Note: For our studies, the difference between the symbols  and  are not crucial,
so we will use  which is more general.
 Complement of a set
Note 1: If you use any other texts or any material on the internet, you should be aware
c
that some authors use A and some use A for the complement of A. You may
use whichever notation you wish. A  Ac  A
c
Note 2: Think of the complement as meaning “not.” Thus, E means “not E.” So, if E
was the set of all even whole numbers in the Universe of all whole numbers, then
E would be the set of all odd whole numbers.
42
2-Math 130
 Union
Note:
Think “or,” but keep in mind that mathematically the word “or” means one or
the other or both, whereas in many everyday usages of the word “or” we mean
an exclusive “or” such as, “Would you like potatoes or rice with dinner?”
 Intersection
Note: Think “AND.” It must be in both sets.
 Disjoint sets
 De Morgan’s Laws
These are two rules having to do with complements, union, and intersection which are
not covered in your text. You may find that they make some problems easier.
The two laws are as follows:
i)  A B   A B
not (A or B) = (not A) and (not B)
ii)  A B   A B
not (A and B) = (not A) or (not B)
Example for De Morgan’s Laws
Let U  a, b, c, d , e, f , g , A  a, b, g, and B  b, d , e, f 
Then
A
and
B   a, b, d , e, f , g   c
A B  c, d , e, f 
a, c, g  c
Section 6.2 Counting Using a Venn Diagram
Venn diagrams are a useful tool for counting elements in a set. We will use them again in the next
chapter on probability, so this is an important technique.
When given the number of elements in the overlapping region (intersection) of two or three sets, I
always start with a Venn diagram as is shown in examples 2 (p. 433) and 6 (pp. 435-6). Start from the
inside (intersection) of the sets and work your way out. Note that on example 4 (p. 434), I would start
with a Venn diagram as illustrated on the bottom of page 434 in figure 6-9. After filling in the Venn
diagram, I would use it to answer the questions (a) through (d).
When the number of elements in the intersection of two sets is unknown, then you should use the
formula as illustrated in example 3 (p. 433). You will not be given any problems involving three sets
unless the number of elements in the intersection of all three sets is also given.  You’re welcome. 
Section 6.3 Basic Counting Principles
The most important concept in this section is the Multiplication Rule on page 443. You will use it
repeatedly for the remainder of this course. While tree diagrams can be useful for visualizing the
Multiplication Rule, their most important use won’t be seen until section 7.6 in lesson 10.
43
2-Math 130
You may find it helpful to think of the multiplication principle as “fill in the slots” problems. For instance,
if I’m going to choose a hat, a scarf, and a coat to wear from my wardrobe which consists of 4 hats, 6
scarves, and 2 coats, then I have 3 “slots”:
. I fill in the “slots” with the number
hats
scarves
coats
of available choices and then multiply those numbers to get the total number of possible outfits:
4
∙
2
∙
6
= 48 possible outfits.
hats
scarves
coats
Notice that we are choosing a hat AND a scarf AND a coat. It is the “AND” which tells us to use the
Multiplication Rule.
Note how this is different from the following problem which uses “OR”. Let’s suppose I’m choosing one
dessert from five types of pies and three types of cakes, then I am choosing a piece of pie OR a piece of
cake and the number of possibilities is 5  3  8.
As always, there is no substitute for practice when studying the different ways that the multiplication
principle can be applied. Here are some more examples to help you with this section.
Example: How many ways can 5 CD’s be arranged on a shelf?
5 CD’s  5 slots  5  4  3  2  1  120
Example: There are 3 Jazz CD’s, 4 rock ‘n roll CD’s, and 3 country &amp; western CD’s.
a) How many ways can they all be arranged on a shelf?
10 CD’s  10 slots  10  9  8  7  6  5  4  3  2  1  3, 628,800
b) How many arrangements are there if the types of music must be grouped together?
This type of problem involves using the multiplication principle more than once.
For the order of the music types there are 3  2 1  6 orders. Then we have to
count the order of the individual CD’s within each group:
3  2  1  4  3  2  1  3  2  1  6  5,184
Jazz
R 'n R
Country
Example: How many three-letter “words” can be formed if
a) repetition of letters is allowed?
b) repetition is not allowed?
26  26  26  17,576
26  25  24  15,600
44
music
type
2-Math 130
Textbook example 8 (pp. 444-5):
Alternate solution: First arrange the three couples: 3 ∙ 2 ∙ 1 but then there are two orders for
each couple to have sat down in (partner one, partner two OR partner two, partner one). So, the
total possible seating orders are: 3 ∙ 2 ∙ 1 ∙ 2 ∙ 2 ∙ 2 = 48
This last example illustrates that it’s sometimes easier to count the number of ways to not do an
arrangement (the complement) and then subtract that number from the total possible arrangements.
We will revisit this method again in section 6.6 and in chapter 7. For now, it should help you do exercise
#54 (p. 450).
Example: How many ways can seven students be assigned seats in a row of seven desks if there are two
particular students who are not allowed to sit next to each other?
i)
The total number of ways to arrange the seven students with no restrictions is
7  6  5  4  3  2 1  5040 .
ii) To find the number of ways to arrange the seven students if the pair in question is
sitting next to each other, first think of the pair as one student, leaving 6 students
and six slots, so 6  5 4  3 2 1 , but then as before there are two orders for the pair.
So, the total number of seating arrangements with the pair next to each other is
6  5  4  3  2 1 2  1440 .
iii) Therefore, the number of arrangements for the particular pair of students to not sit
beside each other is 5040  1440  3600 .
Self-Study Practice Problems
These are not to be handed in. They are for you to check your understanding of the material; to exercise
your math brain. Make sure you attempt each problem without looking in the solutions manual. It is
crucial for you to practice, to do math, and not just read about it.
Section 6.1 (p. 430):
Section 6.2 (p. 436):
Section 6.3 (p. 447):
#1 – 19 odd, 23 – 53 odd, 57 – 71 odd
#3 – 11 odd, 15 – 27 odd, 33, 39
#1 – 11 odd, 15 – 33 odd, 49 – 61 odd
Show all your work in a neat and organized manner or you will receive no credit.
(15 problems)
Section 6.1 (p. 430):
Section 6.2 (p. 436):
Section 6.3 (p. 447):
#42, 44, 52, 70, 72, 74
#12, 14, 20, 22, 24, 28
#50, 52, 54
45
2-Math 130
Lesson 8
Sets and Counting, Part II
Chapter 6, Sets and Counting, Sections 6.4, 6.5, and 6.6
Overview
These next three sections from Chapter 6 Sets and Counting deal with two types of arrangements called
permutations and combinations. These will be used extensively in the chapter on probability.
*A note about the use of calculators: please use a calculator to perform the arithmetic in this section,
but keep in mind that you will not have a calculator on the exam. The numbers on the exam will be of
reasonable magnitude for you to compute by hand. You may need to do some cancelling such as:
10  9  8  7
10 2 9 3 8 27 2  3  2  7 3  2  7



