Skip to content
Textbook Guides

Solution Manual for Linear Algebra, 6th Edition

Solution Manual for Linear Algebra, 6th Edition offers a comprehensive guide to solving every question in your textbook, helping you master the material.

Preview 16 of 484 pages

See the first 16 pages exactly as they appear in the document.

  1. Page 1
    Read page 1 as text

     S OLUTIONS M ANUAL L INEAR A LGEBRA AND I TS A PPLICATIONS S IXTH E DITION David C. Lay University of Maryland–College Park Steven R. Lay Lee University Judi J. McDonald Washington State University

  2. Page 2
  3. Page 3
    Read page 3 as text

    Contents Introduction v Chapter 1 Linear Equations in Linear Algebra 1-1 1.1 Systems of Linear Equations 1-1 1.2 Row Reduction and Echelon Forms 1-8 1.3 Vector Equations 1-16 1.4 The Matrix Equation A x = b 1-25 1.5 Solution Sets of Linear Systems 1-33 1.6 Applications of Linear Systems 1-42 1.7 Linear Independence 1-51 1.8 Introduction to Linear Transformations 1-58 1.9 The Matrix of a Linear Transformation 1-65 1.10 Linear Models in Business, Science, and Engineering 1-71 Supplementary Exercises 1-80 Chapter 2 Matrix Algebra 2-1 2.1 Matrix Operations 2-1 2.2 The Inverse of a Matrix 2-7 2.3 Characterization of Invertible Matrices 2-15 2.4 Partitioned Matrices 2-23 2.5 Matrix Factorizations 2-32 2.6 The Leontief Input-Output Model 2-47 2.7 Applications to Computer Graphics 2-51 2.8 Subspaces of  n 2-58 2.9 Dimension and Rank 2-66 Supplementary Exercises 2-72 Chapter 3 Determinants 3-1 3.1 Introduction to Determinants 3-1 3.2 Properties of Determinants 3-8 3.3 Cramer’s Rule, Volume, and Linear Transformations 3-14 Supplementary Exercises 3-22 Chapter 4 Vector Spaces 4-1 4.1 Vector Spaces and Subspaces 4-1 4.2 Null Spaces, Column Spaces, Row Spaces, and Linear Transformations 4-7 4.3 Linearly Independent Sets; Bases 4-15 4.4 Coordinate Systems 4-23 4.5 The Dimension of a Vector Space 4-30 4.6 Change of Basis 4-36 4.7 Digital Signal Processing 4-40 4.8 Applications to Difference Equations 4-43 Supplementary Exercises 4-52 iii

  4. Page 4
    Read page 4 as text

    Chapter 5 Eigenvalues and Eigenvectors 5-1 5.1 Eigenvalues and Eigenvectors 5-1 5.2 The Characteristic Equation 5-10 5.3 Diagonalization 5-15 5.4 Eigenvalues and Linear Transformations 5-29 5.5 Complex Eigenvalues 5-35 5.6 Discrete Dynamical Systems 5-43 5.7 Applications to Differential Equations 5-49 5.8 Iterative Estimates for Eigenvalues 5-59 5.9 Applications to Markov Chains 5-67 Supplementary Exercises 5-75 Chapter 6 Orthogonality and Least Squares 6-1 6.1 Inner Product, Length, and Orthogonality 6-1 6.2 Orthogonal Sets 6-5 6.3 Orthogonal Projections 6-10 6.4 The Gram-Schmidt Process 6-18 6.5 Least-Squares Problems 6-24 6.6 Machine Learning and Linear Models 6-29 6.7 Inner Product Spaces 6-34 6.8 Applications of Inner Product Spaces 6-38 Supplementary Exercises 6-43 Chapter 7 Symmetric Matrices and Quadratic Forms 7-1 7.1 Diagonalization of Symmetric Matrices 7-1 7.2 Quadratic Forms 7-14 7.3 Constrained Optimization 7-22 7.4 The Singular Value Decomposition 7-27 7.5 Applications to Image Processing and Statistics 7-37 Supplementary Exercises 7-40 Chapter 8 The Geometry of Vector Spaces 8-1 8.1 Affine Combinations 8-1 8.2 Affine Independence 8-5 8.3 Convex Combinations 8-10 8.4 Hyperplanes 8-15 8.5 Polytopes 8-19 8.6 Curves and Surfaces 8-22 Supplementary Exercises 8-27 Chapter 9 Optimization 9-1 9.1 Matrix Games 9-1 9.2 Linear Programming—Geometric Method 9-7 9.3 Linear Programming—Simplex Method 9-11 9.4 Duality 9-15 Supplementary Exercises 9-22 iv

