MATH 244 King Saud University Linear Algebra General Vector Space Worksheet

Description

i want help to solve this homework , it is about math linear algebra ch 4 ( general vector space )

Second homework of 244 Math
Summer semester-1441
Let n=(x1 x2 x3 x4 x5 x6 x7 x8 x9) be your university ID.
Q1- Let
1
x
A  1
x 4

x 7
1
x2
x5
x8
1
x 3 
x6

x9
(i) Find a basis for the row space of A consisting entirely of row
vectors from A. (4 marks)
(ii) Find dim(Col(A)) (in a simple way). (1 mark)
(iii) Find nullity(AT) (in a simple way). (1 mark)
Q2- Find the transition matrix from the standard basis of ℝ2 to
the basis B={(5,6),(x8,x9)}. (4 marks)
Q3- Show that the the set of upper triangular matrices of order
2 is a subspace of M22. (5 marks)
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11 T
H
EDITION
Elementary
Linear
Algebra
Applications Version
H OWA R D
A NT O N
Professor Emeritus, Drexel University
C H R I S
R O R R E S
University of Pennsylvania
VICE PRESIDENT AND PUBLISHER
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Library of Congress Cataloging-in-Publication Data
Anton, Howard, author.
Elementary linear algebra : applications version / Howard Anton, Chris Rorres. — 11th edition.
pages cm
Includes index.
ISBN 978-1-118-43441-3 (cloth)
1. Algebras, Linear–Textbooks. I. Rorres, Chris, author. II. Title.
QA184.2.A58 2013
512′.5–dc23
2013033542
ISBN 978-1-118-43441-3
ISBN Binder-Ready Version 978-1-118-47422-8
Printed in the United States of America
10 9 8 7 6 5 4 3 2 1
ABOUT
THE
AUTHOR
Howard Anton obtained his B.A. from Lehigh University, his M.A. from the
University of Illinois, and his Ph.D. from the Polytechnic University of Brooklyn, all in
mathematics. In the early 1960s he worked for Burroughs Corporation and Avco
Corporation at Cape Canaveral, Florida, where he was involved with the manned space
program. In 1968 he joined the Mathematics Department at Drexel University, where
he taught full time until 1983. Since then he has devoted the majority of his time to
textbook writing and activities for mathematical associations. Dr. Anton was president
of the EPADEL Section of the Mathematical Association of America (MAA), served on
the Board of Governors of that organization, and guided the creation of the Student
Chapters of the MAA. In addition to various pedagogical articles, he has published
numerous research papers in functional analysis, approximation theory, and topology.
He is best known for his textbooks in mathematics, which are among the most widely
used in the world. There are currently more than 175 versions of his books, including
translations into Spanish, Arabic, Portuguese, Italian, Indonesian, French, Japanese,
Chinese, Hebrew, and German. For relaxation, Dr. Anton enjoys travel and
photography.
Chris Rorres earned his B.S. degree from Drexel University and his Ph.D. from the
Courant Institute of New York University. He was a faculty member of the
Department of Mathematics at Drexel University for more than 30 years where, in
addition to teaching, he did applied research in solar engineering, acoustic scattering,
population dynamics, computer system reliability, geometry of archaeological sites,
optimal animal harvesting policies, and decision theory. He retired from Drexel in 2001
as a Professor Emeritus of Mathematics and is now a mathematical consultant. He
also has a research position at the School of Veterinary Medicine at the University of
Pennsylvania where he does mathematical modeling of animal epidemics. Dr. Rorres is
a recognized expert on the life and work of Archimedes and has appeared in various
television documentaries on that subject. His highly acclaimed website on Archimedes
(http://www.math.nyu.edu/~crorres/Archimedes/contents.html) is a virtual book that
has become an important teaching tool in mathematical history for students around
the world.
To:
My wife, Pat
My children, Brian, David, and Lauren
My parents, Shirley and Benjamin
My benefactor, Stephen Girard (1750–1831),
whose philanthropy changed my life
Howard Anton
To:
Billie
Chris Rorres
PREFACE
Summary of Changes in
This Edition
vi
This textbook is an expanded version of Elementary Linear Algebra, eleventh edition, by
Howard Anton. The ﬁrst nine chapters of this book are identical to the ﬁrst nine chapters
of that text; the tenth chapter consists of twenty applications of linear algebra drawn
from business, economics, engineering, physics, computer science, approximation theory,
ecology, demography, and genetics. The applications are largely independent of each
other, and each includes a list of mathematical prerequisites. Thus, each instructor has
the ﬂexibility to choose those applications that are suitable for his or her students and to
incorporate each application anywhere in the course after the mathematical prerequisites
have been satisﬁed. Chapters 1–9 include simpler treatments of some of the applications
covered in more depth in Chapter 10.
This edition gives an introductory treatment of linear algebra that is suitable for a
ﬁrst undergraduate course. Its aim is to present the fundamentals of linear algebra in the
clearest possible way—sound pedagogy is the main consideration. Although calculus
is not a prerequisite, there is some optional material that is clearly marked for students
with a calculus background. If desired, that material can be omitted without loss of
continuity.
Technology is not required to use this text, but for instructors who would like to
use MATLAB, Mathematica, Maple, or calculators with linear algebra capabilities, we
have posted some supporting material that can be accessed at either of the following
companion websites:
www.howardanton.com
www.wiley.com/college/anton
Many parts of the text have been revised based on an extensive set of reviews. Here are
the primary changes:
• Earlier Linear Transformations Linear transformations are introduced earlier (starting
in Section 1.8). Many exercise sets, as well as parts of Chapters 4 and 8, have been
revised in keeping with the earlier introduction of linear transformations.
• New Exercises Hundreds of new exercises of all types have been added throughout
the text.
• Technology Exercises requiring technology such as MATLAB, Mathematica, or Maple
have been added and supporting data sets have been posted on the companion websites
for this text. The use of technology is not essential, and these exercises can be omitted
without affecting the ﬂow of the text.
• Exercise Sets Reorganized Many multiple-part exercises have been subdivided to create
a better balance between odd and even exercise types. To simplify the instructor’s task
of creating assignments, exercise sets have been arranged in clearly deﬁned categories.
• Reorganization In addition to the earlier introduction of linear transformations, the
old Section 4.12 on Dynamical Systems and Markov Chains has been moved to Chapter 5 in order to incorporate material on eigenvalues and eigenvectors.
• Rewriting Section 9.3 on Internet Search Engines from the previous edition has been
rewritten to reﬂect more accurately how the Google PageRank algorithm works in
practice. That section is now Section 10.20 of the applications version of this text.
• Appendix A Rewritten The appendix on reading and writing proofs has been expanded
and revised to better support courses that focus on proving theorems.
• Web Materials Supplementary web materials now include various applications modules, three modules on linear programming, and an alternative presentation of determinants based on permutations.
• Applications Chapter Section 10.2 of the previous edition has been moved to the
websites that accompany this text, so it is now part of a three-module set on Linear
Preface
vii
Programming. A new section on Internet search engines has been added that explains
the PageRank algorithm used by Google.
Hallmark Features
• Relationships Among Concepts One of our main pedagogical goals is to convey to the
student that linear algebra is a cohesive subject and not simply a collection of isolated
deﬁnitions and techniques. One way in which we do this is by using a crescendo of
Equivalent Statements theorems that continually revisit relationships among systems
of equations, matrices, determinants, vectors, linear transformations, and eigenvalues.
To get a general sense of how we use this technique see Theorems 1.5.3, 1.6.4, 2.3.8,
4.8.8, and then Theorem 5.1.5, for example.
• Smooth Transition to Abstraction Because the transition from R n to general vector
spaces is difﬁcult for many students, considerable effort is devoted to explaining the
purpose of abstraction and helping the student to “visualize” abstract ideas by drawing
analogies to familiar geometric ideas.
• Mathematical Precision When reasonable, we try to be mathematically precise. In
keeping with the level of student audience, proofs are presented in a patient style that
is tailored for beginners.
• Suitability for a Diverse Audience This text is designed to serve the needs of students
in engineering, computer science, biology, physics, business, and economics as well as
those majoring in mathematics.
• Historical Notes To give the students a sense of mathematical history and to convey
that real people created the mathematical theorems and equations they are studying, we
have included numerous Historical Notes that put the topic being studied in historical
perspective.
About the Exercises
• Graded Exercise Sets Each exercise set in the ﬁrst nine chapters begins with routine
drill problems and progresses to problems with more substance. These are followed
by three categories of exercises, the ﬁrst focusing on proofs, the second on true/false
exercises, and the third on problems requiring technology. This compartmentalization
is designed to simplify the instructor’s task of selecting exercises for homework.
• Proof Exercises Linear algebra courses vary widely in their emphasis on proofs, so
exercises involving proofs have been grouped and compartmentalized for easy identiﬁcation. Appendix A has been rewritten to provide students more guidance on proving
theorems.
• True/False Exercises The True/False exercises are designed to check conceptual understanding and logical reasoning. To avoid pure guesswork, the students are required
to justify their responses in some way.
• Technology Exercises Exercises that require technology have also been grouped. To
avoid burdening the student with keyboarding, the relevant data ﬁles have been posted
on the websites that accompany this text.
• Supplementary Exercises Each of the ﬁrst nine chapters ends with a set of supplementary exercises that draw on all topics in the chapter. These tend to be more challenging.
Supplementary Materials
for Students
• Student Solutions Manual This supplement provides detailed solutions to most oddnumbered exercises (ISBN 978-1-118-464427).
• Data Files Data ﬁles for the technology exercises are posted on the companion websites
that accompany this text.
• MATLAB Manual and Linear Algebra Labs This supplement contains a set of MATLAB
laboratory projects written by Dan Seth of West Texas A&M University. It is designed
to help students learn key linear algebra concepts by using MATLAB and is available in
PDF form without charge to students at schools adopting the 11th edition of the text.
• Videos A complete set of Daniel Solow’s How to Read and Do Proofs videos is available
to students through WileyPLUS as well as the companion websites that accompany
viii
Preface
this text. Those materials include a guide to help students locate the lecture videos
appropriate for speciﬁc proofs in the text.
Supplementary Materials
for Instructors
• Instructor’s Solutions Manual This supplement provides worked-out solutions to most
exercises in the text (ISBN 978-1-118-434482).
• PowerPoint Presentations PowerPoint slides are provided that display important definitions, examples, graphics, and theorems in the book. These can also be distributed
to students as review materials or to simplify note taking.
• Test Bank Test questions and sample exams are available in PDF or LATEX form.
• WileyPLUS An online environment for effective teaching and learning. WileyPLUS
builds student conﬁdence by taking the guesswork out of studying and by providing a
clear roadmap of what to do, how to do it, and whether it was done right. Its purpose is
to motivate and foster initiative so instructors can have a greater impact on classroom
achievement and beyond.
A Guide for the Instructor
Although linear algebra courses vary widely in content and philosophy, most courses
fall into two categories—those with about 40 lectures and those with about 30 lectures.
Accordingly, we have created long and short templates as possible starting points for
constructing a course outline. Of course, these are just guides, and you will certainly
want to customize them to ﬁt your local interests and requirements. Neither of these
sample templates includes applications or the numerical methods in Chapter 9. Those
can be added, if desired, and as time permits.
Long Template
Chapter 1: Systems of Linear Equations and Matrices
8 lectures
6 lectures
Chapter 2: Determinants
3 lectures
2 lectures
Chapter 3: Euclidean Vector Spaces
4 lectures
3 lectures
10 lectures
9 lectures
Chapter 5: Eigenvalues and Eigenvectors
3 lectures
3 lectures
Chapter 6: Inner Product Spaces
3 lectures
1 lecture
Chapter 7: Diagonalization and Quadratic Forms
4 lectures
3 lectures
Chapter 8: General Linear Transformations
4 lectures
3 lectures
39 lectures
30 lectures
Chapter 4: General Vector Spaces
Total:
Reviewers
Short Template
The following people reviewed the plans for this edition, critiqued much of the content,
and provided me with insightful pedagogical advice:
John Alongi, Northwestern University
Jiu Ding, University of Southern Mississippi
Eugene Don, City University of New York at Queens
John Gilbert, University of Texas Austin
Danrun Huang, St. Cloud State University
Craig Jensen, University of New Orleans
Steve Kahan, City University of New York at Queens
Harihar Khanal, Embry-Riddle Aeronautical University
Firooz Khosraviyani, Texas A&M International University
Y. George Lai, Wilfred Laurier University
Kouok Law, Georgia Perimeter College
Mark MacLean, Seattle University
Preface
ix
Vasileios Maroulas, University of Tennessee, Knoxville
Daniel Reynolds, Southern Methodist University
Qin Sheng, Baylor University
Laura Smithies, Kent State University
Larry Susanka, Bellevue College
Cristina Tone, University of Louisville
Yvonne Yaz, Milwaukee School of Engineering
Ruhan Zhao, State University of New York at Brockport
Exercise Contributions
Special thanks are due to three talented people who worked on various aspects of the
exercises:
Przemyslaw Bogacki, Old Dominion University – who solved the exercises and created
the solutions manuals.
Roger Lipsett, Brandeis University – who proofread the manuscript and exercise solutions for mathematical accuracy.
Daniel Solow, Case Western Reserve University – author of “How to Read and Do Proofs,”
for providing videos on techniques of proof and a key to using those videos in coordination with this text.
Sky Pelletier Waterpeace – who critiqued the technology exercises, suggested improvements, and provided the data sets.
Special Contributions
I would also like to express my deep appreciation to the following people with whom I
worked on a daily basis:
Anton Kaul – who worked closely with me at every stage of the project and helped to write
some new text material and exercises. On the many occasions that I needed mathematical
or pedagogical advice, he was the person I turned to. I cannot thank him enough for his
guidance and the many contributions he has made to this edition.
David Dietz – my editor, for his patience, sound judgment, and dedication to producing
a quality book.
Anne Scanlan-Rohrer – of Two Ravens Editorial, who coordinated the entire project and
brought all of the pieces together.
Jacqueline Sinacori – who managed many aspects of the content and was always there
to answer my often obscure questions.
Carol Sawyer – of The Perfect Proof, who managed the myriad of details in the production
process and helped with proofreading.
Maddy Lesure – with whom I have worked for many years and whose elegant sense of
design is apparent in the pages of this book.
Lilian Brady – my copy editor for almost 25 years. I feel fortunate to have been the beneﬁciary of her remarkable knowledge of typography, style, grammar, and mathematics.
Pat Anton – of Anton Textbooks, Inc., who helped with the mundane chores duplicating,
shipping, accuracy checking, and tasks too numerous to mention.
John Rogosich – of Techsetters, Inc., who programmed the design, managed the composition, and resolved many difﬁcult technical issues.
Brian Haughwout – of Techsetters, Inc., for his careful and accurate work on the illustrations.
Josh Elkan – for providing valuable assistance in accuracy checking.
Howard Anton
Chris Rorres
CONTENTS
C HA PT E R
1
Systems of Linear Equations and Matrices
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
Introduction to Systems of Linear Equations 2
Gaussian Elimination 11
Matrices and Matrix Operations 25
Inverses; Algebraic Properties of Matrices 39
Elementary Matrices and a Method for Finding A−1
More on Linear Systems and Invertible Matrices 61
Diagonal, Triangular, and Symmetric Matrices 67
Matrix Transformations 75
Applications of Linear Systems 84
• Network Analysis (Trafﬁc Flow) 84
• Electrical Circuits 86
• Balancing Chemical Equations 88
• Polynomial Interpolation 91
1.10 Application: Leontief Input-Output Models 96
C HA PT E R
2
Determinants
52
105
2.1 Determinants by Cofactor Expansion 105
2.2 Evaluating Determinants by Row Reduction 113
2.3 Properties of Determinants; Cramer’s Rule 118
C HA PT E R
3
Euclidean Vector Spaces
3.1
3.2
3.3
3.4
3.5
C HA PT E R
4
131
Vectors in 2-Space, 3-Space, and n-Space
Norm, Dot Product, and Distance in Rn
Orthogonality 155
The Geometry of Linear Systems 164
Cross Product 172
General Vector Spaces
131
142
183
4.1 Real Vector Spaces 183
4.2 Subspaces 191
4.3 Linear Independence 202
4.4 Coordinates and Basis 212
4.5 Dimension 221
4.6 Change of Basis 229
4.7 Row Space, Column Space, and Null Space 237
4.8 Rank, Nullity, and the Fundamental Matrix Spaces
4.9 Basic Matrix Transformations in R2 and R3 259
4.10 Properties of Matrix Transformations 270
4.11 Application: Geometry of Matrix Operators on R2
x
248
280
1
Contents
C HA PT E R
5
Eigenvalues and Eigenvectors
5.1
5.2
5.3
5.4
5.5
C HA PT E R
6
C HA PT E R
7
8
C HA PT E R
9
10
Orthogonal Matrices 401
Orthogonal Diagonalization 409
Quadratic Forms 417
Optimization Using Quadratic Forms 429
Hermitian, Unitary, and Normal Matrices
387
401
437
447
General Linear Transformations 447
Compositions and Inverse Transformations 458
Isomorphism 466
Matrices for General Linear Transformations 472
Similarity 481
Numerical Methods
9.1
9.2
9.3
9.4
9.5
C HA PT E R
Inner Products 345
Angle and Orthogonality in Inner Product Spaces 355
Gram–Schmidt Process; QR-Decomposition 364
Best Approximation; Least Squares 378
Application: Mathematical Modeling Using Least Squares
Application: Function Approximation; Fourier Series 394
General Linear Transformations
8.1
8.2
8.3
8.4
8.5
332
345
Diagonalization and Quadratic Forms
7.1
7.2
7.3
7.4
7.5
C HA PT E R
Eigenvalues and Eigenvectors 291
Diagonalization 302
Complex Vector Spaces 313
Application: Differential Equations 326
Application: Dynamical Systems and Markov Chains
Inner Product Spaces
6.1
6.2
6.3
6.4
6.5
6.6
291
491
LU-Decompositions 491
The Power Method 501
Comparison of Procedures for Solving Linear Systems 509
Singular Value Decomposition 514
Application: Data Compression Using Singular Value Decomposition
Applications of Linear Algebra
527
10.1 Constructing Curves and Surfaces Through Speciﬁed Points
10.2 The Earliest Applications of Linear Algebra 533
10.3 Cubic Spline Interpolation 540
528
521
xi
xii
Contents
10.4 Markov Chains 551
10.5 Graph Theory 561
10.6 Games of Strategy 570
10.7 Leontief Economic Models 579
10.8 Forest Management 588
10.9 Computer Graphics 595
10.10 Equilibrium Temperature Distributions 603
10.11 Computed Tomography 613
10.12 Fractals 624
10.13 Chaos 639
10.14 Cryptography 652
10.15 Genetics 663
10.16 Age-Speciﬁc Population Growth 673
10.17 Harvesting of Animal Populations 683
10.18 A Least Squares Model for Human Hearing
10.19 Warps and Morphs 697
10.20 Internet Search Engines 706
APPENDIX A
Working with Proofs
APPENDIX B
Complex Numbers
A1
A5
Answers to Exercises
Index
I1
A13
691
CHAPTER
1
Systems of Linear
Equations and Matrices
CHAPTER CONTENTS
1.1
Introduction to Systems of Linear Equations
1.2
Gaussian Elimination
1.3
Matrices and Matrix Operations
1.4
Inverses; Algebraic Properties of Matrices
1.5
Elementary Matrices and a Method for Finding A−1
1.6
More on Linear Systems and Invertible Matrices
1.7
Diagonal,Triangular, and Symmetric Matrices
11
1.8
MatrixTransformations
1.9
Applications of Linear Systems
•
•
•
•
25
39
52
61
67
75
84
Network Analysis (Trafﬁc Flow) 84
Electrical Circuits 86
Balancing Chemical Equations 88
Polynomial Interpolation 91
1.10 Leontief Input-Output Models
INTRODUCTION
2
96
Information in science, business, and mathematics is often organized into rows and
columns to form rectangular arrays called “matrices” (plural of “matrix”). Matrices
often appear as tables of numerical data that arise from physical observations, but they
occur in various mathematical contexts as well. For example, we will see in this chapter
that all of the information required to solve a system of equations such as
5x + y = 3
2x − y = 4
is embodied in the matrix

