Textbook of Linear Algebra by Jim Hefferon

7/28/2019 Textbook of Linear Algebra by Jim Hefferon

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Jim Hefferonhttp://joshua.smcvt.edu/linearalgebra

inearinearlgebraAlgebra
http://joshua.smcvt.edu/linearalgebrahttp://joshua.smcvt.edu/linearalgebra


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Notation

R, R+, Rn real numbers, reals greater than 0, n-tuples of realsN, C natural numbers: { 0 , 1 , 2 , . . .}, complex numbers

(a . . b), [a . . b] interval (open, closed) of reals between a and b

. . . sequence; like a set but order mattersV,W,U vector spacesv, w, 0, 0V vectors, zero vector, zero vector ofV

B, D, , bases, basis vectorsEn = e1, . . . , en standard basis for Rn

RepB(v) matrix representing the vectorPn set of degree n polynomials

Mnm set ofnmmatrices[S] span of the setS

M N direct sum of subspacesV =W isomorphic spaces

h, g homomorphisms, linear mapsH, G matrices

t, s transformations; maps from a space to itselfT, S square matrices

RepB,D(h) matrix representing the map hhi,j matrix entry from rowi, column j

Znm, Z , Inn, I zero matrix, identity matrix|T| determinant of the matrixT

R(h),N(h) range space and null space of the map hR(h),N(h) generalized range space and null space

Lower case Greek alphabet, with pronounciation

character name character name alpha AL-fuh nu NEW beta BAY-tuh xi KSIGH gamma GAM-muh o omicron OM-uh-CRON delta DEL-tuh pi PIE epsilon EP-suh-lon rho ROW zeta ZAY-tuh sigma SIG-muh

eta AY-tuh tau TOW as in cow theta THAY-tuh upsilon OOP-suh-LON iota eye-OH-tuh phi FEE, or FI as in hi kappa KAP-uh chi KI as in hi lambda LAM-duh psi SIGH, or PSIGH mu MEW omega oh-MAY-guh


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Preface

This book helps students to master the material of a standard US undergraduatefirst course in Linear Algebra.

The material is standard in that the subjects covered are Gaussian reduction,

vector spaces, linear maps, determinants, and eigenvalues and eigenvectors.Another standard is books audience: sophomores or juniors, usually witha background of at least one semester of calculus. The help that it gives tostudents comes from taking a developmental approach this books presentationemphasizes motivation and naturalness, using many examples as well as extensiveand careful exercises.

The developmental approach is what most recommends this book so I willelaborate. Courses at the beginning of a mathematics program focus less ontheory and more on calculating. Later courses ask for mathematical maturity: theability to follow different types of arguments, a familiarity with the themes thatunderlie many mathematical investigations such as elementary set and functionfacts, and a capacity for some independent reading and thinking. Some programshave a separate course devoted to developing maturity and some do not. Ineither case, a Linear Algebra course is an ideal spot to work on this transition.It comes early in a program so that progress made here pays off later but alsocomes late enough that students are serious about mathematics. The materialis accessible, coherent, and elegant. There are a variety of argument styles,including direct proofs, proofs by contradiction, and proofs by induction. And,examples are plentiful.

Helping readers start the transition to being serious students of mathematicsrequires taking the mathematics seriously so all of the results here are proved.On the other hand, we cannot assume that students have already arrived and soin contrast with more advanced texts this book is filled with examples, oftenquite detailed.

Some books that assume a not-yet-sophisticated reader begin with extensivecomputations of linear systems, matrix multiplications, and determinants. Then,when vector spaces and linear maps finally appear and definitions and proofsstart, the abrupt change can bring students to an abrupt stop. While this book


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starts with linear reduction, from the first we do more than compute. Thefirst chapter includes proofs showing that linear reduction gives a correct andcomplete solution set. Then, with the linear systems work as motivation so thatthe study of linear combinations is natural, the second chapter starts with thedefinition of a real vector space. In the schedule below this happens at the start

of the third week.Another example of this books emphasis on motivation and naturalness

is that the third chapter on linear maps does not begin with the definition ofhomomorphism. Instead we start with the definition of isomorphism, which isnatural: students themselves observe that some spaces are the same as others.After that, the next section takes the reasonable step of isolating the operation-preservation idea to define homomorphism. This approach loses mathematicalslickness but it is a good trade because it gives to students a large gain insensibility.

A student progresses most in mathematics while doing exercises. In thisbook problem sets start with simple checks and range up to reasonably involved

proofs. Since instructors usually assign about a dozen exercises I have tried toput two dozen in each set, thereby giving a selection. There are even a few thatare puzzles taken from various journals, competitions, or problems collections.These are marked with a ? and as part of the fun I have retained the originalwording as much as possible.

That is, as with the rest of the book the exercises are aimed to both buildan ability at, and help students experience the pleasure of, doingmathematics.Students should see how the ideas arise and should be able to picture themselvesdoing the same type of work.

Applications and computers. The point of view taken here, that students should

think of Linear Algebra as about vector spaces and linear maps, is not taken tothe complete exclusion of others. Applications and computing are interestingand vital aspects of the subject. Consequently each of this books chapters closeswith a few topics in those areas. They are brief enough that an instructor can doone in a days class or can assign them as independent or small-group projects.Most simply give a reader a taste of the subject, discuss how Linear Algebracomes in, point to some further reading, and give a few exercises. Whether theyfigure formally in a course or not these help readers see for themselves thatLinear Algebra is a tool that a professional must master.

Availability. This book is freely available. In particular, instructors can printcopies for students and sell them out of a college bookstore. Seehttp://joshua.smcvt.edu/linearalgebrafor the license details. That page also contains thisbooks latest version, answers to the exercises, and the LATEX source.

A text is a large and complex project. One of the lessons of softwaredevelopment is that such a project will have errors. I welcome bug reports and Iperiodically issue revisions. My contact information is on the web page.

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If you are reading this on your own. This books emphasis on motivation anddevelopment, and its availability, make it widely used for self-study. If you arean independent student then good for you; I admire your industry. However,you may find some advice helpful.

While an experienced instructor knows what subjects and pace suit theirclass, you may find useful a timetable for a semester. (This is adapted from onecontributed by George Ashline.)

week Monday Wednesday Friday

1 One.I.1 One.I.1, 2 One.I.2, 32 One.I.3 One.III.1 One.III.23 Two.I.1 Two.I.1, 2 Two.I.24 Two.II.1 Two.III.1 Two.III.25 Two.III.2 Two.III.2, 3 Two.III.36 exam Three.I.1 Three.I.17 Three.I.2 Three.I.2 Three.II.1

8 Three.II.1 Three.II.2 Three.II.29 Three.III.1 Three.III.2 Three.IV.1, 2

10 Three.IV.2, 3 Three.IV.4 Three.V.111 Three.V.1 Three.V.2 Four.I.112 exam Four.I.2 Four.III.113 Five.II.1 Thanksgiving break14 Five.II.1, 2 Five.II.2 Five.II.3

This timetable supposes that you already know Section One.II, the elements ofvectors. Note that in addition to the exams and the final exam that is not shown,an important part of the above course is that there are required take-homeproblem sets that include proofs. The computations are important in this course

but so are the proofs.In the table of contents I have marked subsections as optional if some

instructors will pass over them in favor of spending more time elsewhere.

You might pick one or two topics that appeal to you from the end of eachchapter. Youll get more from these if you have access to software for calculations.I recommend Sage, freely available from http://sagemath.org.

My main advice is: do many exercises. I have marked a good sample withs in the margin. For all of them, you must justify your answer either with acomputation or with a proof. Be aware that few inexperienced people can writecorrect proofs; try to find a knowledgeable person to work with you on these.

Finally, a caution for all students, independent or not: I cannot overemphasizehow much the statement, I understand the material but its only that I havetrouble with the problems shows a misconception. Being able to do thingswith the ideas is their entire point. The quotes below express this sentimentadmirably. They capture the essence of both the beauty and the power of

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mathematics and science in general, and of Linear Algebra in particular. (I tookthe liberty of formatting them as poetry).

I know of no better tactic

than the illustration of exciting principles

by well-chosen particulars.

Stephen Jay Gould

If you really wish to learn

then you must mount the machine

and become acquainted with its tricks

by actual trial.

Wilbur Wright

Jim HefferonMathematics, Saint Michaels CollegeColchester, Vermont USA 05439http://joshua.smcvt.edu/linearalgebra

2012-Feb-29

Authors Note. Inventing a good exercise, one that enlightens as well as tests,is a creative act and hard work. The inventor deserves recognition. But textshave traditionally not given attributions for questions. I have changed that herewhere I was sure of the source. I would be glad to hear from anyone who canhelp me to correctly attribute others of the questions.