  5. Page 5
    Read page 5 as text

    Introduction I fell in love with linear algebra when I was an undergraduate student and it has remained a central part of my life since that time. It is an interesting and beautiful subject, with a broad range of applications. In recent years, I consistently hear from industry partners about how much the high-tech industry appreciates individuals having a strong foundation in both technical and theoretical aspects of linear algebra. I hope you will enjoy teaching this course as much as I do. You are also welcome to email me at LLinearAlgebra@gmail.com any time you have comments and suggestions, or just want to talk about linear algebra. There are many ways in which modern technology can support (or hinder) your student’s learning. The interactive figures from the electronic textbook can be used in classroom demonstrations to bring linear algebraic concepts alive and demonstrate numerous examples with the push of a button. Take time to explore with the interactive figures and show your students how to use technology to find key definitions and theorems quickly in the electronic textbook. If your course uses MyLab for homework, there are several things to be aware of. First, for some exercises, your students will enter only a final answer. To get to that answer, they may have half a page or more of calculations. Encourage them to keep a notebook with the exercise statement, worked solutions, and summary notes about what they learned while solving an exercise. As an instructor, you can choose many settings in the program. You can set how many tries students are allowed to solve each question. Most exercises let the student have three tries before MyLab either records an incorrect answer or offers the student a similar question. In my experience, persistence pays off – if students are allowed to continue to work similar exercises, mastery of the skill will result. I have also found that the “View an Example” and “Help Me Solve It” tab help get students going again when they are stuck. At the end of each chapter, we have highlighted some of the projects that are available online, but moved away from updating the toolbox and the computer manuals. I find that when I am trying to code almost anything, I go to the help features for the program or open a search engine and enter some key words. There are still tips in the Student Study Guide about appropriate MATLAB code for various parts of the course. Technology also provides students with easy access to a wealth of videos on linear algebra and solutions for some of the exercises. Please refrain from posting portions of this ’s Solution Manual online, as by doing so you are giving other instructors’ students solutions to the exercises. Some of the open-access online videos are amazing. Others contain errors or introduce the material in a different order from how it is covered in this text, leading to confusion. I try to talk to my students about using technology to learn effectively without it becoming a crutch that leaves them with perfect homework and failed exams. The ’s Solution Manual contains detailed solutions for all the exercises, as well as advice on the exercises themselves. I am interested to hear from you at LLinearAlgebra@gmail.com as to what types of material you would like to use in your course, additional topics you would like to see covered, any typos you find, or just to talk about my favorite subject – linear algebra. —Judi J. McDonald v