5
1
2 −1

3
4
and that the solution of the system can be obtained by performing appropriate
operations on this matrix. This is particularly important in developing computer
programs for solving systems of equations because computers are well suited for
manipulating arrays of numerical information. However, matrices are not simply a
notational tool for solving systems of equations; they can be viewed as mathematical
objects in their own right, and there is a rich and important theory associated with
them that has a multitude of practical applications. It is the study of matrices and
related topics that forms the mathematical ﬁeld that we call “linear algebra.” In this
chapter we will begin our study of matrices.
1
2
Chapter 1 Systems of Linear Equations and Matrices
1.1 Introduction to Systems of Linear Equations
Systems of linear equations and their solutions constitute one of the major topics that we
will study in this course. In this ﬁrst section we will introduce some basic terminology and
discuss a method for solving such systems.
Linear Equations
Recall that in two dimensions a line in a rectangular xy -coordinate system can be represented by an equation of the form
ax + by = c (a, b not both 0)
and in three dimensions a plane in a rectangular xyz-coordinate system can be represented by an equation of the form
ax + by + cz = d (a, b, c not all 0)
These are examples of “linear equations,” the ﬁrst being a linear equation in the variables
x and y and the second a linear equation in the variables x , y , and z. More generally, we
deﬁne a linear equation in the n variables x1 , x2 , . . . , xn to be one that can be expressed
in the form
a1 x1 + a2 x2 + · · · + an xn = b
(1)
where a1 , a2 , . . . , an and b are constants, and the a ’s are not all zero. In the special cases
where n = 2 or n = 3, we will often use variables without subscripts and write linear
equations as
a1 x + a2 y = b (a1 , a2 not both 0)
a1 x + a2 y + a3 z = b (a1 , a2 , a3 not all 0)
(2)
(3)
In the special case where b = 0, Equation (1) has the form
a1 x1 + a2 x2 + · · · + an xn = 0
(4)
which is called a homogeneous linear equation in the variables x1 , x2 , . . . , xn .
E X A M P L E 1 Linear Equations
Observe that a linear equation does not involve any products or roots of variables. All
variables occur only to the ﬁrst power and do not appear, for example, as arguments of
trigonometric, logarithmic, or exponential functions. The following are linear equations:
x + 3y = 7
1
x − y + 3z = −1
2
x1 − 2×2 − 3×3 + x4 = 0
x1 + x2 + · · · + xn = 1
The following are not linear equations:
x + 3y 2 = 4
sin x + y = 0
3x + 2y − xy = 5
√
x1 + 2×2 + x3 = 1
A ﬁnite set of linear equations is called a system of linear equations or, more brieﬂy,
a linear system. The variables are called unknowns. For example, system (5) that follows
has unknowns x and y , and system (6) has unknowns x1 , x2 , and x3 .
5x + y = 3
2x − y = 4
4×1 − x2 + 3×3 = −1
3×1 + x2 + 9×3 = −4
(5–6)
1.1 Introduction to Systems of Linear Equations
The double subscripting on
the coefﬁcients aij of the unknowns gives their location
in the system—the ﬁrst subscript indicates the equation
in which the coefﬁcient occurs,
and the second indicates which
unknown it multiplies. Thus,
a12 is in the ﬁrst equation and
multiplies x2 .
3
A general linear system of m equations in the n unknowns x1 , x2 , . . . , xn can be written
as
a11 x1 + a12 x2 + · · · + a1n xn = b1
a21 x1 + a22 x2 + · · · + a2n xn = b2
..
..
..
..
.
.
.
.
am1 x1 + am2 x2 + · · · + amn xn = bm
(7)
A solution of a linear system in n unknowns x1 , x2 , . . . , xn is a sequence of n numbers
s1 , s2 , . . . , sn for which the substitution
x1 = s1 , x2 = s2 , . . . , xn = sn
makes each equation a true statement. For example, the system in (5) has the solution
x = 1, y = −2
and the system in (6) has the solution
x1 = 1, x2 = 2, x3 = −1
These solutions can be written more succinctly as
(1, −2) and (1, 2, −1)
in which the names of the variables are omitted. This notation allows us to interpret
these solutions geometrically as points in two-dimensional and three-dimensional space.
More generally, a solution
x1 = s1 , x2 = s2 , . . . , xn = sn
of a linear system in n unknowns can be written as
(s1 , s2 , . . . , sn )
which is called an ordered n-tuple. With this notation it is understood that all variables
appear in the same order in each equation. If n = 2, then the n-tuple is called an ordered
pair, and if n = 3, then it is called an ordered triple.
Linear Systems inTwo and
Three Unknowns
Linear systems in two unknowns arise in connection with intersections of lines. For
example, consider the linear system
a1 x + b1 y = c1
a2 x + b2 y = c2
in which the graphs of the equations are lines in the xy-plane. Each solution (x, y) of this
system corresponds to a point of intersection of the lines, so there are three possibilities
(Figure 1.1.1):
1. The lines may be parallel and distinct, in which case there is no intersection and
consequently no solution.
2. The lines may intersect at only one point, in which case the system has exactly one
solution.
3. The lines may coincide, in which case there are inﬁnitely many points of intersection
(the points on the common line) and consequently inﬁnitely many solutions.
In general, we say that a linear system is consistent if it has at least one solution and
inconsistent if it has no solutions. Thus, a consistent linear systemof two equations in
4
Chapter 1 Systems of Linear Equations and Matrices
y
y
y
One solution
No solution
x
x
x
Figure 1.1.1
Infinitely many
solutions
(coincident lines)
two unknowns has either one solution or inﬁnitely many solutions—there are no other
possibilities. The same is true for a linear system of three equations in three unknowns
a1 x + b1 y + c1 z = d1
a2 x + b2 y + c2 z = d2
a3 x + b3 y + c3 z = d3
in which the graphs of the equations are planes. The solutions of the system, if any,
correspond to points where all three planes intersect, so again we see that there are only
three possibilities—no solutions, one solution, or inﬁnitely many solutions (Figure 1.1.2).
No solutions
(three parallel planes;
no common intersection)
No solutions
(two parallel planes;
no common intersection)
No solutions
(no common intersection)
No solutions
(two coincident planes
parallel to the third;
no common intersection)
One solution
(intersection is a point)
Infinitely many solutions
(intersection is a line)
Infinitely many solutions
(planes are all coincident;
intersection is a plane)
Infinitely many solutions
(two coincident planes;
intersection is a line)
Figure 1.1.2
We will prove later that our observations about the number of solutions of linear
systems of two equations in two unknowns and linear systems of three equations in
three unknowns actually hold for all linear systems. That is:
Every system of linear equations has zero, one, or inﬁnitely many solutions. There are
no other possibilities.
1.1 Introduction to Systems of Linear Equations
5
E X A M P L E 2 A Linear System with One Solution
Solve the linear system
x−y =1
2x + y = 6
x from the second equation by adding −2 times the ﬁrst
equation to the second. This yields the simpliﬁed system
Solution We can eliminate
x−y =1
3y = 4
From the second equation we obtain y = 43 , and on substituting this value in the ﬁrst
equation we obtain x = 1 + y = 73 . Thus, the system has the unique solution
x = 73 , y =
4
3
Geometrically, this means that
the
lines represented by the equations in the system
intersect at the single point 73 , 43 . We leave it for you to check this by graphing the
lines.
E X A M P L E 3 A Linear System with No Solutions
Solve the linear system
x+ y=4
3x + 3y = 6
Solution We can eliminate x from the second equation by adding −3 times the ﬁrst
equation to the second equation. This yields the simpliﬁed system
x+y =
4
0 = −6
The second equation is contradictory, so the given system has no solution. Geometrically,
this means that the lines corresponding to the equations in the original system are parallel
and distinct. We leave it for you to check this by graphing the lines or by showing that
they have the same slope but different y -intercepts.
E X A M P L E 4 A Linear System with Inﬁnitely Many Solutions
Solve the linear system
4x − 2y = 1
16x − 8y = 4
Solution We can eliminate x from the second equation by adding −4 times the ﬁrst
equation to the second. This yields the simpliﬁed system
4 x − 2y = 1
0=0
The second equation does not impose any restrictions on x and y and hence can be
omitted. Thus, the solutions of the system are those values of x and y that satisfy the
single equation
4x − 2y = 1
(8)
Geometrically, this means the lines corresponding to the two equations in the original
system coincide. One way to describe the solution set is to solve this equation for x in
terms of y to obtain x = 41 + 21 y and then assign an arbitrary value t (called a parameter)
6
Chapter 1 Systems of Linear Equations and Matrices
In Example 4 we could have
also obtained parametric
equations for the solutions
by solving (8) for y in terms
of x and letting x = t be
the parameter. The resulting
parametric equations would
look different but would
deﬁne the same solution set.
to y . This allows us to express the solution by the pair of equations (called parametric
equations)
x=
1
4
+ 21 t, y = t
We can obtain speciﬁc numerical solutions from these equations by substituting
1 numerical values for the parameter
t
.
For
example,
t
=
0
yields
the
solution
,0 , t = 1
4

yields the solution 43 , 1 , and t = −1 yields the solution − 41 , −1 . You can conﬁrm
that these are solutions by substituting their coordinates into the given equations.
E X A M P L E 5 A Linear System with Inﬁnitely Many Solutions
Solve the linear system
x − y + 2z = 5
2x − 2y + 4z = 10
3x − 3y + 6z = 15
Solution This system can be solved by inspection, since the second and third equations
are multiples of the ﬁrst. Geometrically, this means that the three planes coincide and
that those values of x , y , and z that satisfy the equation
x − y + 2z = 5
(9)
automatically satisfy all three equations. Thus, it sufﬁces to ﬁnd the solutions of (9).
We can do this by ﬁrst solving this equation for x in terms of y and z, then assigning
arbitrary values r and s (parameters) to these two variables, and then expressing the
solution by the three parametric equations
x = 5 + r − 2s, y = r, z = s
Speciﬁc solutions can be obtained by choosing numerical values for the parameters r
and s . For example, taking r = 1 and s = 0 yields the solution (6, 1, 0).
Augmented Matrices and
Elementary Row Operations
As the number of equations and unknowns in a linear system increases, so does the
complexity of the algebra involved in ﬁnding solutions. The required computations can
be made more manageable by simplifying notation and standardizing procedures. For
example, by mentally keeping track of the location of the +’s, the x ’s, and the =’s in the
linear system
a11 x1 + a12 x2 + · · · + a1n xn = b1
a21 x1 + a22 x2 + · · · + a2n xn = b2
..
..
..
..
.
.
.
.
am1 x1 + am2 x2 + · · · + amn xn = bm
we can abbreviate the system by writing only the rectangular array of numbers
⎡
a11
As noted in the introduction
to this chapter, the term “matrix” is used in mathematics to
denote a rectangular array of
numbers. In a later section
we will study matrices in detail, but for now we will only
be concerned with augmented
matrices for linear systems.
⎢
⎢a21
⎢ .
⎣ ..
am1
a12
· · · a1n
a22
..
.
· · · a2 n
..
.
am2
· · · amn
b1
⎤
⎥
b2 ⎥
.. ⎥
. ⎦
bm
This is called the augmented matrix for the system. For example, the augmented matrix
for the system of equations
⎡
x1 + x2 + 2×3 = 9
2×1 + 4×2 − 3×3 = 1
3×1 + 6×2 − 5×3 = 0
is
1
⎤
⎢
⎣2
1
2
9
4
−3
1⎦
3
6
−5
0
⎥
1.1 Introduction to Systems of Linear Equations
7
The basic method for solving a linear system is to perform algebraic operations on
the system that do not alter the solution set and that produce a succession of increasingly
simpler systems, until a point is reached where it can be ascertained whether the system
is consistent, and if so, what its solutions are. Typically, the algebraic operations are:
1. Multiply an equation through by a nonzero constant.
2. Interchange two equations.
3. Add a constant times one equation to another.
Since the rows (horizontal lines) of an augmented matrix correspond to the equations in
the associated system, these three operations correspond to the following operations on
the rows of the augmented matrix:
1. Multiply a row through by a nonzero constant.
2. Interchange two rows.
3. Add a constant times one row to another.
These are called elementary row operations on a matrix.
In the following example we will illustrate how to use elementary row operations and
an augmented matrix to solve a linear system in three unknowns. Since a systematic
procedure for solving linear systems will be developed in the next section, do not worry
about how the steps in the example were chosen. Your objective here should be simply
to understand the computations.
E X A M P L E 6 Using Elementary Row Operations
In the left column we solve a system of linear equations by operating on the equations in
the system, and in the right column we solve the same system by operating on the rows
of the augmented matrix.
⎡
x + y + 2z = 9
1
1
2
⎤
9
2x + 4y − 3z = 1
⎢
⎣2
4
−3
1⎦
3x + 6y − 5z = 0
3
6
−5
0
Add −2 times the ﬁrst equation to the second
to obtain
x + y + 2z =
9
2y − 7z = −17
3x + 6y − 5z =
Maxime Bôcher
(1867–1918)
0
⎥
Add −2 times the ﬁrst row to the second to
obtain
⎡
1
⎢
⎣0
1
2
2
3
6
−7
−5
9
⎤
⎥
−17⎦
0
Historical Note The ﬁrst known use of augmented matrices appeared
between 200 B.C. and 100 B.C. in a Chinese manuscript entitled Nine
Chapters of Mathematical Art. The coefﬁcients were arranged in
columns rather than in rows, as today, but remarkably the system was
solved by performing a succession of operations on the columns. The
actual use of the term augmented matrix appears to have been introduced by the American mathematician Maxime Bôcher in his book Introduction to Higher Algebra, published in 1907. In addition to being an
outstanding research mathematician and an expert in Latin, chemistry,
philosophy, zoology, geography, meteorology, art, and music, Bôcher
was an outstanding expositor of mathematics whose elementary textbooks were greatly appreciated by students and are still in demand
today.
[Image: Courtesy of the American Mathematical Society
www.ams.org]
8
Chapter 1 Systems of Linear Equations and Matrices
Add −3 times the ﬁrst equation to the third to
obtain
Add −3 times the ﬁrst row to the third to obtain
⎡
x + y + 2z =
9
2y − 7z = −17
3y − 11z = −27
Multiply the second equation by
1
2
x + y + 2z =
to obtain
⎢
⎣0
1
2
2
−7
0
3
−11
Multiply the second row by
⎡
9
1
= − 172
⎢
⎣0
3y − 11z = −27
0
y−
7
z
2
Add −3 times the second equation to the third
to obtain
x + y + 2z =
⎡
1
9
− 21 z = − 23
Multiply the third equation by −2 to obtain
x + y + 2z =
y−
11
z
2
7
z
2
=
=
z=
3
y
−11
1
− 27
0
0
− 21
0
⎥
−17⎦
−27
1
2
to obtain
9
⎤
⎥
− 172 ⎦
−27
⎤
9
⎥
− 172 ⎥
⎦
− 23
1
2
1
− 27
0
1
⎤
9
⎥
− 172 ⎦
3
Add −1 times the second row to the ﬁrst to
obtain
⎡
⎢
⎢0
⎣
0
1
11
2
− 27
0
0
1
1
Add −11
times the third equation to the ﬁrst
2
and 27 times the third equation to the second to
obtain
x
3
⎢
⎢0
⎣
⎢
⎣0
3
35
2
− 172
− 27
2
1
Add −1 times the second equation to the ﬁrst
to obtain
+
1
1
⎡
y − 27 z = − 172
x
2
⎤
Multiply the third row by −2 to obtain
9
z=
1
9
Add −3 times the second row to the third to
obtain
y − 27 z = − 172
The solution in this example
can also be expressed as the ordered triple (1, 2, 3) with the
understanding that the numbers in the triple are in the
same order as the variables in
the system, namely, x, y, z.
1
⎤
35
2
⎥
− 172 ⎥
⎦
3
Add − 11
times the third row to the ﬁrst and
2
times the third row to the second to obtain
⎡
=1
=2
⎤
⎢
⎣0
0
0
1
1
0
2⎦
0
0
1
3
z=3
1
7
2
⎥
The solution x = 1, y = 2, z = 3 is now evident.
Exercise Set 1.1
1. In each part, determine whether the equation is linear in x1 ,
x2 , and x3 .
(a) x1 + 5×2 −
√
2 x3 = 1
(c) x1 = −7×2 + 3×3
(e)
3/5
x1
− 2×2 + x3 = 4
2. In each part, determine whether the equation is linear in x
and y .
(b) x1 + 3×2 + x1 x3 = 2
(a) 21/3 x +
(d) x1−2 + x2 + 8×3 = 5
(c) cos
(f ) πx1 −
(e) xy = 1
√
2 x2 = 7
1/3
π
7
√
3y = 1
x − 4y = log 3
√
(b) 2x 1/3 + 3 y = 1
(d)
π
7
cos x − 4y = 0
(f ) y + 7 = x
1.1 Introduction to Systems of Linear Equations
3. Using the notation of Formula (7), write down a general linear
system of
(d)
(a) two equations in two unknowns.
(b) three equations in three unknowns.
(c) two equations in four unknowns.
4. Write down the augmented matrix for each of the linear systems in Exercise 3.
In each part of Exercises 5–6, ﬁnd a linear system in the unknowns x1 , x2 , x3 , . . . , that corresponds to the given augmented
matrix.
⎡
2
⎢
5. (a) ⎣3
0