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Contents

Chapter One: Linear Systems

I Solving Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . 1I.1 Gauss Method . . . . . . . . . . . . . . . . . . . . . . . . . . 2

I.2 Describing the Solution Set . . . . . . . . . . . . . . . . . . . 11I.3 General=Particular+Homogeneous. . . . . . . . . . . . . . 20

II Linear Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . 32II.1 Vectors in Space* . . . . . . . . . . . . . . . . . . . . . . . . 32II.2 Length and Angle Measures* . . . . . . . . . . . . . . . . . . 39

III Reduced Echelon Form . . . . . . . . . . . . . . . . . . . . . . . . 46III.1 Gauss-Jordan Reduction. . . . . . . . . . . . . . . . . . . . . 46III.2 The Linear Combination Lemma . . . . . . . . . . . . . . . . 51

Topic: Computer Algebra Systems . . . . . . . . . . . . . . . . . . . 59Topic: Input-Output Analysis . . . . . . . . . . . . . . . . . . . . . . 61Topic: Accuracy of Computations . . . . . . . . . . . . . . . . . . . . 65

Topic: Analyzing Networks. . . . . . . . . . . . . . . . . . . . . . . . 69

Chapter Two: Vector Spaces

I Definition of Vector Space . . . . . . . . . . . . . . . . . . . . . . 76I.1 Definition and Examples . . . . . . . . . . . . . . . . . . . . 76I.2 Subspaces and Spanning Sets . . . . . . . . . . . . . . . . . . 87

II Linear Independence . . . . . . . . . . . . . . . . . . . . . . . . . 97II.1 Definition and Examples . . . . . . . . . . . . . . . . . . . . 97

III Basis and Dimension . . . . . . . . . . . . . . . . . . . . . . . . . 109III.1 Basis. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109III.2 Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

III.3 Vector Spaces and Linear Systems . . . . . . . . . . . . . . . 121III.4 Combining Subspaces*. . . . . . . . . . . . . . . . . . . . . . 128Topic: Fields. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137Topic: Crystals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139Topic: Voting Paradoxes . . . . . . . . . . . . . . . . . . . . . . . . . 143Topic: Dimensional Analysis . . . . . . . . . . . . . . . . . . . . . . . 149


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Chapter Three: Maps Between Spaces

I Isomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157

I.1 Definition and Examples . . . . . . . . . . . . . . . . . . . . 157

I.2 Dimension Characterizes Isomorphism . . . . . . . . . . . . . 166

II Homomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . 173II.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173

II.2 Range space and Null space . . . . . . . . . . . . . . . . . . . 180

III Computing Linear Maps . . . . . . . . . . . . . . . . . . . . . . . 191

III.1 Representing Linear Maps with Matrices . . . . . . . . . . . 191

III.2 Any Matrix Represents a Linear Map* . . . . . . . . . . . . . 201

IV Matrix Operations . . . . . . . . . . . . . . . . . . . . . . . . . . 208

IV.1 Sums and Scalar Products. . . . . . . . . . . . . . . . . . . . 208

IV.2 Matrix Multiplication . . . . . . . . . . . . . . . . . . . . . . 211

IV.3 Mechanics of Matrix Multiplication . . . . . . . . . . . . . . 218

IV.4 Inverses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227

V Change of Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235V.1 Changing Representations of Vectors . . . . . . . . . . . . . . 235

V.2 Changing Map Representations . . . . . . . . . . . . . . . . . 239

VI Projection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247

VI.1 Orthogonal Projection Into a Line* . . . . . . . . . . . . . . 247

VI.2 Gram-Schmidt Orthogonalization* . . . . . . . . . . . . . . . 251

VI.3 Projection Into a Subspace*. . . . . . . . . . . . . . . . . . . 256

Topic: Line of Best Fit . . . . . . . . . . . . . . . . . . . . . . . . . . 265

Topic: Geometry of Linear Maps . . . . . . . . . . . . . . . . . . . . 270

Topic: Magic Squares . . . . . . . . . . . . . . . . . . . . . . . . . . . 276

Topic: Markov Chains . . . . . . . . . . . . . . . . . . . . . . . . . . 281Topic: Orthonormal Matrices . . . . . . . . . . . . . . . . . . . . . . 287

Chapter Four: Determinants

I Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294

I.1 Exploration* . . . . . . . . . . . . . . . . . . . . . . . . . . . 294

I.2 Properties of Determinants . . . . . . . . . . . . . . . . . . . 299

I.3 The Permutation Expansion . . . . . . . . . . . . . . . . . . 303

I.4 Determinants Exist* . . . . . . . . . . . . . . . . . . . . . . . 311

II Geometry of Determinants . . . . . . . . . . . . . . . . . . . . . . 318

II.1 Determinants as Size Functions . . . . . . . . . . . . . . . . . 318

III Laplaces Expansion . . . . . . . . . . . . . . . . . . . . . . . . . 324III.1 Laplaces Expansion Formula* . . . . . . . . . . . . . . . . . 324

Topic: Cramers Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . 329

Topic: Speed of Calculating Determinants . . . . . . . . . . . . . . . 332

Topic: Projective Geometry . . . . . . . . . . . . . . . . . . . . . . . 335


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Chapter Five: Similarity

I Complex Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . 347I.1 Factoring and Complex Numbers; A Review* . . . . . . . . . 348I.2 Complex Representations . . . . . . . . . . . . . . . . . . . . 350

II Similarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352

II.1 Definition and Examples . . . . . . . . . . . . . . . . . . . . 352II.2 Diagonalizability . . . . . . . . . . . . . . . . . . . . . . . . . 354II.3 Eigenvalues and Eigenvectors . . . . . . . . . . . . . . . . . . 358

III Nilpotence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 367III.1 Self-Composition* . . . . . . . . . . . . . . . . . . . . . . . . 367III.2 Strings* . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371

IV Jordan Form. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 381IV.1 Polynomials of Maps and Matrices* . . . . . . . . . . . . . . 381IV.2 Jordan Canonical Form*. . . . . . . . . . . . . . . . . . . . . 389

Topic: Method of Powers . . . . . . . . . . . . . . . . . . . . . . . . . 402Topic: Stable Populations . . . . . . . . . . . . . . . . . . . . . . . . 406

Topic: Page Ranking . . . . . . . . . . . . . . . . . . . . . . . . . . . 408Topic: Linear Recurrences . . . . . . . . . . . . . . . . . . . . . . . . 412

Appendix

Propositions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A-1Quantifiers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A-3Techniques of Proof . . . . . . . . . . . . . . . . . . . . . . . . . . A-4Sets, Functions, and Relations . . . . . . . . . . . . . . . . . . . . . A-6

Starred subsections are optional.


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Chapter One

Linear Systems

I Solving Linear Systems

Systems of linear equations are common in science and mathematics. These twoexamples from high school science[Onan]give a sense of how they arise.

The first example is from Statics. Suppose that we have three objects, onewith a mass known to be 2 kg and we want to find the unknown masses. Supposefurther that experimentation with a meter stick produces these two balances.

ch 2

15

40 50

c h2

25 50

25

For the masses to balance we must have that the sum of moments on the leftequals the sum of moments on the right, where the moment of an object is itsmass times its distance from the balance point. That gives a system of twoequations.

40h+ 15c= 100

25c= 50 + 50h

The second example of a linear system is from Chemistry. We can mix,under controlled conditions, toluene C7H8 and nitric acid HNO3 to producetrinitrotoluene C7H5O6N3 along with the byproduct water (conditions have

to be very well controlled trinitrotoluene is better known as TNT). In whatproportion should we mix them? The number of atoms of each element presentbefore the reaction

x C7H8 + y HNO3 zC7H5O6N3 + w H2O


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2 Chapter One. Linear Systems

must equal the number present afterward. Applying that in turn to the elementsC, H, N, and O gives this system.

7x= 7z

8x + 1y= 5z+ 2w

1y= 3z

3y= 6z+ 1w

Both examples come down to solving a system of equations. In each system,the equations involve only the first power of each variable. This chapter showshow to solve any such system.

I.1 Gauss Method

1.1 Definition A linear combinationofx1, . . . , xn has the form

a1x1+ a2x2+ a3x3+ + anxn

where the numbers a1, . . . , an Rare the combinations coefficients. A linearequationin the variables x1, . . . , xn has the form a1x1+ a2x2+ a3x3+ +anxn = d where d Ris the constant.

An n-tuple (s1, s2, . . . , sn) Rn is a solutionof, or satisfies, that equationif substituting the numbers s1, . . . , sn for the variables gives a true statement:a1s1+ a2s2+ + ansn = d. Asystem of linear equations

a1,1x1+ a1,2x2+

+ a1,nxn = d1

a2,1x1+ a2,2x2+ + a2,nxn = d2...

am,1x1+ am,2x2+ + am,nxn = dmhas the solution(s1, s2, . . . , sn)if thatn-tuple is a solution of all of the equationsin the system.

