Estudiante
Páginas: 249 (62245 palabras)
Publicado: 22 de octubre de 2012
Introductory Linear Algebra: An Applied First Course, 8/E - Prentice Hall Catalog
Select a Discipline
Enter Zip Code
by Keyword
Welcome, Muhammad Yousaf | Sign out
Advanced Search
View Cart
ABOUT THIS PRODUCT
Description Table of Contents Features New To This Edition
PACKAGE OPTIONS
Introductory Linear Algebra: An Applied First Course, 8/E
Bernard Kolman, DrexelUniversity David R. Hill, Temple University
ISBN: 0-13-143740-2 Publisher: Prentice Hall Copyright: 2005 Format: Cloth; 768 pp
Valuepack(s)
View Larger Image RESOURCES
Our Price: $124.80 Status: Instock Published: 08/02/2004 Add to Cart
Student Instructor Course-Specific Discipline-Specific First Days Of Class
RELATED TITLES
Add to Exam Copy Bookbag
Print Product InformationInstructor Resource Center (Please scroll down to see all available instructor resources)
Help downloading Instructor Resources Instructor's Manual, 8/E by Dennis Kletzing © 2005 | Paper; 214 pages | ISBN: 0131437410 | Status: Instock Chapter 1--Instructor's Solutions Manual (.2 MB | pdf file | Type: Manuals/Guides)
Introductory Linear Algebra (Mathematics) Linear Algebra (Engineering: Mechanical)Chapter 2--Instructor's Solutions Manual (.1 MB | pdf file | Type: Manuals/Guides)
Chapter 3--Instructor's Solutions Manual (.1 MB | pdf file | Type: Manuals/Guides) More files available...
Copyright 2006 by Pearson Education - Legal Notice - Privacy Policy - Permissions Technical Support: Pearson 24/7
http://vig.prenhall.com/catalog/academic/product/0,1144,0131437402-IS,00.html7/7/20061:20:01 PM
Chapter 1
Linear Equations and Matrices
Section 1.1, p. 8
2. x = 1, y = 2, z = −2. 4. No solution. 6. x = 13 + 10z, y = −8 − 8z, z = any real number. 8. No solution. 10. x = 2, y = −1. 12. No solution. 14. x = −1, y = 2, z = −2. 16. (c) Yes. (d) Yes.
18. x = 2, y = 1, z = 0. 20. There is no such value of r. 22. Zero, infinitely many, zero. 24. 1.5 tons of regular and 2.5 tonsof special plastic. 26. 20 tons of 2-minute developer and a total of 40 tons of 6-minute and 9-minute developer. 28. $7000, $14,000, $3000. T.1. The same numbers sj satisfy the system when the pth equation is written in place of the qth equation and vice versa. T.2. If s1 , s2 , . . . , sn is a solution to (2), then the ith equation of (2) is satisfied: ai1 s1 +ai2 s2 +· · ·+ain sn = bi . Then forany r = 0, rai1 s1 + rai2 s2 + · · · + rain sn = rbi . Hence s1 , s2 , . . . , sn is a solution to the new system. Conversely, for any solution s1 , s2 , . . . , sn to the new system, rai1 s1 +· · ·+rain sn = rbi , and dividing both sides by nonzero r we see that s1 , . . . , sn must be a solution to the original linear system. T.3. If s1 , s2 , . . . , sn is a solution to (2), then the pth andqth equations are satisfied: ap1 s1 + · · · apn = bp aq1 s1 + · · · aqn = bq .
2 Thus, for any real number r, (ap1 + raq1 )s1 + · · · + (apn + raqn )sn = bp + rbq
Chapter 1
and so s1 , . . . , sn is a solution to the new system. Conversely, any solution to the new system is also a solution to the original system (2). T.4. Yes; x = 0, y = 0 is a solution for any values of a, b, c, and d.Section 1.2, p. 19
2. a = 3, b = 1, c = 8, d = −2. ⎡ ⎤ 5 −5 8 7 −7 2 9⎦. 4. (a) C + E = E + C = ⎣4 (b) Impossible. (c) . 0 1 5 3 4 ⎤ ⎡ ⎡ ⎤ −9 3 −9 0 10 −9 (d) ⎣−12 −3 −15⎦. (e) ⎣ 8 −1 −2⎦. (f) Impossible. −6 −3 −9 −5 −4 3 ⎡ ⎤ ⎡ ⎤ 1 2 5 4 5 1 2 3 1⎦, (AT )T = 2 3⎦. . (b) ⎣−5 (c) 6. (a) AT = ⎣2 2 1 4 3 4 8 9 4 ⎡ ⎤ 3 4 0 −4 17 2 3⎦. (d) . (e) ⎣6 (f) . 4 0 −16 6 9 10 1 0 1 0 3 0 +1 = . 0 1 0 0 0 2 ⎡⎤ λ − 1 −2 −3 10. ⎣ −6 λ + 2 −3 ⎦. −2 λ − 4 −5 8. Yes: 2 12. (a) 0 0 . 1 1 (b) 1 1 . 1 0 (c) 1 1 . 0 1 (d) 0 1 . 0 1 (e) 1 1 . 0 1
−6 11
10 . 17
14. v = 0 0 0 0 . T.1. Let A and B each be diagonal n × n matrices. Let C = A + B, cij = aij + bij . For i = j, aij and bij are each 0, so cij = 0. Thus C is diagonal. If D = A − B, dij = aij − bij , then dij = 0. Therefore D is diagonal. T.2....