6. (a)
⎤
0
−4
1
0
3
−1
5
2
0
⎡
3
⎢−4
⎢
(b) ⎢
⎣−1
0
⎡
0
⎥
0⎦
1
0
0
3
0
3
⎢
(b) ⎣7
0
−1
−3
−4
1
4
0
0
1
−2
−1
−1
−6

0
1
−2
−2
4
1
⎤
5
⎥
−3⎦
7
(c)
x3
(d)
2
, 25 , 2

,
10 2
,
7 7

(e)
5

7
, 87 , 0
7
, 227 , 2
5
(c) (5, 8, 1)

11. In each part, solve the linear system, if possible, and use the
result to determine whether the lines represented by the equations in the system have zero, one, or inﬁnitely many points of
intersection. If there is a single point of intersection, give its
coordinates, and if there are inﬁnitely many, ﬁnd parametric
equations for them.
(a) 3x − 2y = 4
6x − 4 y = 9
(b) 2x − 4y = 1
4 x − 8y = 2
(c) x − 2y = 0
x − 4y = 8
12. Under what conditions on a and b will the following linear
system have no solutions, one solution, inﬁnitely many solutions?
2 x − 3y = a
4x − 6y = b
(d) 3v − 8w + 2x − y + 4z = 0
14. (a) x + 10y = 2
(b) x1 + 3×2 − 12×3 = 3
(c) 4×1 + 2×2 + 3×3 + x4 = 20
(d) v + w + x − 5y + 7z = 0
In Exercises 15–16, each linear system has inﬁnitely many solutions. Use parametric equations to describe its solution set.
(b) 2×1
+ 2×3 = 1
3×1 − x2 + 4×3 = 7
6×1 + x2 − x3 = 0
=1
=2
=3
2×1 − 4×2 − x3 = 1
x1 − 3×2 + x3 = 1
3×1 − 5×2 − 3×3 = 1
13
7
5
(b)
(c) −8×1 + 2×2 − 5×3 + 6×4 = 1
9. In each part, determine whether the given 3-tuple is a solution
of the linear system
(a) (3, 1, 1)