1.2 Example The combination 3x1+ 2x2ofx1and x2is linear. The combination3x21+ 2 sin(x2)is not linear, nor is 3x

21+ 2x2.

1.3 Example The ordered pair (1, 5)is a solution of this system.

3x1+ 2x2= 7

x1+ x2= 6

In contrast, (5, 1)is not a solution.

Finding the set of all solutions is solvingthe system. We dont need guessworkor good luck; there is an algorithm that always works. This algorithm is Gaussmethod (orGaussian eliminationor linear elimination).


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Section I. Solving Linear Systems 3

1.4 Example To solve this system

3x3= 9

x1+ 5x2 2x3= 213

x1+ 2x2 = 3

we transform it, step by step, until it is in a form that we can easily solve.The first transformation rewrites the system by interchanging the first and

third row.

swap row 1 with row 313

x1+ 2x2 = 3

x1+ 5x2 2x3= 2

3x3= 9

The second transformation rescales the first row by multiplying both sides ofthe equation by 3.

multiply row 1 by 3

x1+ 6x2 = 9

x1+ 5x2 2x3= 23x3= 9

The third transformation is the only nontrivial one in this example. We mentallymultiply both sides of the first row by 1, mentally add that to the second row,and write the result in as the new second row.

add 1 times row 1 to row 2x1+ 6x2 = 9

x2 2x3= 7

3x3= 9

The point of these steps is that weve brought the system to a form where we can

easily find the value of each variable. The bottom equation shows that x3 = 3.Substituting 3for x3 in the middle equation shows that x2 =1. Substitutingthose two into the top equation gives that x1 = 3. Thus the system has a uniquesolution; the solution set is { (3,1,3) }.

Most of this subsection and the next one consists of examples of solvinglinear systems by Gauss method. We will use it throughout the book. It is fastand easy. But before we do those examples we will first show that this methodis also safe in that it never loses solutions or picks up extraneous solutions.

1.5 Theorem (Gauss method) If a linear system is changed to another by one ofthese operations

(1) an equation is swapped with another(2) an equation has both sides multiplied by a nonzero constant

(3) an equation is replaced by the sum of itself and a multiple of another

then the two systems have the same set of solutions.


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Each of the three Gauss method operations has a restriction. Multiplyinga row by 0is not allowed because obviously that can change the solution setof the system. Similarly, adding a multiple of a row to itself is not allowedbecause adding 1times the row to itself has the effect of multiplying the rowby0. Finally, we disallow swapping a row with itself to make some results in

the fourth chapter easier to state and remember, and also because its pointless.Proof We will cover the equation swap operation here. The other two casesare Exercise31.

Consider the swap of row i with row j. The tuple (s1, . . . , sn)satisfies thesystem before the swap if and only if substituting the values for the variables,the ss for the xs, gives a conjunction of true statements: a1,1s1+ a1,2s2 + + a1,nsn= d1and .. . ai,1s1+ ai,2s2+ + ai,nsn = diand .. . aj,1s1+aj,2s2+ + aj,nsn= dj and .. . am,1s1+ am,2s2+ + am,nsn = dm.

In a list of statements joined with and we can rearrange the order of thestatements. Thus this requirement is met if and only ifa1,1s1+ a1,2s2+ +a1,nsn = d1and .. . aj,1s1 + aj,2s2 +

+ aj,nsn = djand .. . ai,1s1 + ai,2s2 +

+ ai,nsn = diand .. . am,1s1 + am,2s2 + + am,nsn = dm. This is exactlythe requirement that (s1, . . . , sn)solves the system after the row swap. QED

1.6 Definition The three operations from Theorem1.5 are the elementary re-duction operations, or row operations, or Gaussian operations. They areswapping, multiplying by a scalar (orrescaling), and row combination.

When writing out the calculations, we will abbreviate rowi by i. Forinstance, we will denote a row combination operation by ki+ j, with the rowthat changes written second. To save writing we will often combine additionsteps when they use the same i; see the next example.

1.7 Example Gauss method systematically applies the row operations to solve asystem. Here is a typical case.

x + y = 0

2x y + 3z= 3

x 2y z= 3

We begin by using the first row to eliminate the 2x in the second row and the xin the third. To get rid of the 2x, we mentally multiply the entire first row by2, add that to the second row, and write the result in as the new second row.To eliminate the x leading the third row, we multiply the first row by 1, addthat to the third row, and write the result in as the new third row.

21+21+3

x + y = 03y + 3z= 3

3y z= 3

To finish we transform the second system into a third system, where the lastequation involves only one unknown. We use the second row to eliminate y from


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the third row.

2+3x + y = 0

3y + 3z= 3

4z= 0

Now the systems solution is easy to find. The third row shows that z = 0.Substitute that back into the second row to get y = 1 and then substituteback into the first row to get x = 1.

1.8 Example For the Physics problem from the start of this chapter, Gaussmethod gives this.

40h+ 15c = 100

50h+ 25c = 50

5/41+2 40h+ 15c = 100(175/4)c = 175

So c= 4, and back-substitution gives that h= 1. (We will solve the Chemistryproblem later.)

1.9 Example The reduction

x + y + z= 92x +4y 3z= 1

3x + 6y 5z= 0

21+231+3

x + y + z= 92y 5z= 17

3y 8z= 27

(3/2)2+3x + y + z= 9

2y 5z= 17

(1/2)z= (3/2)

shows that z= 3, y = 1, and x = 7.

As illustrated above, the point of Gauss method is to use the elementaryreduction operations to set up back-substitution.

1.10 Definition In each row of a system, the first variable with a nonzero coefficient

is the rows leading variable. A system is in echelon form if each leadingvariable is to the right of the leading variable in the row above it (except for theleading variable in the first row).

1.11 Example The prior three examples only used the operation of row combina-tion. This linear system requires the swap operation to get it into echelon formbecause after the first combination

x y = 0

2x 2y + z+ 2w =4

y + w = 0

2z+ w = 5

21+2x y = 0

z+ 2w =4

y + w = 0

2z+ w = 5

the second equation has no leading y. To get one, we put in place a lower-downrow that has a leading y.

23x y = 0

y + w = 0

z+ 2w =4

2z+ w = 5


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(Had there been more than one suitable row below the second then we couldhave swapped in any one.) With that, Gauss method proceeds as before.

23+4

x y = 0

y + w = 0

z+ 2w = 43w = 3

Back-substitution gives w = 1, z= 2 , y= 1, and x = 1.

The row rescaling operation is not needed, strictly speaking, to solve linearsystems. But we will use it later in this chapter as part of a variation on Gaussmethod, the Gauss-Jordan method.

All of the systems seen so far have the same number of equations as unknowns.All of them have a solution, and for all of them there is only one solution. Wefinish this subsection by seeing some other things that can happen.

1.12 Example This system has more equations than variables.

x + 3y = 1

2x + y = 3

2x + 2y = 2

Gauss method helps us understand this system also, since this

21+221+3

x + 3y = 1

5y = 5

4y = 4

shows that one of the equations is redundant. Echelon form

(4/5)2+3x + 3y = 1

5y = 5

0 = 0

gives that y = 1 and x = 2. The 0= 0 reflects the redundancy.

Gauss method is also useful on systems with more variables than equations.Many examples are in the next subsection.

Another way that linear systems can differ from the examples shown earlieris that some linear systems do not have a unique solution. This can happen intwo ways.

The first is that a system can fail to have any solution at all.1.13 Example Contrast the system in the last example with this one.

x + 3y = 1

2x + y = 3

2x + 2y = 0

21+221+3

x + 3y = 1

5y = 5

4y = 2


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Here the system is inconsistent: no pair of numbers satisfies all of the equationssimultaneously. Echelon form makes this inconsistency obvious.

(4/5)2+3x + 3y = 1

5y = 5

0 = 2

The solution set is empty.

1.14 Example The prior system has more equations than unknowns, but thatis not what causes the inconsistency Example1.12has more equations thanunknowns and yet is consistent. Nor is having more equations than unknownsnecessary for inconsistency, as we see with this inconsistent system that has thesame number of equations as unknowns.

x + 2y = 8

2x +4y = 8

21+2 x + 2y = 80 = 8

The other way that a linear system can fail to have a unique solution, besideshaving no solutions, is to have many solutions.

1.15 Example In this system

x + y =4

2x + 2y = 8

any pair of real numbers (s1, s2)satisfying the first equation also satisfies thesecond. The solution set {(x, y)

x+y = 4} is infinite; some of its membersare (0, 4), (1, 5), and (2.5, 1.5).

The result of applying Gauss method here contrasts with the prior examplebecause we do not get a contradictory equation.

21+2 x +y =40 = 0

Dont be fooled by the final system in that example. A 0= 0 equation itnot the signal that a system has many solutions.

