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Fourth Edition LINEAR ALGEBRA AND ITS APPLICATIONS Gilbert Strang THOMSON BROOKS/COLE Linear Algebra and Its Applications, Fourth Edition Gilbert Strang Permissions Editor: Audrey Petrengill Production Editor: Rozi Harris, ICC Text Designer: Kim Rokusek. Illustrator: Brett Coonley/ICC Cover Designer: Laurie Albrecht Cover Image: Judith Laurel Harkness Compositor: interactive Composition Corporation Cover/Interior Printer: RR. Donnelley/Crawfordsy Acquisitions Editor: Johm-Paul Ramin Assistant Editor: Katherine Brayton Editorial Assistant: Leata Holloway Marketing Manager: Tom Ziolkowski Marketing Assistant: Jennifer Velasquez Marketing Communications Manager: Bryan Vamn Senior Project Manager, Editorial Production: Janet Bit Senior Art Director: Vernon Boes Print Buyer: Liso Claudeanos €2006 Thomson Brooks/Cole, a part of The Thomson Corporation. Thomson, the Star logo, and Brooks/Cole are trademarks used berein under ticense. 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Library of Congress Control Number: 2005923623 Student Edition: ISBN 0-03-010567-6 Thomson Higher Education 10 Davis Drive Belmont, CA 94002-3098 USA Asia (including Índia) Thomson Learning 5 Shenton Way 01-01 UIC Building Singapore 068808 Australia/New Zealand Thomson Learning Australia 102 Dodás Street Southbank. Victoria 3006 Australia Canada Thomson Nelson 1120 Birchmount Road Toronto, Ontario MIK 5G4 Canada UK /Europe/ Middle East/Africa Thomson Learning High Holborn House 50/51 Bedford Row London WCIR 4LR United Kingdom Latin America Thomson Learning Seneca, 53 Colonia Polanco 11560 Mexico D.F. Mexico Spain (including Portugal) Thomson Paraninfo Calle Magailanes, 25 28015 Madrid, Spain Iv Table of Contents E Chapter 5 EIGENVALUES AND EFIGENVECTORS 233 5.1 Introduction 233 5.2 Diagonalization of a Matrix 245 5.3 Difference Equations and Powers 4? 254 5.4 Differential Equations ande” 266 5.5 Complex Matrices 280 5.6 Similasity Transformations 293 Review Exercises: Chapter 5 307 EE Chapter 6 POSITIVE DEFINITE MATRICES 311 6.1 Minima, Maxima, and Saddle Points 311 6.2 — Tests for Positive Definiteness 318 6.3 Singular Value Decomposition 331 6.4 Minimum Principles 339 6.5 The Finité Element Method 346 BE Chapter 7 COMPUTATIONS WITH MATRICES 351 71 Introduction 351 7.2 Matrix Norm-and-Condition Number 352 7.3 Computation of Eigenvalues 359 74 Terative Methods Tor Ax =b 367 E Chapter 8 LINEAR PROGRAMIENG AND GAME THEORY 8.1 Linear Inegualities 377 8.2 The Simplex Method 382 8.3 The Dual Problem 392 84 Network Models 401 8.5 Game Theory 408 Fi Appendix A E Appendix B INTERSECTION, SUM, AND PRODUCT OF SPACES THE JORDAN FORM 422 Seclutions to Selected Exercises 428 Matrix Factorizations 474 Glossary 476 MATLAB Teaching Codes 481 Index 482 Linear Algebra in a Nutshell 488 415 Preface Revising this textbook has been a special challenge, for a very nice reason. So many people have read this book, and taught from it, and even loved it. The spirit of the book could never change. This text was written to help our teaching of linear algebra keep up with the enormous importance of this subject--which just continues to grow. One step was certainly possible and desirable-—to add new problems. Teaching for all these years required hundreds of new exam questions (especially with quizzes going onto the web). 1 think you will approve of the extended choice of problems. The questions are still a mixture of explain and compute-—the two complementary approaches to learning this beautiful subject. 1 personally believe that many more people need lincar algebra than calculus. Isaac Newton might not agree ! But he isn't teaching mathematics in the 21st century (and maybe he wasn't a great teacher, but we will give him the benefit of the doubt). Certainly the laws of physics are well expressed by differential equations. Newton needed calculus-—quite right. But the scope of science and engineering and management (and life) is now so much wider, and linear algebra has moved into a central place. May Isay a little more, because many universities have not yet adjusted the balance toward linear algebra. Working with curved lines and curved surfaces, the first step is always to linearize. Replace the curve by its tangent line, fit the surface by a plane, and the problem becomes linear. The power of this subject comes when you have ten variables, or 1000 variables, instead of two. You might think 1 am exaggerating to use the word “beautiful” for a basic course in mathematics. Not at all. This subject begins with two vectors v and w, pointing in different directions. The key step is to take their linear combinations. We multiply to get 3v and 4w, and we add to get the particular combination 3v + 4%w. That new vector is in the same plane as v and w. When we take all