, 87 , 1
(b) 3×1 − 5×2 + 4×3 = 7
(b) 6×1 − x2 + 3×3 = 4
5×2 − x3 = 1
8. (a) 3×1 − 2×2 = −1
4×1 + 5×2 = 3
7×1 + 3×2 = 2
x2
7
13. (a) 7x − 5y = 3
2×2
− 3×4 + x5 = 0
−3×1 − x2 + x3
= −1
6×1 + 2×2 − x3 + 2×4 − 3×5 = 6
(c) x1
5
In each part of Exercises 13–14, use parametric equations to
describe the solution set of the linear equation.
⎤
3
−3 ⎥
⎥
⎥
−9 ⎦
−2
In each part of Exercises 7–8, ﬁnd the augmented matrix for
the linear system.
7. (a) −2×1 = 6
3×1 = 8
9×1 = −3
(a)
9
(b) (3, −1, 1)
(c) (13, 5, 2)
(e) (17, 7, 5)
10. In each part, determine whether the given 3-tuple is a solution
of the linear system
x + 2y − 2z = 3
3x − y + z = 1
−x + 5y − 5z = 5
15. (a) 2x − 3y = 1
6 x − 9y = 3
(b)
x1 + 3×2 − x3 = −4
3×1 + 9×2 − 3×3 = −12
−x1 − 3×2 + x3 =
4
16. (a) 6×1 + 2×2 = −8
3×1 + x2 = −4
(b)
2x − y + 2z = −4
6x − 3y + 6z = −12
−4 x + 2 y − 4 z =
8
In Exercises 17–18, ﬁnd a single elementary row operation that
will create a 1 in the upper left corner of the given augmented matrix and will not create any fractions in its ﬁrst row.
⎡
−3
17. (a) ⎣ 2
0
⎡
2
18. (a) ⎣ 7
−5
−1
−3
2
4
1
4
⎤
2
3
−3
4
2⎦
1
−6
8
3⎦
7
4
2
⎤
⎡
0
(b) ⎣2
1
⎡
7
(b) ⎣ 3
−6
−1
−9
⎤
−5
0
2⎦
3
3
−3
4
−4
−1
3
−2
8
−1
⎤
2
1⎦
4
10
Chapter 1 Systems of Linear Equations and Matrices
In Exercises 19–20, ﬁnd all values of k for which the given
augmented matrix corresponds to a consistent linear system.
19. (a)
1
4
k
−4
8
2
20. (a)
3
−6
−4
k
8
5
(b)
(b)
1
4
k
k
1
−1
4
8
−1
−4
−2
2
21. The curve y = ax 2 + bx + c shown in the accompanying ﬁgure passes through the points (x1 , y1 ), (x2 , y2 ), and (x3 , y3 ).
Show that the coefﬁcients a , b, and c form a solution of the
system of linear equations whose augmented matrix is
⎡
x12
⎢ 2
⎣x2
x32
y
y1
⎤
x1
1
x2
1
⎥
y2 ⎦
x3
1
y3
Let x, y, and z denote the number of ounces of the ﬁrst, second, and third foods that the dieter will consume at the main
meal. Find (but do not solve) a linear system in x, y, and z
whose solution tells how many ounces of each food must be
consumed to meet the diet requirements.
26. Suppose that you want to ﬁnd values for a, b, and c such that
the parabola y = ax 2 + bx + c passes through the points
(1, 1), (2, 4), and (−1, 1). Find (but do not solve) a system
of linear equations whose solutions provide values for a, b,
and c. How many solutions would you expect this system of
equations to have, and why?
27. Suppose you are asked to ﬁnd three real numbers such that the
sum of the numbers is 12, the sum of two times the ﬁrst plus
the second plus two times the third is 5, and the third number
is one more than the ﬁrst. Find (but do not solve) a linear
system whose equations describe the three conditions.
True-False Exercises
y = ax2 + bx + c
TF. In parts (a)–(h) determine whether the statement is true or
false, and justify your answer.
(x3, y3)
(x1, y1)
(a) A linear system whose equations are all homogeneous must
be consistent.
(x2, y2)
x
Figure Ex-21
22. Explain why each of the three elementary row operations does
not affect the solution set of a linear system.
23. Show that if the linear equations
x1 + kx2 = c
and
x1 + lx2 = d
have the same solution set, then the two equations are identical
(i.e., k = l and c = d ).
24. Consider the system of equations
ax + by = k
cx + dy = l
ex + fy = m
Discuss the relative positions of the lines ax + by = k ,
cx + dy = l , and ex + fy = m when
(a) the system has no solutions.
(b) the system has exactly one solution.
(b) Multiplying a row of an augmented matrix through by zero is
an acceptable elementary row operation.
(c) The linear system
x− y =3
2x − 2y = k
cannot have a unique solution, regardless of the value of k .
(d) A single linear equation with two or more unknowns must
have inﬁnitely many solutions.
(e) If the number of equations in a linear system exceeds the number of unknowns, then the system must be inconsistent.
(f ) If each equation in a consistent linear system is multiplied
through by a constant c, then all solutions to the new system
can be obtained by multiplying solutions from the original
system by c.
(g) Elementary row operations permit one row of an augmented
matrix to be subtracted from another.
(h) The linear system with corresponding augmented matrix
(c) the system has inﬁnitely many solutions.
25. Suppose that a certain diet calls for 7 units of fat, 9 units of
protein, and 16 units of carbohydrates for the main meal, and
suppose that an individual has three possible foods to choose
from to meet these requirements:
Food 1: Each ounce contains 2 units of fat, 2 units of
protein, and 4 units of carbohydrates.
Food 2: Each ounce contains 3 units of fat, 1 unit of
protein, and 2 units of carbohydrates.
Food 3: Each ounce contains 1 unit of fat, 3 units of
protein, and 5 units of carbohydrates.
2
0
−1
0
4
−1
is consistent.
Working withTechnology
T1. Solve the linear systems in Examples 2, 3, and 4 to see how
your technology utility handles the three types of systems.
T2. Use the result in Exercise 21 to ﬁnd values of a , b, and c
for which the curve y = ax 2 + bx + c passes through the points
(−1, 1, 4), (0, 0, 8), and (1, 1, 7).
1.2 Gaussian Elimination
11
1.2 Gaussian Elimination
In this section we will develop a systematic procedure for solving systems of linear
equations. The procedure is based on the idea of performing certain operations on the rows
of the augmented matrix that simplify it to a form from which the solution of the system
can be ascertained by inspection.
Considerations in Solving
Linear Systems
When considering methods for solving systems of linear equations, it is important to
distinguish between large systems that must be solved by computer and small systems
that can be solved by hand. For example, there are many applications that lead to
linear systems in thousands or even millions of unknowns. Large systems require special
techniques to deal with issues of memory size, roundoff errors, solution time, and so
forth. Such techniques are studied in the ﬁeld of numerical analysis and will only be
touched on in this text. However, almost all of the methods that are used for large
systems are based on the ideas that we will develop in this section.
Echelon Forms
In Example 6 of the last section, we solved a linear system in the unknowns x , y , and z
by reducing the augmented matrix to the form
⎡
1
⎢0
⎣
0
0
1
0
0
0
1
⎤
1
2⎥
⎦
3
from which the solution x = 1, y = 2, z = 3 became evident. This is an example of a
matrix that is in reduced row echelon form. To be of this form, a matrix must have the
following properties:
1. If a row does not consist entirely of zeros, then the ﬁrst nonzero number in the row
is a 1. We call this a leading 1.
2. If there are any rows that consist entirely of zeros, then they are grouped together at
the bottom of the matrix.
3. In any two successive rows that do not consist entirely of zeros, the leading 1 in the
lower row occurs farther to the right than the leading 1 in the higher row.
4. Each column that contains a leading 1 has zeros everywhere else in that column.
A matrix that has the ﬁrst three properties is said to be in row echelon form. (Thus,
a matrix in reduced row echelon form is of necessity in row echelon form, but not
conversely.)
E X A M P L E 1 Row Echelon and Reduced Row Echelon Form
The following matrices are in reduced row echelon form.
⎡
1
⎢
⎣0
0
0
1
0
0
0
1
⎤
⎡
4
1
⎥ ⎢
7⎦ , ⎣0
0
−1
0
1
0
⎤
0
⎥
0⎦ ,
1
⎡
0
⎢0
⎢
⎢
⎣0
0
1
0
0
0
−2
0
0
0
0
1
0
0
⎤
1
3⎥
⎥
⎥,
0⎦
0
0
0
0
0
The following matrices are in row echelon form but not reduced row echelon form.
⎡
1
⎢
⎣0
0
4
1
0
−3
6
1
⎤
⎡
7
1
⎥ ⎢
2⎦ , ⎣0
5
0
1
1
0
⎤
⎡
0
0
⎥ ⎢
0⎦ , ⎣0
0
0
1
0
0
2
1
0
6
−1
0
⎤
0
⎥
0⎦
1
12
Chapter 1 Systems of Linear Equations and Matrices
E X A M P L E 2 More on Row Echelon and Reduced Row Echelon Form
As Example 1 illustrates, a matrix in row echelon form has zeros below each leading 1,
whereas a matrix in reduced row echelon form has zeros below and above each leading
1. Thus, with any real numbers substituted for the ∗’s, all matrices of the following types
are in row echelon form:
⎡
1
⎢0
⎢
⎢
⎣0
0
⎤
∗ ∗ ∗
1 ∗ ∗⎥
⎥
⎥,
0 1 ∗⎦
⎡
1
⎢0
⎢
⎢
⎣0
0
0 0 1
⎤
∗ ∗ ∗
1 ∗ ∗⎥
⎥
⎥,
0 1 ∗⎦
⎡
1
⎢0
⎢
⎢
⎣0
0
0 0 0
⎤
∗ ∗ ∗
1 ∗ ∗⎥
⎥
⎥,
0 0 0⎦
0 0 0
⎡
0
⎢0
⎢
⎢
⎢0
⎢
⎣0
0
1
0
0
0
0
⎤
∗
∗⎥
⎥
⎥
∗⎥
⎥
∗⎦
0 0 0 0 0 0 1 ∗
∗ ∗ ∗ ∗ ∗
0 1 ∗ ∗ ∗
0 0 1 ∗ ∗
0 0 0 1 ∗
∗
∗
∗
∗
All matrices of the following types are in reduced row echelon form:
⎡
⎤
⎡
⎤
⎡
1 0 0 0
1 0 0 ∗
1 0
⎢0 1 0 0⎥ ⎢0 1 0 ∗⎥ ⎢0 1
⎢
⎥ ⎢
⎥ ⎢
⎢
⎥, ⎢
⎥, ⎢
⎣0 0 1 0⎦ ⎣0 0 1 ∗⎦ ⎣0 0
0 0 0 1
0 0
0 0 0 0
⎡
⎤
0
∗ ∗
⎢0
⎢
∗ ∗⎥
⎥ ⎢
⎥ , ⎢0
⎢
0 0⎦
⎣0
0 0
0
1
0
0
0
0
∗ 0 0 0 ∗
0 1 0 0 ∗
0 0 1 0 ∗
0 0 0 1 ∗
∗
∗
∗
∗
0
0
0
0
0 0 0 0 0 0 1
∗
∗
∗
∗
⎤
∗
∗⎥
⎥
⎥
∗⎥
⎥
∗⎦
∗
If, by a sequence of elementary row operations, the augmented matrix for a system of
linear equations is put in reduced row echelon form, then the solution set can be obtained
either by inspection or by converting certain linear equations to parametric form. Here
are some examples.
E X A M P L E 3 Unique Solution
Suppose that the augmented matrix for a linear system in the unknowns x1 , x2 , x3 , and
x4 has been reduced by elementary row operations to
⎡
1
⎢0
⎢
⎢
⎣0
0
0
1
0
0
0
0
1
0
0
0
0
1
⎤
3
−1⎥
⎥
⎥
0⎦
5
This matrix is in reduced row echelon form and corresponds to the equations
x1
In Example 3 we could, if
desired, express the solution
more succinctly as the 4-tuple
(3, −1, 0, 5).
x2
x3
= 3
= −1
= 0
x4 = 5
Thus, the system has a unique solution, namely, x1 = 3, x2 = −1, x3 = 0, x4 = 5.
E X A M P L E 4 Linear Systems in Three Unknowns
In each part, suppose that the augmented matrix for a linear system in the unknowns
x , y , and z has been reduced by elementary row operations to the given reduced row
echelon form. Solve the system.
⎡
1
⎢
(a) ⎣0
0
0
1
0
0
2
0
⎤
0
⎥
0⎦
1
⎡
1
⎢
(b) ⎣0
0
0
1
0
3
−4
0
⎤
−1
⎥
2⎦
0
⎡
1
⎢
(c) ⎣0
0
−5
0
0
1
0
0
⎤
4
⎥
0⎦
0
1.2 Gaussian Elimination
13
Solution (a) The equation that corresponds to the last row of the augmented matrix is
0x + 0y + 0z = 1
Since this equation is not satisﬁed by any values of x , y , and z, the system is inconsistent.
Solution (b) The equation that corresponds to the last row of the augmented matrix is
0x + 0y + 0z = 0
This equation can be omitted since it imposes no restrictions on x , y , and z; hence, the
linear system corresponding to the augmented matrix is
+ 3z = −1
y − 4z = 2
x
Since x and y correspond to the leading 1’s in the augmented matrix, we call these
the leading variables. The remaining variables (in this case z) are called free variables.
Solving for the leading variables in terms of the free variables gives
x = −1 − 3z
y = 2 + 4z
From these equations we see that the free variable z can be treated as a parameter and
assigned an arbitrary value t , which then determines values for x and y . Thus, the
solution set can be represented by the parametric equations
x = −1 − 3t, y = 2 + 4t, z = t
By substituting various values for t in these equations we can obtain various solutions
of the system. For example, setting t = 0 yields the solution
x = −1, y = 2, z = 0
and setting t = 1 yields the solution
x = −4, y = 6, z = 1
Solution (c) As explained in part (b), we can omit the equations corresponding to the
zero rows, in which case the linear system associated with the augmented matrix consists
of the single equation
x − 5y + z = 4
(1)
We will usually denote parameters in a general solution
by the letters r, s, t, . . . , but
any letters that do not conﬂict with the names of the
unknowns can be used. For
systems with more than three
unknowns, subscripted letters
such as t1 , t2 , t3 , . . . are convenient.
from which we see that the solution set is a plane in three-dimensional space. Although
(1) is a valid form of the solution set, there are many applications in which it is preferable
to express the solution set in parametric form. We can convert (1) to parametric form
by solving for the leading variable x in terms of the free variables y and z to obtain