1.16 Example The absence of a 0 = 0 does not keep a system from havingmany different solutions. This system is in echelon form has no 0= 0, but hasinfinitely many solutions.

x +y + z= 0

y + z= 0

Some solutions are: (0,1, 1), (0, 1/2, 1/2), (0,0,0), and (0, , ). There areinfinitely many solutions because any triple whose first component is 0 andwhose second component is the negative of the third is a solution.

Nor does the presence of a 0 = 0 mean that the system must have manysolutions. Example1.12shows that. So does this system, which does not have


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any solutions at all despite that in echelon form it has a 0= 0 row.

2x 2z= 6

y + z= 1

2x + y z= 7

3y + 3z= 0

1+32x 2z= 6

y + z= 1

y + z= 1

3y + 3z= 0

2+332+4

2x 2z= 6

y + z= 1

0 = 0

0 = 3

We will finish this subsection with a summary of what weve seen so farabout Gauss method.

Gauss method uses the three row operations to set a system up for backsubstitution. If any step shows a contradictory equation then we can stop withthe conclusion that the system has no solutions. If we reach echelon form withouta contradictory equation, and each variable is a leading variable in its row, thenthe system has a unique solution and we find it by back substitution. Finally,if we reach echelon form without a contradictory equation, and there is not aunique solution that is, at least one variable is not a leading variable thenthe system has many solutions.

The next subsection deals with the third case. We will see that such a systemmust have infinitely many solutions and we will describe the solution set.

Note For all exercises, you must justify your answer. For instance, if a

question asks whether a system has a solution then you must justify a

yes response by producing the solution and must justify a no response by

showing that no solution exists.

Exercises

1.17 Use Gauss method to find the unique solution for each system.

(a) 2x+3y= 13

x y= 1 (b)

x z=0

3x+y =1

x+y+z=4

1.18 Use Gauss method to solve each system or conclude many solutions or no

solutions.(a) 2x+2y=5

x4y=0

(b) x+y=1

x+y=2

(c) x3y+ z= 1

x+ y+2z=14

(d) x y=1

3x3y=2

(e) 4y+z=20

2x2y+z= 0

x +z= 5

x+ yz=10

(f) 2x + z+w= 5

y w= 1

3x zw= 0

4x+y+2z+w= 9

1.19 We can solve linear systems by methods other than Gauss method. One oftentaught in high school is to solve one of the equations for a variable, then substitutethe resulting expression into other equations. Then we repeat that step until thereis an equation with only one variable. From that we get the first number in the

solution and then we get the rest with back-substitution. This method takes longer
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than Gauss method, since it involves more arithmetic operations, and is also morelikely to lead to errors. To illustrate how it can lead to wrong conclusions, we willuse the system

x+3y= 1

2x+ y= 3

2x+2y= 0from Example1.13.

(a) Solve the first equation for x and substitute that expression into the second

equation. Find the resulting y.(b) Again solve the first equation for x, but this time substitute that expression

into the third equation. Find this y.What extra step must a user of this method take to avoid erroneously concluding asystem has a solution?

1.20 For which values of k are there no solutions, many solutions, or a unique

solution to this system?x y= 1

3x3y=k

1.21 This system is not linear, in some sense,2 sin cos +3 tan= 3

4 sin +2 cos 2 tan=106 sin 3 cos + tan= 9

and yet we can nonetheless apply Gauss method. Do so. Does the system have asolution?

1.22 [Anton] What conditions must the constants, the bs, satisfy so that each ofthese systems has a solution? Hint. Apply Gauss method and see what happens

to the right side.(a) x3y=b1

3x+ y=b2x+7y=b3

2x+

4y=

b4

(b) x1+ 2x2+3x3=b12x1+ 5x2+3x3=b2

x1 +8x3=b3

1.23 True or false: a system with more unknowns than equations has at least one

solution. (As always, to say true you must prove it, while to say false you mustproduce a counterexample.)

1.24 Must any Chemistry problem like the one that starts this subsection a balancethe reaction problem have infinitely many solutions?

1.25 Find the coefficients a, b, and c so that the graph off(x) =ax2 + bx + cpassesthrough the points (1, 2), (1, 6), and (2, 3).

1.26 After Theorem1.5 we note that multiplying a row by 0 is not allowed becausethat could change a solution set. Give an example of a system with solution set S0where after multiplying a row by 0 the new system has a solution set S1 and S0 isa proper subset ofS1. Give an example where S0 = S1.

1.27 Gauss method works by combining the equations in a system to make newequations.(a) Can we derive the equation 3x 2y = 5 by a sequence of Gaussian reductionsteps from the equations in this system?

x+y=1

4xy=6
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(b)Can we derive the equation 5x 3y = 2 with a sequence of Gaussian reductionsteps from the equations in this system?

2x+2y=5

3x+ y=4

(c) Can we derive 6x 9y + 5z= 2 by a sequence of Gaussian reduction steps

from the equations in the system?2x+ yz=4

6x3y+z=5

1.28 Prove that, where a, b, . . . , e are real numbers and a =0, ifax + by= c

has the same solution set asax + dy= e

then they are the same equation. What ifa = 0?

1.29 Show that ifad bc =0 thenax+by= j

cx+dy=k

has a unique solution. 1.30 In the system

ax+by=c

dx+ ey = f

each of the equations describes a line in the xy-plane. By geometrical reasoning,

show that there are three possibilities: there is a unique solution, there is no

solution, and there are infinitely many solutions.

1.31 Finish the proof of Theorem1.5.

1.32 Is there a two-unknowns linear system whose solution set is all ofR2?

1.33 Are any of the operations used in Gauss method redundant? That is, can wemake any of the operations from a combination of the others?

1.34 Prove that each operation of Gauss method is reversible. That is, show that if

two systems are related by a row operation S1 S2 then there is a row operationto go back S2 S1.?1.35 [Anton]A box holding pennies, nickels and dimes contains thirteen coins with

a total value of83 cents. How many coins of each type are in the box?

?1.36 [Con. Prob. 1955] Four positive integers are given. Select any three of theintegers, find their arithmetic average, and add this result to the fourth integer.

Thus the numbers 29, 23, 21, and 17 are obtained. One of the original integers

is:(a) 19 (b) 21 (c) 23 (d) 29 (e) 17

? 1.37 [Am. Math. Mon., Jan. 1935] Laugh at this: AHAHA+TEHE= TEHAW. Itresulted from substituting a code letter for each digit of a simple example in

addition, and it is required to identify the letters and prove the solution unique.

?1.38 [Wohascum no. 2]The Wohascum County Board of Commissioners, which has

20 members, recently had to elect a President. There were three candidates (A, B,andC); on each ballot the three candidates were to be listed in order of preference,with no abstentions. It was found that 11 members, a majority, preferredA over

B (thus the other 9 preferred B over A). Similarly, it was found that 12 members

preferred C over A. Given these results, it was suggested that B should withdraw,to enable a runoff election between A and C. However, B protested, and it was
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then found that 14 members preferred B over C! The Board has not yet recoveredfrom the resulting confusion. Given that every possible order ofA, B, C appearedon at least one ballot, how many members voted for B as their first choice?

?1.39 [Am. Math. Mon., Jan. 1963] This system of n linear equations with n un-knowns, said the Great Mathematician, has a curious property.

Good heavens! said the Poor Nut, What is it?Note, said the Great Mathematician, that the constants are in arithmeticprogression.

Its all so clear when you explain it! said the Poor Nut. Do you mean like

6x + 9y= 12 and 15x + 18y= 21?Quite so, said the Great Mathematician, pulling out his bassoon. Indeed,

the system has a unique solution. Can you find it?Good heavens! cried the Poor Nut, I am baffled.Are you?

I.2 Describing the Solution Set

A linear system with a unique solution has a solution set with one element. Alinear system with no solution has a solution set that is empty. In these casesthe solution set is easy to describe. Solution sets are a challenge to describe onlywhen they contain many elements.

2.1 Example This system has many solutions because in echelon form

2x + z= 3

x y z= 1

3x y =4

(1/2)1+2(3/2)1+3

2x + z= 3

y (3/2)z= 1/2

y (3/2)z= 1/2

2+32x + z= 3

y (3/2)z= 1/2

0 = 0

not all of the variables are leading variables. Theorem1.5shows that an (x,y,z)satisfies the first system if and only if it satisfies the third. So we can describethe solution set { (x,y,z)

2x + z= 3 and x y z= 1 and 3x y= 4 }in thisway.

{(x,y,z)2x + z= 3 and y 3z/2= 1/2 } ()

This description is better because it has two equations instead of three but it isnot optimal because it still has some hard to understand interactions among thevariables.

To improve it, use the variable that does not lead any equation, z, to describethe variables that do lead, xandy. The second equation givesy = (1/2)(3/2)zand the first equation givesx = (3/2)(1/2)z. Thus we can describe the solutionset in this way.

