Diablo Valley College Application of Linear Algebra in Engineering Paper

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Index of Applications
BIOLOGY AND LIFE SCIENCES
Age distribution vector, 378, 391, 392, 395
Age progression software, 180
Age transition matrix, 378, 391, 392, 395
Agriculture, 37, 50
Cosmetic surgery results simulation, 180
Duchenne muscular dystrophy, 365
Galloping speeds of animals, 276
Genetics, 365
Health care expenditures, 146
Heart rhythm analysis, 255
Hemophilia A, 365
Hereditary baldness, 365
Nutrition, 11
Population
of deer, 37
of laboratory mice, 91
of rabbits, 379
of sharks, 396
of small fish, 396
Population age and growth over time, 331
Population genetics, 365
Population growth, 378, 379, 391, 392,
395, 396, 398
Predator-prey relationship, 396
Red-green color blindness, 365
Reproduction rates of deer, 103
Sex-linked inheritance, 365
Spread of a virus, 91, 93
Vitamin C content, 11
Wound healing simulation, 180
X-linked inheritance, 365
BUSINESS AND ECONOMICS
Airplane allocation, 91
Borrowing money, 23
Demand, for a rechargeable power drill, 103
Demand matrix, external, 98
Economic system, 97, 98
of a small community, 103
Finance, 23
Fundraising, 92
Gasoline sales, 105
Industrial system, 102, 107
Input-output matrix, 97
Leontief input-output model(s), 97, 98, 103
Major League Baseball salaries, 107
Manufacturing
labor and material costs, 105
models and prices, 150
production levels, 51, 105
Net profit, Microsoft, 32
Output matrix, 98
Petroleum production, 292
Profit, from crops, 50
Purchase of a product, 91
Revenue
fast-food stand, 242
General Dynamics Corporation, 266, 276
Google, Inc., 291
telecommunications company, 242
software publishers, 143
Sales, 37
concession area, 42
stocks, 92
Wal-Mart, 32
Sales promotion, 106
Satellite television service, 85, 86, 147
Software publishing, 143
ENGINEERING AND TECHNOLOGY
Aircraft design, 79
Circuit design, 322
Computer graphics, 338
Computer monitors, 190
Control system, 314
Controllability matrix, 314
Cryptography, 94–96, 102, 107
Data encryption, 94
Decoding a message, 96, 102, 107
Digital signal processing, 172
Electrical network analysis, 30, 31, 34, 37,
150
Electronic equipment, 190
Encoding a message, 95, 102, 107
Encryption key, 94
Engineering and control, 130
Error checking
digit, 200
matrix, 200
Feed horn, 223
Global Positioning System, 16
Google’s Page Rank algorithm, 86
Image morphing and warping, 180
Information retrieval, 58
Internet search engine, 58
Ladder network, 322
Locating lost vessels at sea, 16
Movie special effects, 180
Network analysis, 29–34, 37
Radar, 172
Sampling, 172
Satellite dish, 223
Smart phones, 190
Televisions, 190
Wireless communications, 172
MATHEMATICS AND GEOMETRY
Adjoint of a matrix, 134, 135, 142, 146, 150
Collinear points in the xy-plane, 139, 143
Conic section(s), 226, 229
general equation, 141
rotation of axes, 221–224, 226, 229,
   383–385, 392, 395
Constrained optimization, 389, 390, 392,
  395
Contraction in R2, 337, 341, 342
Coplanar points in space, 140, 143
Cramer’s Rule, 130, 136, 137, 142, 143, 146
Cross product of two vectors, 277–280,
288, 289, 294
Differential equation(s)
linear, 218, 225, 226, 229
second order, 164
system of first order, 354, 380, 381,
   391, 392, 395, 396, 398
Expansion in R2, 337, 341, 342, 345
Fibonacci sequence, 396
Fourier approximation(s), 285–287, 289, 292
Geometry of linear transformations in R2,
336–338, 341, 342, 345
Hessian matrix, 375
Jacobian, 145
Lagrange multiplier, 34
Laplace transform, 130
Least squares approximation(s), 281–284, 289
linear, 282, 289, 292
quadratic, 283, 289, 292
Linear programming, 47
Magnification in R2, 341, 342
Mathematical modeling, 273, 274, 276
Parabola passing through three points, 150
Partial fraction decomposition, 34, 37
Polynomial curve fitting, 25–28, 32, 34, 37
Quadratic form(s), 382–388, 392, 395, 398
Quadric surface, rotation of, 388, 392
Reflection in R2, 336, 341, 342, 345, 346
Relative maxima and minima, 375
Rotation
in R2, 303, 343, 393, 397
in R3, 339, 340, 342, 345
Second Partials Test for relative extrema, 375
Shear in R2, 337, 338, 341, 342, 345
Taylor polynomial of degree 1, 282
Three-point form of the equation of a plane,
141, 143, 146
Translation in R2, 308, 343
Triple scalar product, 288
Two-point form of the equation of a line,
139, 143, 146, 150
Unit circle, 253
Wronskian, 219, 225, 226, 229
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PHYSICAL SCIENCES
Acoustical noise levels, 28
Airplane speed, 11
Area
of a parallelogram using cross product,
   279, 280, 288, 294
of a triangle
using cross product, 289
using determinants, 138, 142, 146,
  150
Astronomy, 27, 274
Balancing a chemical equation, 4
Beam deflection, 64, 72
Chemical
changing state, 91
mixture, 37
reaction, 4
Comet landing, 141
Computational fluid dynamics, 79
Crystallography, 213
Degree of freedom, 164
Diffusion, 354
Dynamical systems, 396
Earthquake monitoring, 16
Electric and magnetic flux, 240
Flexibility matrix, 64, 72
Force
matrix, 72
to pull an object up a ramp, 157
Geophysics, 172
Grayscale, 190
Hooke’s Law, 64
Kepler’s First Law of Planetary Motion, 141
Kirchhoff’s Laws, 30, 322
Lattice of a crystal, 213
Mass-spring system, 164, 167
Mean distance from the sun, 27, 274
Natural frequency, 164
Newton’s Second Law of Motion, 164
Ohm’s Law, 322
Pendulum, 225
Planetary periods, 27, 274
Primary additive colors, 190
RGB color model, 190
Stiffness matrix, 64, 72
Temperature, 34
Torque, 277
Traffic flow, 28, 33
Undamped system, 164
Unit cell, 213
end-centered monoclinic, 213
Vertical motion, 37
Volume
of a parallelepiped, 288, 289, 292
of a tetrahedron, 114, 140, 143
Water flow, 33
Wind energy consumption, 103
Work, 248
SOCIAL SCIENCES AND
DEMOGRAPHICS
Caribbean Cruise, 106
Cellular phone subscribers, 107
Consumer preference model, 85, 86, 92, 147
Final grades, 105
Grade distribution, 92
Master’s degrees awarded, 276
Politics, voting apportionment, 51
Population
of consumers, 91
regions of the United States, 51
of smokers and nonsmokers, 91
United States, 32
world, 273
Population migration, 106
Smokers and nonsmokers, 91
Sports
activities, 91
Super Bowl I, 36
Television watching, 91
Test scores, 108
STATISTICS AND PROBABILITY
Canonical regression analysis, 304
Least squares regression
analysis, 99–101, 103, 107, 265, 271–276
cubic polynomial, 276
line, 100, 103, 107, 271, 274, 276, 296
quadratic polynomial, 273, 276
Leslie matrix, 331, 378
Markov chain, 85, 86, 92, 93, 106
absorbing, 89, 90, 92, 93, 106
Multiple regression analysis, 304
Multivariate statistics, 304
State matrix, 85, 106, 147, 331
Steady state probability vector, 386
Stochastic matrices, 84–86, 91–93, 106, 331
MISCELLANEOUS
Architecture, 388
Catedral Metropolitana Nossa Senhora
Aparecida, 388
Chess tournament, 93
Classified documents, 106
Determining directions, 16
Dominoes, A2
Flight crew scheduling, 47
Sudoku, 120
Tips, 23
U.S. Postal Service, 200
ZIP + 4 barcode, 200
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Elementary Linear Algebra
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Elementary Linear Algebra
8e
Ron Larson
The Pennsylvania State University
The Behrend College
Australia • Brazil • Mexico • Singapore • United Kingdom • United States
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Elementary Linear Algebra
Eighth Edition
Ron Larson
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Contents
1
Systems of Linear Equations
1.1
1.2
1.3
2
3
2
13
25
35
38
38
Matrices
39
2.1
2.2
2.3
2.4
2.5
2.6
40
52
62
74
84
94
104
108
108
Operations with Matrices
Properties of Matrix Operations
The Inverse of a Matrix
Elementary Matrices
Markov Chains
More Applications of Matrix Operations
Review Exercises
Project 1 Exploring Matrix Multiplication
Project 2 Nilpotent Matrices
Determinants
3.1
3.2
3.3
3.4
4
Introduction to Systems of Linear Equations
Gaussian Elimination and Gauss-Jordan Elimination
Applications of Systems of Linear Equations
Review Exercises
Project 1 Graphing Linear Equations
Project 2 Underdetermined and Overdetermined Systems
1
The Determinant of a Matrix
Determinants and Elementary Operations
Properties of Determinants
Applications of Determinants
Review Exercises
Project 1 Stochastic Matrices
Project 2 The Cayley-Hamilton Theorem
Cumulative Test for Chapters 1–3
Vector Spaces
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
Rn
Vectors in
Vector Spaces
Subspaces of Vector Spaces
Spanning Sets and Linear Independence
Basis and Dimension
Rank of a Matrix and Systems of Linear Equations
Coordinates and Change of Basis
Applications of Vector Spaces
Review Exercises
Project 1 Solutions of Linear Systems
Project 2 Direct Sum
109
110
118
126
134
144
147
147
149
151
152
161
168
175
186
195
208
218
227
230
230
v
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vi
Contents
5
Inner Product Spaces
5.1
5.2
5.3
5.4
5.5
6
Linear Transformations
6.1
6.2
6.3
6.4
6.5
7
Introduction to Linear Transformations
The Kernel and Range of a Linear Transformation
Matrices for Linear Transformations
Transition Matrices and Similarity
Applications of Linear Transformations
Review Exercises
Project 1 Reflections in R 2 (I)
Project 2 Reflections in R 2 (II)
Eigenvalues and Eigenvectors
7.1
7.2
7.3
7.4
8
Length and Dot Product in R n
Inner Product Spaces
Orthonormal Bases: Gram-Schmidt Process
Mathematical Models and Least Squares Analysis
Applications of Inner Product Spaces
Review Exercises
Project 1 The QR-Factorization
Project 2 Orthogonal Matrices and Change of Basis
Cumulative Test for Chapters 4 and 5
Eigenvalues and Eigenvectors
Diagonalization
Symmetric Matrices and Orthogonal Diagonalization
Applications of Eigenvalues and Eigenvectors
Review Exercises
Project 1 Population Growth and Dynamical Systems (I)
Project 2 The Fibonacci Sequence
Cumulative Test for Chapters 6 and 7
231
232
243
254
265
277
290
293
294
295
297
298
309
320
330
336
343
346
346
347
348
359
368
378
393
396
396
397
Complex Vector Spaces (online)*
8.1
8.2
8.3
8.4
8.5
Complex Numbers
Conjugates and Division of Complex Numbers
Polar Form and DeMoivre’s Theorem
Complex Vector Spaces and Inner Products
Unitary and Hermitian Matrices
Review Exercises
Project 1 The Mandelbrot Set
Project 2 Population Growth and Dynamical Systems (II)
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Contents
9
Linear Programming (online)*
9.1
9.2
9.3
9.4
9.5
10
vii
Systems of Linear Inequalities
Linear Programming Involving Two Variables
The Simplex Method: Maximization
The Simplex Method: Minimization
The Simplex Method: Mixed Constraints
Review Exercises
Project 1 Beach Sand Replenishment (I)
Project 2 Beach Sand Replenishment (II)
Numerical Methods (online)*
10.1
10.2
10.3
10.4
Gaussian Elimination with Partial Pivoting
Iterative Methods for Solving Linear Systems
Power Method for Approximating Eigenvalues
Applications of Numerical Methods
Review Exercises
Project 1 The Successive Over-Relaxation (SOR) Method
Project 2 United States Population
Appendix
A1
Mathematical Induction and Other Forms of Proofs
Answers to Odd-Numbered Exercises and Tests
Index
Technology Guide*
*Available online at CengageBrain.com.