x = 4 + 5y − z
From this equation we see that the free variables can be assigned arbitrary values, say
y = s and z = t , which then determine the value of x . Thus, the solution set can be
expressed parametrically as
x = 4 + 5s − t, y = s, z = t
(2)
Formulas, such as (2), that express the solution set of a linear system parametrically
have some associated terminology.
DEFINITION 1 If a linear system has inﬁnitely many solutions, then a set of parametric
equations from which all solutions can be obtained by assigning numerical values to
the parameters is called a general solution of the system.
14
Chapter 1 Systems of Linear Equations and Matrices
Elimination Methods
We have just seen how easy it is to solve a system of linear equations once its augmented
matrix is in reduced row echelon form. Now we will give a step-by-step elimination
procedure that can be used to reduce any matrix to reduced row echelon form. As we
state each step in the procedure, we illustrate the idea by reducing the following matrix
to reduced row echelon form.
⎡
0
0
⎢
⎣2
4
2
4
−2
−10
−5
0
7
6
12
6
12
⎤
⎥
28⎦
−5 −1
Step 1. Locate the leftmost column that does not consist entirely of zeros.
⎡
0
⎢
2
⎣
2
0
4
4
2
10
5
0
6
6
7
12
5
⎤
12
⎥
28⎦
1
Leftmost nonzero column
Step 2. Interchange the top row with another row, if necessary, to bring a nonzero entry
to the top of the column found in Step 1.
⎡
2
⎢
⎣0
2
−10
−2
0
−5
4
4
6
12
0
7
6
−5
⎤
28
⎥
12⎦
The ﬁrst and second rows in the preceding
matrix were interchanged.
−1
Step 3. If the entry that is now at the top of the column found in Step 1 is a , multiply
the ﬁrst row by 1/a in order to introduce a leading 1.
⎡
1
⎢
⎣0
2
−5
0 −2
4 −5
2
⎤
3
6
14
0
7
12⎦
6
−5
⎥
The ﬁrst row of the preceding matrix was
multiplied by 21 .
−1
Step 4. Add suitable multiples of the top row to the rows below so that all entries below
the leading 1 become zeros.
⎡
1
⎢
⎣0
0
⎤
−5
0 −2
3
6
14
0
7
12⎦
0
0
2
5
⎥
−17 −29
−2 times the ﬁrst row of the preceding
matrix was added to the third row.
Step 5. Now cover the top row in the matrix and begin again with Step 1 applied to the
submatrix that remains. Continue in this way until the entire matrix is in row
echelon form.
⎡
1
⎢
⎣0
0
2
0
0
5
2
5
3
0
0
6
7
17
⎤
14
⎥
12 ⎦
29
Leftmost nonzero column
in the submatrix
⎡
1
⎢
⎣0
2
5
3
6
0
1
0
7
2
0
0
5
0
17
14
⎤
⎥
6⎦
29
The first row in the submatrix was
multiplied by 1 to introduce a
2
leading 1.
1.2 Gaussian Elimination
⎡
1
⎢0
⎣
2
5
3
6
0
1
0
0
0
0
7
2
1
2
0
⎡
1
⎢
⎣0
2
5
3
6
0
1
0
0
0
0
0
7
2
1
2
⎡
1
⎢
⎣0
0
⎤
14
⎥
6⎦
1
14
15
⎤
⎥
6⎦
1
–5 times the first row of the submatrix
was added to the second row of the
submatrix to introduce a zero below
the leading 1.
The top row in the submatrix was
covered, and we returned again to
Step 1.
Leftmost nonzero column
in the new submatrix
2
5
3
6
0
0
1
0
0
0
7
2
1
14
⎤
⎥
6⎦
2
The first (and only) row in the new
submatrix was multiplied by 2 to
introduce a leading 1.
The entire matrix is now in row echelon form. To ﬁnd the reduced row echelon form we
need the following additional step.
Step 6. Beginning with the last nonzero row and working upward, add suitable multiples
of each row to the rows above to introduce zeros above the leading 1’s.
⎡
1
⎢
⎣0
0
2
0
0
−5
1
⎢
⎣0
0
2
0
0
−5
1
⎢
⎣0
0
2
0
0
⎡
⎡
⎤
3
0
0
6
0
1
14
⎥
1⎦
2
7
times the third row of the preceding
2
matrix was added to the second row.
1
0
3
0
0
0
0
1
2
⎥
1⎦
2
−6 times the third row was added to the
ﬁrst row.
0
1
0
3
0
0
0
0
1
7
⎥
1⎦
2
5 times the second row was added to the
ﬁrst row.
1
0
⎤
⎤
The last matrix is in reduced row echelon form.
The procedure (or algorithm) we have just described for reducing a matrix to reduced
row echelon form is called Gauss–Jordan elimination. This algorithm consists of two
parts, a forward phase in which zeros are introduced below the leading 1’s and a backward
phase in which zeros are introduced above the leading 1’s. If only theforward phase is
Carl Friedrich Gauss
(1777–1855)
Wilhelm Jordan
(1842–1899)
Historical Note Although versions of Gaussian elimination were known much
earlier, its importance in scientiﬁc computation became clear when the great
German mathematician Carl Friedrich Gauss used it to help compute the orbit
of the asteroid Ceres from limited data. What happened was this: On January 1,
1801 the Sicilian astronomer and Catholic priest Giuseppe Piazzi (1746–1826)
noticed a dim celestial object that he believed might be a “missing planet.” He
named the object Ceres and made a limited number of positional observations
but then lost the object as it neared the Sun. Gauss, then only 24 years old,
undertook the problem of computing the orbit of Ceres from the limited data
using a technique called “least squares,” the equations of which he solved by
the method that we now call “Gaussian elimination.” The work of Gauss created a sensation when Ceres reappeared a year later in the constellation Virgo
at almost the precise position that he predicted! The basic idea of the method
was further popularized by the German engineer Wilhelm Jordan in his book
on geodesy (the science of measuring Earth shapes) entitled Handbuch der Vermessungskunde and published in 1888.
[Images: Photo Inc/Photo Researchers/Getty Images (Gauss);
Leemage/Universal Images Group/Getty Images (Jordan)]
16
Chapter 1 Systems of Linear Equations and Matrices
used, then the procedure produces a row echelon form and is called Gaussian elimination.
For example, in the preceding computations a row echelon form was obtained at the end
of Step 5.
E X A M P L E 5 Gauss–Jordan Elimination
Solve by Gauss–Jordan elimination.
x1 + 3×2 − 2×3
+ 2×5
2×1 + 6×2 − 5×3 − 2×4 + 4×5 − 3×6
5×3 + 10×4
+ 15×6
2×1 + 6×2
+ 8×4 + 4×5 + 18×6
= 0
= −1
= 5
= 6
Solution The augmented matrix for the system is
⎡
1
⎢2
⎢
⎢
⎣0
2
3
6
0
6
−2
−5
5
0
0
−2
10
8
2
4
0
4
0
−3
15
18
⎤
0
−1⎥
⎥
⎥
5⎦
6
Adding −2 times the ﬁrst row to the second and fourth rows gives
⎡
1
⎢0
⎢
⎢
⎣0
0
3
0
0
0
−2
−1
5
4
0
−2
10
8
2
0
0
0
0
−3
15
18
⎤
0
−1⎥
⎥
⎥
5⎦
6
Multiplying the second row by −1 and then adding −5 times the new second row to the
third row and −4 times the new second row to the fourth row gives
⎡
1
⎢0
⎢
⎢
⎣0
0
3
0
0
0
−2
1
0
0
0
2
0
0
2
0
0
0
⎤
0
3
0
6
0
1⎥
⎥
⎥
0⎦
2
Interchanging the third and fourth rows and then multiplying the third row of the resulting matrix by 16 gives the row echelon form
⎡
1
⎢0
⎢
3
0
−2
0
0
0
⎢
⎣0
⎤
1
0
2
2
0
0
3
0
1⎥
⎥
0
0
0
0
0
0
1
0
0
1⎥
⎦
3
This completes the forward phase since
there are zeros below the leading 1’s.
Adding −3 times the third row to the second row and then adding 2 times the second
row of the resulting matrix to the ﬁrst row yields the reduced row echelon form
⎡
3
0
0
1
4
2
2
0
0
0
0
0⎥
⎥
0
0
0
0
0
0
0
0
0
1
0
0
⎢
⎣0
Note that in constructing the
linear system in (3) we ignored
the row of zeros in the corresponding augmented matrix.
Why is this justiﬁed?
⎤
1
⎢0
⎢
1⎥
⎦
3
This completes the backward phase since
there are zeros above the leading 1’s.
The corresponding system of equations is
x1 + 3×2
+ 4×4 + 2×5
x3 + 2×4
=0
=0
x6 =
1
3
(3)
1.2 Gaussian Elimination
17
Solving for the leading variables, we obtain
x1 = −3×2 − 4×4 − 2×5
x3 = −2×4
x6 =
1
3
Finally, we express the general solution of the system parametrically by assigning the
free variables x2 , x4 , and x5 arbitrary values r, s , and t , respectively. This yields
x1 = −3r − 4s − 2t, x2 = r, x3 = −2s, x4 = s, x5 = t, x6 =
Homogeneous Linear
Systems
1
3
A system of linear equations is said to be homogeneous if the constant terms are all zero;
that is, the system has the form
a11 x1 + a12 x2 + · · · + a1n xn = 0
a21 x1 + a22 x2 + · · · + a2n xn = 0
..
..
..
..
.
.
.
.
am1 x1 + am2 x2 + · · · + amn xn = 0
Every homogeneous system of linear equations is consistent because all such systems
have x1 = 0, x2 = 0, . . . , xn = 0 as a solution. This solution is called the trivial solution;
if there are other solutions, they are called nontrivial solutions.
Because a homogeneous linear system always has the trivial solution, there are only
two possibilities for its solutions:
• The system has only the trivial solution.
• The system has inﬁnitely many solutions in addition to the trivial solution.
In the special case of a homogeneous linear system of two equations in two unknowns,
say
a1 x + b1 y = 0 (a1 , b1 not both zero)
a2 x + b2 y = 0 (a2 , b2
not both zero)
the graphs of the equations are lines through the origin, and the trivial solution corresponds to the point of intersection at the origin (Figure 1.2.1).
y
y
a1x + b1y = 0
x
a 2 x + b2 y = 0
Only the trivial solution
Figure 1.2.1
x
a1x + b1y = 0
and
a 2 x + b2 y = 0
Infinitely many
solutions
There is one case in which a homogeneous system is assured of having nontrivial
solutions—namely, whenever the system involves more unknowns than equations. To
see why, consider the following example of four equations in six unknowns.
18
Chapter 1 Systems of Linear Equations and Matrices
E X A M P L E 6 A Homogeneous System
Use Gauss–Jordan elimination to solve the homogeneous linear system
x1 + 3×2 − 2×3
+ 2 x5
2×1 + 6×2 − 5×3 − 2×4 + 4×5 − 3×6
+ 15×6
5×3 + 10×4
+ 8×4 + 4×5 + 18×6
2×1 + 6×2
=0
=0
=0
=0
(4)
Solution Observe ﬁrst that the coefﬁcients of the unknowns in this system are the same
as those in Example 5; that is, the two systems differ only in the constants on the right
side. The augmented matrix for the given homogeneous system is
⎡
1
⎢2
⎢
⎢
⎣0
2
3
6
0
6
−2
−5
0
−2
10
8
5
0
2
4
0
4
⎤
0
−3
15
18
0
0⎥
⎥
⎥
0⎦
0
(5)
which is the same as the augmented matrix for the system in Example 5, except for zeros
in the last column. Thus, the reduced row echelon form of this matrix will be the same
as that of the augmented matrix in Example 5, except for the last column. However,
a moment’s reﬂection will make it evident that a column of zeros is not changed by an
elementary row operation, so the reduced row echelon form of (5) is
⎡
1
⎢0
⎢
⎢
⎣0
0
3
0
0
0
0
1
0
0
4
2
0
0
2
0
0
0
0
0
1
0
⎤
0
0⎥
⎥
⎥
0⎦
0
(6)
The corresponding system of equations is
x1 + 3×2
+ 4×4 + 2×5
x3 + 2×4
=0
=0
x6 = 0
Solving for the leading variables, we obtain
x1 = −3×2 − 4×4 − 2×5
x3 = −2×4
x6 = 0
(7)
If we now assign the free variables x2 , x4 , and x5 arbitrary values r , s , and t , respectively,
then we can express the solution set parametrically as
x1 = −3r − 4s − 2t, x2 = r, x3 = −2s, x4 = s, x5 = t, x6 = 0
Note that the trivial solution results when r = s = t = 0.
Free Variables in
Homogeneous Linear
Systems
Example 6 illustrates two important points about solving homogeneous linear systems:
1. Elementary row operations do not alter columns of zeros in a matrix, so the reduced
row echelon form of the augmented matrix for a homogeneous linear system has
a ﬁnal column of zeros. This implies that the linear system corresponding to the
reduced row echelon form is homogeneous, just like the original system.
1.2 Gaussian Elimination
19
2. When we constructed the homogeneous linear system corresponding to augmented
matrix (6), we ignored the row of zeros because the corresponding equation
0x1 + 0x2 + 0x3 + 0x4 + 0x5 + 0x6 = 0
does not impose any conditions on the unknowns. Thus, depending on whether or
not the reduced row echelon form of the augmented matrix for a homogeneous linear
system has any rows of zero, the linear system corresponding to that reduced row
echelon form will either have the same number of equations as the original system
or it will have fewer.
Now consider a general homogeneous linear system with n unknowns, and suppose
that the reduced row echelon form of the augmented matrix has r nonzero rows. Since
each nonzero row has a leading 1, and since each leading 1 corresponds to a leading
variable, the homogeneous system corresponding to the reduced row echelon form of
the augmented matrix must have r leading variables and n − r free variables. Thus, this
system is of the form