  6. Page 6
  7. Page 7
    Read page 7 as text

    1-1 1.1 - Systems Of Linear Equations Notes : The key exercises are 7 (or 11 or 12), 23–26, and 35. For brevity, the symbols R1, R2,…, stand for row 1 (or equation 1), row 2 (or equation 2), and so on. Additional notes are at the end of the section. In Exercises 15–18, students are asked to check their answers to Exercises 11–14; checking that solutions are correct, or at least reasonable, is an important skill in the high-tech industry. 1 . 1 2 1 2 5 7 2 7 5 x x x x + = − − = − 1 5 7 2 7 5     − − −   Replace R2 by R2 + (2)R1 and obtain: 1 2 2 5 7 3 9 x x x + = = 1 5 7 0 3 9       Scale R2 by 1/3: 1 2 2 5 7 3 x x x + = = 1 5 7 0 1 3       Replace R1 by R1 + (–5)R2: 1 2 8 3 x x = − = 1 0 8 0 1 3 −       The solution is ( x 1 , x 2 ) = (–8, 3), or simply (–8, 3). 2 . 1 2 1 2 2 4 4 5 7 11 x x x x + = − + = 2 4 4 5 7 11 −       Scale R1 by 1/2 and obtain: 1 2 1 2 2 2 5 7 11 x x x x + = − + = 1 2 2 5 7 11 −       Replace R2 by R2 + (–5)R1: 1 2 2 2 2 3 21 x x x + = − − = 1 2 2 0 3 21 −     −   Scale R2 by –1/3: 1 2 2 2 2 7 x x x + = − = − 1 2 2 0 1 7 −     −   Replace R1 by R1 + (–2)R2: 1 2 12 7 x x = = − 1 0 12 0 1 7     −   The solution is ( x 1 , x 2 ) = (12, –7), or simply (12, –7). 3 . The point of intersection satisfies the system of two linear equations:

  8. Page 8
    Read page 8 as text

    1-2 Chapter 1  Linear Equations in Linear Algebra 1 2 1 2 5 7 2 2 x x x x + = − = − 1 5 7 1 2 2     − −   Replace R2 by R2 + (–1)R1 and obtain: 1 2 2 5 7 7 9 x x x + = − = − 1 5 7 0 7 9     − −   Scale R2 by –1/7: 1 2 2 5 7 9/7 x x x + = = 1 5 7 0 1 9/7       Replace R1 by R1 + (–5)R2: 1 2 4/7 9/7 x x = = 1 0 4/7 0 1 9/7       The point of intersection is ( x 1 , x 2 ) = (4/7, 9/7). 4 . The point of intersection satisfies the system of two linear equations: 1 2 1 2 5 1 3 7 5 x x x x − = − = 1 5 1 3 7 5 −     −   Replace R2 by R2 + (–3)R1 and obtain: 1 2 2 5 1 8 2 x x x − = = 1 5 1 0 8 2 −       Scale R2 by 1/8: 1 2 2 5 1 1/4 x x x − = = 1 5 1 0 1 1/4 −       Replace R1 by R1 + (5)R2: 1 2 9/4 1/4 x x = = 1 0 9/4 0 1 1/4       The point of intersection is ( x 1 , x 2 ) = (9/4, 1/4). 5 . The system is already in “triangular” form. The fourth equation is x 4 = –5, and the other equations do not contain the variable x 4 . The next two steps should be to use the variable x 3 in the third equation to eliminate that variable from the first two equations. In matrix notation, that means to replace R2 by its sum with 3 times R3, and then replace R1 by its sum with –5 times R3. 6 . One more step will put the system in triangular form. Replace R4 by its sum with –3 times R3, which produces 1 6 4 0 1 0 2 7 0 4 0 0 1 2 3 0 0 0 5 15 − −     −     −   −   . After that, the next step is to scale the fourth row by –1/5. 7 . Ordinarily, the next step would be to interchange R3 and R4, to put a 1 in the third row and third column. But in this case, the third row of the augmented matrix corresponds to the equation 1 2 3 0 0 0 1 x x x + + = , or simply, 0 = 1. A system containing this condition has no solution. Further row operations are unnecessary once an equation such as 0 = 1 is evident. The solution set is empty. 8 . The standard row operations begin by multiplying R3 by ½. Next, replace R2 by R2+ (–7)R3 and R1 by R1+ (–2)R3. Finally, replace R1 by R1+ (–1)R2.