combinations, we are filling in the whole plane. If I draw v and w on this page, their combinations cv + dw fill the page (and beyond), but they don't go up from the page. In the language of linear equations, 1 can solve cu + dw = b exactly when the vector b lies in the same plane as v and w. Matrices 1 will keep going a little more to convert combinations of three-dimensional vectors into linear algebra. IH the vectors are v = (1,2,3) and w = (1,3, 4), put them into the columns of a matrix: 11 matrix = |2 3 34 Preface vii voiceover is low, and FlashPlayer is freely available. This offers a quick review (with active experiment), and the full lectures can be downloaded. I hope professors and students worldwide will find these web pages helpful. My goal is to make this book as useful as possible with all the course material 1 can provide. Other Supporiing Materials Student Solutions Manual 0-495-01325-0 The Student Solutions Manual provides solutions to the odd-numbered problems in the text. instructor's Solutions Manual 1-030-40568-4 The Instructor's Solutions Manual has teaching notes for each chapter and solutions to all of the problems in the text. Structure of the Course The two fundamental problems are Ax = band Ax = Ax for square matrices A. The first problem Ax = b has a solution when A has independent columns. The second problem 4x = Ax looks for independent eigenvectors. A crucial part of this course is to learn what “independence” means. Tbelieve that most of us learn first from examples. You can see that 1 1 A=|1 2 does not have independent columns. 3 + to O Column 1 plus colurm 2 equals column 3. A wonderful theorem of linear algebra says that the three rows are not independent either. The third row must lie in the same plane as the first two rows. Some combination of rows 1 and 2 will produce row 3. You might find that combination quickly (1 didn"t). In the end 1 had to use elimination to discover that the right combination uses 2 times row 2, minus row 1. Elimination is the simple and natural way to understand a matrix by producing a lot of zero entries. So the course starts there. But don't stay there too long ! You have to get from combinations of the rows. to independence of the rows, to “dimension of the row space.” That is a key goal, to see whole spaces of vectors: the row space and the column space and the nullspace. A further goal is to understand how the matrix acts. When A multíplies x itproduces the new vector Ax. The whole space of vectors moves —itis “transformed” by A. Special transformations come from particular matrices, and those are the foundation stones of linear algebra: diagonal matrices, orthogonal matrices, triangular matrices, symmetric matrices, The eigenvalues of those matrices are special too. I think 2 by 2 matrices provide terrific examples of the information that eigenvalues À can give. Sections 5.1 and 5.2 are worth careful reading, to see how Ax = Ax is useful, Here is a case in which small matrices allow tremendous insight. Overall, the beauty of linear algebra is seen in so many different ways: 1. Visualization. Combinations of vectors. Spaces of vectors. Rotation and reflection and projection of vectors. Perpendicutar vectors. Four fundamental subspaces. Preface 2 Abstraction. Independence of vectors. Basis and dimension of a vector space. Linear transformations. Singular value decomposition and the best basis. 3 Compntation. Elimination to produce zero entries. Gram-Schmidt to produce orthogonal vectors. Eigenvalues to solve differential and difference equations, 4. Applications. Least-squares solution when Ax = b has too many equations. Dif- ference equations approximating differential equations. Markov probability matrices (the basis for Google!). Orthogonal eigenvectors as principal axes (and more ...). To go further with those applications, may I mention the books published by Wellesley- Cambridge Press. They are all linear algebra in disguise, applied to signal processing and partial differential equations and scientific computing (and even GPS). If you look athtip://www.wellesleycambridge.com, you will see part of the reason that linear algebra is so widely used. After this preface, the book will speak for itself. You will see the spirit right away. The emphasis is on understanding—T try to explain rather than to deduce. This is a book about real mathematics, not endless drill, In class, 1 am constantly working with examples to teach what students need, Acknowledgments Tenjoyed writing this book, and 1 certainly hope you enjoy reading it. A big part of the pleasure comes from working with friends. J had wonderful help from Brett Coonley and Cordula Robinson and Erin Maneri. They created the HTgX files and drew all the figures. Without Brett's steady support I would never have completed this new