 42
5  4  3  2  1 5  4  3  2 1
1
2 1
Section 6.4 Permutations
A permutation is a special case usage of the multiplication rule where each element is chosen from the
same set, repetition of elements is not allowed, and the order of choice is important. Here are two more
examples to accompany the ten examples given in the text.
Example: How many ways are there for five people to get in line?
“In line”  order matters  P(5, 5)
or 5!
or 5  4  3  2  1  120
Sometimes there’s more than one way to think about a problem.
Example: How many ways are there to elect two officers, a president and vice president, from a group
of five eligible people?
“officers”  order matters  P(5, 2)
or 5  4  20
46
2-Math 130
Section 6.5 Combinations
A combination differs from a permutation only in the fact that the order of choice is not important.
Keep the concept of a committee (a group of people whose order does not matter) in mind when
dealing with combinations. You may skip the material on the Binomial Theorem, text pages 470 – 472.
The first twelve examples, along with the four presented here sufficiently cover combinations.
Example: How many ways are there to choose a three-person committee from a group of five people?
“committee”  order is irrelevant  C (5,3)
or P(5,3)
3!
or 5  4  3  10
3!
Example: How many different pizzas can be made with three toppings if there are six toppings
available?
“pizza toppings”  mathematically it makes no difference as to the order we put the
toppings on a pizza, so  C (6,3)

654
 20
3!
Example: Twelve students apply for three assistantships. How many ways can these be awarded if
a) no candidate is given preference?
b) there are seven men and five women and at least one woman must be given a position?
“assistantships”  similar to a committee
(a) C (12,3)  12  11  10  220
3!
(b) three possibilities:
(three women and no men) OR (two women and one man) OR (one woman and
two men)
C (5,3)  C (7, 0)  C (5, 2)  C (7,1)  C (5,1)  C (7, 2)
5  4  3
5  4
7  6