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A7
A41
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Preface
Welcome to Elementary Linear Algebra, Eighth Edition. I am proud to present to you this new edition. As with
all editions, I have been able to incorporate many useful comments from you, our user. And while much has
changed in this revision, you will still find what you expect—a pedagogically sound, mathematically precise, and
comprehensive textbook. Additionally, I am pleased and excited to offer you something brand new— a companion
website at LarsonLinearAlgebra.com. My goal for every edition of this textbook is to provide students with the
tools that they need to master linear algebra. I hope you find that the changes in this edition, together with
LarsonLinearAlgebra.com, will help accomplish just that.
New To This Edition
NEW LarsonLinearAlgebra.com
This companion website offers multiple tools and
resources to supplement your learning. Access to
these features is free. Watch videos explaining
concepts from the book, explore examples, download
data sets and much more.
5.2
In Exercises 85 and 86, determine
whether each statement is true or false. If a statement
is true, give a reason or cite an appropriate statement
from the text. If a statement is false, provide an example
that shows the statement is not true in all cases or cite an
appropriate statement from the text.
true or False?
85. (a) The dot product is the only inner product that can be
defined in Rn.
(b) A nonzero vector in an inner product can have a
norm of zero.
86. (a) The norm of the vector u is the angle between u and
the positive x-axis.
(b) The angle θ between a vector v and the projection
of u onto v is obtuse when the scalar a < 0 and acute when a > 0, where av = projvu.
87. Let u = (4, 2) and v = (2, −2) be vectors in R2 with
the inner product 〈u, v〉 = u1v1 + 2u2v2.
(a) Show that u and v are orthogonal.
(b) Sketch u and v. Are they orthogonal in the Euclidean
sense?
88. Proof Prove that
u + v2 + u − v2 = 2u2 + 2v2
for any vectors u and v in an inner product space V.
89. Proof Prove that the function is an inner product on Rn.
〈u, v〉 = c1u1v1 + c2u2v2 + . . . + cnunvn, ci > 0
90. Proof Let u and v be nonzero vectors in an inner
product space V. Prove that u − projvu is orthogonal
to v.
91. Proof Prove Property 2 of Theorem 5.7: If u, v,
and w are vectors in an inner product space V, then
〈u + v, w〉 = 〈u, w〉 + 〈v, w〉.
92. Proof Prove Property 3 of Theorem 5.7: If u and v
are vectors in an inner product space V and c is any real
number, then 〈u, cv〉 = c〈u, v〉.
93. guided Proof Let W be a subspace of the inner
product space V. Prove that the set
W⊥ = { v ∈ V: 〈v, w〉 = 0 for all w ∈ W }
is a subspace of V.
Getting Started: To prove that W⊥ is a subspace of
V, you must show that W⊥ is nonempty and that the
closure conditions for a subspace hold (Theorem 4.5).
(i) Find a vector in W⊥ to conclude that it is nonempty.
(ii) To show the closure of W⊥ under addition, you
need to show that 〈v1 + v2, w〉 = 0 for all w ∈ W
and for any v1, v2 ∈ W⊥. Use the properties of
inner products and the fact that 〈v1, w〉 and 〈v2, w〉
are both zero to show this.
(iii) To show closure under multiplication by a scalar,
proceed as in part (ii). Use the properties of inner
products and the condition of belonging to W⊥.
9781305658004_0502.indd 253
253
Exercises
94. Use the result of Exercise 93 to find W⊥ when W is the
span of (1, 2, 3) in V = R3.
95. guided Proof Let 〈u, v〉 be the Euclidean inner
product on Rn. Use the fact that 〈u, v〉 = uTv to prove
that for any n × n matrix A,
(a) 〈ATAu, v〉 = 〈u, Av〉
and
(b) 〈ATAu, u〉 = Au2.
Getting Started: To prove (a) and (b), make use of both
the properties of transposes (Theorem 2.6) and the
properties of the dot product (Theorem 5.3).
(i) To prove part (a), make repeated use of the property
〈u, v〉 = uTv and Property 4 of Theorem 2.6.
(ii) To prove part (b), make use of the property
〈u, v〉 = uTv, Property 4 of Theorem 2.6, and
Property 4 of Theorem 5.3.
96. CAPSTONE
(a) Explain how to determine whether a function
defines an inner product.
(b) Let u and v be vectors in an inner product space V,
such that v ≠ 0. Explain how to find the orthogonal
projection of u onto v.
In Exercises 97–100,
find c1 and c2 for the inner product of R2,
〈u, v〉 = c1u1v1 + c2u2v2
such that the graph represents a unit circle as shown.
y
y
97.
98.
Finding Inner Product Weights
4
3
2
||u|| = 1
−3 − 2
2 3
−3
−2
−3
1
3
y
100.
5
6
4
||u|| = 1
1
−5 − 3
−1
x
−4
y
99.
||u|| = 1
1
x
1
3
5
x
−6
||u|| = 1
6
x
−4
−5
−6
101. Consider the vectors
u = (6, 2, 4) and v = (1, 2, 0)
from Example 10. Without using Theorem 5.9, show
that among all the scalar multiples cv of the vector
v, the projection of u onto v is the vector closest to
u—that is, show that d(u, projvu) is a minimum.
REVISED Exercise Sets
The exercise sets have been carefully and extensively
examined to ensure they are rigorous, relevant, and
cover all the topics necessary to understand the
fundamentals of linear algebra. The exercises are
ordered and titled so you can see the connections
between examples and exercises. Many new skillbuilding, challenging, and application exercises have
been added. As in earlier editions, the following
pedagogically-proven types of exercises are included.
•
•
•
•
•
True or False Exercises
Proofs
Guided Proofs
Writing Exercises
Technology Exercises (indicated throughout the
text with
)
Exercises utilizing electronic data sets are indicated
by
and found at CengageBrain.com.
8/18/15 10:21 AM
ix
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
x
Preface
2
Table of Contents Changes
Based on market research and feedback from users,
Section 2.5 in the previous edition (Applications of
Matrix Operations) has been expanded from one section
to two sections to include content on Markov chains.
So now, Chapter 2 has two application sections:
Section 2.5 (Markov Chains) and Section 2.6 (More
Applications of Matrix Operations). In addition,
Section 7.4 (Applications of Eigenvalues and
Eigenvectors) has been expanded to include content
on constrained optimization.
2.1
2.2
2.3
2.4
2.5
2.6
Trusted Features
Matrices
Operations with Matrices
Properties of Matrix Operations
The Inverse of a Matrix
Elementary Matrices
Markov Chains
More Applications of Matrix Operations
Data Encryption (p. 94)
Computational Fluid Dynamics (p. 79)
®
For the past several years, an independent website—
CalcChat.com—has provided free solutions to all
odd-numbered problems in the text. Thousands of
students have visited the site for practice and help
with their homework from live tutors. You can also
use your smartphone’s QR Code® reader to scan the
icon
at the beginning of each exercise set to
access the solutions.
Beam Deflection (p. 64)
Information Retrieval (p. 58)
Flight Crew Scheduling (p. 47)
62
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Chapter 2
2.3 The Inverse of a Matrix
Chapter Openers
Use properties of inverse matrices.
Each Chapter Opener highlights five real-life
applications of linear algebra found throughout the
chapter. Many of the applications reference the
Linear Algebra Applied feature (discussed on the
next page). You can find a full list of the
applications in the Index of Applications on the
inside front cover.
Use an inverse matrix to solve a system of linear equations.
Matrices and their inverses
Section 2.2 discussed some of the similarities between the algebra of real numbers and
the algebra of matrices. This section further develops the algebra of matrices to include
the solutions of matrix equations involving matrix multiplication. To begin, consider
the real number equation ax = b. To solve this equation for x, multiply both sides of
the equation by a−1 (provided a ≠ 0).
ax = b
(a−1a)x = a−1b
(1)x = a−1b
x = a−1b
The number a−1 is the multiplicative inverse of a because a−1a = 1 (the identity
element for multiplication). The definition of the multiplicative inverse of a matrix is
similar.
Section Objectives
definition of the inverse of a Matrix
A bulleted list of learning objectives, located at
the beginning of each section, provides you the
opportunity to preview what will be presented
in the upcoming section.
An n × n matrix A is invertible (or nonsingular) when there exists an n × n
matrix B such that
AB = BA = In
where In is the identity matrix of order n. The matrix B is the (multiplicative)
inverse of A. A matrix that does not have an inverse is noninvertible (or
singular).
Nonsquare matrices do not have inverses. To see this, note that if A is of size
m × n and B is of size n × m (where m ≠ n), then the products AB and BA are of
different sizes and cannot be equal to each other. Not all square matrices have inverses.