edition. Earlier help with the Teaching Codes came from Steven Lee and Cleve Moler. Those foilow the steps described in the book; MATLAB and Maple and Mathematica are faster for large matrices. All can be used (optionally) in this course. I could have added “Factorization” to that list above, as a fifth avenue to the understanding of matrices: IL,U,P] = Iu(A) for linear equations IQ,R] = qr(A) to make the columns orthogonal [S,E] = eig(A) to find eigenvectors and eigenvalues. In giving thanks, I never forget the first dedication of this textbook. years ago. That was a special chance to thank my parents for so many unselfish gifts. Their example is an inspiration for my life. And I thank the reader too, hoping you like this book. Gilbert Strang Chapter 1 Matrices and Gaussian Elimination That could seem a little mysterious, unless you already know about 2 by 2 determinants. They gave the same answer y = 2, coming from the same ratio of —6 to —3. If we stay with determinants (which we don't plan to do), there will be a similar formula to compute the other unknown, x: tr as O BS asa A |" ricas Caro 6) Let me compare those two approaches, looking ahead to real problems when n is much larger (n = 1000 is a very moderate size in scientific computing). The truth is that direct use of the determinant formula for 1000 equations would be a total disaster. E would use the million numbers on the left sides correctly, but not efficiently. We will find that formula (Cramer's Rule) in Chapter 4, but we want a good method to solve] 1000 equations in Chapter 1. That good method is Gaussian Elimination. This is the algorithm that is constantly used to solve large systems of equations. From the examples in a textbook (n = 3 is close to the upper limit on the patience of the author and reader) you might not see much difference. Equations (2) and (4) used essentially the same steps to find y = 2. Certain x came faster by the back-substitution in equation (3) than the ratio in (5). For larger n there is absolutely no question. Elimination wins (and this is even the best way to compute determinants). - The idea of elimination is deceptively simple—you will master it after a few exam" ples. It will become the basis for half of this book, simplifying a matrix so that we can understand it. Together with the mechanies of the algorithm, we want to explain four deeper aspects in this chapter. They are: gar mM 1. Linear eguations lead to geometry of planes. It is not easy to visualize a nine- dimensional plane in ten-dimensional space. It is harder to see ten of those planes, intersecting at the solution to ten equations—but somehow this is almost possible. Our example has two lines in Figure 1.1, meeting at the point (x,y) = (—1,2). Linear algebra moves that picture into ten dimensions, where the intuition has to imagine the geometry (and gets it right). 2. We move to matrix notation, writing the n unknowns as a vector x and the n equations as 4x = b. We multiply A by “elimination matrices” to reach an up- per triangular matrix U. Those steps factor A into L times U, where L is lower y y y E x42y=3 x+2y=3 x+2y=3 Bed aa 7 ENA + E 4x +5y=6 4x+8y=6 4x +8y =12 olution (x,y) = (1,2) Parallel: No solution Whole line of solutions Figure £.1 The example has one solution (Singular cases have none or too many. us 1.2 The Geometry of Linear Equations 3 triangular. 'will write down A and its factors for our example, and explain them at the right time: 45 Ho —3 First we have to introduce matrices and vectors and the rules for multiplication. Every matrix has a transpose A?. This matrix has an inverse A”1. 3. Inmostcaseselimination goes forward without difficulties. The matrix has an inverse and the system Ax = b has one solution. In exceptional cases the method will break down-—either the equations were written in the wrong order, which is easily fixed by exchanging them, or the equations don't have a unique solution. That singular case will appear if 8 replaces 5 in our example: Factorization A= Ê :| = lá '| Ê à] =LitimesU. (6) Singular case lx + 2y = 3 7 Two parallel lines 4 + 8y = 6. (7) Elimination still innocently subtracts 4 times the first equation from the second. But look at the result! (equation 2) — d(equation 1) 0=-—6. This singular case has no solution. Other singular cases have infinitely many solu- tions. (Change 6 to 12 in the example, and elimination will lead to O = 0. Now y can have any value.) When elimination breaks down, we want to find every possible solution. 