1 
7  5
3!
2!
2!
 10  70  105  185
Notice and recall “AND”  multiply
47
2-Math 130
Example: From a group of ten students, how many ways are there to
a) choose at least nine of them?
b) choose at least two of them?
c) choose no more than eight students?
(a) Choosing: nine students OR all ten
C (10,9)  C (10,10)
(b) Choosing: two students OR three students OR…all ten. There must be an easier way!
(Review the last example from Lesson 7 Sets and Counting, Part I in this study
guide.)
 at least 2 students 
C
  no student OR one student 
Now, think of each student as either being chosen or not being chosen.
So, the number of ways to choose at least two students is
possible
of ways
Number of ways
  Number
Totalchoices
to choose none   to choose one 
210  C (10, 0)  C (10,1)
(c) Similarly, it is easier to count the complement in this case.
possible
of ways
Number of ways
Number of ways
  Number
Totalchoices
to choose 8   to choose 9   to choose all 10 
210  C (10,8)  C(10,9)  C(10,10)
Section 6.6 A Mixture of Counting Problems
This section will give you practice in distinguishing between permutations and combinations. Also, you
will see some problems where a permutation and a combination are used jointly. Here is one more
example to help you with the multiplication principle, permutations, combinations, “AND”, and “OR”.
Example: There are seven math, eight science and five literature books in a pile. How many ways are
there to:
a) line them all up?
b) line them all up, if the subjects must be grouped together?
c) choose any three books?
d) choose one book from each subject?
e) choose two books from each subject?
f) choose exactly two books from one subject?
(a) 20  19  18  1 or 20!
(b) 7!  8!  5!  3!
subject
order
(c)
(d)
(e)
(f)
C (20,3)
7  8  5 or C (7,1)  C (8,1)  C (5,1)
C (7,2)  C (8,2)  C (5,2)
two math OR two science OR two literature
C (7,2)  C (8,2)  C (5,2)
48
2-Math 130
Self-Study Practice Problems
These are not to be handed in. They are for you to check your understanding of the material; to exercise
your math brain. Make sure you attempt each problem without looking in the solutions manual. It is
crucial for you to practice, to do math, and not just read about it.
Section 6.4 (p. 459):
Section 6.5 (p. 473):
Section 6.6 (p. 482):
#1 – 41 odd, 45, 57, 59
#1 – 11 odd, 19, 23, 25, 29, 31, 41, 43, 45, 47
#1, 3, 5, 9, 11, 13, 17, 19, 21, 23, 31
Show all your work in a neat and organized manner or you will receive no credit.
(17 problems)
Section 6.4 (p. 459):
Section 6.5 (p. 473):
Section 6.6 (p. 482):
#8, 38, 60, 62
#34, 36, 38, 42
#10, 12, 14, 16, 18, 28, 30, 32, 34
49
2-Math 130
Lesson 9
Probability, Part I
Chapter 7, Probability, Sections 7.1, 7.2, and 7.3
Overview
As you are about to start the last chapter for this course, if you haven’t given yourself a congratulatory
treat in a while, do so now. It’s important to celebrate the small victories and prepare for the work
Sections 7.1, 7.2, 7.3 Basic Probability
The first three sections in this chapter contain a lot of new vocabulary and formulas. Instead of outlining
each section separately, I’m going to list the most important terms and formulas so you’ll have them
together as a reference and then conclude with a few examples to supplement those given in the text.
7.1
7.1
7.1
7.1
Experiment – An activity that has observable results.
Sample space, S – The set of all possible outcomes of an experiment.
Event, E – Any subset of a sample space.
P( E ) – The relative frequency of the event.
7.1 0  P( E )  1
7.1 P   0
7.1 PS  1  Ps1  Ps2   . . .  Psn , where S  s1 , s2 , . . . , sn
number of ways E can occur
, if the simple outcomes of E are
equally likely to occur
total number of possible outcomes
7.3 PE  1  PE
7.2 P( E ) 
7.3 Mutually exclusive events – Two events, E and F, are mutually exclusive if, as sets, they are
disjoint. i.e., E F    and PE F  0
7.3 PE or F  PE
F   PE  PF   PE
F
Note that if E and F are mutually exclusive then PE
F   PE  PF  . However, be
very, very careful to not assume mutually exclusive events without the proper
information. Peoples’ intuition is often wrong in these cases.
Example: Experiment: Roll two distinguishable dice. (This is the same as rolling a single die twice.)
a) Find the probability of rolling at least one four.
b) Find the probability of rolling a sum of at least five.
a) E  1, 4, 2, 4, 3, 4, 4, 4, 5, 4, 6, 4, 4, 1, 4, 2, 4, 3, 4, 5, 4, 6
11
(See the table on page 512 [Figure 7-5] of the textbook.)
36
b) E  1, 4, 2, 3, 3, 2, 4, 1
PE 
50
2-Math 130
PE 
4
1

36 9
Example: Experiment: Roll one 20-sided die.
Find the probability of rolling a multiple of 3 or a multiple of 5.
Let T = rolling a multiple of 3 and let F = rolling a multiple of 5.
Then nT   6 and nF  4 .
Caution here, the answer is not 10 ! Why not? Because T and F are not mutually
20
exclusive events. There is one number common to both events. The number 15 is a
multiple of both 3 and 5.
T   3, 6, 9, 12, 15, 18  and F   5, 10, 15, 20 
So, Pmultiple of 3 OR multiple of 5  PT or F   PT
6
4
1
9
 PT   PF   PT F  



20 20 20
20
F
Example: Experiment: There are five red, four white, and one black marble in a sack.
a) If one marble is chosen without looking, what is the probability that it is not white?
b) If two marbles are chosen without looking, what is the probability that they are the same color?
a)
Pnot white  Pred OR black 
5 1
6
3


10
10 5
Or Pnot white  Pred  Pblack  Pred and black  5  1  0  6  3
10 10 10 10 5
4
6
3
Or 1  Pwhite  1 


10 10 5
C5, 2  C4, 2 5241  4213 10  6 16
b) Psame color  Prr OR ww 
 109 

C10, 2
45
45
2 1
Example: Experiment: A family is planning on having three children. What is the probability of having:
a) Exactly two girls?
b) At least two girls?
We make the assumption in probability that a baby is equally likely to be born male or female. Because
of this assumption, you can use a tree diagram, similar to the one in the textbook in example 5 of
section 7.2 (pp. 513-4) to find all the possible gender outcomes.
3
a) Pexactly 2 girls  Pggb or gbg or bgg 
8
51
2-Math 130
4 1