  9. Page 9
    Read page 9 as text

    1.1 - Systems Of Linear Equations 1-3 1 1 2 0 1 1 2 0 1 1 0 2 1 0 0 5 0 1 7 0 ~ 0 1 7 0 ~ 0 1 0 7 ~ 0 1 0 7 0 0 2 2 0 0 1 1 0 0 1 1 0 0 1 1 −                                 − − − −         The solution set contains one solution: ( ) 5, 7, 1 − − . 9 . The system has already been reduced to triangular form. Notice that in the last row, the only nonzero entry is in the last column. The corresponding system of equations does not have a solution. The solution set is the empty set. 10 . The system has already been reduced to triangular form. Use the 1 in the fourth row to change the –4 and 3 above it to zeros. That is, replace R2 by R2 + (4)R4 and replace R1 by R1 + (–3)R4. For the final step, replace R1 by R1 + (2)R2. 1 2 0 3 0 1 2 0 0 0 1 0 0 0 0 0 1 0 4 0 0 1 0 0 0 0 1 0 0 0 ~ ~ 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 − −             −                         The solution set contains one solution: (0, 0, 0, 0). 11 . First, swap R1 and R2. Then replace R3 by R3 + (–3)R1. To reach triangular form, replace R3 by R3 + (2)R2. Next, divide the last row by 4. 0 1 4 4 1 3 3 2 1 3 3 2 1 3 3 2 1 3 3 2 1 3 3 2 ~ 0 1 4 4 ~ 0 1 4 4 ~ 0 1 4 4 ~ 0 1 4 4 3 7 5 6 3 7 5 6 0 2 4 12 0 0 4 4 0 0 1 1 − − − − −                     − − − − −                     − −           To continue, subtract (4)R3 from R2 and subtract (3)R3 from R1. Finally, replace R1 by R1+(–3)R2. 1 3 0 5 1 0 0 19 0 1 0 8 0 1 0 8 0 0 1 1 0 0 1 1 −         − −              The solution set contains one solution: (19, –8, 1). 12 . Replace R2 by R2 + (–3)R1 and replace R3 by R3 + (4)R1. To avoid fractions, it is easier to replace R3 by (–1/6)R3, then interchange R2 and R3. 1 3 4 4 1 3 4 4 1 3 4 4 1 3 4 4 3 7 7 8 ~ 0 2 5 4 ~ 0 2 5 4 ~ 0 1 3 2 4 6 2 4 0 6 18 12 0 1 3 2 0 2 5 4 − − − − − − − −                 − − − − −                 − − − − −         Next, subtract 2R2 from R3 to complete the forward phase. Replacing R1 by R1+(–4)R3 and R2 by R2+(3)R3 creates zeros above the 1 in the third column. Finally, replace R1 by R1+(3)R2 1 3 4 4 1 3 0 4 1 0 0 2 0 1 3 2 0 1 0 2 ~ 0 1 0 2 0 0 1 0 0 0 1 0 0 0 1 0 − − − −             −                    The solution set contains one solution: (2, 2, 0).

  10. Page 10
    Read page 10 as text

    1-4 Chapter 1  Linear Equations in Linear Algebra 13 . Replace R2 by R2 + (–2)R1. Then interchange R2 and R3. Next replace R3 by R3 + (–2)R2. Then divide R3 by 5. Finally, replace R1 by R1 + (–2)R3. 1 0 3 8 1 0 3 8 1 0 3 8 1 0 3 8 2 2 9 7 ~ 0 2 15 9 ~ 0 1 5 2 ~ 0 1 5 2 0 1 5 2 0 1 5 2 0 2 15 9 0 0 5 5 − − − −                 − − −                 − − − −         1 0 3 8 ~ 0 1 5 2 0 0 1 1 −     −    −    1 0 0 5 ~ 0 1 0 3 0 0 1 1        −    . The solution is (5, 3, –1). 14 . Replace R2 by R2 + R1. Then interchange R2 and R3. Next replace R3 by R3 + 2R2. Then divide R3 by 7. Next replace R2 by R2 + (–1)R3. Finally, replace R1 by R1 + 3R2. 1 3 0 5 1 3 0 5 1 3 0 5 1 3 0 5 1 1 5 2 ~ 0 2 5 7 ~ 0 1 1 0 ~ 0 1 1 0 0 1 1 0 0 1 1 0 0 2 5 7 0 0 7 7 − − − −                 − −                 −         1 3 0 5 ~ 0 1 1 0 0 0 1 1 −           1 3 0 5 1 0 0 2 ~ 0 1 0 1 ~ 0 1 0 1 0 0 1 1 0 0 1 1 −         − −             . The solution is (2, –1, 1). 15 . The solution found in Exercise 11 is (19, –8, 1). Substituting into the equations 2 3 1 2 3 1 2 3 4 4 3 3 2 3 7 5 6 x x x x x x x x + = − + + = − + + = results in ( 8) 4(1) 4 19 3( 8) 3(1) 2 3(19) 7( 8) 5(1) 6 − + = − + − + = − + − + = , which establishes that the solution found to Exercise 11 is indeed correct. 16 . The solution found in Exercise 12 is (2, 2, 0). Substituting into the equations 1 2 3 1 2 3 1 2 3 3 4 4 3 7 7 8 4 6 2 4 x x x x x x x x x − + = − − + = − − + + = results in 2 3(2) 4(0) 4 3(2) 7(2) 7(0) 8 4(2) 6(2) 2(0) 4 − + = − − + = − − + + = , which establishes that the answer found in Exercise 12 is indeed correct.