4. Weneedarough count of the number of elimination steps required to solve a system of size n. The computing cost often determines the accuracy in the model. A hundred equations require a third of a million steps (multiplications and sabtractions). The computer can do those quickly, but not many trillions. And already after a million steps, roundoff error could be significant. (Some problems are sensitive; others are not.) Without trying for full detail, we wantto see large systems that arise in practice, and how they are actually solved, The final result of this chapter will be an elimination algorithm that is about as efficient as possible. It is essentially the algorithm that is in constant use in a tremendous variety of applications. And at the same time, understanding it in terms of matrices—the coefficient matrix A, the matrices E for elimination and P for row exchanges, and the final factors L and U —is an essential foundation for the theory. [hope you will enjoy this book and this course. 1.2 THE GEOMETRY OF LINEAR EQUATIONS The way to understand this subject is by example. We begin with two extremely humble equations, recognizing that you could solve them without a course in linear algebra. Nevertheless I hope you will give Gauss a chance: 2x-y=1 x+y=5. We can look at that system by rows or by columns. We want to see them both. 1.2 The Geometry of Linear Eguations 5 2u+v +w = 5 (sloping plane) 4u — 6u = —2 (vertical plane) ] (1,1,2) = point of intersection with third plane = solution line of intersection: first two planes Figure 1.3 The row picture: three intersecting planes from three linear equations. the center pointu = 0, v = 0, w = O. Changing the right side moves the plane parallel to itself, and the plane 2u + v + w = O goes through the origin. The second plane is 4u — Gu = —2. Itis drawn vertically, because ww can take any value. The coefficient of w is zero, but this remains a plane in 3-space. (The equation 4u = 3, or even the extreme case u = 0, would still describe a plane.) The figure shows the intersection of the second plane with the first. That intersection is a line. Jn three dimensions a line requires two eguations; in n dimensions it will require n — 1. Finally the third plane intersects this line in a point. The plane (not drawn) represents the third equation —2u + 7v + 2w =9, anditcrosses the lineatu = L,v=|,w=2, That triple intersection point (1, 1, 2) solves the linear system. How does this row picture extend into n dimensions? The n equations will contain n unknowns. The first equation still determines a “plane” Tt is no longer a two-dimensional plane in 3-space; somehow it has “dimension”n — 1. It must be flat and extremely thin within n-dimensional space, although it would look solid to us. TÉ time is the fourth dimension, then the plane + = 0 cuts through four-dimensional space and produces the three-dimensional universe we live in-(or rather, the universe as it was at! = 0). Another plane is z = 0, which is also three-dimensional; it is the ordinary x-y plane taken over all time. Those three-dimensional planes will intersect! They share the ordinary x-y plane at t = O. We are down to two dimensions, and the next plane leaves a line. Finally a fourth plane leaves a single point. It is the intersection point of 4 planes in 4 dimensions, and it solves the 4 underlying equations. I will be in trouble if that example from relativity goes any further. The point is that linear algebra can operate with any number of equations. The first equation produces an (n — 1)-dimensional plane in n dimensions. The second plane intersects it (we hope) in pter 1 Matrices and Gaussian Elimination a smaller set of “dimension n — 2” Assuming all goes well, every new plane (every new equation) reduces the dimension by one. At the end, when all n planes are accounted for, the intersection has dimension zero. It is a point, it lies on all the planes, and its coordinates satisfy all n equations, It is the solution! Column Vectors and Linear Combinations We tum to the columns. This time the vector equation (the same equation as (1)) is 2 1 i 5 Column form ul 4|+o|-6| +w |0|=|--2) =b. (2) —2, 7 2 9 Those are three-dimensional column vectors. The vector b is identified with the point whose coordinates are 5, —2,9. Every point in three-dimensional space is matched to a vector, and vice versa. That was the idea of Descartes, who turned geometry into algebra by working with the coordinates of the point. We can write the vector in a column, or we can list its components as b = (5, —2,9), or we can represent it geometrically by an atrow from the origin. You can choose the arrow, or the point, or the three numbers. In six dimensions it is probably easiest to choose the six numbers. We use parentheses and commas when the components ate listed horizontally, and square brackets (with no commas) when a column vector is printed vertically. What really matters is addition of vectors and multiplication by a scalar (a number). In Figure 1.4a you see a vector addition, component by component: 5 0 (o) 5 Vector addition O +i-2) + 10] = |-21. 