8 2
Example: Experiment: Toss a coin ten times and observe the number of heads. Find the probability of
getting:
C10, 2  C8, 8 10219  1
45
a) P2H  P2H and 8T 
 10 
 0.0439
10
2
2
1024
b) P at least 2H  1  P 1H OR 0H  1  P 1H AND 9T OR 0H AND 10T
b)
Pat least 2 girls  Pggb or gbg or bgg or ggg 




 

C10, 1  C9, 9  C10, 0  C10, 10
1

210
10  1 1024
11
1013
1



 0.9893
10
2
1024 1024 1024
Example: A club is comprised of twelve females and eight males. Find the probability that a random
committee of five contains:
a) Two males
b) At least one male
c) At least two males.
n2M and 3F
C8, 2  C12, 3
6160
a) P2M and 3F 


 0.3973
ncommittees of 5
C20, 5
15,504
b) Pat least one male  1  Pno males  1  Pall females
1
c)
C8, 0  C12, 5
C20, 5
1
792
 0.9489
15,504
Pat least 2 males  1  P1 male or 0 males
C8, 1  C12, 4  C8, 0  C12, 5
3960  792
1
1
 0.6935
C20, 5
15,504
Self-Study Practice Problems
These are not to be handed in. They are for you to check your understanding of the material; to exercise
your math brain. Make sure you attempt each problem without looking in the solutions manual. It is
crucial for you to practice, to do math, and not just read about it.
Section 7.1 (p. 506):
Section 7.2 (p. 515):
Section 7.3 (p. 528):
#1, 5 – 19 odd, 23
#1, 3, 7, 9, 13, 17, 19, 21, 23, 25, 27, 37, 39, 41, 43, 55
#1 - 15 odd, 19, 21, 25, 27, 29, 33, 35, 37
Show all your work in a neat and organized manner or you will receive no credit.
(15 problems)
Section 7.1 (p. 506):
#10, 14, 24
52
2-Math 130
Section 7.2 (p. 515):
Section 7.3 (p. 528):
#6, 10, 18, 20, 24, 26, 30, 58
#16, 18, 22, 34
53
2-Math 130
Lesson 10
Probability, Part II
Chapter 7, Probability, Sections 7.4, 7.5, and 7.6
Overview
These next three sections will finish the topics that are considered basics of probability.
Section 7.4 Conditional Probability
I’m going to start right in with an example to reinforce the textbook’s introduction to the concept of
conditional probability.
Example: A sack contains two red and three black chips which are also numbered as shown here:
1
2
3
4
5
The experiment is to randomly choose one chip. Let E be the event that an even number is on the chip
and let B be the event that the chip is black. Furthermore, let’s assume we have an accomplice who
gives us some partial information about the chip. (Silly, I know, but there is a real concrete use for this
concept, as you will see soon enough.)
1
Then, P  chip is even numbered, given that we know it's black   P  E B  
3
1
and P  chip is black, given that we know it is even   P  B E  
2
In general, P  E F  is said as, “the probability of E, given F”. That is, the conditional probability of E
occurring given that F has already occurred. In essence, a conditional probability reduces the number of
elements in the sample space. Also, notice that P  E F  does not necessarily equal P  E F  .
Example: Textbook exercise #18 (p. 543):
a)
P  4th grader  
678
1205
b) P  4th grader AND above grade level   216
1205

324  98 422

1205
1205
120
225
120
P  scoring below grade level, given a 4th grader  
678
d) P  4th grader, given scoring below grade level  
e)
54
2-Math 130
When the information doesn’t fit in a nice chart, as in the above example, the formula presented in the
textbook can be used:
PE F 
PE F  
PF 
Pay close attention to the textbook examples #4 (p. 536), 6 (pp. 538-9), and 8 (p. 540).
Section7.5 Independent Events
Two events E and F are called independent events if any one of the following is true:
i) P  E F   P  E 
ii) P  F E   P  F 
iii) P  E F   P  E   P  F 
You must be very careful to NOT use your everyday understanding of “independent”; intuition is often
wrong here. You must verify, mathematically, that one of the above three conditions is true. Also, be
careful to not confuse mutually exclusive events and independent events. Exercise #1 of section 7.5 (p.
559) presents you with examples to show that two sets may be:
mutually exclusive and independent
mutually exclusive and dependent
not mutually exclusive and independent
not mutually exclusive and dependent.
Section 7.6 Bayes’ Rule
For the problems in this section, I require you to draw tree diagrams. I have redone the textbook’s
example #1 below to show you how to use the tree diagram as the first step in solving the problem.
Example: Textbook example #1 (pp. 567-8)
D  P 
D    0.65 0.05  0.0325
0.05
I
0.65
0.95
0.35
D′
 P   D    0.35 0.15  0.0525
D
0.15
II
0.85
P  I D 
D′
PI
D
P  D