Select a Discipline
Enter Zip Code
by Keyword
Welcome, Muhammad Yousaf | Sign out
Advanced Search
View Cart
ABOUT THIS PRODUCT
Description Table of Contents Features New To This Edition
PACKAGE OPTIONS
Introductory Linear Algebra: An Applied First Course, 8/E
Bernard Kolman, DrexelUniversity David R. Hill, Temple University
ISBN: 0-13-143740-2 Publisher: Prentice Hall Copyright: 2005 Format: Cloth; 768 pp
Valuepack(s)
View Larger Image RESOURCES
Our Price: $124.80 Status: Instock Published: 08/02/2004 Add to Cart
Student Instructor Course-Specific Discipline-Specific First Days Of Class
RELATED TITLES
Add to Exam Copy Bookbag
Print Product InformationInstructor Resource Center (Please scroll down to see all available instructor resources)
Help downloading Instructor Resources Instructor's Manual, 8/E by Dennis Kletzing © 2005 | Paper; 214 pages | ISBN: 0131437410 | Status: Instock Chapter 1--Instructor's Solutions Manual (.2 MB | pdf file | Type: Manuals/Guides)
Introductory Linear Algebra (Mathematics) Linear Algebra (Engineering: Mechanical)Chapter 2--Instructor's Solutions Manual (.1 MB | pdf file | Type: Manuals/Guides)
Chapter 3--Instructor's Solutions Manual (.1 MB | pdf file | Type: Manuals/Guides) More files available...
Copyright 2006 by Pearson Education - Legal Notice - Privacy Policy - Permissions Technical Support: Pearson 24/7
http://vig.prenhall.com/catalog/academic/product/0,1144,0131437402-IS,00.html7/7/20061:20:01 PM
Chapter 1
Linear Equations and Matrices
Section 1.1, p. 8
2. x = 1, y = 2, z = −2. 4. No solution. 6. x = 13 + 10z, y = −8 − 8z, z = any real number. 8. No solution. 10. x = 2, y = −1. 12. No solution. 14. x = −1, y = 2, z = −2. 16. (c) Yes. (d) Yes.
18. x = 2, y = 1, z = 0. 20. There is no such value of r. 22. Zero, infinitely many, zero. 24. 1.5 tons of regular and 2.5 tonsof special plastic. 26. 20 tons of 2-minute developer and a total of 40 tons of 6-minute and 9-minute developer. 28. $7000, $14,000, $3000. T.1. The same numbers sj satisfy the system when the pth equation is written in place of the qth equation and vice versa. T.2. If s1 , s2 , . . . , sn is a solution to (2), then the ith equation of (2) is satisfied: ai1 s1 +ai2 s2 +· · ·+ain sn = bi . Then forany r = 0, rai1 s1 + rai2 s2 + · · · + rain sn = rbi . Hence s1 , s2 , . . . , sn is a solution to the new system. Conversely, for any solution s1 , s2 , . . . , sn to the new system, rai1 s1 +· · ·+rain sn = rbi , and dividing both sides by nonzero r we see that s1 , . . . , sn must be a solution to the original linear system. T.3. If s1 , s2 , . . . , sn is a solution to (2), then the pth andqth equations are satisfied: ap1 s1 + · · · apn = bp aq1 s1 + · · · aqn = bq .
2 Thus, for any real number r, (ap1 + raq1 )s1 + · · · + (apn + raqn )sn = bp + rbq
Chapter 1
and so s1 , . . . , sn is a solution to the new system. Conversely, any solution to the new system is also a solution to the original system (2). T.4. Yes; x = 0, y = 0 is a solution for any values of a, b, c, and d.Section 1.2, p. 19
2. a = 3, b = 1, c = 8, d = −2. ⎡ ⎤ 5 −5 8 7 −7 2 9⎦. 4. (a) C + E = E + C = ⎣4 (b) Impossible. (c) . 0 1 5 3 4 ⎤ ⎡ ⎡ ⎤ −9 3 −9 0 10 −9 (d) ⎣−12 −3 −15⎦. (e) ⎣ 8 −1 −2⎦. (f) Impossible. −6 −3 −9 −5 −4 3 ⎡ ⎤ ⎡ ⎤ 1 2 5 4 5 1 2 3 1⎦, (AT )T = 2 3⎦. . (b) ⎣−5 (c) 6. (a) AT = ⎣2 2 1 4 3 4 8 9 4 ⎡ ⎤ 3 4 0 −4 17 2 3⎦. (d) . (e) ⎣6 (f) . 4 0 −16 6 9 10 1 0 1 0 3 0 +1 = . 0 1 0 0 0 2 ⎡⎤ λ − 1 −2 −3 10. ⎣ −6 λ + 2 −3 ⎦. −2 λ − 4 −5 8. Yes: 2 12. (a) 0 0 . 1 1 (b) 1 1 . 1 0 (c) 1 1 . 0 1 (d) 0 1 . 0 1 (e) 1 1 . 0 1
−6 11
10 . 17
14. v = 0 0 0 0 . T.1. Let A and B each be diagonal n × n matrices. Let C = A + B, cij = aij + bij . For i = j, aij and bij are each 0, so cij = 0. Thus C is diagonal. If D = A − B, dij = aij − bij , then dij = 0. Therefore D is diagonal. T.2....
Leer documento completo
Regístrate para leer el documento completo.