{ (x,y,z) = ((3/2) (1/2)z, (1/2) (3/2)z, z)z R} ()
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Compared with (), the advantage of () is that zcan be any real number.This makes the job of deciding which tuples are in the solution set much easier.For instance, taking z= 2 shows that (1/2, 5/2, 2)is a solution.

2.2 Definition In an echelon form linear system the variables that are not leading

are free.

2.3 Example Reduction of a linear system can end with more than one variablefree. On this system Gauss method

x + y + z w = 1

y z+ w = 1

3x + 6z 6w = 6

y + z w = 1

31+3x + y + z w = 1

y z+ w = 1

3y + 3z 3w = 3

y + z w = 1

32+3

2+4

x +y + zw = 1

y z+w = 1

0 = 0

0 = 0

leaves xand y leading, and both zand wfree. To get the description that weprefer we work from the bottom. We first express the leading variable y in termsofzand w, with y= 1 + zw. Moving up to the top equation, substitutingfory givesx + (1 + zw) + zw= 1 and solving forx leavesx = 2 2z+ 2w.The solution set

{ (2 2z+ 2w, 1 + zw, z, w)z, w R} ()

has the leading variables in terms of the free variables.

2.4 Example The list of leading variables may skip over some columns. Afterthis reduction

2x 2y = 0

z+ 3w = 2

3x 3y = 0

x y + 2z+ 6w =4

(3/2)1+3(1/2)1+4

2x 2y = 0

z+ 3w = 2

0 = 0

2z+ 6w =4

22+42x 2y = 0

z+ 3w = 2

0 = 0

0 = 0

xand zare the leading variables, not x and y. The free variables are y and wand so we can describe the solution set as { (y, y, 2 3w,w)

y, w R }. Forinstance, (1,1,2,0)satisfies the systemtake y = 1 and w = 0. The four-tuple(1,0,5,4)is not a solution since its first coordinate does not equal its second.

A variable that we use to describe a family of solutions is a parameter. Wesay that the solution set in the prior example is parametrizedwith y and w.


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(The terms parameter and free variable do not mean the same thing. In theprior example y and w are free because in the echelon form system they do notlead while they are parameters because of how we used them to describe the setof solutions. Had we instead rewritten the second equation as w = 2/3 (1/3)zthen the free variables would still be yand wbut the parameters would be y

and z.)In the rest of this book we will solve linear systems by bringing them to

echelon form and then using the free variables as parameters in the descriptionof the solution set.

2.5 Example This is another system with infinitely many solutions.

x + 2y = 1

2x + z = 2

3x + 2y + zw =4

21+231+3

x + 2y = 1

4y + z = 0

4y + zw = 1

2+3

x + 2y = 1

4y + z = 0

w = 1

The leading variables are x, y, and w. The variable z is free. Notice that,although there are infinitely many solutions, the value ofwdoesnt vary but isconstant at 1. To parametrize, write w in terms ofzwith w= 1 + 0z. Theny = (1/4)z. Substitute for y in the first equation to get x = 1 (1/2)z. Thesolution set is { (1 (1/2)z, (1/4)z,z, 1)

z R}.Parametrizing solution sets shows that systems with free variables have

infinitely many solutions. In the prior example, ztakes on all real number values,each associated with an element of the solution set, and so there are infinitelymany such elements.

We finish this subsection by developing a streamlined notation for linearsystems and their solution sets.

2.6 Definition An mn matrix is a rectangular array of numbers with mrowsand n columns. Each number in the matrix is an entry.

Matrices are usually named by upper case roman letters such as A. Forinstance,

A=

1 2.2 5

3 4 7

has 2 rows and 3 columns and so is a 23 matrix. Read that aloud as two-

by-three; the number of rows is always first. We denote entries with thecorresponding lower-case letter so that ai,j is the number in row i and column jof the array. The entry in the second row and first column is a2,1 =3. Notethat the order of the subscripts matters: a1,2= a2,1 since a1,2 = 2.2. (Theparentheses around the array are so that when two matrices are adjacent thenwe can tell where one ends and the next one begins.)


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Matrices occur throughout this book. We shall use Mnm to denote thecollection ofnmmatrices.2.7 Example We can abbreviate this linear system

x + 2y =4

y z= 0

x + 2z=4

with this matrix. 1 2 0 40 1 1 0

1 0 2 4

The vertical bar just reminds a reader of the difference between the coefficientson the systems left hand side and the constants on the right. With a bar, thisis an augmentedmatrix. In this notation the clerical load of Gauss methodthe copying of variables, the writing of+s and =sis lighter.

1 2 0 40 1 1 01 0 2 4

1+3

1 2 0 40 1 1 0

0 2 2 0

22+3

1 2 0 40 1 1 0

0 0 0 0

The second row stands for y z= 0 and the first row stands for x + 2y= 4 sothe solution set is { (4 2z,z,z)

z R}.We will also use the matrix notation to clarify the descriptions of solution sets.

The description { (2 2z+ 2w, 1 + zw, z, w)z, w R}from Example2.3

is hard to read. We will rewrite it to group all the constants together, all thecoefficients ofztogether, and all the coefficients ofw together. We will write

them vertically, in one-column matrices.

{

2

1

0

0

+

2

1

1

0

z+

2

1

0

1

w

z, w R}

For instance, the top line says that x = 2 2z+2w and the second line saysthaty = 1 + zw. The next section gives a geometric interpretation that willhelp us picture the solution sets.

2.8 Definition A vector (or column vector) is a matrix with a single column.

A matrix with a single row is a row vector. The entries of a vector are itscomponents. A column or row vector whose components are all zeros is a zerovector.

Vectors are an exception to the convention of representing matrices withcapital roman letters. We use lower-case roman or greek letters overlined with an


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arrow: a, b, . . . or , , ... (boldface is also common: aor ). For instance,this is a column vector with a third component of7.

v=

1

3

7

A zero vector is denoted 0. There are many different zero vectors, e.g., theone-tall zero vector, the two-tall zero vector, etc. Nonetheless we will usuallysay the zero vector, expecting that the size will be clear from the context.

2.9 Definition The linear equationa1x1 + a2x2 + + anxn = d with unknownsx1, . . . , xn issatisfiedby

s=

s1...

sn

ifa1s1+ a2s2+ + ansn = d. A vector satisfies a linear system if it satisfieseach equation in the system.

The style of description of solution sets that we use involves adding thevectors, and also multiplying them by real numbers, such as the zand w. Weneed to define these operations.

2.10 Definition The vector sumof uand vis the vector of the sums.

u+ v=

u1...

un

+

v1...

vn

=

u1+ v1...

un+ vn

Note that the vectors must have the same number of entries for the additionto be defined. This entry-by-entry addition works for any pair of matrices, not

just vectors, provided that they have the same number of rows and columns.

2.11 Definition The scalar multiplicationof the real number rand the vector vis the vector of the multiples.

r v= r

v1...

vn

=

rv1...

rvn

As with the addition operation, the entry-by-entry scalar multiplicationoperation extends beyond just vectors to any matrix.

We write scalar multiplication in either order, as r vor v r, or without the symbol: rv. (Do not refer to scalar multiplication as scalar product becausewe use that name for a different operation.)


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2.12 Example

2

3

1

+

3

1

4

=

2 + 3

3 1

1 +4

=

5

2

5

7

1

4

1

3

=

7

28

7

21

Notice that the definitions of vector addition and scalar multiplication agreewhere they overlap, for instance, v + v= 2v.

With the notation defined, we can now solve systems in the way that we willuse from now on.

2.13 Example This system

2x +y w =4

y + w +u=4

x z+ 2w = 0

reduces in this way.2 1 0 1 0 40 1 0 1 1 4

1 0 1 2 0 0

(1/2)1+3

2 1 0 1 0 40 1 0 1 1 4

0 1/2 1 5/2 0 2

(1/2)2+3

2 1 0 1 0 40 1 0 1 1 4

0 0 1 3 1/2 0

The solution set is { (w + (1/2)u, 4 w u, 3w + (1/2)u, w, u)w, u R}. We

write that in vector form.

{

x

y

z

w

u

=

04

0

0

0

+

11

3

1

0

w +

1/21

1/2

0

1

uw, u R }

Note how well vector notation sets off the coefficients of each parameter. Forinstance, the third row of the vector form shows plainly that ifuis fixed then zincreases three times as fast as w. Another thing shown plainly is that settingboth w and uto zero gives that this vector

x

yz

w

u

=

0

40

0

0

is a particular solution of the linear system.


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2.14 Example In the same way, this system

x y + z= 1

3x + z= 3

5x 2y + 3z= 5

reduces1 1 1 13 0 1 3

5 2 3 5

31+2

51+3

1 1 1 10 3 2 0

0 3 2 0

2+3

1 1 1 10 3 2 0

0 0 0 0

to a one-parameter solution set.