0.0325
0.0325 325 13



0.0325  0.0525
0.085
850 34
55
2-Math 130
Self-Study Practice Problems
These are not to be handed in. They are for you to check your understanding of the material; to exercise
your math brain. Make sure you attempt each problem without looking in the solutions manual. It is
crucial for you to practice, to do math, and not just read about it.
Section 7.4 (p. 541):
Section 7.5 (p. 559):
Section 7.6 (p. 574):
Section 7.6 (p. 574):
#3, 5, 9 – 17 odd, 23 – 31 odd, 35, 37, 39
#1 – 15 odd, 19 – 27 odd
#1, 9, 11, 13, 15, 23, 31 [First draw a tree diagram to solve each of these.]
#25
Show all your work in a neat and organized manner or you will receive no credit.
(19 problems)
Section 7.4 (p. 541):
Section 7.5 (p. 559):
Section 7.6 (p. 574):
Section 7.6 (p.574):
#4, 6, 10, 12, 14, 16, 28, 36
#6, 10, 24, 26
#2, 6, 18, 24, 38 [First draw a tree diagram to solve each of these.]
#26
56
2-Math 130
Exam 3 Information
Make arrangements with your proctor to schedule Exam 3.
Prior to taking this exam:
• You must submit lessons 7, 8, 9, and 10 to your instructor before taking this exam.
• Please do not take this exam until you have received graded lessons 7, 8, 9, and 10 back from your
instructor.
• Do not submit any subsequent lessons until you have taken this exam.
 Below, there is a test outline containing textbook exercises that are similar to what will be on the
test. You should do all of these problems to prepare for the test. You may find that you need to do
them more than once. Keep track of the problems you had to look up in the textbook, in this course
guide, or on your past homework before you could solve the problems. Do those over again until
you can do every problem without any references.
Exam components:
• This exam covers lessons 7, 8, 9, and 10.
• This is a closed-book, closed-notes exam.
• Calculators are NOT allowed.
 The format and questions on this exam are similar to the textbook problems.
• This exam is worth 100 points.
• Time limit: 75 minutes.
Items to take to the exam:
• Photo identification
• V number
• Pencil and eraser
• Graded exams will not be returned to you. However, arrangements can be made to view graded
Exam 3 Topic Outline (Refer to your lessons for page numbers)
1. Set Operations [6.1 #43 – 53 odd]
2. Counting with Venn diagrams [6.2 #3 – 11 odd, 23, 28]
3. Multiplication Rule [6.3 #5, 7, 9, 11, 17, 29, 49, 53]
4. Permutations [6.4 #7, 11, 17, 23b, 29, 37, 43, 57]
5. Combinations [6.5 #19, 23, 27, 41, 43]
6. Mixed counting practice [6.6 #3, 5, 11, 13, 17, 21]
7. Basic Probability [7.1 #1; 7.2 #1, 3, 7, 9, 13, 19, 21, 23, 27, 39, 41]
8. Compound Events [7.3 #5, 7, 9, 21, 25, 29, 33, 35]
9. Conditional Probability [7.4 #5, 9, 17]
10. Independence [7.5 #1, 3, 5, 7, 9, 11]
11. Bayes’ Rule [7.6 #9, 11, 13, 25]
57
2-Math 130
Extra Practice Problems are on the next page.
Extra Practice – These problems do not necessarily cover all the topics. Make sure you can do the exam
outline problems above. It’s also a good idea to make sure you can do the examples that were
presented in this study guide.
pp. 491 – 494 #5, 9, 11, 15, 17, 21, 23, 25, 27, 29, 35, 43, 51
pp. 593 – 596 #3, 5, 7, 15, 21, 27, 29, 33, 35, 37
58
2-Math 130
Lesson 11 [Self-study]
Markov Chains
Section 7.7
Overview
Section 7.7 Markov Chains
This last section is going to show you an application that uses both matrices and probabilities. Are you
excited? Go ahead, you can admit it, no one will know. 
A Markov chain, also called a Markov sequence or process, is a mathematical model used to describe an
experiment or measurement that is repeated at regular time intervals where the possible outcomes for
each trial are the same and the probabilities for one trial depends solely on the immediately preceding
trial.
Here is one more example to accompany the ones presented in the textbook:
Suppose the data shows that 80% of the daughters of working women go on to be working mothers
while only 30% of the daughters of non-working women go on to become working mothers. (Note that
“working” here implies working outside of the home.)
This data can be represented by the following matrix:
not
working working
 0.8
T 
 0.3
0.2 
0.7 
working
not working
Suppose that at generation 0, 40% of women work. This can be represented by the row matrix:
d0  0.4 0.6
We wish to find the percent of women working in the next two generations.
0.8 0.2 
d1  d0  T   0.4 0.6 
  0.5 0.5
 0.3 0.7 
50% of women will be working.
0.8 0.2 
d 2  d1  T   0.5 0.5 
  0.55 0.45
0.3 0.7 
55% of women will be working.
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2-Math 130
The following table shows the data after many generations:
Generation % working
0
40.00
1
50.00
2
55.00
3
57.50
4
58.75
5.
59.38
.
.
.
.
.
10
11
12
13
14
59.98
59.99
60.00
60.00
60.00
.
.
.
.
.
.
The long term trend is for 60% of women to work. Of course, we would like to find the long term trend
without doing the repeated matrix multiplications. We can do that as follows:
Let  x
y  be the stable distribution.
0.8 0.2 
y 
  x y
0.3 0.7 
0.8 x  0.3 y  x  0.2 x  0.3 y  0
So we have