0 0 9 9 5 E = linear combination equals b EE 2column 3) ENC Duma columns 1+2 (a) Add vectors along axes (b) Add columns 1+4+24-(343) Figure 1.4 The column picture: linear combination of columns equals b. 1 Matrices and Gaussian Elimination third plane is not parallei to the other planes, but it is parallel to their line of intersection. This corresponds to a singular system with b = (2,5, 6): utv+ w=2 No solution, as in Figure 1.5b Zu +3w=5 (3) Ju +v+ Sw =6. The first two Jeft sides add up to the third. On the right side that fails: 2 +5 *£ 6. Equation 1 plus equation 2 minus equation 3 is the impossible statement O = 1. Thus the equations are inconsistent, as Gaussian elimination will systematically discover. Another singular system, close to this one, has an infinity of solutions. When the 6 in the last equation becomes 7, the three equations combine to give O = 0, Now the third equation is the sum of the first two. In that case the three planes have a whole line in common (Figure 1.5c). Changing the right sides will move the planes in Figure 1.5b parallel to themselves, and for b = (2,5, 7) the figure is suddenly different. The lowest plane moved up to meet the others, and there is a line of solutions. Problem 1.5c is still singular, but now it suffers from too many solutions instead of too few. The extreme case is three parallel planes. For most right sides there is no solution (Figure 1.5d). For special right sides (like b = (0,0,0)!) there is a whole plane of solutions—because the three paraliel planes move over to become the same. What happens to the columa picture when the system is singular? Ithas to go wrong; the question is how. There are still three columns on the left side of the equations, and we try to combine them to produce b. Stay with equation (3): Singular case: Column picture 1 1 1 Three columns in the same plane u|2l +02 |0| +wij3j =b. (4) Solvable only for b in that plane 3 1 44 For b = (2,5, 7) this was possible; for b = (2,5, 6) it was not. The reason is that those three columns lie in a plane. Then every combination is also in the plane (which goes through the origin). If the vector b is not in that plane, no solution is possible (Figure 1.6). That is by far the most likely event; a singular system generally has no solution. But | [3 columns | in a plane t 3 columns | in a plane ) | Figure 1.6 Singular cases: b outside or inside the plane with all three columns. (a) no solution 1.2 The Geometry of Linear Equations 9 there is a chance that b does lie in the plane of the colupans. In that case there are too many solutions; the three columns can be combined in infinitely many ways to produce b. That column picture in Figure 1.6b corresponds to the row picture in Figure 1.5c. How do we know that the three columns lie in the same plane? One answer is to find a combination of the columns that adds to zero. After some calculation, itis u = 3, v = —1,w = —2, Three times column 1 equals column 2 plus twice column 3. Column 1 is in the plane of columns 2 and 3. Only two columns are independent. The vector b = (2,5,7) is in that plane of the columns—it is column 1 plus column 3—so (1,0,1) is a solution. We can add any multiple of the combination (3, —1, —2) that gives b = 0. So there is a whole line of solutions-—as we know from the row picture. The truth is that we knew the columns would combine to give zero, because the rows did. That is a fact of mathematics, not of computation—and it remains true in dimension n. If the n planes have no point in common, or infinitely many points, then the n columns lie in the same plane. TÉ the row picture breaks down, so does the column picture. That brings out the difference between Chapter 1 and Chapter 2, This chapter studies the most important probtem-—the nonsingular case-—where there is one solution and it has to be found. Chapter 2 studies the general case, where there may be many solutions or none. In both cases we cannot continue without a decent notation (Gnatrix notation) and a decent algoritim (elimination). After the exercises, we start with elimination. Problem Set 1.2 1. For the equations x + y = 4,2x — 2y = 4, draw the row picture (two intersecting lines) and the column picture (combination of two columns equal to the column vector (4, 4) on the right side). 2. Solve to find a combination of the columns that equals b: u-v—w= bh “ Triangular system v+tw=b w = ba. 3. (Recommended) Describe the intersection of the three planes u+v+Hw+z = 6 andutw+z=4andu-tHy = 2 (allinfou-dimensional space). Is it a line or a point or an empty set? What is the intersection if the fourth plane u = —1 is included? Find a fourth equation that leaves us with no solution. 4. Sketch these three lines and decide if the equations are solvable: x+2y=2 3by 2 system x— y=2 y=5 What happens if all right-hand sides are zero? Is there any nonzero choice of right- hand sides that allows the three lines to intersect at the same point? 