{

10

0

+

1/32/3

1

zz R }

As in the prior example, the vector not associated with the parameter10

0

is a particular solution of the system.

Before the exercises, we will consider what we have accomplished and whatwe have yet to do.

So far we have done the mechanics of Gauss method. Except for one result,Theorem1.5 which we did because it says that the method gives the rightanswers we have not stopped to consider any of the interesting questions that

arise.For example, can we prove that we can always describe solution sets as above,

with a particular solution vector added to an unrestricted linear combinationof some other vectors? Weve noted that the solution sets we described in thisway have infinitely many solutions so an answer to this question would tell usabout the size of solution sets. It will also help us understand the geometry ofthe solution sets.

Many questions arise from our observation that we can do Gauss method inmore than one way (for instance, when swapping rows we may have a choice ofmore than one row). Theorem1.5says that we must get the same solution setno matter how we proceed but if we do Gauss method in two ways must we get

the same number of free variables in each echelon form system? Must those bethe same variables, that is, is solving a problem one way to get y and wfreeand solving it another way to get y and zfree impossible?

In the rest of this chapter we will answer these questions. The answer toeach is yes. We do the first one, the proof about the description of solution sets,in the next subsection. Then, in the chapters second section, we will describe


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the geometry of solution sets. After that, in this chapters final section, we willsettle the questions about the parameters. When we are done we will not onlyhave a solid grounding in the practice of Gauss method, we will also have a solidgrounding in the theory. We will know exactly what can and cannot happen ina reduction.

Exercises

2.15 Find the indicated entry of the matrix, if it is defined.

A=

1 3 1

2 1 4

(a) a2,1 (b) a1,2 (c) a2,2 (d) a3,1

2.16 Give the size of each matrix.

(a)

1 0 4

2 1 5

(b)

1 11 1

3 1

(c) 5 10

10 5

2.17 Do the indicated vector operation, if it is defined.

(a)21

1

+ 304

(b) 5 41

(c)

151

311

(d) 721

+9

35

(e)

1

2

+

12

3

(f) 6

31

1

4

20

3

+ 2

11

5

2.18 Solve each system using matrix notation. Express the solution using vec-

tors.(a) 3x+6y=18

x+2y= 6

(b) x+y= 1

xy= 1

(c) x1 + x3= 4

x1 x2+ 2x3= 5

4x1 x2+ 5x3= 17

(d) 2a+bc=2

2a +c=3

ab =0

(e) x+2yz =3

2x+ y +w=4

x y+z+w=1

(f) x +z+w=4

2x+y w=2

3x+y+z =7 2.19 Solve each system using matrix notation. Give each solution set in vector

notation.(a) 2x+yz=1

4xy =3

(b) x z =1

y+2zw=3

x+2y+3zw=7

(c) x y+ z =0

y +w=0

3x 2y+3z+w=0

y w=0

(d) a+2b+3c+de=1

3a b+ c+d+e=3

2.20 The vector is in the set. What value of the parameters produces that vec-

tor?

(a)

5

5

, {

1

1

k

k R }

(b)

121

, {210

i + 301

j i, j R}

(c)

04

2

, {

11

0

m+

20

1

n m, n R}
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2.21 Decide if the vector is in the set.

(a)

3

1

, {

6

2

kk R }

(b)

5

4

, {

5

4

jj R}

(c) 21

1

, { 037

+ 113

r r R }

(d)

10

1

, {

20

1

j +

31

1

kj, k R }

2.22 [Cleary]A farmer with 1200 acres is considering planting three different crops,corn, soybeans, and oats. The farmer wants to use all 1200acres. Seed corn costs$20 per acre, while soybean and oat seed cost $50 and $12 per acre respectively.

The farmer has $40 000available to buy seed and intends to spend it all.(a) Use the information above to formulate two linear equations with three

unknowns and solve it.(b) Solutions to the system are choices that the farmer can make. Write down

two reasonable solutions.(c) Suppose that in the fall when the crops mature, the farmer can bring in

revenue of $100per acre for corn, $300 per acre for soybeans and $80per acre

for oats. Which of your two solutions in the prior part would have resulted in alarger revenue?

2.23 Parametrize the solution set of this one-equation system.

x1+ x2+ + xn = 0 2.24 (a)Apply Gauss method to the left-hand side to solve

x+2y w=a

2x +z =b

x+ y +2w= c

for x, y, z, and w, in terms of the constants a, b, and c.(b) Use your answer from the prior part to solve this.

x+2y w= 3

2x +z = 1

x+ y +2w= 2

2.25 Why is the comma needed in the notation ai,j for matrix entries?

2.26 Give the 4 4matrix whose i, j-th entry is(a) i +j; (b) 1to the i +jpower.

2.27 For any matrix A, the transpose of A, written Atrans, is the matrix whose

columns are the rows ofA. Find the transpose of each of these.

(a)

1 2 3

4 5 6

(b)

2 3

1 1

(c)

5 10

10 5

(d)

11

0

2.28 (a) Describe all functionsf(x) =ax2 + bx + csuch that f(1) =2 and f(1) =6.

(b) Describe all functionsf(x) =ax2 + bx + csuch that f(1) =2.2.29 Show that any set of five points from the plane R2 lie on a common conic section,

that is, they all satisfy some equation of the form ax2 + by2 + cxy + dx + ey + f= 0where some ofa, . . . , f are nonzero.

2.30 Make up a four equations/four unknowns system having(a) a one-parameter solution set;
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(b)a two-parameter solution set;(c)a three-parameter solution set.

?2.31 [Shepelev]This puzzle is from a Russian web-site http://www.arbuz.uz/, andthere are many solutions to it, but mine uses linear algebra and is very naive.

Theres a planet inhabited by arbuzoids (watermeloners, if to translate from

Russian). Those creatures are found in three colors: red, green and blue. Thereare 13 red arbuzoids, 15 blue ones, and 17 green. When two differently colored

arbuzoids meet, they both change to the third color.The question is, can it ever happen that all of them assume the same color?

?2.32 [USSR Olympiad no. 174](a)Solve the system of equations.

ax+ y=a2

x+ay= 1

For what values ofa does the system fail to have solutions, and for what valuesofa are there infinitely many solutions?

(b)Answer the above question for the system.ax+ y=a3

x+ay= 1

?2.33 [Math. Mag., Sept. 1952] In air a gold-surfaced sphere weighs 7588 grams. Itis known that it may contain one or more of the metals aluminum, copper, silver,or lead. When weighed successively under standard conditions in water, benzene,

alcohol, and glycerin its respective weights are 6588, 6688, 6778, and 6328grams.

How much, if any, of the forenamed metals does it contain if the specific gravitiesof the designated substances are taken to be as follows?

Aluminum 2.7 Alcohol 0.81Copper 8.9 Benzene 0.90Gold 19.3 Glycerin 1.26Lead 11.3 Water 1.00Silver 10.8

I.3 General = Particular + Homogeneous

In the prior subsection the descriptions of solution sets all fit a pattern. Theyhave a vector that is a particular solution of the system added to an unre-stricted combination of some other vectors. The solution set from Example2.13illustrates.

{

0

4

0

0

0

particular

solution

+w

1

1

3

1

0

+u

1/2

1

1/2

0

1

unrestrictedcombination

w, u R}

The combination is unrestricted in that wand ucan be any real numbers there is no condition like such that 2w u= 0 that would restrict which pairsw, uwe can use.
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That example shows an infinite solution set fitting the pattern. The othertwo kinds of solution sets also fit. A one-element solution set fits because it hasa particular solution and the unrestricted combination part is trivial. (That is,instead of being a combination of two vectors or of one vector, it is a combinationof no vectors. We are using the convention that the sum of an empty set of

vectors is the vector of all zeros.) A zero-element solution set fits the patternbecause there is no particular solution and so there are no sums of that form.

3.1 Theorem Any linear systems solution set has the form

{p + c11+ + ckkc1, . . . , ck R }

where p is any particular solution and where the number of vectors 1, . . . ,k equals the number of free variables that the system has after a Gaussianreduction.

The solution description has two parts, the particular solution p and the

unrestricted linear combination of the s. We shall prove the theorem in twocorresponding parts, with two lemmas.

We will focus first on the unrestricted combination. For that we considersystems that have the vector of zeroes as a particular solution so that we canshorten p + c11+ + ckk to c11+ + ckk.

3.2 Definition A linear equation is homogeneousif it has a constant of zero, sothat it can be written as a1x1+ a2x2+ + anxn = 0.

3.3 Example With any linear system like

3x +4y =3

2x y =1

we associate a system of homogeneous equations by setting the right side tozeros.

3x +4y =0

2x y =0

Our interest in this comes from comparing the reduction of the original system

3x +4y =3

2x y =1

(2/3)1+2 3x + 4y =3(11/3)y = 1

with the reduction of the associated homogeneous system.