(See Example 4.) The next theorem, however, states that if a matrix does have an
inverse, then that inverse is unique.
theoreM 2.7
Theorems, Definitions, and
Properties
Presented in clear and mathematically precise
language, all theorems, definitions, and properties
are highlighted for emphasis and easy reference.
Uniqueness of an inverse Matrix
If A is an invertible matrix, then its inverse is unique. The inverse of A is
denoted by A−1.
proof
If A is invertible, then it has at least one inverse B such that
AB = I = BA.
Proofs in Outline Form
Assume that A has another inverse C such that
In addition to proofs in the exercises, some
proofs are presented in outline form. This omits
the need for burdensome calculations.
AC = I = CA.
Demonstrate that B and C are equal, as shown on the next page.
QR Code is a registered trademark of Denso Wave Incorporated
9/10/15 10:21 AM
9781305658004_0201.indd 39
Find the inverse of a matrix (if it exists).
9781305658004_0203.indd 62
39
Matrices
8/18/15 11:34 AM
Copyright 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
[
∑
Using the Discovery feature helps you develop
or
an intuitive understanding of mathematical
det(A) = ∣A∣ = ∑ a C
concepts and relationships.
j=1
ij
i=1
[
−2
1
2
4
1
−1
A=
0
3
]
3
0
0
0
using Gauss-Jordan
elimination to obtain
the transition matrix
P −1 when the change
of basis is from a
nonstandard basis to
a standard basis.
2 2 1
0
2
.
3
−2
(
TeChnology
Many graphing utilities and
software programs can
find the determinant of
a square matrix. If you use
a graphing utility, then you may
see something similar to the
screen below for Example 4.
The Technology guide at
CengageBrain.com can help
you use technology to find a
determinant.
3
0
0
0
cn1
cn2 . . . cnn
⋮
C13 = (−1)
1+3
=
∣
−1
0
3
∣
−1
0
3
1
2
4
∣
∣
1
2
4
2
3
−2
.
xi
Finding a transition Matrix
[
]
[
]
3
1
[
1
0
1
0
−1
2
2
3
{(
√2 √2
]
)(
[
6
6
3
) (3
2
1
0
0
1
0
0 −1
3
0
1
0
0
1
0
3
First show that the three vectors are mutually orthogonal.
−5
0
0
1
0
0
1
1
SOlutiOn
1 1
v1 ∙ v2 = −that
+ +transition
0=0
From this, you can conclude
matrix from B to B′ is
6 the
6
z
P−1 =
)
(
)
[
−1
3
1
3 3
4
−7
−2
)}
2
−3
−1
]
2
2
v14∙ v3 =2
−
+0=0
−7 −33√
. 2 3√2
√2
√2
2√2
−2
−1
v2 ∙ v3 = −
−
+
=0
]
2
3
−2
C13 = (0)(−1)2+1
matrices in R3 to designate the locations of atoms in a
unit cell. For example, the figure below shows the unit
an Orthonormal
Basis for P3
cell known as end-centered
monoclinic.
Delete 1st row and 3rd column.
In P3 , with the inner product
∣
〈 p, q〉 = a0b0 + a1b1 + a2b2 + a3b3
the standard basis B = { 1, x, x2, x3 } is orthonormal. The verification of this is left as an
exercise. (See Exercise 17.)
2
−1
+ (2)(−1)2+2
−2
3
= 0 + 2(1)(−4) + 3(−1)(−7)
= 13.
called a lattice. The simplest
repeating unit in a lattice is a
3
so you know that they span R . By Theorem 4.12, they form a (nonstandard)
applied5.11),
unitbasis
cell.for
Crystallographers
can use bases and coordinate
orthonormal
R3.
Simplify.
∣ ∣
1
4
)
(
∣
∣
2
−1
+ (3)(−1)2+3
−2
3
∣
1
4
One possible coordinate matrix for the top end-centered
Time-frequency
analysis
of irregular physiological signals,
T
linear
(blue) atom such
is [xas
]B′ beat-to-beat
= [12 12 1]cardiac
.
rhythm variations (also known
algeBra as heart rate variability or HRV), canBrazhnykov
be difficult.
This is
Andriy/Shutterstock.com
applied
because the structure of a signal can include multiple
periodic, nonperiodic, and pseudo-periodic components.
Researchers have proposed and validated a simplified HRV
analysis method called orthonormal-basis partitioning and
time-frequency representation (OPTR). This method can
detect both abrupt and slow changes in the HRV signal’s
structure, divide a nonstationary HRV signal into segments
8/18/15
11:58 AM
that are “less nonstationary,” and determine patterns
in the
HRV. The researchers found that although it had poor time
resolution with signals that changed gradually, the OPTR
method accurately represented multicomponent and abrupt
changes in both real-life and simulated HRV signals.
You obtain
∣A∣ = 3(13)
9781305658004_0407.indd 213
= 39.
Chapter 2
]
Show that the set is an orthonormal basis for R .
Then form the matrix
[B′ B] and use Gauss-Jordan elimination to rewrite [B′ B] as
1
1
√2 √2 2√2
2 2 1
[I3 P−1].
,
,
S = {v , v , v } =
,
,0 , −
, ,− ,
Figure 5.11
Expanding by cofactors in the second row yields
0 ]
2 ]
3 ]
-2]]
39
2
⋮
2 , 2of
9
9
9
, 2 2
−some
Notice that three of the entries in the third column are zeros. So, ,to− eliminate
,
3
6 6
3 3 3
Multiply P−1 byNow,
the coordinate
of 1x because
= [1 2 −1]T to see that the result is the
k
the work in the expansion, use the third column.
each vector ismatrix
of length
v2
same
as
that
obtained
in
Example
3.
v3
v1 = √v1 ∙ v1 = √12 + 12 + 0 = 1
∣A∣ = 3(C13) + 0(C23) + 0(C333) + 0(C43)
i
1
1
v1 j
v2 = √v2 ∙ v2 = √18
+ 18
+ 89 = 1
The cofactors C23, C33, and C43 have zero coefficients, so you need only find the
4
4
1
y
v3Crystallography
= √v3 ∙ v3 = √9 is
+ the
= 1. of atomic and molecular
9 + 9science
1 , linear
1the
cofactor C13. To do this, delete the first row and third columnx of A and evaluate
,0
structure. In a crystal, atoms are in a repeating pattern
2
2
So, S is an orthonormal set. The three vectors do not lie in the same plane (see Figure
determinant of the resulting matrix.
algeBra
det A
108
⋮
Find the transition matrix from B to B′ for the bases for R3 below.
soluTion
-2
1
2
4
1
⋮
jth column
See LarsonLinearAlgebra.com for an interactive version of this type of example.
expansion
. . .+a C .
ij = a1jC1j + a2jC2j +
nj nj
Technology notes show how you can use
D I S C O V E RY
graphing utilities and software programs
appropriately in the problem-solving process.
and B′ = {(1, 0), (0, 1)}.
the matrix
Many of the Technology notes reference the The Determinant Form
of order 4
[of
B′ aBmatrix
].
2.
Make
a conjecture
Technology Guide at CengageBrain.com.
Find the determinant of
about the necessity of
[[1
[-1
[0
[3
. . .
c12 . . . c1n
c22 . . . c2n
When expanding by cofactors, you do not need to find cofactors of zero entries,
B = {(1, 0, 0), (0, 1, 0), (0, 0, 1)} and B′ = {(1, 0, 1), (0, −1, 2), (2, 3, −5)}
because zero times its cofactor is zero.
solution
5.3 Orthonormal Bases: Gram-Schmidt Process
255
aijCij = (0)Cij
First use the vectors in the two bases to form the matrices B and B′.
=0
Example 1 describes another nonstandard orthonormal basis for R3.
1
0
0
1
0
2
The row (or column) containing the most zeros is usually the best choice for expansion
B= 0
1
0 and B′ = 0 −1
3
a nonstandard Orthonormal Basis for R 3
Let B = {(1, 0), (1, 2)}
by cofactors. The next example demonstrates1.
this.
0
0
1
1
2 −5
Technology Notes
A
0
c11
c21
ith row
expansion
aijCij = ai1Ci1 + ai2Ci2 + . . . + ainCin
n
0
⋮ ⋮
0
0
In the next example, you will apply this procedure to the change of basis problem
from Example 3.
Let A be a square matrix of order n. Then the determinant of A is
n
. . .
. . .
Preface
Theorem 3.1 expansion by Cofactors
det(A) = ∣A∣ =
0
1
By 113
the lemma following Theorem 4.20, however, the right-hand side of this matrix
is Q = P−1, which implies that the matrix has the form [I P−1], which proves the
theorem.
3.1 The Determinant of a Matrix
Discovery
1
0
(Source: Orthonormal-Basis Partitioning and Time-Frequency
Representation of Cardiac Rhythm Dynamics, Aysin, Benhur, et al,
IEEE Transactions on Biomedical Engineering, 52, no. 5)
Matrices
Sebastian Kaulitzki/Shutterstock.com
9781305658004_0301.indd 113
8/18/15 2:14 PM
Projects
1 Exploring Matrix Multiplication
Test 1
Test 2
Anna
84
96
Bruce
56
72
Chris
78
83
David
82
91
The table shows the first two test scores for Anna, Bruce, Chris, and David. Use the
table to create a matrix M to represent the data. Input M into a software program or
a graphing utility and use it to answer the questions below.
1. Which test was more difficult? Which was easier? Explain.
2. How would you rank the performances of the four students?
1
0
3. Describe the meanings of the matrix products M
and M
.
0
1
[]
9781305658004_0503.indd 255
Linear Algebra Applied
The Linear Algebra Applied feature describes a real-life
application of concepts discussed in a section. These
applications include biology and life sciences, business
and economics, engineering and technology, physical
sciences, and statistics and probability.
[]
4. Describe the meanings of the matrix products [1 0 0 0]M and [0 0 1 0]M.
1
1
5. Describe the meanings of the matrix products M
and 12M
.
1
1
6. Describe the meanings of the matrix products [1 1 1 1]M and 14 [1 1 1 1]M.
1
7. Describe the meaning of the matrix product [1 1 1 1]M
.
1
8. Use matrix multiplication to find the combined overall average score on
both tests.
9. How could you use matrix multiplication to scale the scores on test 1 by a
factor of 1.1?
[]
[]
[]
Capstone Exercises
2 Nilpotent Matrices
The Capstone is a conceptual problem that synthesizes
key topics to check students’ understanding of the
section concepts. I recommend it.