0.2 x  0.7 y  y 
0.2 x  0.3 y  0
Then  x
y T   x
y   x
Notice that both of these equations are equivalent to 2x  3 y  0 . We need another equation in order to
get a unique solution. Notice that for the stable distribution x  y  1.
So,
x  y 1
1 1 R2  R2 2 R1  1
1
1 R1 5 R1  R2  5
1
 
  0 5 2  0
2 x  3 y  0
2

3
0





0
5
3
1 0
R1  15 R1


1
2 R2  5 R2 0 1
3
5
2
5
So, x  3  60%
5
Now, you are due a huge congratulations! And, good luck on the final exam.
Self-Study Practice Problems
These are not to be handed in. They are for you to check your understanding of the material; to exercise
your math brain. Make sure you attempt each problem without looking in the solutions manual. It is
crucial for you to practice, to do math, and not just read about it.
Section 7.7 (p. 589):
#1, 3, 5, 7, 9, 13, 17, 21, 39
The material will be covered on the final exam so be sure and do the self-study problems.
There is no graded assignment for this section. Do not send anything in for this section.
60



2-Math 130
Final Exam Information
Make arrangements with your proctor to schedule the Final Exam.
Prior to taking this exam:
• You must submit lessons 1-10 to your instructor before taking this exam.
 On the next pages there are two sample final exams, followed by their answers, which are similar to
the final exam that you will take. Before you even try these, go back to lesson 1 and review all of the
examples in this study guide. Then work whichever problems you can without looking at the
answers. This is hard to do, I know, but if you resist you will get a better indicator of which topics
you need to review the most. Keep track of the problems you had to look up in the textbook, in this
guide, or on your past homework before you could solve them. You will know those are the topics
that you need to go back and review the most.
Exam components:
• This exam covers lessons 1-11.
• This is a closed-book, closed-notes exam.
• Calculators are NOT allowed.
 The format and questions on this exam are similar to the textbook problems.
• This exam is worth 100 points.
• Time limit: 120 minutes (2 hours).
Items to take to the exam:
• Photo identification
• V number
• Pencil and eraser
• Graded exams will not be returned to you. However, arrangements can be made to view graded
Final Exam reviews, and answers, continue on the following pages
61
2-Math 130
Final Exam Review #1
1. Find the equation of the line that passes through (–3, 4) and (2, 24).
x  y  2z   1
2. Use the Gauss-Jordan Method to solve: 3 x  2 y  z 
6
 x  y  4 z  11
v  4w
3. Find all the solutions to the following system of linear equations:
v  4w
2
4. Compute the inverse of A  
 3y
1
x 2y
2
 3y  z  5
6
 in two ways.
 3 10 
1 0 0 
5. (a) Use the Gauss-Jordan Method to compute the inverse of the matrix: A  0 1 2 


2 3 4 
x
 1
y  2z 
(b) Use the answer in Part (a) to solve:
6
2 x  3 y  4 z  11
6. Use the Geometric (graphical) method to solve the linear programming problem:
Minimize C  3x  5 y
Subject to: 2 x  y  8
x  2 y  10
x, y  0
7. A mining company owns two mines which produce three grades of ore. Each day, mine A produces
1 ton of high-grade ore, 2 tons of medium-grade ore, and 3 tons of low-grade ore. Mine B produces
1 ton of high-grade ore, 1 ton of medium-grade ore, and 2 tons of low-grade ore daily. An order is
placed for 70 tons of high-grade ore, 40 tons of medium-grade ore, and 60 tons of low-grade ore. If
the operating costs for production are \$500 for mine A and \$350 for mine B, find the number of
days each mine should be used to fill this order at the lowest cost.
(a) Define the variables you will use.
(b) Set up the LP problem (do not solve).
62
2-Math 130
8. Use the simplex method to solve the linear programming problem:
Maximize: P  2 x1  3x2  6 x3
Subject to:
2 x1  3 x2  x3  10
2 x2  3x3  6
x1  0, x2  0, x3  0
9. For the given linear programming problem, set up the initial simplex tableau for the dual problem,
and label each column with the appropriate variable. Do not solve.
Minimize: C  30 x1  60 x2  50 x3
5 x1  3 x2  x3  2
Subject to:
x1  2 x2  x3  3
x1  0, x2  0, x3  0
10. Find the solution to an original minimum problem if the final tableau is:
y1
y2
y3
x1
x2
P
1
35
0
13
 17
8
35
0
8
20
16
1
92
39
35
1
0
1
7
 32
35
0
1
96
0
0