  11. Page 11
    Read page 11 as text

    1.1 - Systems Of Linear Equations 1-5 17 . The solution found in Exercise 13 is (5, 3, –1). Substituting into the equations 1 3 1 2 3 2 3 3 8 2 2 9 7 5 2 x x x x x x x − = + + = + = − results in 5 3( 1) 8 2(5) 2(3) 9( 1) 7 (3) 5( 1) 2 − − = + + − = + − = − , which establishes that the answer found in Exercise 13 is indeed correct. 18 . The solution found in Exercise 14 is (2, (–1, 1). Substituting into the equations 1 2 1 2 3 2 3 3 5 5 2 0 x x x x x x x − = − + + = + = results in 2 3( 1) 5 (2) ( 1) 5(1) 2 ( 1) (1) 0 − − = − + − + = − + = , which establishes that the answer found in Exercise 14 is indeed correct. 19 . First, replace R4 by R4 + (–3)R1, then replace R3 by R3 + (2)R2, and finally replace R4 by R4 + (3)R3. 1 0 3 0 2 1 0 3 0 2 0 1 0 3 3 0 1 0 3 3 ~ 0 2 3 2 1 0 2 3 2 1 3 0 0 7 5 0 0 9 7 11         − −         − −     − − −     1 0 3 0 2 1 0 3 0 2 0 1 0 3 3 0 1 0 3 3 ~ ~ 0 0 3 4 7 0 0 3 4 7 0 0 9 7 11 0 0 0 5 10         − −         − −     − − −     The resulting triangular system indicates that a solution exists. In fact, using the argument from Example 2, one can see that the solution is unique. 20 . First replace R4 by R4 + (2)R1 and replace R4 by R4 + (–3/2)R2. (One could also scale R2 before adding to R4, but the arithmetic is rather easy keeping R2 unchanged.) Finally, replace R4 by R4 + R3. 1 0 0 2 3 1 0 0 2 3 0 2 2 0 0 0 2 2 0 0 ~ 0 0 1 3 1 0 0 1 3 1 2 3 2 1 5 0 3 2 3 1 − − − −                     − − −     1 0 0 2 3 1 0 0 2 3 0 2 2 0 0 0 2 2 0 0 ~ ~ 0 0 1 3 1 0 0 1 3 1 0 0 1 3 1 0 0 0 0 0 − − − −                     − − −     The system is now in triangular form and has a solution. The next section discusses how to continue with this type of system. 21 . Row reduce the augmented matrix corresponding to the given system of three equations:

  12. Page 12
    Read page 12 as text

    1-6 Chapter 1  Linear Equations in Linear Algebra 1 4 1 1 4 1 1 4 1 2 1 3 ~ 0 7 5 ~ 0 7 5 1 3 4 0 7 5 0 0 0 − − −             − − − −             − − −       The system is consistent, and using the argument from Example 2, there is only one solution. So the three lines have only one point in common. 22 . Row reduce the augmented matrix corresponding to the given system of three equations: 1 2 1 4 1 2 1 4 1 2 1 4 0 1 1 1 ~ 0 1 1 1 ~ 0 1 1 1 1 3 0 0 0 1 1 4 0 0 0 5             − − −             − − −       The third equation, 0 = –5, shows that the system is inconsistent, so the three planes have no point in common. 23 . 1 4 1 4 ~ 3 6 8 0 6 3 4 h h h         − −     Write c for 6 – 3 h . If c = 0, that is, if h = 2, then the system has no solution, because 0 cannot equal –4. Otherwise, when h ≠ 2, the system has a solution. 24 . 1 3 1 3 ~ . 2 4 6 0 4 2 0 h h h − −         − +     Write c for 4 + 2 h . Then the second equation cx 2 = 0 has a solution for every value of c. So the system is consistent for all h . 25 . 1 3 2 1 3 2 ~ . 4 8 0 12 0 h h − −         − +     Write c for h + 12. Then the second equation cx 2 = 0 has a solution for every value of c . So the system is consistent for all h . 26 . 2 3 2 3 ~ . 6 9 5 0 0 5 3 h h h − −         − +     The system is consistent if and only if 5 + 3 h = 0, that is, if and only if h = –5/3. 27 . True. See the remarks following the box titled “Elementary Row Operations” . 28 . True. See the box preceding the subsection titled “Existence and Uniqueness Questions” . 29 . False. A 5 × 6 matrix has five rows. 30 . False. The definition of row equivalent requires that there exist a sequence of row operations that transforms one matrix into the other. 31 . False. The description given applies to a single solution. The solution set consists of all possible solutions. Only in special cases does the solution set consist of exactly one solution. Mark a statement True only if the statement is always true. 32 . False. By definition, an inconsistent system has no solution. 33 . True. See the box before Example 2. 34 . True. This definition of equivalent systems is in the second paragraph after equation (2) in Section 1.1 the text.

  13. Page 13
    Read page 13 as text

    1.1 - Systems Of Linear Equations 1-7 35 . 1 4 7 1 4 7 1 4 7 0 3 5 ~ 0 3 5 ~ 0 3 5 2 5 9 0 3 5 2 0 0 0 2 g g g h h h k k g k g h − − −             − − −             − − − + + +       Let b denote the number k + 2 g + h . Then the third equation represented by the augmented matrix above is 0 = b . This equation is possible if and only if b is zero. So the original system has a solution if and only if k + 2 g + h = 0. 36 . A basic principle of this section is that row operations do not affect the solution set of a linear system. Begin with a simple augmented matrix for which the solution is obviously (–2, 1, 0), and then perform any elementary row operations to produce other augmented matrices. Here are three examples. The fact that they are all row equivalent proves that they all have the solution set given by (–2, 1, 0). 1 0 0 2 1 0 0 2 1 0 0 2 0 1 0 1 ~ 2 1 0 3 ~ 2 1 0 3 0 0 1 0 0 0 1 0 2 0 1 4 − − −             − −             −       37 . Study the augmented matrix for the given system, replacing R2 by R2 + (– c )R1: 1 3 1 3 ~ 0 3 f f c d g d c g cf         − −     This shows that shows d – 3 c must be nonzero, since f and g are arbitrary. Otherwise, for some choices of f and g the second row would correspond to an equation of the form 0 = b , where b is nonzero. Thus d ≠ 3 c . 38 . Row reduce the augmented matrix for the given system. Scale the first row by 1/ a , which is possible since a is nonzero. Then replace R2 by R2 + (– c )R1. 1 / / 1 / / ~ ~ 0 ( / ) ( / ) a b f b a f a b a f a c d g c d g d c b a g c f a             − −       The quantity d – c ( b / a ) must be nonzero, in order for the system to be consistent when the quantity g – c ( f / a ) is nonzero (which can certainly happen). The condition that d – c ( b / a ) ≠ 0 can also be written as ad – bc ≠ 0, or ad ≠ bc . 39 . Swap R1 and R2; swap R1 and R2. 40 . Multiply R2 by –1/2; multiply R2 by –2. 41 . Replace R3 by R3 + (–4)R1; replace R3 by R3 + (4)R1. 42 . Replace R3 by R3 + (3)R2; replace R3 by R3 + (–3)R2. 43 . The first equation was given. The others are: 2 1 3 2 1 3 ( 20 40 )/4, or 4 60 T T T T T T = + + + − − = 3 4 2 3 4 2 ( 40 30)/4, or 4 70 T T T T T T = + + + − − = 4 1 3 4 1 3 (10 30)/4, or 4 40 T T T T T T = + + + − − = Rearranging,