Let A be a nonzero square matrix. Is it possible that a positive integer k exists such
that Ak = O? For example, find A3 for the matrix
[
0
A= 0
0
1
0
0
]
2
1 .
0
A square matrix A is nilpotent of index k when A ≠ O, A2 ≠ O, . . . , Ak−1 ≠ O,
but Ak = O. In this project you will explore nilpotent matrices.
Chapter Projects
1. The matrix in the example above is nilpotent. What is its index?
2. Use a software program or a graphing utility to determine which matrices below
are nilpotent and find their indices.
0
1
0
1
0
0
(a)
(b)
(c)
0
0
1
0
1
0
[
(d)
[
]
1
1
]
0
0
[
[
0
(e) 0
0
]
0
0
0
[
1
0
0
]
[
0
(f) 1
1
Two per chapter, these offer the opportunity for group
activities or more extensive homework assignments,
and are focused on theoretical concepts or applications.
Many encourage the use of technology.
]
0
0
1
0
0
0
]
3. Find 3 × 3 nilpotent matrices of indices 2 and 3.
4. Find 4 × 4 nilpotent matrices of indices 2, 3, and 4.
5.
6.
7.
8.
Find a nilpotent matrix of index 5.
Are nilpotent matrices invertible? Prove your answer.
When A is nilpotent, what can you say about AT? Prove your answer.
Show that if A is nilpotent, then I − A is invertible.
Supri Suharjoto/Shutterstock.com
9781305658004_020R.indd 108
8/18/15 4:07 PM
9/8/15 8:41 AM
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Instructor Resources
Media
Instructor’s Solutions Manual
The Instructor’s Solutions Manual provides worked-out solutions for all even-numbered
exercises in the text.
Cengage Learning Testing Powered by Cognero (ISBN: 978-1-305-65806-6)
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Student Resources
Print
Student Solutions Manual
ISBN-13: 978-1-305-87658-3
The Student Solutions Manual provides complete worked-out solutions to all
odd-numbered exercises in the text. Also included are the solutions to all
Cumulative Test problems.
Media
MindTap for Larson’s Elementary Linear Algebra
MindTap is a digital representation of your course that provides you with the tools
you need to better manage your limited time, stay organized and be successful.
You can complete assignments whenever and wherever you are ready to learn with
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To access additional course materials and companion resources, please visit
CengageBrain.com. At the CengageBrain.com home page, search for the ISBN
of your title (from the back cover of your book) using the search box at the top of
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Acknowledgements
I would like to thank the many people who have helped me during various stages
of writing this new edition. In particular, I appreciate the feedback from the dozens
of instructors who took part in a detailed survey about how they teach linear algebra.
I also appreciate the efforts of the following colleagues who have provided valuable
suggestions throughout the life of this text:
Michael Brown, San Diego Mesa College
Nasser Dastrange, Buena Vista University
Mike Daven, Mount Saint Mary College
David Hemmer, University of Buffalo, SUNY
Wai Lau, Seattle Pacific University
Jorge Sarmiento, County College of Morris.
I would like to thank Bruce H. Edwards, University of Florida, and
David C. Falvo, The Pennsylvania State University, The Behrend College, for
their contributions to previous editions of Elementary Linear Algebra.
On a personal level, I am grateful to my spouse, Deanna Gilbert Larson, for
her love, patience, and support. Also, a special thanks goes to R. Scott O’Neil.
Ron Larson, Ph.D.
Professor of Mathematics
Penn State University
www.RonLarson.com
xiv
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1
1.1
1.2
1.3
Systems of Linear
Equations
Introduction to Systems of Linear Equations
Gaussian Elimination and Gauss-Jordan Elimination
Applications of Systems of Linear Equations
Traffic Flow (p. 28)
Electrical Network Analysis (p. 30)
Global Positioning System (p. 16)
Airspeed of a Plane (p. 11)
Balancing Chemical Equations (p. 4)
Clockwise from top left, Rafal Olkis/Shutterstock.com; michaeljung/Shutterstock.com;
Fernando Jose V. Soares/Shutterstock.com; Alexander Raths/Shutterstock.com; edobric/Shutterstock.com
Copyright 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1
2
Chapter 1
Systems of Linear Equations
1.1 Introduction to Systems of Linear Equations
Recognize a linear equation in n variables.
Find a parametric representation of a solution set.
Determine
whether a system of linear equations is consistent or

inconsistent.
Use
back-substitution and Gaussian elimination to solve a system

of linear equations.
Linear Equations in n Variables
The study of linear algebra demands familiarity with algebra, analytic geometry,
and trigonometry. Occasionally, you will find examples and exercises requiring a
knowledge of calculus, and these are marked in the text.
Early in your study of linear algebra, you will discover that many of the solution
methods involve multiple arithmetic steps, so it is essential that you check your work. Use
software or a calculator to check your work and perform routine computations.
Although you will be familiar with some material in this chapter, you should
carefully study the methods presented. This will cultivate and clarify your intuition for
the more abstract material that follows.
Recall from analytic geometry that the equation of a line in two-dimensional space
has the form
a1x + a2y = b,   a1, a2, and b are constants.
This is a linear equation in two variables x and y. Similarly, the equation of a plane
in three-dimensional space has the form
a1x + a2 y + a3z = b,   a1, a2, a3, and b are constants.
This is a linear equation in three variables x, y, and z. A linear equation in n variables
is defined below.
Definition of a Linear Equation in n Variables
A linear equation in n variables x1, x2, x3, . . . , xn has the form
a1x1 + a2 x2 + a3 x3 + . . . + an xn = b.
The coefficients a1, a2, a3, . . . , an are real numbers, and the constant term b
is a real number. The number a1 is the leading coefficient, and x1 is the
leading variable.
Linear equations have no products or roots of variables and no variables involved
in trigonometric, exponential, or logarithmic functions. Variables appear only to the
first power.
Linear and Nonlinear Equations
Each equation is linear.
a. 3x + 2y = 7
b. 12x + y − πz = √2
c. (sin π )x1 − 4×2 = e2
Each equation is not linear.
a. xy + z = 2
b. e x − 2y = 4
c. sin x1 + 2×2 − 3×3 = 0
Copyright 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1.1
Introduction to Systems of Linear Equations
3
Solutions and Solution Sets
A solution of a linear equation in n variables is a sequence of n real numbers s1, s2,
s3, . . . , sn that satisfy the equation when you substitute the values
x1 = s1,  x 2 = s2,  x 3 = s3,   . . . ,   xn = sn
into the equation. For example, x1 = 2 and x 2 = 1 satisfy the equation x1 + 2×2 = 4.
Some other solutions are x1 = −4 and x 2 = 4, x1 = 0 and x 2 = 2, and x1 = −2 and
x 2 = 3.
The set of all solutions of a linear equation is its solution set, and when you have
found this set, you have solved the equation. To describe the entire solution set of a
linear equation, use a parametric representation, as illustrated in Examples 2 and 3.
Parametric Representation of a Solution Set
Solve the linear equation x1 + 2×2 = 4.
solution
To find the solution set of an equation involving two variables, solve for one of the
variables in terms of the other variable. Solving for x1 in terms of x2, you obtain
x1 = 4 − 2×2.
In this form, the variable x2 is free, which means that it can take on any real value. The
variable x1 is not free because its value depends on the value assigned to x2. To represent
the infinitely many solutions of this equation, it is convenient to introduce a third variable
t called a parameter. By letting x2 = t, you can represent the solution set as
x1 = 4 − 2t,  x2 = t,  t is any real number.
To obtain particular solutions, assign values to the parameter t. For instance, t = 1
yields the solution x1 = 2 and x2 = 1, and t = 4 yields the solution x1 = −4
and x2 = 4.
To parametrically represent the solution set of the linear equation in Example 2
another way, you could have chosen x1 to be the free variable. The parametric
representation of the solution set would then have taken the form
x1 = s,  x2 = 2 − 12s,  s is any real number.
For convenience, when an equation has more than one free variable, choose the
variables that occur last in the equation to be the free variables.
Parametric Representation of a Solution Set
Solve the linear equation 3x + 2y − z = 3.
solution
Choosing y and z to be the free variables, solve for x to obtain
3x = 3 − 2y + z
x = 1 − 23y + 13z.
Letting y = s and z = t, you obtain the parametric representation
x = 1 − 23s + 13t,  y = s,  z = t
where s and t are any real numbers. Two particular solutions are
x = 1, y = 0, z = 0  and  x = 1, y = 1, z = 2.
Copyright 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
4
Chapter 1
Systems of Linear Equations
SyStEmS oF LInEar EquatIonS
A system of m linear equations in n variables is a set of m equations, each of which
is linear in the same n variables:
rEmarK
The double-subscript notation
indicates aij is the coefficient
of xj in the ith equation.
a11x1 + a12x2 + a13x3 + . . . + a1n xn = b1
a21x1 + a22x2 + a23x3 + . . . + a2n xn = b2
a31x1 + a32x2 + a33x3 + . . . + a3n xn = b3
⋮
am1x1 + am2x2 + am3x3 + . . . + amn xn = bm.
A system of linear equations is also called a linear system. A solution of a linear
system is a sequence of numbers s1, s2, s3, . . . , sn that is a solution of each equation
in the system. For example, the system
3×1 + 2×2 = 3
−x1 + x2 = 4
has x1 = −1 and x2 = 3 as a solution because x1 = −1 and x2 = 3 satisfy both
equations. On the other hand, x1 = 1 and x2 = 0 is not a solution of the system because
these values satisfy only the first equation in the system.
DI S C O VERY
1.
Graph the two lines
3x − y = 1
2x − y = 0
2.
in the xy-plane. Where do they intersect? How many solutions does
this system of linear equations have?
Repeat this analysis for the pairs of lines
3x − y = 1
3x − y = 1
and
3x − y = 0
6x − 2y = 2.
3.
What basic types of solution sets are possible for a system of two
linear equations in two variables?
See LarsonLinearAlgebra.com for an interactive version of this type of exercise.
LInEar
aLgEbra
aPPLIED
In a chemical reaction, atoms reorganize in one or more
substances. For example, when methane gas (CH4 )
combines with oxygen (O2) and burns, carbon dioxide
(CO2 ) and water (H2O) form. Chemists represent this
process by a chemical equation of the form
(x1)CH4 + (x2)O2 → (x3)CO2 + (x4)H2O.
A chemical reaction can neither create nor destroy atoms.