5. Find two points on the line of intersection of the three planes + = O and z = O and x+y+z+t = 1 in four-dimensional space. 1.3 An Example of Gaussian Elimination 1 17. The first of these equations plus the second equals the third: x+ y+ 7=2 x+2y+ 2=3 2x+3y +27 =5. The first two planes meet along a line. The third plane contains that line, because if x, y,z satisfy the first two equations then they also... The equations have infinitely many solutions (the whole line L). Find three solutions. 18. Move the third plane in Problem 17 to a parallel plane 2x + 3y + 27 = 9. Now the three equations have no solution—wAy not? The first two planes meet along the line L, but the third plane doesn't that line. 19, Tn Problem 17 the columns are (1, 1, 2) and (1,2, 3) and (1, 1, 2). This is a “singular case” because the third column is - Find two combinations of the columns that give b = (2,3, 5). This is only possible forb= (4,6, )Jifc= 20, Normally 4 “planes” in four-dimensional space meet at a - Normally 4 col- umn vectors in four-dimensional space can combine to produce b. What combination of (1,0,0,0), (1, 1,0,0), (1,1,1,0), (1. 1,1, 1) produces b = (3,3,3,2)? What 4 eguations for x, y, z, t are you solving? 21. When equation 1 is added to equation 2, which of these are changed: the planes in the row picture, the column picture, the coefficient matrix, the solution? 22. If (a, b) is a multiple of (c, d) with abed + 0, show that (a, c) is a multiple of (b, d). This is surprisingly important: call it a challenge question. You could use numbers first to see how a, b, c, and d are related. The question will lead to: FA= [: à] has dependent rows then it has dependent columns. 23. In these equations, the third column (multiplying w) is the same as the right side b. The column form of the equations immediately gives what solution for (4, v, w)? GU+Tv+B8w=8 Au + Sv+ PM =9 Qu —- 2 + Tw=T. 1.3 ANEXAMPLE OF GAUSSIAN ELIMINATION The way to understand elimination is by example. We begin in three dimensions: U+ v+ w= 5 Original system 4u — bu =—2 (1) QU ++ = 9. The problem is to find the unknown values of u, v, and w, and we shall apply Gaussian elimination. (Gauss is recognized as the greatest of all mathematicians, but certainly not because of this invention, which probably took him ten minutes. Ironically, rt Matrices and Gaussian Elimination it is the most frequently used of all the ideas that bear his name.) The method starts by subtracting multiples of the first equation from the other equations. The goal is to eliminate u from the last two equations. This requires that we (a) subtract 2 times the first equation from the second (b) subtract —1 times the first equation from the third. M+ vt w= 5 Equivalent system — 82 —2wy = -—12 (2) 8Bvr+Iw= 14, The coefficient 2 is the first pivot. Elimination is constantly dividing the pivot into the numbers underneath it, to find out the right multipliers. The pivot for the second stage of elimination is —8. We now ignore the first equation. A multiple of the second equation will be subtracted from the remaining equations (in this case there is only the third one) so as to eliminate v. We add the second equation to the third or, in other words, we (c) subtract —1 times the second equation from the third. The elimination process is now complete, at least in the “forward” direction: lU+ vt w= ís5 Triangular system — Bv —2w =-12 (3) lw= 2 This system is solved backward, bottom to top. The last equation gives w = 2. Substituting into the second equation, we find v = 1. Then the first equation gives u = 1, This process is called back-substitution. To repeat: Forward elimination produced the pivots 2, —8, 1. It subtracted multiples of each row from the rows beneath. It reached the “triangular” system (3), which is solved in reverse order: Substitute each newly computed value into the equations that are waiting. Remark One good way to write down the forward elimination steps is to include the right-hand side as an extra column. There is no need to copy u and v and w and = at every step, so we are left with the bare minimum: 2 1 1 5 2 1 1 5 2 1 1 5 4-6 0 2)>/0 -8 20 —12)->1]0 -8 2 12. -2 7 2 9 0 8 3 14 0-0 1 2 At the end is the triangular system, ready for back-substitution. You may prefer this arrangement, which guarantees that operations on the left-hand side of the equations are also done on the right-hand side—because both sides are there together. In a larger problem, forward elimination takes most of the effort. We use multiples of the first equation to produce zeros below the first pivot. Then the second column is cleared out below the second pivot. The forward step is finished when the system is triangular; equation n contains only the last unknown multiplied by the last pivot. 