  14. Page 14
    Read page 14 as text

    1-8 Chapter 1  Linear Equations in Linear Algebra 1 2 4 1 2 3 2 3 4 1 3 4 4 30 4 60 4 70 4 40 T T T T T T T T T T T T − − = − + − = − + − = − − + = 44 . Begin by interchanging R1 and R4, then create zeros in the first column: 4 1 0 1 30 1 0 1 4 40 1 0 1 4 40 1 4 1 0 60 1 4 1 0 60 0 4 0 4 20 ~ ~ 0 1 4 1 70 0 1 4 1 70 0 1 4 1 70 1 0 1 4 40 4 1 0 1 30 0 1 4 15 190 − − − − − −             − − − − −             − − − − − −       − − − − − −       Scale R1 by –1 and R2 by 1/4, create zeros in the second column, and replace R4 by R4 + R3: 1 0 1 4 40 1 0 1 4 40 1 0 1 4 40 0 1 0 1 5 0 1 0 1 5 0 1 0 1 5 ~ ~ ~ 0 1 4 1 70 0 0 4 2 75 0 0 4 2 75 0 1 4 15 190 0 0 4 14 195 0 0 0 12 270 − − − − − −             − − −             − − − −       − − −       Scale R4 by 1/12, use R4 to create zeros in column 4, and then scale R3 by 1/4: 1 0 1 4 40 1 0 1 0 50 1 0 1 0 50 0 1 0 1 5 0 1 0 0 27.5 0 1 0 0 27.5 ~ ~ ~ 0 0 4 2 75 0 0 4 0 120 0 0 1 0 30 0 0 0 1 22.5 0 0 0 1 22.5 0 0 0 1 22.5 − −             −             −             The last step is to replace R1 by R1 + (–1)R3: 1 0 0 0 20.0 0 1 0 0 27.5 ~ . 0 0 1 0 30.0 0 0 0 1 22.5             The solution is (20, 27.5, 30, 22.5). Notes : The Study Guide includes a “Mathematical Note” about statements, “If … , then … .” This early in the course, students typically use single row operations to reduce a matrix. As a result, even the small grid for Exercise 44 leads to about 25 multiplications or additions (not counting operations with zero). This exercise should give students an appreciation for matrix programs such as MATLAB. Exercise 14 in Section 1.10 returns to this problem and states the solution in case students have not already solved the system of equations. Exercise 31 in Section 2.5 uses this same type of problem in connection with an LU factorization. For instructors who wish to use technology in the course, the Study Guide provides boxed MATLAB notes at the ends of many sections. Parallel notes for Maple, Mathematica, and some calculators appear in separate appendices at the end of the Study Guide . 1.2 - Row Reduction and Echelon Forms Notes : The key exercises are 1–24 and 35–40. (Students should work at least four or five from Exercises 7–14, in preparation for Section 1.5.)