So, all of the atoms represented on the left side of the
arrow must also be on the right side of the arrow. This
is called balancing the chemical equation. In the above
example, chemists can use a system of linear equations
to find values of x1, x2, x3, and x4 that will balance the
chemical equation.
Elnur/Shutterstock.com
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1.1
5
Introduction to Systems of Linear Equations
It is possible for a system of linear equations to have exactly one solution,
infinitely many solutions, or no solution. A system of linear equations is consistent
when it has at least one solution and inconsistent when it has no solution.
Systems of Two Equations in Two Variables
Solve and graph each system of linear equations.
a. x + y = 3
x − y = −1
b.
x+ y=3
2x + 2y = 6
c. x + y = 3
x+y=1
solution
a. T
 his system has exactly one solution, x = 1 and y = 2. One way to obtain
the solution is to add the two equations to give 2x = 2, which implies x = 1
and so y = 2. The graph of this system is two intersecting lines, as shown in
Figure 1.1(a).
b. T
 his system has infinitely many solutions because the second equation is the result
of multiplying both sides of the first equation by 2. A parametric representation of
the solution set is
x = 3 − t,  y = t,  t is any real number.
The graph of this system is two coincident lines, as shown in Figure 1.1(b).
c. T
 his system has no solution because the sum of two numbers cannot be 3 and 1
simultaneously. The graph of this system is two parallel lines, as shown in
Figure 1.1(c).
y
y
4
y
3
3
3
2
2
2
1
1
1
−1
1
2
3
a. Two intersecting lines:
x+y= 3
x − y = −1
x
x
1
2
3
b. Two coincident lines:
x+ y=3
2x + 2y = 6
−1
1
2
3
x
−1
c. Two parallel lines:
x+y=3
x+y=1
Figure 1.1
Example 4 illustrates the three basic types of solution sets that are possible for a
system of linear equations. This result is stated here without proof. (The proof is
provided later in Theorem 2.5.)
Number of Solutions of a System of Linear Equations
For a system of linear equations, precisely one of the statements below is true.
1. The system has exactly one solution (consistent system).
2. The system has infinitely many solutions (consistent system).
3. The system has no solution (inconsistent system).
Copyright 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
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6
Chapter 1
Systems of Linear Equations
Solving a System of Linear Equations
Which system is easier to solve algebraically?
x − 2y + 3z = 9
−x + 3y
= −4
2x − 5y + 5z = 17
x − 2y + 3z = 9
y + 3z = 5
z=2
The system on the right is clearly easier to solve. This system is in row‑echelon form,
which means that it has a “stair-step” pattern with leading coefficients of 1. To solve
such a system, use back-substitution.
Using Back-Substitution in Row-Echelon Form
Use back-substitution to solve the system.
x − 2y = 5
y = −2
Equation 1
Equation 2
solution
From Equation 2, you know that y = −2. By substituting this value of y into Equation 1,
you obtain
x − 2(−2) = 5
x = 1.
Substitute −2 for y.
Solve for x.
The system has exactly one solution: x = 1 and y = −2.
The term back-substitution implies that you work backwards. For instance,
in Example 5, the second equation gives you the value of y. Then you substitute
that value into the first equation to solve for x. Example 6 further demonstrates this
procedure.
Using Back-Substitution in Row-Echelon Form
Solve the system.
x − 2y + 3z = 9
y + 3z = 5
z=2
Equation 1
Equation 2
Equation 3
solution
From Equation 3, you know the value of z. To solve for y, substitute z = 2 into
Equation 2 to obtain
y + 3(2) = 5
y = −1.
Substitute 2 for z.
Solve for y.
Then, substitute y = −1 and z = 2 in Equation 1 to obtain
x − 2(−1) + 3(2) = 9
x = 1.
Substitute −1 for y and 2 for z.
Solve for x.
The solution is x = 1, y = −1, and z = 2.
Two systems of linear equations are equivalent when they have the same solution
set. To solve a system that is not in row-echelon form, first rewrite it as an equivalent
system that is in row-echelon form using the operations listed on the next page.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1.1
Introduction to Systems of Linear Equations
7
operations that Produce Equivalent Systems
Each of these operations on a system of linear equations produces an equivalent
system.
1. Interchange two equations.
2. Multiply an equation by a nonzero constant.
3. Add a multiple of an equation to another equation.
Rewriting a system of linear equations in row-echelon form usually involves
a chain of equivalent systems, using one of the three basic operations to obtain
each system. This process is called Gaussian elimination, after the German
mathematician Carl Friedrich Gauss (1777–1855).
Carl Friedrich gauss
(1777–1855)
German mathematician
Carl Friedrich Gauss is
recognized, with Newton
and Archimedes, as one
of the three greatest
mathematicians in history.
Gauss used a form of what
is now known as Gaussian
elimination in his research.
Although this method was
named in his honor, the
Chinese used an
almost identical
method some
2000 years prior
to Gauss.
x − 2y +3z = 9
2x − 5y + 5z = 17
− x + 3y = − 4
z
(1, −1, 2)
y
x
using Elimination to rewrite
a System in row-Echelon Form
See LarsonLinearAlgebra.com for an interactive version of this type of example.
Solve the system.
x − 2y + 3z = 9
−x + 3y
= −4
2x − 5y + 5z = 17
SoLutIon
Although there are several ways to begin, you want to use a systematic procedure
that can be applied to larger systems. Work from the upper left corner of the
system, saving the x at the upper left and eliminating the other x-terms from the
first column.
x − 2y + 3z = 9
y + 3z = 5
2x − 5y + 5z = 17
Adding the first equation to
the second equation produces
a new second equation.
x − 2y + 3z = 9
y + 3z = 5
−y − z = −1
Adding −2 times the first
equation to the third equation
produces a new third equation.
Now that you have eliminated all but the first x from the first column, work on the
second column.
x − 2y + 3z = 9
y + 3z = 5
2z = 4
Adding the second equation to
the third equation produces
a new third equation.
x − 2y + 3z = 9
y + 3z = 5
z=2
Multiplying the third equation
by 12 produces a new third
equation.
This is the same system you solved in Example 6, and, as in that example, the solution is
x = 1,
Figure 1.2
y = −1,
z = 2.
Each of the three equations in Example 7 represents a plane in a three-dimensional
coordinate system. The unique solution of the system is the point (x, y, z) = (1, −1, 2),
so the three planes intersect at this point, as shown in Figure 1.2.
Nicku/Shutterstock.com
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
8
Chapter 1
Systems of Linear Equations
Many steps are often required to solve a system of linear equations, so it is
very easy to make arithmetic errors. You should develop the habit of checking your
solution by substituting it into each equation in the original system. For instance,
in Example 7, check the solution x = 1, y = −1, and z = 2 as shown below.
Equation 1: (1) − 2(−1) + 3(2) = 9
= −4
Equation 2: − (1) + 3(−1)
Equation 3: 2(1) − 5(−1) + 5(2) = 17
Substitute the solution
into each equation of the
original system.
The next example involves an inconsistent system—one that has no solution.
The key to recognizing an inconsistent system is that at some stage of the Gaussian
elimination process, you obtain a false statement such as 0 = −2.
An Inconsistent System
Solve the system.
x1 − 3×2 + x3 = 1
2×1 − x2 − 2×3 = 2
x1 + 2×2 − 3×3 = −1
solution
x1 − 3×2 + x3 = 1
5×2 − 4×3 = 0
x1 + 2×2 − 3×3 = −1
  equation to the second equation
x1 − 3×2 + x3 = 1
5×2 − 4×3 = 0
5×2 − 4×3 = −2
  equation to the third equation
Adding −2 times the first
produces a new second equation.
Adding −1 times the first
produces a new third equation.
(Another way of describing this operation is to say that you subtracted the first
equation from the third equation to produce a new third equation.)
x1 − 3×2 + x3 = 1
5×2 − 4×3 = 0
0 = −2
Subtracting the second equation
  from the third equation produces
a new third equation.
The statement 0 = −2 is false, so this system has no solution. Moreover, this system
is equivalent to the original system, so the original system also has no solution.
As in Example 7, the three equations in
Example 8 represent planes in a three-dimensional
coordinate system. In this example, however, the
system is inconsistent. So, the planes do not have a
point in common, as shown at the right.
x1 + 2×2 − 3×3 = −1
x1 − 3×2 + x3 = 1
x3
x1
x2
2×1 − x2 − 2×3 = 2
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1.1
Introduction to Systems of Linear Equations
9
This section ends with an example of a system of linear equations that has infinitely
many solutions. You can represent the solution set for such a system in parametric
form, as you did in Examples 2 and 3.
a System with Infinitely many Solutions
Solve the system.
x2 − x3 = 0
x1
− 3×3 = −1
−x1 + 3×2
= 1
SoLutIon
Begin by rewriting the system in row-echelon form, as shown below.
− 3×3 = −1
x2 − x3 = 0
−x1 + 3×2
= 1
x1
x1
− 3×3 = −1
x2 − x3 = 0
3×2 − 3×3 = 0
x1
− 3×3 = −1
x2 − x3 = 0
0= 0
Interchange the first
two equations.
Adding the first equation to the
third equation produces a new
third equation.
Adding −3 times the second
equation to the third equation
eliminates the third equation.
The third equation is unnecessary, so omit it to obtain the system shown below.
x1
− 3×3 = −1
x2 − x3 = 0
To represent the solutions, choose x3 to be the free variable and represent it by the
parameter t. Because x2 = x3 and x1 = 3×3 − 1, you can describe the solution set as
x1 = 3t − 1,
x2 = t,
x3 = t, t is any real number.
D ISCO VERY
1.
Graph the two lines represented by the system of equations.
x − 2y = 1
−2x + 3y = −3
2.
Use Gaussian elimination to solve this system as shown below.
x − 2y = 1
−1y = −1
rEmarK
You are asked to repeat this
graphical analysis for other
systems in Exercises 91
and 92.
x − 2y = 1
y=1
x=3
y=1
Graph the system of equations you obtain at each step of this
process. What do you observe about the lines?
See LarsonLinearAlgebra.com for an interactive version of this type of exercise.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
10
Chapter 1
Systems of Linear Equations
1.1 Exercises
See CalcChat.com for worked-out solutions to odd-numbered exercises.
Linear Equations In Exercises 1–6, determine whether
the equation is linear in the variables x and y.
1. 2x − 3y = 4  2. 3x − 4xy = 0
3.
3 2
+ − 1 = 0  4. x 2 + y2 = 4
y
x
5. 2 sin x − y = 14  6. (cos 3)x + y = −16
Parametric Representation In Exercises 7–10, find
a parametric representation of the solution set of the
linear equation.