xk1
+ ()=0
+
xk2
..

()=0
..
.
.

x kr + ( ) = 0
(8)
where in each equation the expression ( ) denotes a sum that involves the free variables,
if any [see (7), for example]. In summary, we have the following result.
THEOREM 1.2.1 Free Variable Theorem for Homogeneous Systems
If a homogeneous linear system has n unknowns, and if the reduced row echelon form
of its augmented matrix has r nonzero rows, then the system has n − r free variables.
Note that Theorem 1.2.2 applies only to homogeneous
systems—a nonhomogeneous
system with more unknowns
than equations need not be
consistent. However, we will
prove later that if a nonhomogeneous system with more
unknowns then equations is
consistent, then it has inﬁnitely many solutions.
Theorem 1.2.1 has an important implication for homogeneous linear systems with
more unknowns than equations. Speciﬁcally, if a homogeneous linear system has m
equations in n unknowns, and if m
(d)
⎤
0
1
0
⎡
0
⎥
0⎦
1
1
⎢
(b) ⎣0
0
1
0
3
1
0
1
2
4
⎡
0
⎢
(f ) ⎣0
0
⎡
0
⎢0
⎢
⎣0
0
⎥
0⎦
(g)
0
⎤
2
0
1
0⎦
0
0
⎡
⎥
⎤
5
1
−3
0
0
0
⎥
1⎦
⎡
0
⎢
(c) ⎣0
0
⎤
1
0
0
0
0
⎥
1⎦
0
⎤
2
0
3
0
1
1
0
0
0
⎥
1⎦
0
0
0
0
−7
5
5
0
1
3
2
⎤
1
0
0
1
0⎦
2
0
⎥
⎡
1
⎢
(e) ⎣0
0
⎡