11. A committee of 4 people is to be selected from a group of 7 men and 3 women. In how many ways
can the committee be selected if (a) 2 men and 2 women are selected? (b) at least 2 women are
selected?
12. A campus radio station surveyed 190 students. The survey determined that 114 liked rock music, 50
liked country music, and 41 liked classical music. Moreover, 14 liked rock and country, 15 liked rock
and classical, 11 liked country and classical, and 5 liked all three types. How many students (a) do
not like any of the three types of music? (b) like only country music?
what is the probability that the driver had also received a speeding ticket?
14. E and F are events from an experiment with P( E )  0.6, P  F   0.3, and P( E
F )  0.4.
Compute P( F ), P( F E ), and P( E F ).
Are E and F independent events? Justify mathematically.
15. A company manufactures calculators at three different production lines. 50% are produced in line 1,
and of these 5% are defective. 20% are produced in line 2 of which 3% are defective, and 30% are
produced in line 3 of which 1% are defective. First draw a tree diagram and then answer the
following. What is the probability that (a) a calculator is defective? (b) given a defective calculator,
it was produced in line 2?
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2-Math 130
16. The matrix given below represents the transition of college students in a big city university between
living on campus (C) and living at home (H) at the end of a semester. Find the percentage of those
students that live at home when the stable distribution is achieved.
From
C
C 0.9
To
H
0.1
H 0.4 0.6 
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2-Math 130
Final Exam Review #2
1. Find the point of intersection of the lines:
3x  4 y  10
2x  y  8
x  2y
 1
2. Use the Gauss-Jordan Method to solve:  2 x  5 y  z  3
3x 
7z  2
w x
3. Use the Gauss-Jordan Method to find all the solutions for:
5w  x  2 y
w
 2
 4
z 1
1 4
4. Compute the inverse of the matrix A  
 .
2 6 
 1 3
2 1 6 
5. Let A  
, and B  


3 5 2
2 4
(a) Find AB if possible. If not possible, state why.
(b) Find BA if possible. If not possible, state why.
 1 23 
8 3
6. (a) Show that the matrices 
and 
 are inverses of each other.
6 2 


 3 4 
(b) Use part (a) to solve the system of equations:
x
3
y6
2
3 x  4 y  8
1 0 5
 6 10 5
7. Use the fact that the inverse of the matrix A  1 1 0  is the matrix A1   6 9 5 




3 2 6 
 1 2 1 
x
to solve
x y
5z  2
 1
3x  2 y  6 z  1
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2-Math 130
8. Use the Geometric (graphical) method to solve the linear programming problem:
Maximize: P  x  y
Subject to: 2 x  y  12
2 x  3 y  24
x  0, y  0
9. Use the simplex method to solve the linear programming problem:
Maximize: M  x1  2 x2  3x3
Subject to: x1  2 x2  x3  40
x1  x2  x3  30
x1  0, x2  0, x3  0
10. For the given linear programming problem, set up the initial simplex tableau for the dual problem,
and label each column with the appropriate variable. Do not solve.
Minimize: C  20 x1  16 x2
Subject to: 7 x1  2 x2  12
3 x1  5 x2  8
6 x1  10 x2  9
x1  0, x2  0
11. At the left below is the primal problem and on the right is the final simplex tableau for the dual
problem. Give the solution to the original primal problem.
Minimize: C  8 x1  12 x2
y1
y2
Subject to: x1  3 x2  2
0
1
2 x1  2 x2  3
1
x1  0, x2  0
0
0
0
x1

x2
P
3
4
 14
0
3
1
2
1
2
0
2
5
4
1
4
1
13
12. Set up the following problem. Do not solve.
A farmer is going to plant at least 100 acres with 2 crops, wheat and lentils. Each acre of wheat costs
\$40 and takes 2 hours to plant. Each acre of lentils costs \$30 and takes 3 hours to plant. The farmer
has at most \$12,000 and 240 hours to spend on planting. If she expects to make \$50 for each acre
of wheat and \$75 for each acre of lentils, how many acres of each crop should she plant in order to
maximize her profits?
(a) Identify the unknowns.
(b) Continue by forming the objective function and a system of constraints to model this problem.
13. To help plan the number of meals to be prepared in a college cafeteria, a survey was conducted and
the following data were obtained: 130 students ate breakfast, 180 students ate lunch, 275 students
ate dinner, 68 students ate breakfast and lunch, 112 students ate breakfast and dinner, 90 students
ate lunch and dinner, and 58 students ate all three meals. How many students ate (a) only dinner in
the cafeteria? (b) exactly two meals in the cafeteria?
66
2-Math 130
14. In how many ways can four married couples attending a concert be seated to a row of eight chairs if:
(b) each married couple is seated together? Do not simplify your answer.
15. An organization is comprised of 12 men and 10 women. They need to form a committee of 2 men
and 3 women to plan a party. In how many ways can the committee be formed? Do not simplify
16. Let E and F be events with P( E )  0.4, P( F )  0.3, and P  E F   0.9 .
(a) Find P( F ), P( E F ), P( E F ), and P( E F ).
(b) Are E and F independent events? Justify mathematically.
(c) Are E and F mutually exclusive? Justify mathematically.
17. The attendant at a toll booth of an airport parking lot has been noticing the types of vehicles and
drivers that pass through his station. He has determined that the probability that the vehicle is a
pickup is 0.4 and the probability that the driver is a male is 0.6. The probability that both occur is
0.3.
(a) Find the probability that either the vehicle is a pickup or that the driver is a male.
(b) If he sees that the next vehicle is a pickup, what is the conditional probability that the driver is a
male?
18. A professor has two assistants who grade for him. Mr. Goofy grades 70% of the papers, while Mr.
Sloppy grades the remaining 30% of the papers. Of the papers Mr. Goofy grades, 10% are graded
(a) Draw a tree diagram for this situation.
(b) A grading error is found in a paper. What is the probability that it was graded by Mr. Sloppy?
19. Birds migrate between habits I and II each year as follows: Of the birds in habitat I, 70% will migrate
to habitat II the next year and 30% will remain in habitat I. Of the birds in habitat II, 50% will migrate
to habitat I and 50% will remain in habitat II.
(a) Find the transition matrix for this Markov process.
(b) If in one year 80% of the birds are in habitat I and 20% are in habitat II, give the percentages of
birds in each habitat the following year.
67
2-Math 130
1.
y 4 
28
( x  3)
5
2.
x  2, y  1, z  2
3.
v  1  4w  3 y 
w  any number 