7. 2x − 4y = 0  8. 3x − 12 y = 9
9. x + y + z = 1
10. 12×1 + 24×2 − 36×3 = 12
In Exercises 11–24, graph the
system of linear equations. Solve the system and
interpret your answer.
11. 2x + y = 4
12. x + 3y = 2
x−y=2
−x + 2y = 3
Graphical Analysis
13. −x + y = 1
3x − 3y = 4
15. 3x − 5y = 7
2x + y = 9
17. 2x − y = 5
5x − y = 11
19.
x+3 y−1
+
= 1
4
3
2x − y = 12
21. 0.05x − 0.03y = 0.07
0.07x + 0.02y = 0.16
23.
x
y
+ =1
4 6
x−y=3
1
− 3y = 1
−2x + 43 y = −4
16. −x + 3y = 17
4x + 3y = 7
14.
18.
20.
1
2x
32.
33. 2x − 8y = 3
1
2x + y = 0
34. 9x − 4y = 5
1
1
2x + 3 y = 0
35.
4x − 8y = 9
0.8x − 1.6y = 1.8
4x − 5y = 3
−8x + 10y = 14
36. −14.7x + 2.1y = 1.05
44.1x − 6.3y = −3.15
System of Linear Equations In Exercises 37–56, solve
the system of linear equations.
37. x1 − x2 = 0
38. 3x + 2y = 2
3×1 − 2×2 = −1
6x + 4y = 14
x − 5y = 21
6x + 5y = 21
41. 9x − 3y = −1
1
2
1
5x + 5y = − 3
42. 23×1 + 16×2 = 0
4×1 + x2 = 0
x−1 y+2
+
=4
2
3
x − 2y = 5
43.
y
2
2x
+ =
3
6 3
4x + y = 4
In Exercises 25–30, use backsubstitution to solve the system.
26. 2×1 − 4×2 = 6
25. x1 − x2 = 2
x2 = 3
3×2 = 9
29. 5×1 + 2×2 + x3 = 0
2×1 + x2
=0
31. −3x − y = 3
6x + 2y = 1
40.
Back-Substitution
27. −x + y − z = 0
2y + z = 3
1
2z = 0
(a) Use a graphing utility to graph the system.
(b) Use the graph to determine whether the system is
consistent or inconsistent.
(c) If the system is consistent, approximate the solution.
(d) Solve the system algebraically.
(e) 
Compare the solution in part (d) with the
approximation in part (c). What can you conclude?
39. 3u + v = 240
u + 3v = 240
22. 0.2x − 0.5y = −27.8
0.3x − 0.4y = 68.7
24.
Graphical Analysis In Exercises 31–36, complete parts
(a)–(e) for the system of equations.
28. x − y
= 5
3y + z = 11
4z = 8
30. x1 + x2 + x3 = 0
x2
=0
44.
x1 − 2×2 = 0
6×1 + 2×2 = 0
x−2 y−1
+
= 2
4
3
x − 3y = 20
x1 + 4 x2 + 1
+
= 1
3
2
3×1 − x2 = −2
45. 0.02×1 − 0.05×2 = −0.19
0.03×1 + 0.04×2 = 0.52
46. 0.05×1 − 0.03×2 = 0.21
0.07×1 + 0.02×2 = 0.17
47. x − y − z = 0
x + 2y − z = 6
2x
−z=5
48. x + y + z = 2
−x + 3y + 2z = 8
4x + y
=4
49. 3×1 − 2×2 + 4×3 = 1
x1 + x2 − 2×3 = 3
2×1 − 3×2 + 6×3 = 8
The symbol     indicates an exercise in which you are instructed to use a
graphing utility or software program.
Copyright 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1.1
50. 5×1 − 3×2 + 2×3 = 3
2×1 + 4×2 − x3 = 7
x1 − 11×2 + 4×3 = 3
51. 2×1 + x2 − 3×3 = 4
4×1
+ 2×3 = 10
−2×1 + 3×2 − 13×3 = −8
52.
x1
+ 4×3 = 13
7
4×1 − 2×2 + x3 =
2×1 − 2×2 − 7×3 = −19
x − 3y + 2z = 18
5x − 15y + 10z = 18
54. x1 − 2×2 + 5×3 = 2
3×1 + 2×2 − x3 = −2
55.
x+ y+z+ w=6
2x + 3y
− w=0
−3x + 4y + z + 2w = 4
x + 2y − z + w = 0
53.
56. −x1
+ 2×4 = 1
4×2 − x3 − x4 = 2
x2
− x4 = 0
3×1 − 2×2 + 3×3
=4
System of Linear Equations In Exercises 57–62, use
a software program or a graphing utility to solve the
system of linear equations.
57. 123.5x + 61.3y − 32.4z = −262.74
54.7x − 45.6y + 98.2z = 197.4
42.4x − 89.3y + 12.9z =
33.66
58. 120.2x + 62.4y − 36.5z = 258.64
56.8x − 42.8y + 27.3z = −71.44
88.1x + 72.5y − 28.5z = 225.88
59.
x1 + 0.5×2 + 0.33×3 + 0.25×4 = 1.1
0.5×1 + 0.33×2 + 0.25×3 + 0.21×4 = 1.2
0.33×1 + 0.25×2 + 0.2×3 + 0.17×4 = 1.3
0.25×1 + 0.2×2 + 0.17×3 + 0.14×4 = 1.4
60. 0.1x − 2.5y + 1.2z
2.4x + 1.5y − 1.8z
0.4x − 3.2y + 1.6z
1.6x + 1.2y − 3.2z
1
3
2
− 0.75w = 108
+ 0.25w = −81
− 1.4w = 148.8
+ 0.6w = −143.2
349
61. 2×1 − 7×2 + 9×3 = 630
2
4
2
19
3 x1 + 9 x2 − 5 x3 = − 45
4
1
4
139
5 x1 − 8 x2 + 3 x3 = 150
62. 18 x − 17 y + 16 z
1
1
1
7x + 6y − 5z
1
1
1
6x − 5y + 4z
1
1
1
5x + 4y − 3z
− 15 w = 1
+ 14 w = 1
− 13 w = 1
+ 12 w = 1
Exercises
11
Number of Solutions In Exercises 63–66, state why
the system of equations must have at least one solution.
Then solve the system and determine whether it has
exactly one solution or infinitely many solutions.
63. 4x + 3y + 17z = 0
=0
64. 2x + 3y
5x + 4y + 22z = 0
4x + 3y − z = 0
8x + 3y + 3z = 0
4x + 2y + 19z = 0
65. 5x + 5y − z = 0
66. 16x + 3y + z = 0
16x + 2y − z = 0
10x + 5y + 2z = 0
5x + 15y − 9z = 0
67. N
 utrition One eight-ounce glass of apple juice and
one eight-ounce glass of orange juice contain a total of
227 milligrams of vitamin C. Two eight-ounce glasses
of apple juice and three eight-ounce glasses of orange
juice contain a total of 578 milligrams of vitamin C.
How much vitamin C is in an eight-ounce glass of each
type of juice?
68. A
 irplane Speed Two planes start from Los Angeles
International Airport and fly in opposite directions. The
second plane starts 12 hour after the first plane, but its
speed is 80 kilometers per hour faster. Two hours after
the first plane departs, the planes are 3200 kilometers
apart. Find the airspeed of each plane.
True or False? In Exercises 69 and 70, determine
whether each statement is true or false. If a statement
is true, give a reason or cite an appropriate statement
from the text. If a statement is false, provide an example
that shows the statement is not true in all cases or cite an
appropriate statement from the text.
69. (a) A system of one linear equation in two variables is
always consistent.
(b) A system of two linear equations in three variables
is always consistent.
(c) If a linear system is consistent, then it has infinitely
many solutions.
70. (a) A linear system can have exactly two solutions.
(b) 
Two systems of linear equations are equivalent
when they have the same solution set.
(c) A system of three linear equations in two variables
is always inconsistent.
71. F
 ind a system of two equations in two variables, x1 and
x2, that has the solution set given by the parametric
representation x1 = t and x2 = 3t − 4, where t is any
real number. Then show that the solutions to the system
can also be written as
x1 =
4
t
+   and  x2 = t.
3 3
The symbol
indicates that electronic data sets for these exercises are available
at LarsonLinearAlgebra.com. The data sets are compatible with MATLAB,
Mathematica, Maple, TI-83 Plus, TI-84 Plus, TI-89, and Voyage 200.
Copyright 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
12
Chapter 1
Systems of Linear Equations
72. F
ind a system of two equations in three variables,
x1, x2, and x3, that has the solution set given by the
parametric representation
x1 = t,  x2 = s,  and  x3 = 3 + s − t
86. C
 APSTONE Find values of a, b, and c such
that the system of linear equations has (a) exactly
one solution, (b) infinitely many solutions, and
(c) no solution. Explain.
where s and t are any real numbers. Then show that the
solutions to the system can also be written as
x + 5y + z = 0
x + 6y − z = 0
2x + ay + bz = c
x1 = 3 + s − t,  x2 = s,  and  x3 = t.
In Exercises 73–76, solve the system
of equations by first letting A = 1x, B = 1y, and
C = 1z.
12 12
3 2
73.
74. + = −1
−
=7
x
y
x
y
17
3 4
2 3
+ =0
− =−
x
y
x
y
6
Substitution
75.
2 1
3
+
− = 4
x
y
z
4
2
+ = 10
x
z
2 3 13
− + −
= −8
x
y
z
76.
2 1 2
+ − = 5
x
y
z
3 4
−
= −1
x
y
2 1 3
+ + = 0
x
y
z
Trigonometric Coefficients In Exercises 77 and 78,
solve the system of linear equations for x and y.
77. (cos θ )x + (sin θ )y = 1
(−sin θ )x + (cos θ )y = 0
78. (cos θ )x + (sin θ )y = 1
(−sin θ )x + (cos θ )y = 1
In Exercises 79–84, determine the
value(s) of k such that the sys­tem of linear equations has
the indicated number of solutions.
79. No solution
80. Exactly one solution
Coefficient Design
x + ky = 2
x + ky = 0
kx + y = 4
kx + y = 0
81. Exactly one solution
82. No solution
x + 2y + kz = 6
kx + 2ky + 3kz = 4k
x + y + z = 0 3x + 6y + 8z = 4
2x − y + z = 1
83. Infinitely many solutions
4x + ky = 6
kx + y = −3
84. Infinitely many solutions
87. W
 riting Consider the system of linear equations in x
and y.
a1 x + b1 y = c1
a2 x + b2 y = c2
a3 x + b3 y = c3

Describe the graphs of these three equations in the
xy-plane when the system has (a) exactly one solution,
(b) infinitely many solutions, and (c) no solution.