⎤
3
4
0
1⎦
0
0
2
3
0
0⎦
0
1
⎥
⎥
⎤
2
3
4
5
0
7
1
0
0
0
1⎦
0
0
0
0

3⎥
⎥
⎥
(g)
1
−2
0
1
0
0
1
−2

In Exercises 3–4, suppose that the augmented matrix for a linear system has been reduced by row operations to the given row
echelon form. Solve the system.
⎤
1
−3
4
7
3. (a) ⎣0
0
1
2
2⎦
0
1
5
1
0
8
6
(b) ⎣0
0
1
4
−5
−9
0
1
1
2
1
⎢0
⎢
(c) ⎢
⎣0
0
7
0
0
0
−2
−8
1
0
0
0
1
1
0
1
⎢
(d) ⎣0
0
−3
7
4
0
1
⎥
0⎦
1
⎡
⎢
1
⎤
0
⎢
(c) ⎣0
0
⎢
⎢1
⎢
(f ) ⎢
⎣0
⎡
0⎥
⎥
1
(b) ⎣0
0
⎢
⎢
(d) ⎣0
1
1
(e) ⎢

1
⎡
0
⎥
0⎦
0
⎡

⎤
2. (a) ⎣0
0
⎢
0
1
0
⎤
1
⎡
⎡
1
0
⎥
⎤
⎤
⎥
3⎦
6
3
0
⎤
−3
5⎥
⎥
⎥
9⎦
0
1.2 Gaussian Elimination
⎡
1
⎤
−3
⎥
0⎦
17. 3×1 + x2 + x3 + x4 = 0
5×1 − x2 + x3 − x4 = 0
0
0
1
0
0
0
1
1
⎢
(b) ⎣0
0
0
1
0
0
0
1
−7
3
1
8
⎥
2⎦
−5
1
⎢0
⎢
(c) ⎢
⎣0
0
−6
0
1
0
0
0
0
1
0
3
4
5
0
1
−3
0
0
0
0
1
0
0
⎥
0⎦
1
⎢
4. (a) ⎣0
⎡
⎡
⎡
⎢
(d) ⎣0
0
0
0
7
⎤
−2
⎤
7⎥
⎥
⎥
8⎦
0
In Exercises 5–8, solve the linear system by Gaussian elimination.
5.
x1 + x2 + 2×3 = 8
−x1 − 2×2 + 3×3 = 1
3×1 − 7×2 + 4×3 = 10
7.
x − y + 2z − w
2x + y − 2z − 2w
−x + 2y − 4z + w
− 3w
3x
8.
− 2b + 3c = 1
3a + 6b − 3c = −2
6a + 6b + 3c = 5
2u + v − 4w + 3x = 0
2u + 3v + 2w − x = 0
−4u − 3v + 5w − 4x = 0
6.
2×1 + 2×2 + 2×3 = 0
−2×1 + 5×2 + 2×3 = 1
8×1 + x2 + 4×3 = −1
20. x1 + 3×2
x1 + 4×2
− 2×2
2×1 − 4×2
x1 − 2×2
+ 4z = 0
− 3z = 0
+ z=0
− 2z = 0
+ 2×3
− 2×3
+ x3
− x3
+ x4 = 0
=0
− x4 = 0
+ x4 = 0
+ x4 = 0
21. 2I1 − I2 + 3I3
I1
− 2I3
3I1 − 3I2 + I3
2I1 + I2 + 4I3
22.
= −1
= −2
= 1
= −3
+ 2y
− y
+ y
+ 3y
2x
w
2 w + 3x
−2 w + x
+ 4I4
+ 7I4
+ 5I4
+ 4I4
v + 3w − 2 x = 0
18.
⎤
19.
23
= 9
= 11
= 8
= 10
Z3 + Z4
−Z1 − Z2 + 2Z3 − 3Z4
Z1 + Z2 − 2Z3
2Z1 + 2Z2 − Z3
+ Z5
+ Z5
− Z5
+ Z5
=0
=0
=0
=0
In each part of Exercises 23–24, the augmented matrix for a
linear system is given in which the asterisk represents an unspeciﬁed real number. Determine whether the system is consistent,
and if so whether the solution is unique. Answer “inconclusive” if
there is not enough information to make a decision.
In Exercises 9–12, solve the linear system by Gauss–Jordan
elimination.
9. Exercise 5
10. Exercise 6
11. Exercise 7
12. Exercise 8
In Exercises 13–14, determine whether the homogeneous system has nontrivial solutions by inspection (without pencil and
paper).
⎡
1
23. (a) ⎣0
0
⎡
1
(c) ⎣0
0
⎡
1
24. (a) ⎣0
0
13. 2×1 − 3×2 + 4×3 − x4 = 0
7×1 + x2 − 8×3 + 9×4 = 0
2×1 + 8×2 + x3 − x4 = 0
⎡
1
(c) ⎣1
1
14. x1 + 3×2 − x3 = 0
x2 − 8×3 = 0
4 x3 = 0
∗
1
0
∗
1
0
∗
1
0
∗
∗
⎤
∗
∗⎦
∗
⎤
∗
∗⎦
0
1
∗
∗
⎤
∗
∗⎦
1
1
∗
∗
1
⎤
0
0
0
0
0
1⎦
∗
∗
∗
⎡
1
(b) ⎣0
0
⎡
1
(d) ⎣0
0
⎡
∗
1
0
∗
0
0
∗
∗
⎤
∗
∗⎦
0
0
∗
∗
⎤
∗
0⎦
∗
1
1
0
1
∗
⎡
∗
0
0
1
∗
∗
0
0
0
0
(b) ⎣∗
1
(d) ⎣1
1
⎤
∗
∗⎦
∗
⎤
∗
1⎦
1
In Exercises 15–22, solve the given linear system by any
method.
In Exercises 25–26, determine the values of a for which the
system has no solutions, exactly one solution, or inﬁnitely many
solutions.
15. 2×1 + x2 + 3×3 = 0
x1 + 2×2
=0
x2 + x3 = 0
25. x + 2y −
3z =
4
5z =
2
3x − y +
4x + y + (a 2 − 14)z = a + 2
16. 2x − y − 3z = 0
−x + 2y − 3z = 0
x + y + 4z = 0
24
Chapter 1 Systems of Linear Equations and Matrices
26. x + 2y +
z=2
3z = 1
2x − 2y +
x + 2y − (a 2 − 3)z = a
36. Solve the following system for x, y, and z.
1
x
2
In Exercises 27–28, what condition, if any, must a , b, and c
satisfy for the linear system to be consistent?
27. x + 3y − z = a
x + y + 2z = b
2 y − 3z = c
28.
x + 3y + z = a
−x − 2y + z = b
3x + 7y − z = c
In Exercises 29–30, solve the following systems, where a , b,
and c are constants.
29. 2x + y = a
x
−
+
y
3
y
9
y
−
4
+
8
+
10
20
(0, 10)
2
7
(1, 7)
⎡
2
(3, –11)
–20
1
⎢
⎣0
−2
3
4
(4, –14)
Figure Ex-37
This exercise shows that a matrix can have multiple row echelon forms.
32. Reduce
=5
x

3
z
=0
6
31. Find two different row echelon forms of
1
z
=1
y
–2

z
y = ax 3 + bx 2 + cx + d.
+ 2×3 = b
3×2 + 3×3 = c
2×1
+
2
37. Find the coefﬁcients a, b, c, and d so that the curve shown
in the accompanying ﬁgure is the graph of the equation
30. x1 + x2 + x3 = a
3x + 6y = b
1
x
+
3
⎤
⎥
−29⎦
38. Find the coefﬁcients a, b, c, and d so that the circle shown in
the accompanying ﬁgure is given by the equation
ax 2 + ay 2 + bx + cy + d = 0.
y
(–2, 7)
(–4, 5)
5
to reduced row echelon form without introducing fractions at
any intermediate stage.
x
33. Show that the following nonlinear system has 18 solutions if
0 ≤ α ≤ 2π , 0 ≤ β ≤ 2π , and 0 ≤ γ ≤ 2π .
sin α + 2 cos β + 3 tan γ = 0
2 sin α + 5 cos β + 3 tan γ = 0
− sin α − 5 cos β + 5 tan γ = 0
[Hint: Begin by making the substitutions x = sin α ,
y = cos β , and z = tan γ .]
34. Solve the following system of nonlinear equations for the unknown angles α , β , and γ , where 0 ≤ α ≤ 2π , 0 ≤ β ≤ 2π ,
and 0 ≤ γ = 0, then the reduced row echelon
form of