3x +4y =02x y =0

(2/3)1+2 3x + 4y =0(11/3)y =0

Obviously the two reductions go in the same way. We can study how to reducea linear systems by instead studying how to reduce the associated homogeneoussystem.


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Studying the associated homogeneous system has a great advantage overstudying the original system. Nonhomogeneous systems can be inconsistent.But a homogeneous system must be consistent since there is always at least onesolution, the zero vector.

3.4 Example Some homogeneous systems have the zero vector as their onlysolution.

3x + 2y + z= 0

6x +4y = 0

y + z= 0

21+23x + 2y + z= 0

2z= 0

y + z= 0

233x + 2y + z= 0

y + z= 0

2z= 0

3.5 Example Some homogeneous systems have many solutions. One example isthe Chemistry problem from the first page of this book.

7x 7z = 0

8x + y 5z 2w = 0

y 3z = 0

3y 6z w = 0

(8/7)1+27x 7z = 0

y + 3z 2w = 0

y 3z = 0

3y 6z w = 0

2+332+4

7x 7z = 0

y + 3z 2w = 0

6z+ 2w = 0

15z+ 5w = 0

(5/2)3+47x 7z = 0

y + 3z 2w = 0

6z+ 2w = 0

0 = 0

The solution set

{

1/3

1

1/3

1

ww R }

has many vectors besides the zero vector (if we interpret w as a number ofmolecules then solutions make sense only when wis a nonnegative multiple of3).

3.6 Lemma For any homogeneous linear system there exist vectors 1, . . . , ksuch that the solution set of the system is

{ c11+ + ckk c1, . . . , ck R }

wherekis the number of free variables in an echelon form version of the system.

We will make two points before the proof. The first point is that the basicidea of the proof is straightforward. Consider a system of homogeneous equations


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in echelon form.x +y + 2z+ s + t = 0

y + z+ s t = 0

s + t = 0

Start at the bottom, expressing its leading variable in terms of the free variableswith s = t. For the next row up, substitute the expression giving s as acombination of free variables y+ z+ (t) t = 0 and solve for its leadingvariable y = z+ 2t. Iterate: on the next row up, substitute expressions derivedfrom prior rows x+ (z+ 2t) + 2z+ (t) + t = 0 and solve for the leadingvariable x = z 2t. Now to finish, write the solution in vector notation

x

y

z

s

t

=

1

1

1

0

0

z+

2

2

0

1

1

t z, t R

and recognize that the 1 and 2 of the lemma are the vectors associated withthe free variables zand t.

The prior paragraph is a sketch, not a proof; for instance, a proof would haveto hold no matter how many equations are in the system.

The second point we will make about the proof concerns its style. Theabove sketch moves row-by-row up the system, using the equations derived forthe earlier rows to do the next row. This suggests a proof by mathematicalinduction. Induction is an important and non-obvious proof technique that weshall use a number of times in this book.

We prove a statement by mathematical induction using two steps, a basestep and an inductive step. In the base step we establish that the statement istrue for some first instance, here that for the bottom equation we can write theleading variable in terms of the free variables. In the inductive step we mustverify an implication, that if the statement is true for all prior cases then itfollows for the present case also. Here we will argue that if we can express theleading variables from the bottom-most t rows in terms of the free variables thenwe can express the leading variable of the next row up thet + 1-th row fromthe bottom in terms of the free variables. Those two steps together prove thestatement because by the base step it is true for the bottom equation, and bythe inductive step the fact that it is true for the bottom equation shows thatit is true for the next one up. Then another application of the inductive stepimplies that it is true for the third equation up, etc.

Proof Apply Gauss method to get to echelon form. We may get some 0= 0equations (if the entire system consists of such equations then the result is triviallytrue) but because the system is homogeneous we cannot get any contradictoryequations. We will use induction to show this statement: each leading variable

More information on mathematical induction is in the appendix.


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can be expressed in terms of free variables. That will finish the proof becausewe can then use the free variables as the parameters and the s are the vectorsof coefficients of those free variables.

For the base step, consider the bottommost equation that is not 0 = 0. Callit equation mso we have

am,m xm+ am,m+1xm+1+ + am,nxn = 0

wheream,m=0. (The means leading so that xm is the leading variablein row m.) This is the bottom row so any variables xm+1, . . . after the leadingvariable in this equation must be free variables. Move these to the right sideand divide by am,m

xm = (am,m+1/am,m )xm+1+ + (am,n/am,m )xnto express the leading variable in terms of free variables. (If there are no variablesto the right ofxlm then xm =0; see the tricky point following this proof.)

For the inductive step assume that for the m-th equation, and the (m 1)-thequation, etc., up to and including the (m t)-thequation (where 0 t < m),we can express the leading variable in terms of free variables. We must verify thatthis statement also holds for the next equation up, the (m (t + 1))-th equation.As in the earlier sketch, take each variable that leads in a lower-down equationxm , . . . ,xmt and substitute its expression in terms of free variables. (We onlyneed do this for the leading variables from lower-down equations because thesystem is in echelon form and so in this equation none of the variables leadinghigher up equations appear.) The result has the form

am(t+1),m(t+1)xm(t+1)+a linear combination of free variables =0

witham(t+1),m(t+1)=0. Move the free variables to the right side and dividebyam(t+1),m(t+1) to end with xm(t+1) expressed in terms of free variables.Because we have shown both the base step and the inductive step, by the

principle of mathematical induction the proposition is true. QED

We say that the set { c11+ + ckkc1, . . . , ck R }is generated byor

spanned bythe set of vectors { 1, . . . , k }.There is a tricky point to this. We rely on the convention that the sum of an

empty set of vectors is the zero vector. In particular, we need this in the casewhere a homogeneous system has a unique solution. Then the homogeneouscase fits the pattern of the other solution sets: in the proof above, we derive thesolution set by taking the cs to be the free variables and if there is a uniquesolution then there are no free variables.

Note that the proof shows, as discussed after Example2.4,that we can alwaysparametrize solution sets using the free variables.

The next lemma finishes the proof of Theorem3.1 by considering the partic-ular solution part of the solution sets description.


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3.7 Lemma For a linear system, where pis any particular solution, the solutionset equals this set.

{p + h

hsatisfies the associated homogeneous system}

Proof We will show mutual set inclusion, that any solution to the system is inthe above set and that anything in the set is a solution to the system .

For set inclusion the first way, that if a vector solves the system then it is inthe set described above, assume that ssolves the system. Then s psolves theassociated homogeneous system since for each equation index i,

ai,1(s1 p1) + + ai,n(snpn)= (ai,1s1+ + ai,nsn) (ai,1p1+ + ai,npn) =di di = 0

wherepj and sj are the j-th components of pand s. Express sin the requiredp + hform by writing s pas h.

For set inclusion the other way, take a vector of the form p+ h, where psolves the system and hsolves the associated homogeneous system and notethat p + hsolves the given system: for any equation index i,

ai,1(p1+ h1) + + ai,n(pn+ hn)= (ai,1p1+ + ai,npn) + (ai,1h1+ + ai,nhn) =di+ 0= di

wherepj and hj are the j-th components of pand h. QED

The two lemmas above together establish Theorem3.1. We remember thattheorem with the slogan, General=Particular+Homogeneous.

3.8 Example This system illustrates Theorem3.1.x + 2y z= 1

2x +4y = 2

y 3z= 0

Gauss method

21+2x + 2y z= 1

2z= 0

y 3z= 0

23x + 2y z= 1

y 3z= 0

2z= 0

shows that the general solution is a singleton set.

{

10

0

}

More information on equality of sets is in the appendix.


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That single vector is obviously a particular solution. The associated homogeneoussystem reduces via the same row operations

x + 2y z= 0

2x +4y = 0

y 3z= 0

21+2 23x + 2y z= 0

y 3z= 0

2z= 0

to also give a singleton set.

{

00

0

}

So, as discussed at the start of this subsection, in this single-solution case thegeneral solution results from taking the particular solution and adding to it theunique solution of the associated homogeneous system.

3.9 Example Also discussed at the start of this subsection is that the casewhere the general solution set is empty also fits the General = Particular+Homogeneous pattern. This system illustrates. Gauss method

x + z+ w = 1

2x y + w = 3

x +y + 3z+ 2w = 1

21+21+3

x + z+w = 1

y 2zw = 5

y + 2z+w = 2

shows that it has no solutions because the final two equations are in conflict.The associated homogeneous system has a solution, because all homogeneoussystems have at least one solution.

x + z+ w = 0

2x y + w = 0

x +y + 3z+ 2w = 0

21+21+3

2+3x + z+w = 0

y 2zw = 0

0 = 0

In fact the solution set of this homogeneous system is infinite.

{

1

2

1

0

z+

1

1

0

1

w

z, w R}

However, because no particular solution of the original system exists, the generalsolution set is empty there are no vectors of the form p + hbecause there areno p s.