88. Writing Explain why the system of linear equations
in Exercise 87 must be consistent when the constant
terms c1, c2, and c3 are all zero.
89. 
Show that if ax 2 + bx + c = 0 for all x, then
a = b = c = 0.
90. Consider the system of linear equations in x and y.
ax + by = e
cx + dy = f
Under what conditions will the system have exactly one
solution?
Discovery In Exercises 91 and 92, sketch the lines
represented by the system of equations. Then use
Gaussian elimination to solve the system. At each step of
the elimination process, sketch the corresponding lines.
What do you observe about the lines?
91.
x − 4y = −3
5x − 6y = 13
2x − 3y =
7
−4x + 6y = −14
Writing In Exercises 93 and 94, the graphs of the
two equations appear to be parallel. Solve the system
of equations algebraically. Explain why the graphs are
misleading.
93. 100y − x = 200
99y − x = −198
94. 21x − 20y = 0
13x − 12y = 120
y
kx + y = 16
3x − 4y = −64
85. Determine the values of k such that the system of linear
equations does not have a unique solution.
x + y + kz = 3
x + ky + z = 2
kx + y + z = 1
92.
y
20
4
3
10
1
−3
−1
1 2 3 4
x
−10
10
20
x
−3
−4
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1.2
13
Gaussian Elimination and Gauss-Jordan Elimination
1.2 Gaussian Elimination and Gauss-Jordan Elimination
Determine the size of a matrix and write an augmented or
coefficient matrix from a system of linear equations.
Use
matrices and Gaussian elimination with back-substitution

to solve a system of linear equations.
Use
matrices and Gauss-Jordan elimination to solve a system

of linear equations.
Solve a homogeneous system of linear equations.
Matrices
Section 1.1 introduced Gaussian elimination as a procedure for solving a system of
linear equations. In this section, you will study this procedure more thoroughly,
beginning with some definitions. The first is the definition of a matrix.
REMARK
The plural of matrix is matrices.
When each entry of a matrix is
a real number, the matrix
is a real matrix. Unless stated
otherwise, assume all matrices
in this text are real matrices.
Definition of a Matrix
If m and n are positive integers, then an m
array
Column 1
Row 1
Row 2
Row 3
⋮
Row m
[
×
n (read “m by n”) matrix is a rectangular
Column 2
Column 3
. . .
a11
a21
a31
a12
a22
a32
a13
a23
a33
…
…
…
a1n
a2n
a3n
am1
am2
am3
…
amn
⋮
⋮
⋮
Column n
⋮
]
in which each entry, aij, of the matrix is a number. An m × n matrix has m rows
and n columns. Matrices are usually denoted by capital letters.
The entry aij is located in the ith row and the jth column. The index i is called the
row subscript because it identifies the row in which the entry lies, and the index j is
called the column subscript because it identifies the column in which the entry lies.
A matrix with m rows and n columns is of size m × n. When m = n, the matrix is
square of order n and the entries a11, a22, a33, . . . , ann are the main diagonal entries.
Sizes of Matrices
Each matrix has the indicated size.
a. [2] Size: 1 × 1
REMARK
Begin by aligning the variables
in the equations vertically. Use
0 to show coefficients of zero
in the matrix. Note the fourth
column of constant terms in
the augmented matrix.
b.
[00 00] Size: 2 × 2
c.
[πe
2
√2
−7
Size: 2 × 3
4
]
One common use of matrices is to represent systems of linear equations. The
matrix derived from the coefficients and constant terms of a system of linear equations
is the augmented matrix of the system. The matrix containing only the coefficients of
the system is the coefficient matrix of the system. Here is an example.
System
x − 4y + 3z = 5
−x + 3y − z = −3
2x
− 4z = 6
Augmented Matrix
[
1
−1
2
−4
3
0
3
−1
−4
5
−3
6
Cofficient Matrix
] [
1
−1
2
−4
3
0
3
−1
−4
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
]
14
Chapter 1
Systems of Linear Equations
Elementary Row Operations
In the previous section, you studied three operations that produce equivalent systems
of linear equations.
1. Interchange two equations.
2. Multiply an equation by a nonzero constant.
3. Add a multiple of an equation to another equation.
In matrix terminology, these three operations correspond to elementary row operations.
An elementary row operation on an augmented matrix produces a new augmented matrix
corresponding to a new (but equivalent) system of linear equations. Two matrices are
row-equivalent when one can be ­
obtained from the other by a finite sequence of
elementary row operations.
Elementary Row Operations
1. Interchange two rows.
2. Multiply a row by a nonzero constant.
3. Add a multiple of a row to another row.
technology
Many graphing utilities and
software programs can perform
elementary row operations on
matrices. If you use a graphing
utility, you may see something
similar to the screen below for
Example 2(c). The Technology
Guide at CengageBrain.com
can help you use technology
to perform elementary row
operations.
A
[[1 2
[0 3
[2 1
-4 3 ]
-2 -1 ]
5 -2 ]]
mRAdd(-2, A, 1, 3)
[[1 2 -4 3 ]
[0 3 -2 -1 ]
[0 -3 13 -8]]
Although elementary row operations are relatively simple to perform, they can
involve a lot of arithmetic, so it is easy to make a mistake. Noting the elementary row
operations performed in each step can make checking your work easier.
Solving some systems involves many steps, so it is helpful to use a shorthand
method of notation to keep track of each elementary row operation you perform. The
next example introduces this notation.
Elementary Row Operations
a. Interchange the first and second rows.
Original Matrix
[
0
−1
2
1
2
−3
3
0
4
New Row-Equivalent Matrix
4
3
1
] [
−1
0
2
2
1
−3
0
3
4
3
4
1
]
Notation
R1 ↔ R2
b. Multiply the first row by 12 to produce a new first row.
Original Matrix
[
2
1
5
−4
3
−2
6
−3
1
−2
0
2
New Row-Equivalent Matrix
]
[
1
1
5
−2
3
−2
3
−3
1
−1
0
2
]
Notation
(12 )R1 → R1
c. Add −2 times the first row to the third row to produce a new third row.
Original Matrix
[
1
0
2
2
3
1
−4
−2
5
3
−1
−2
New Row-Equivalent Matrix
]
[
1
0
0
2
3
−3
−4
−2
13
3
−1
−8
]
Notation
R3 + (−2)R1 → R3
Notice that adding −2 times row 1 to row 3 does not change row 1.
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1.2
Gaussian Elimination and Gauss-Jordan Elimination
15
In Example 7 in Section 1.1, you used Gaussian elimination with back‑substitution
to solve a system of linear equations. The next example demonstrates the matrix
version of Gaussian elimination. The two methods are essentially the same. The basic
difference is that with matrices you do not need to keep writing the variables.
Using Elementary Row Operations
to Solve a System
Linear System
x − 2y + 3z = 9
−x + 3y
= −4
2x − 5y + 5z = 17
Add the first equation to the second
equation.
x − 2y + 3z = 9
y + 3z = 5
2x − 5y + 5z = 17
Add −2 times the first equation to the
third equation.
x − 2y + 3z = 9
y + 3z = 5
−y − z = −1
Add the second equation to the third
equation.
x − 2y + 3z = 9
y + 3z = 5
2z = 4
Associated Augmented Matrix
[
1
−1
2
−2
3
−5
3
0
5
9
−4
17
]
Add the first row to the second row to
produce a new second row.
[
1
0
2
−2
1
−5
3
3
5
]
9
5   R2 + R1 → R2
17
Add −2 times the first row to the third
row to produce a new third row.
[
1
0
0
−2
1
−1
3
3
−1
]
9
5   
−1
R3 + (−2)R1 → R3
Add the second row to the third row to
produce a new third row.
[
1
0
0
−2
1
0
3
3
2
]
9
5   
4
R3 + R2 → R 3
Multiply the third equation by 12. Multiply the third row by 12 to produce
a new third row.
REMARK
The term echelon refers to the
stair-step pattern formed by
the nonzero elements of the
matrix.
x − 2y + 3z = 9
y + 3z = 5
z=2
[
1
0
0
−2
1
0
3
3
1
]
9
5   
2 (12 )R3 → R3
Use back‑substitution to find the solution, as in Example 6 in Section 1.1. The solution
is x = 1, y = −1, and z = 2.
The last matrix in Example 3 is in row-echelon form. To be in this form, a matrix
must have the properties listed below.
Row-Echelon Form and Reduced Row-Echelon Form
A matrix in row-echelon form has the properties below.
1. Any rows consisting entirely of zeros occur at the bottom of the matrix.
2. For each row that does not consist entirely of zeros, the first nonzero entry
is 1 (called a leading 1).
3. For two successive (nonzero) rows, the leading 1 in the higher row is farther
to the left than the leading 1 in the lower row.
A matrix in row-echelon form is in reduced row-echelon form when every column
that has a leading 1 has zeros in every position above and below its leading 1.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
16
Chapter 1
Systems of Linear Equations
row-Echelon Form
tEchnoloGy
Use a graphing utility or
a software program to find
the row-echelon forms of the
matrices in Examples 4(b)
and 4(e) and the reduced
row-echelon forms of the
matrices in Examples 4(a),
4(b), 4(c), and 4(e). The
technology Guide at
CengageBrain.com can help
you use technology to find
the row-echelon and reduced
row-echelon forms of a matrix.
Similar exercises and projects
are also available on the
website.
Determine whether each matrix is in row-echelon form. If it is, determine whether the
matrix is also in reduced row-echelon form.
[
2
1
0
−1
0
1
[
4
3
−2
1
0
0
0
−5
0
0
0
2
1
0
0
−1
3
1
0
1
e. 0
0
[
2
2
0
−3
1
1
4
−1
−3
1
a. 0
0
c.
]
]
3
−2
4
1
]
[
]
2
0
1
−1
0
2
2
0
−4
1
0
0
0
0
0
1
0
−1
2
3
0
0
0
0
0
1
0
5
3
0
1
b. 0
0
d.
[
0
1
0
0
f.
[
1
0
0
]
]
solUtion
The matrices in (a), (c), (d), and (f ) are in row-echelon form. The matrices in (d) and (f)
are in reduced row-echelon form because every column that has a leading 1 has zeros in
every position above and below its leading 1. The matrix in (b) is not in row-echelon form
because the row of all zeros does not occur at the bottom of the matrix. The matrix in (e)
is not in row-echelon form because the first nonzero entry in Row 2 is not 1.