a b
1 0
is
c d
0 1
(b) Use the result in part (a) to prove that if ad − bc = 0, then
the linear system
ax + by = k
cx + dy = l
has exactly one solution.
25
(d) A homogeneous linear system in n unknowns whose corresponding augmented matrix has a reduced row echelon form
with r leading 1’s has n − r free variables.
(e) All leading 1’s in a matrix in row echelon form must occur in
different columns.
(f ) If every column of a matrix in row echelon form has a leading
1, then all entries that are not leading 1’s are zero.
(g) If a homogeneous linear system of n equations in n unknowns
has a corresponding augmented matrix with a reduced row
echelon form containing n leading 1’s, then the linear system
has only the trivial solution.
(h) If the reduced row echelon form of the augmented matrix for
a linear system has a row of zeros, then the system must have
inﬁnitely many solutions.
(i) If a linear system has more unknowns than equations, then it
must have inﬁnitely many solutions.
Working withTechnology
T1. Find the reduced row echelon form of the augmented matrix
for the linear system:
+ 4×4 = −3
6×1 + x2
−9×1 + 2×2 + 3×3 − 8×4 = 1
− 4×3 + 5×4 = 2
7×1
Use your result to determine whether the system is consistent and,
if so, ﬁnd its solution.
True-False Exercises
TF. In parts (a)–(i) determine whether the statement is true or
false, and justify your answer.
(a) If a matrix is in reduced row echelon form, then it is also in
row echelon form.
(b) If an elementary row operation is applied to a matrix that is
in row echelon form, the resulting matrix will still be in row
echelon form.
(c) Every matrix has a unique row echelon form.
T2. Find values of the constants A, B , C , and D that make the
following equation an identity (i.e., true for all values of x ).
C
D
3x 3 + 4x 2 − 6x
Ax + B
+
+
= 2
(x 2 + 2x + 2)(x 2 − 1)
x + 2x + 2 x − 1 x + 1
[Hint: Obtain a common denominator on the right, and then
equate corresponding coefﬁcients of the various powers of x in
the two numerators. Students of calculus will recognize this as a
problem in partial fractions.]
1.3 Matrices and Matrix Operations
Rectangular arrays of real numbers arise in contexts other than as augmented matrices for
linear systems. In this section we will begin to study matrices as objects in their own right
by deﬁning operations of addition, subtraction, and multiplication on them.
Matrix Notation and
Terminology
In Section 1.2 we used rectangular arrays of numbers, called augmented matrices, to
abbreviate systems of linear equations. However, rectangular arrays of numbers occur
in other contexts as well. For example, the following rectangular array with three rows
and seven columns might describe the number of hours that a student spent studying
three subjects during a certain week:
26
Chapter 1 Systems of Linear Equations and Matrices
Math
History
Language
Mon.
Tues.
Wed.
Thurs.
Fri.
Sat.
Sun.
2
0
4
3
3
1
2
1
3
4
4
1
1
3
0
4
2
0
2
2
2
If we suppress the headings, then we are left with the following rectangular array of
numbers with three rows and seven columns, called a “matrix”:
⎡
2
⎢
⎣0
4
3
3
1
2
1
3
4
4
1
1
3
0
⎤
4
2
0
2
⎥
2⎦
2
More generally, we make the following deﬁnition.
DEFINITION 1 A matrix is a rectangular array of numbers. The numbers in the array
are called the entries in the matrix.
E X A M P L E 1 Examples of Matrices
Matrix brackets are often
omitted from 1 × 1 matrices, making it impossible to
tell, for example, whether the
symbol 4 denotes the number “four” or the matrix [4].
This rarely causes problems
because it is usually possible
to tell which is meant from the
context.
Some examples of matrices are
⎡
1
⎣ 3
−1
⎡
⎤
2
0⎦, [2
4
1
e
⎢
− 3], ⎣0
0
0
√ ⎤
− 2
⎥
1 ⎦,
π
1
2
0
0
1
, [4]
3
The size of a matrix is described in terms of the number of rows (horizontal lines)
and columns (vertical lines) it contains. For example, the ﬁrst matrix in Example 1 has
three rows and two columns, so its size is 3 by 2 (written 3 × 2). In a size description,
the ﬁrst number always denotes the number of rows, and the second denotes the number
of columns. The remaining matrices in Example 1 have sizes 1 × 4, 3 × 3, 2 × 1, and
1 × 1, respectively.
A matrix with only one row, such as the second in Example 1, is called a row vector
(or a row matrix), and a matrix with only one column, such as the fourth in that example,
is called a column vector (or a column matrix). The ﬁfth matrix in that example is both
a row vector and a column vector.
We will use capital letters to denote matrices and lowercase letters to denote numerical quantities; thus we might write
A=
2
3
1
4
7
2
or C =
a
d
b
e
c
f
When discussing matrices, it is common to refer to numerical quantities as scalars. Unless
stated otherwise, scalars will be real numbers; complex scalars will be considered later in
the text.
The entry that occurs in row i and column j of a matrix A will be denoted by aij .
Thus a general 3 × 4 matrix might be written as
1.3 Matrices and Matrix Operations
⎡
a11
⎢
A = ⎣a21
a31
and a general m × n matrix as
⎡
a11
⎢a
⎢ 21
A=⎢ .
⎣ ..
am1
A matrix with n rows and n
columns is said to be a square
matrix of order n.
a12
a22
a32
a13
a23
a33
a12
a22
..
.
···
···
am2
27
⎤
a14
⎥
a24 ⎦
a34
···
⎤
a1n
a2 n ⎥
⎥
.. ⎥
. ⎦
amn
(1)
When a compact notation is desired, the preceding matrix can be written as
[aij ]m×n or [aij ]
the ﬁrst notation being used when it is important in the discussion to know the size,
and the second when the size need not be emphasized. Usually, we will match the letter
denoting a matrix with the letter denoting its entries; thus, for a matrix B we would
generally use bij for the entry in row i and column j , and for a matrix C we would use
the notation cij .
The entry in row i and column j of a matrix A is also commonly denoted by the
symbol (A)ij . Thus, for matrix (1) above, we have
(A)ij = aij
and for the matrix
2 −3
7
0
we have (A)11 = 2, (A)12 = −3, (A)21 = 7, and (A)22 = 0.
Row and column vectors are of special importance, and it is common practice to
denote them by boldface lowercase letters rather than capital letters. For such matrices,
double subscripting of the entries is unnecessary. Thus a general 1 × n row vector a and
a general m × 1 column vector b would be written as
A=
⎤
b1
⎢b ⎥
⎢ 2⎥
· · · an ] and b = ⎢ .. ⎥
⎣ . ⎦
bm
⎡
a = [a1 a2
A matrix A with n rows and n columns is called a square matrix of order n, and the
shaded entries a11 , a22 , . . . , ann in (2) are said to be on the main diagonal of A.
⎡
⎢
⎢
⎢
⎣
Operations on Matrices
a11
a21
..
.
an1
a12
a22
..
.
an2
···
···
a1n
a2n
..
.
· · · ann
⎤
⎥
⎥
⎥
⎦
(2)
So far, we have used matrices to abbreviate the work in solving systems of linear equations. For other applications, however, it is desirable to develop an “arithmetic of matrices” in which matrices can be added, subtracted, and multiplied in a useful way. The
remainder of this section will be devoted to developing this arithmetic.
DEFINITION 2 Two matrices are deﬁned to be equal if they have the same size and
their corresponding entries are equal.
28
Chapter 1 Systems of Linear Equations and Matrices
E X A M P L E 2 Equality of Matrices
The equality of two matrices
Consider the matrices
A=
A = [aij ] and B = [bij ]
of the same size can be expressed either by writing
(A)ij = (B)ij
or by writing
aij = bij
where it is understood that the
equalities hold for all values of
i and j .
2
3
1
, B=
x
2
3
1
2
, C=
5
3
1
4
0
0
If x = 5, then A = B , but for all other values of x the matrices A and B are not equal,
since not all of their corresponding entries are equal. There is no value of x for which
A = C since A and C have different sizes.
DEFINITION 3 If A and B are matrices of the same size, then the sum A + B is the
matrix obtained by adding the entries of B to the corresponding entries of A, and
the difference A − B is the matrix obtained by subtracting the entries of B from the
corresponding entries of A. Matrices of different sizes cannot be added or subtracted.
In matrix notation, if A = [aij ] and B = [bij ] have the same size, then
(A + B)ij = (A)ij + (B)ij = aij + bij and (A − B)ij = (A)ij − (B)ij = aij − bij
E X A M P L E 3 Addition and Subtraction
Consider the matrices
⎡
Then
⎤
⎡
1
0
−2
0
2
7
−4
3
⎥
⎢
4⎦, B = ⎣ 2
0
3
−2
4
2
0
5
2
3
2
⎢
A = ⎣−1
4
⎡
⎢
A+B =⎣ 1
7
3
2
2
⎤
5
0
−4
⎤
1
1
⎥
−1⎦, C =
2
5
⎡
4
6
⎥
⎢
3⎦ and A − B = ⎣−3
5
1
−2
−2
−4
−5
2
11
1
2
⎤
2
⎥
5⎦
−5
The expressions A + C , B + C , A − C , and B − C are undeﬁned.
DEFINITION 4 If A is any matrix and c is any scalar, then the product cA is the matrix
obtained by multiplying each entry of the matrix A by c. The matrix cA is said to be
a scalar multiple of A.
In matrix notation, if A = [aij ], then
(cA)ij = c(A)ij = caij
E X A M P L E 4 Scalar Multiples
For the matrices
2
1
A=
3
3
4
0
, B=
−1
1
2
3
7
9
, C=
−5
3
−6
3
12
0
we have
2A =
4
2
6
6
8
0
, (−1)B =
2
1
−2 −7
,
−3
5
It is common practice to denote (−1)B by −B .
1
C
3
=
3
1
−2
0
1
4
1.3 Matrices and Matrix Operations
29
Thus far we have deﬁned multiplication of a matrix by a scalar but not the multiplication of two matrices. Since matrices are added by adding corresponding entries
and subtracted by subtracting corresponding entries, it would seem natural to deﬁne
multiplication of matrices by multiplying corresponding entries. However, it turns out
that such a deﬁnition would not be very useful for most problems. Experience has led
mathematicians to the following more useful deﬁnition of matrix multiplication.
A is an m × r matrix and B is an r × n matrix, then the product
AB is the m × n matrix whose entries are determined as follows: To ﬁnd the entry in
row i and column j of AB , single out row i from the matrix A and column j from
the matrix B . Multiply the corresponding entries from the row and column together,
DEFINITION 5 If
and then add up the resulting products.
E X A M P L E 5 Multiplying Matrices
Consider the matrices
A=
⎡
1
2
2
6
4
4
⎢
, B = ⎣0
0
2
1
−1
7
⎤
4
3
5
3
⎥
1⎦
2
Since A is a 2 × 3 matrix and B is a 3 × 4 matrix, the product AB is a 2 × 4 matrix.
To determine, for example, the entry in row 2 and column 3 of AB , we single out row 2
from A and column 3 from B . Then, as illustrated below, we multiply corresponding
entries together and add up these products.
⎡
4
1 2 4 ⎢
⎣0
2 6 0
2

1
1
7
4
3
5
⎤ ⎡
3
⎥ ⎢
1⎦ = ⎣
2
⎤
⎥
⎦
26
(2 · 4) + (6 · 3) + (0 · 5) = 26
The entry in row 1 and column 4 of AB is computed as follows:

⎡
4
1 2 4 ⎢
⎣0
2 6 0
2
1
1
7
4
3
5
⎤ ⎡
3
⎥ ⎢
1⎦ = ⎣
2
⎤
13 ⎥
⎦
(1 · 3) + (2 · 1) + (4 · 2) = 13
The computations for the remaining entries are
(1 · 4) + (2 · 0) + (4 · 2) = 12
(1 · 1) − (2 · 1) + (4 · 7) = 27
(1 · 4) + (2 · 3) + (4 · 5) = 30
(2 · 4) + (6 · 0) + (0 · 2) = 8
(2 · 1) − (6 · 1) + (0 · 7) = −4
(2 · 3) + (6 · 1) + (0 · 2) = 12
AB =
12
8
27
−4
30
26
13
12
The deﬁnition of matrix multiplication requires that the number of columns of the
ﬁrst factor A be the same as the number of rows of the second factor B in order to form
the product AB . If this condition is not satisﬁed, the product is undeﬁned. A convenient
30
Chapter 1 Systems of Linear Equations and Matrices
way to determine whether a product of two matrices is deﬁned is to write down the size
of the ﬁrst factor and, to the right of it, write down the size of the second factor. If, as in
(3), the inside numbers are the same, then the product is deﬁned. The outside numbers
then give the size of the product.
A
m × r
B
r × n =
AB
m × n
(3)
Inside
Outside
E X A M P L E 6 Determining Whether a Product Is Deﬁned
Suppose that A, B , and C are matrices with the following sizes:
A
B
C
3×4
4×7
7×3
Then by (3), AB is deﬁned and is a 3 × 7 matrix; BC is deﬁned and is a 4 × 3 matrix; and
CA is deﬁned and is a 7 × 4 matrix. The products AC , CB , and BA are all undeﬁned.
In general, if A = [aij ] is an m × r matrix and B = [bij ] is an r × n matrix, then, as
illustrated by the shading in the following display,
⎡
a11
⎢a
⎢ 21
⎢ .
⎢ ..
AB = ⎢
⎢ ai 1
⎢ .
⎢ .
⎣ .
a12
a22
..
.
ai 2
..
.
am1
am2
···
···
···
···
a1r
a2r
..
.
air
..
.
⎤
⎥ ⎡b
⎥ 11
⎥⎢
⎥ ⎢b21
⎥⎢ .
⎥⎣ .
⎥ .
⎥
⎦ br 1
b12
b22
..
.
· · · b1 j
· · · b2 j
..
.
br 2
br j
···
⎤
· · · b1n
· · · b2n ⎥
⎥
.. ⎥
. ⎦
· · · br n
(4)
amr
the entry (AB)ij in row i and column j of AB is given by
(AB)ij = ai 1 b1j + ai 2 b2j + ai 3 b3j + · · · + air brj
(5)
Formula (5) is called the row-column rule for matrix multiplication.
Partitioned Matrices
A matrix can be subdivided or partitioned into smaller matrices by inserting horizontal
and vertical rules between selected rows and columns. For example, the following are
three possible partitions of a general 3 × 4 matrix A—the ﬁrst is a partition of A into
Gotthold Eisenstein
(1823–1852)
Historical Note The concept of matrix multiplication is due to the German mathematician Gotthold
Eisenstein, who introduced the idea around 1844 to
simplify the process of making substitutions in linear systems. The idea was then expanded on and
formalized by Cayley in his Memoir on the Theory
of Matrices that was published in 1858. Eisenstein
was a pupil of Gauss, who ranked him as the equal
of Isaac Newton and Archimedes. However, Eisenstein, suffering from bad health his entire life, died
at age 30, so his potential was never realized.
[Image: http://www-history.mcs.st-andrews.ac.uk/
Biographie…
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