3.10 Corollary Solution sets of linear systems are either empty, have one element,

or have infinitely many elements.

Proof Weve seen examples of all three happening so we need only prove thatthere are no other possibilities.

First, notice a homogeneous system with at least one non-0solution vhasinfinitely many solutions. This is because the set of multiples of vis infinite if


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s, t Rare unequal then sv =tvbecause sv tv= (s t)vis non-0, since anynon-0component of vwhen rescaled by the non-0 factor s twill give a non-0value.

Now apply Lemma3.7to conclude that a solution set

{p +

h

hsolves the associated homogeneous system }is either empty (if there is no particular solution p), or has one element (if thereis a pand the homogeneous system has the unique solution 0), or is infinite (ifthere is a pand the homogeneous system has a non-0solution, and thus by theprior paragraph has infinitely many solutions). QED

This table summarizes the factors affecting the size of a general solution.

number of solutions of thehomogeneous system

particularsolution

exists?

one infinitely many

yes unique

solution

infinitely many

solutionsno

nosolutions

nosolutions

The dimension on the top of the table is the simpler one. When we performGauss method on a linear system, ignoring the constants on the right side andso paying attention only to the coefficients on the left-hand side, we either endwith every variable leading some row or else we find that some variable does notlead a row, that is, we find that some variable is free. (We formalize ignoringthe constants on the right by considering the associated homogeneous system.)

A notable special case is systems having the same number of equations asunknowns. Such a system will have a solution, and that solution will be unique,

if and only if it reduces to an echelon form system where every variable leads itsrow (since there are the same number of variables as rows), which will happen ifand only if the associated homogeneous system has a unique solution.

3.11 Definition A square matrix is nonsingularif it is the matrix of coefficientsof a homogeneous system with a unique solution. It is singular otherwise, thatis, if it is the matrix of coefficients of a homogeneous system with infinitely manysolutions.

3.12 Example The first of these matrices is nonsingular while the second issingular

1 23 4

1 23 6

because the first of these homogeneous systems has a unique solution while thesecond has infinitely many solutions.

x + 2y = 0

3x +4y = 0

x + 2y = 0

3x + 6y = 0


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We have made the distinction in the definition because a system with the samenumber of equations as variables behaves in one of two ways, depending onwhether its matrix of coefficients is nonsingular or singular. A system where thematrix of coefficients is nonsingular has a unique solution for any constants onthe right side: for instance, Gauss method shows that this system

x + 2y = a

3x +4y =b

has the unique solution x = b 2aand y= (3a b)/2. On the other hand, asystem where the matrix of coefficients is singular never has a unique solution it has either no solutions or else has infinitely many, as with these.

x + 2y = 1

3x + 6y = 2

x + 2y = 1

3x + 6y = 3

We use the word singular because it means departing from general expecta-

tion and people often, naively, expect that systems with the same number ofvariables as equations will have a unique solution. Thus, we can think of theword as connoting troublesome, or at least not ideal. (That singular appliesthose systems that do not have one solution is ironic, but it is the standardterm.)

3.13 Example The systems from Example 3.3,Example3.4,and Example3.8each have an associated homogeneous system with a unique solution. Thus thesematrices are nonsingular.

3 4

2 1

3 2 1

6 4 0

0 1 1

1 2 1

2 4 0

0 1 3

The Chemistry problem from Example3.5 is a homogeneous system with morethan one solution so its matrix is singular.

7 0 7 0

8 1 5 2

0 1 3 0

0 3 6 1

The above table has two dimensions. We have considered the one on top: wecan tell into which column a given linear system goes solely by considering thesystems left-hand side the constants on the right-hand side play no role in

this factor.The tables other dimension, determining whether a particular solution exists,

is tougher. Consider these two

3x + 2y = 5

3x + 2y = 5

3x + 2y = 5

3x + 2y =4


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with the same left sides but different right sides. The first has a solution while thesecond does not, so here the constants on the right side decide if the system hasa solution. We could conjecture that the left side of a linear system determinesthe number of solutions while the right side determines if solutions exist butthat guess is not correct. Compare these two systems

3x + 2y = 5

4x + 2y =4

3x + 2y = 5

3x + 2y =4

with the same right sides but different left sides. The first has a solution but thesecond does not. Thus the constants on the right side of a system dont decidealone whether a solution exists; rather, it depends on some interaction betweenthe left and right sides.

For some intuition about that interaction, consider this system with one ofthe coefficients left as the parameter c.

x + 2y + 3z= 1

x + y + z= 1cx + 3y +4z= 0

Ifc = 2 then this system has no solution because the left-hand side has thethird row as a sum of the first two, while the right-hand does not. If c= 2then this system has a unique solution (try it with c = 1). For a system tohave a solution, if one row of the matrix of coefficients on the left is a linearcombination of other rows, then on the right the constant from that row mustbe the same combination of constants from the same rows.

More intuition about the interaction comes from studying linear combinations.That will be our focus in the second chapter, after we finish the study of Gaussmethod itself in the rest of this chapter.

Exercises

3.14 Solve each system. Express the solution set using vectors. Identify the particularsolution and the solution set of the homogeneous system.

(a) 3x+6y=18

x+2y= 6

(b) x+y= 1

xy= 1

(c) x1 + x3= 4

x1x2+ 2x3= 5

4x1x2+ 5x3= 17

(d) 2a+bc=2

2a +c=3

ab =0

(e) x+2yz =3

2x+ y +w=4

x y+z+w=1

(f) x +z+w=4

2x+y w=2

3x+y+z =7

3.15 Solve each system, giving the solution set in vector notation. Identify the

particular solution and the solution of the homogeneous system.

(a) 2x+yz=14xy =3

(b) x z =1y+2zw=3

x+2y+3zw=7

(c) x y+ z =0y +w=0

3x 2y+3z+w=0

y w=0

(d) a+2b+3c+de=1

3a b+ c+d+e=3
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3.16 For the system

2x y w= 3

y+z+2w= 2

x2yz = 1

which of these can be used as the particular solution part of some general solu-

tion?

(a)

0

3

5

0

(b)

2

1

1

0

(c)

1

4

8

1

3.17 Lemma3.7says that we can use any particular solution for p. Find, if possible,a general solution to this system

x y +w=4

2x+3yz =0

y+z+w=4

that uses the given vector as its particular solution.

(a)

0

00

4

(b)

5

17

10

(c)

2

11

1

3.18 One is nonsingular while the other is singular. Which is which?

(a)

1 3

4 12

(b)

1 3

4 12

3.19 Singular or nonsingular?

(a)

1 2

1 3

(b)

1 2

3 6

(c)

1 2 1

1 3 1

(Careful!)

(d)

1 2 1

1 1 3

3 4 7

(e)

2 2 1

1 0 5

1 1 4

3.20 Is the given vector in the set generated by the given set?

(a)

2

3

, {

1

4

,

1

5

}

(b)

10

1

, {

21

0

,

10

1

}

(c)

13

0

, {

10

4

,

21

5

,

33

0

,

42

1

}

(d)

1

0

1

1

, {

2

1

0

1

,

3

0

0

2

}

3.21 Prove that any linear system with a nonsingular matrix of coefficients has a

solution, and that the solution is unique.

3.22 In the proof of Lemma3.6,what happens if there are no non-0= 0 equations?

3.23 Prove that if s and t satisfy a homogeneous system then so do these vec-

tors.
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(a) s +t (b) 3s (c) ks + mtfor k, m RWhats wrong with this argument: These three show that if a homogeneous systemhas one solution then it has many solutions any multiple of a solution is anothersolution, and any sum of solutions is a solution also so there are no homogeneoussystems with exactly one solution.?

3.24 Prove that if a system with only rational coefficients and constants has asolution then it has at least one all-rational solution. Must it have infinitely many?
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II Linear Geometry

If you have seen the elements of vectors before then this section is an

optional review. However, later work will refer to this material so if this is

not a review then it is not optional.

In the first section, we had to do a bit of work to show that there are onlythree types of solution sets singleton, empty, and infinite. But in the specialcase of systems with two equations and two unknowns this is easy to see with apicture. Draw each two-unknowns equation as a line in the plane and then thetwo lines could have a unique intersection, be parallel, or be the same line.

Unique solution

3x+2y= 7x y= 1

No solutions

3x+2y=73x+2y=4

Infinitely many

solutions

3x+2y= 76x+4y=14

These pictures arent a short way to prove the results from the prior section,because those apply to any number of linear equations and any number ofunknowns. But they do help us understand those results. This section developsthe ideas that we need to express our results geometrically. In particular, whilethe two-dimensional case is familiar, to extend to systems with more than twounknowns we shall need some higher-dimensional geometry.

II.1 Vectors in Space

Higher-dimen

Textbook of Linear Algebra by Jim Hefferon

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