 or
y  any number 

z5

x  2  2y
(1  4 j  3k , j, 2  2k , k , 5)
4. The two methods are the Gauss-Jordan method and the short-cut for 2x2’s. In either case, you
 5
should get A1   3
 2
3
1 0 
which you can check by showing A  A1  

.
1
0 1 
 1 0 0
5. (a) A   2 2
1


3
1
 1


2
2
1
(b) Recall A  X  B  X  A1  B
x  1, y  21, z 
27
2
6. Your feasible set should have 3 corners.
The minimum value is C  26 when x  2 and y  4 .
7. (a) Let x1  number of days mine A should be used
and x2  number of days mine B should be used
(b) Minimize: C  500 x1  350 x2
Subject to:
x1  x2  70
2 x1  x2  40
3 x1  2 x2  60
x1  0, x2  0
68
2-Math 130
8.
The maximum is P  20 when x1  4, x2  0, and x3  2.
9.
y1
y2
x1
5
1
1
3
2
1
2
x2
x3
P
0
0
0
30
0
1
0
0
60
1
0
0
1
0
50
3
0
0
0
1
0
10. The minimum is C = 92 when x1  20 and x2  16.
11. (a) C  7,2   C  3,2  
7  6 3 2

 63
2! 2!
(b) C  3,2   C  7,2   C  3,3  C  7,1 
3 2 7  6

 1  7  70
2! 2!
12. From a Venn diagram:
(a) 20
(b) 30
13. P  S P  
14. (a) 0.7,
2
3
P S P  9

P  P
37
,
4
7
(b) No, since P(F)  P(F E)
15. (a) 0.034
16.
1
5
(b)
6
34

3
17
 20%
69
2-Math 130
1. (2, 4)
2.
x  3, y  2, z  1
3.
x  2  w
y  3  2w
z  1 w
w  any number
4.
 3 2 
 1  1

2
5. (a)  11 14
 8 22

12 
4 
(b) Not possible: Number of columns of B  number of rows of A.
1 0 
.
0 1 
6. (a) Show that the product of the two matrices is 
8 3 6
(b) Multiply 
    to get x  24 and y  20 .
6 2 8
 2
7. Use A   1 to get x  3, y  2, and z  1 .
 
 1
1
8. The feasible set should have 3 corners.
Maximum is P  12 when x  12 and y  0 .
9. Maximum is M  90 when x1  0, x2  0, and x3  30 .
10.
y1
y2
y3
x1
x2
P
7
3
6
1
0
0
20
2
5
10
0
1
0
16
12
8
9
0
0
1
0
11. Minimum is C  13 when x1 
5
4
and x2 
1
4
70
2-Math 130
12. (a) x  number of acres of wheat
y  number of acres of lentils
(b) Maximize P  50 x  75 y
Subject to:
x  y  100
40 x  30 y  12000
2 x  3 y  240
x  0, y  0
13. Use a Venn diagram to find (a) 131 and (b) 96.
14. (a) 8! (b) 2∙2∙2∙2∙4!
15. C(12, 2)  C(10,3)
16. (a) 0.7, 0.2,
2
7
, 0.2
(b) No, P( E F )  P( E )  P( F )or P( E F )  P( E )
(c) No, P( E F )  0
17. (a) 0.7
(b)
3
4
I
18. (a)
0.1
(b) P ( S I ) 
6
13
G
0.7
0.9
C
I
0.2
0.3
S
0.8
0.7 0.3

0.5 0.5
19. (a) T  
C
(b) 66% in habitat I and 34% in habitat II
71
2-Math 130
72
2-Math 130
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