Every matrix is row-equivalent to a matrix in row-echelon form. For instance, in
Example 4(e), multiplying the second row in the matrix by 12 changes the matrix to
row-echelon form.
The procedure for using Gaussian elimination with back-substitution is
summarized below.
Gaussian Elimination with Back-substitution
1. Write the augmented matrix of the system of linear equations.
2. Use elementary row operations to rewrite the matrix in row-echelon form.
3. Write the system of linear equations corresponding to the matrix in
row-echelon form, and use back-substitution to find the solution.
Gaussian elimination with back-substitution works well for solving systems of linear
equations by hand or with a computer. For this algorithm, the order in which you perform
the elementary row operations is important. Operate from left to right by columns, using
elementary row operations to obtain zeros in all entries directly below the leading 1’s.
linEar
alGEBra
appliED
The Global Positioning System (GPS) is a network of
24 satellites originally developed by the U.S. military as a
navigational tool. Today, GPS technology is used in a wide
variety of civilian applications, such as package delivery,
farming, mining, surveying, construction, banking, weather
forecasting, and disaster relief. A GPS receiver works by using
satellite readings to calculate its location. In three dimensions,
the receiver uses signals from at least four satellites to
“trilaterate” its position. In a simplified mathematical model,
a system of three linear equations in four unknowns (three
dimensions and time) is used to determine the coordinates
of the receiver as functions of time.
edobric/Shutterstock.com
Copyright 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1.2
Gaussian Elimination and Gauss-Jordan Elimination
17
Gaussian Elimination with Back-Substitution
Solve the system.
x2 + x3 − 2×4 = −3
2
x1 + 2×2 − x3
=
+
+
x
−
=
−2
2×1 4×2
3×4
3
x1 − 4×2 − 7×3 − x4 = −19
solution
The augmented matrix for this system is
[
0
1
2
1
1
2
4
−4
1
−1
1
−7
]
−2 −3
0
2
.
−3 −2
−1 −19
Obtain a leading 1 in the upper left corner and zeros elsewhere in the first column.
[
[
[
1
0
2
1
2
1
4
−4
−1
1
1
−7
0
2
−2 −3
−3 −2
−1 −19
1
0
0
1
2
1
0
−4
−1
1
3
−7
0
2
−2 −3
−3 −6
−1 −19
1
0
0
0
2
1
0
−6
−1
1
3
−6
0
2
−2 −3
−3 −6
−1 −21
]
]
]
Interchange the first
two rows.
R1 ↔ R2
Adding −2 times the
first row to the third
row produces a new
third row.
R3 + (−2)R1 → R3
Adding −1 times the
first row to the fourth
row produces a new
fourth row.
R4 + (−1)R1 → R4
Now that the first column is in the desired form, change the second column as shown
below.
[
1
0
0
0
2
1
0
0
−1
0
2
1 −2 −3
3 −3 −6
0 −13 −39
]
Adding 6 times the
second row to the fourth
row produces a new
fourth row.
R4 + (6)R2 → R4
To write the third and fourth columns in proper form, multiply the third row by 13 and
1
the fourth row by − 13
.
[
1
0
0
0
2
1
0
0
−1
1
1
0
0
−2
−1
1
2
−3
−2
3
]
Multiplying the third
row by 13 and the fourth
1
row by − 13
produces new
third and fourth rows.
(13 )R3 → R3
(− 131 )R4 → R4
The matrix is now in row-echelon form, and the corresponding system is shown below.
x1 + 2×2 − x3
x2 + x3 − 2×4
x3 − x4
x4
= 2
= −3
= −2
= 3
Use back-substitution to find that the solution is x1 = −1, x2 = 2, x3 = 1, and x4 = 3.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
18
Chapter 1
Systems of Linear Equations
When solving a system of linear equations, remember that it is possible for the
system to have no solution. If, in the elimination process, you obtain a row of all zeros
except for the last entry, then it is unnecessary to continue the process. Simply conclude
that the system has no solution, or is inconsistent.
A System with No Solution
Solve the system.
x1 − x2 + 2×3
x1
+ x3
2×1 − 3×2 + 5×3
3×1 + 2×2 − x3
=4
=6
=4
=1
solution
The augmented matrix for this system is
[
1
1
2
3
−1
0
−3
2
2
1
5
−1
]
4
6
.
4
1
Apply Gaussian elimination to the augmented matrix.
[
[
[
[
1
0
2
3
−1
1
−3
2
2
−1
5
−1
4
2
4
1
1
0
0
3
−1
1
−1
2
2
−1
1
−1
4
2
−4
1
1
0
0
0
−1
1
−1
5
2
4
−1
2
1 −4
−7 −11
1
0
0
0
−1
1
0
5
2
4
−1
2
0 −2
−7 −11
]
]
]
]
R2 + (−1)R1 → R2
R3 + (−2)R1 → R3
R4 + (−3)R1 → R4
R3 + R2 → R 3
Note that the third row of this matrix consists entirely of zeros except for the last entry.
This means that the original system of linear equations is inconsistent. To see why this
is true, convert back to a system of linear equations.
4
x1 − x2 + 2×3 =
2
x2 − x3 =
0 = −2
5×2 − 7×3 = −11
The third equation is not possible, so the system has no solution.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1.2
Gaussian Elimination and Gauss-Jordan Elimination
19
GaUss-JorDan EliMination
With Gaussian elimination, you apply elementary row operations to a matrix to
obtain a (row-equivalent) row-echelon form. A second method of elimination, called
Gauss-Jordan elimination after Carl Friedrich Gauss and Wilhelm Jordan (1842–1899),
continues the reduction process until a reduced row-echelon form is obtained.
Example 7 demonstrates this procedure.
Gauss-Jordan Elimination
See LarsonLinearAlgebra.com for an interactive version of this type of example.
Use Gauss-Jordan elimination to solve the system.
x − 2y + 3z = 9
−x + 3y
= −4
2x − 5y + 5z = 17
solUtion
In Example 3, you used Gaussian elimination to obtain the row-echelon form
[
1
0
0
−2
1
0
3
3
1
]
9
5 .
2
Now, apply elementary row operations until you obtain zeros above each of the leading
1’s, as shown below.
[
[
[
1
0
0
0
1
0
9
3
1
19
5
2
1
0
0
0
1
0
9
0
1
19
−1
2
1
0
0
0
1
0
0
0
1
1
−1
2
]
]
]
R1 + (2)R2 → R1
R2 + (−3)R3 → R2
R1 + (−9)R3 → R1
The matrix is now in reduced row-echelon form. Converting back to a system of linear
equations, you have
x= 1
y = −1
z = 2.
The elimination procedures described in this section can sometimes result in
fractional coefficients. For example, in the elimination procedure for the system
rEMarK
No matter which elementary
row operations or order you
use, the reduced row-echelon
form of a matrix is the same.
2x − 5y + 5z = 14
9
3x − 2y + 3z =
−3x + 4y
= −18
you may be inclined to first multiply Row 1 by 12 to produce a leading 1, which will
result in working with fractional coefficients. Sometimes, judiciously choosing which
elementary row operations you apply, and the order in which you apply them, enables
you to avoid fractions.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
20
Chapter 1
Systems of Linear Equations
DI S CO VERY
1.
2.
Without performing any row operations, explain why the system of
linear equations below is consistent.
2×1 + 3×2 + 5×3 = 0
−5×1 + 6×2 − 17×3 = 0
7×1 − 4×2 + 3×3 = 0
The system below has more variables than equations. Why does it
have an infinite number of solutions?
2×1 + 3×2 + 5×3 + 2×4 = 0
−5×1 + 6×2 − 17×3 − 3×4 = 0
7×1 − 4×2 + 3×3 + 13×4 = 0
The next example demonstrates how Gauss-Jordan elimination can be used to
solve a system with infinitely many solutions.
a system with infinitely Many solutions
Solve the system of linear equations.
2×1 + 4×2 − 2×3 = 0
3×1 + 5×2
=1
solUtion
The augmented matrix for this system is
[23
4
5
−2
0
]
0
.
1
Using a graphing utility, a software program, or Gauss-Jordan elimination, verify that
the reduced row-echelon form of the matrix is
[10
0
1
5
−3
]
2
.
−1
The corresponding system of equations is
x1
+ 5×3 = 2
x2 − 3×3 = −1.
Now, using the parameter t to represent x3, you have
x1 = 2 − 5t,
x2 = −1 + 3t,
x3 = t, t is any real number.
Note in Example 8 that the arbitrary parameter t represents the nonleading
variable x3. The variables x1 and x2 are written as functions of t.
You have looked at two elimination methods for solving a system of linear
equations. Which is better? To some degree the answer depends on personal preference.
In real-life applications of linear algebra, systems of linear equations are usually
solved by computer. Most software uses a form of Gaussian elimination, with
special emphasis on ways to reduce rounding errors and minimize storage of data. The
examples and exercises in this text focus on the underlying concepts, so you should
know both elimination methods.
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1.2
Gaussian Elimination and Gauss-Jordan Elimination
21
Homogeneous Systems of Linear Equations
Systems of linear equations in which each of the constant terms is zero are called
homogeneous. A homogeneous system of m equations in n variables has the form
REMARK
A homogeneous system of
three equations in the three
variables x1, x2, and x3 has the
trivial solution x1 = 0, x2 = 0,
and x3 = 0.
a11x1 + a12x2 + a13x3 + . . . + a1nxn = 0
a21x1 + a22x2 + a23x3 + . . . + a2nxn = 0
⋮
am1x1 + am2x2 + am3x3 + . . . + amnxn = 0.
A homogeneous system must have at least one solution. Specifically, if all variables in
a homogeneous system have the value zero, then each of the equations is satisfied. Such
a solution is trivial (or obvious).
Solving a Homogeneous System
of Linear Equations
Solve the system of linear equations.
x1 − x2 + 3×3 = 0
2×1 + x2 + 3×3 = 0
solution
Applying Gauss-Jordan elimination to the augmented matrix
[12
−1
1
3
3
]
0
0
yields the matrices shown below.
[10
−1
3
3
−3
0
0
[10
−1
1
3
−1
0
0
[10
0
1
2
−1
0
0
]
R2 + (−2)R1 → R2
]
(13 )R2 → R2
]
R1 + R2 → R 1
The system of equations corresponding to this matrix is
x1
+ 2×3 = 0
x2 − x3 = 0.
Using…