  15. Page 15
    Read page 15 as text

    1.2 - Row Reduction and Echelon Forms 1-9 1 . Reduced echelon form: a and c . Echelon form: b and d . Not echelon: none. 2 . Reduced echelon form: a and b . Echelon form: d . Not echelon: c . 3 . 1 2 3 4 1 2 3 4 1 2 3 4 4 5 6 7 ~ 0 3 6 9 ~ 0 1 2 3 6 7 8 9 0 5 10 15 0 5 10 15             − − −             − − − − − −       1 2 3 4 1 0 1 2 ~ 0 1 2 3 ~ 0 1 2 3 0 0 0 0 0 0 0 0 − −                     . Pivot cols 1 and 2. 1 2 3 4 4 5 6 7 6 7 8 9           4 . 1 3 5 7 1 3 5 7 1 3 5 7 1 3 5 7 3 5 7 9 ~ 0 4 8 12 ~ 0 1 2 3 ~ 0 1 2 3 5 7 9 1 0 8 16 34 0 8 16 34 0 0 0 10                 − − −                 − − − − − − −         1 3 5 7 1 3 5 0 1 0 1 0 ~ 0 1 2 3 ~ 0 1 2 0 ~ 0 1 2 0 0 0 0 1 0 0 0 1 0 0 0 1 −                               . Pivot cols 1, 2, and 4 1 3 5 7 3 5 7 9 5 7 9 1           5 . . * * 0 , , 0 0 0 0 0                       6 . * * 0 0 , 0 0 , 0 0 0 0 0 0 0 0                                   7 . 1 3 4 7 1 3 4 7 1 3 4 7 1 3 0 5 ~ ~ ~ 3 9 7 6 0 0 5 15 0 0 1 3 0 0 1 3 −                 − −         Corresponding system of equations: 1 2 3 3 5 3 x x x + = − = The basic variables (corresponding to the pivot positions) are x 1 and x 3 . The remaining variable x 2 is free. Solve for the basic variables in terms of the free variable. The general solution is 1 2 2 3 5 3 is free 3 x x x x = − −     =  8 . 1 4 0 7 1 4 0 7 1 4 0 7 1 0 0 5 ~ ~ ~ 2 7 0 11 0 1 0 3 0 1 0 3 0 1 0 3 −                 − −         Corresponding system of equations: 1 2 5 3 x x = − = The basic variables (corresponding to the pivot positions) are x 1 and x 2 . The remaining variable x 3 is free. Solve for the basic variables in terms of the free variable. In this particular problem, the basic variables do not depend on the value of the free variable.

  16. Page 16
    Read page 16 as text

    1-10 Chapter 1  Linear Equations in Linear Algebra General solution: 1 2 3 5 3 is free x x x = −   =    Note : A common error in Exercise 8 is to assume that x 3 is zero. To avoid this, identify the basic variables first. Any remaining variables are free . (This type of computation will arise in Chapter 5.) 9 . 0 1 6 5 1 2 7 4 1 0 5 6 ~ ~ 1 2 7 4 0 1 6 5 0 1 6 5 − − − −             − − − −       Corresponding system: 1 3 2 3 5 6 6 5 x x x x − = − = Basic variables: x 1 , x 2 ; free variable: x 3 . General solution: 1 3 2 3 3 6 5 5 6 is free x x x x x = +   = +    10 . 1 2 1 3 1 2 1 3 1 2 0 4 ~ ~ 3 6 2 2 0 0 1 7 0 0 1 7 − − − − − −             − − − −       Corresponding system: 1 2 3 2 4 7 x x x − = − = − Basic variables: x 1 , x 3 ; free variable: x 2 . General solution: 1 2 2 3 4 2 is free 7 x x x x = − +     = −  11 . 3 4 2 0 3 4 2 0 1 4/ 3 2/ 3 0 9 12 6 0 ~ 0 0 0 0 ~ 0 0 0 0 6 8 4 0 0 0 0 0 0 0 0 0 − − −             − −             − −       Corresponding system: 1 2 3 4 2 0 3 3 0 0 0 0 x x x − + = = = Basic variable: x 1 ; free variables x 2 , x 3 . General solution: 1 2 3 2 3 4 2 3 3 is free is free x x x x x  = −       12 . 1 7 0 6 5 1 7 0 6 5 1 7 0 6 5 0 0 1 2 3 ~ 0 0 1 2 3 ~ 0 0 1 2 3 1 7 4 2 7 0 0 4 8 12 0 0 0 0 0 − − −             − − − − − −             − − −      

468 more pages in the full document

Unlock it once and it stays in your library, ready to chat with or turn into flashcards.

Unlock the full document

More Mathematics study materials

See all

Study this document with CramX

Once it is in your library, every tool can work from it.