Elementary linear algebra, applications version: student solutions manual

274 Pages·2010·16.798 MB·English

by  Howard Anton, Chris Rorres

TENTH Elementary
EDITION
Linear Algebra
APPLICATIONS VERSION
/
I
HOWARD ANTON / CHRIS RORRES
STUDENT SOLUTIONS MANUAL
C Trimble & Associates
I
STUDENT SOLUTIONS MANUAL
C Trimble & Associates
ELEMENTARY
LINEAR ALGEBRA
APPLICATIONS VERSION
Tenth Edition
Howard Anton
Professor Emeritus, Drexei University
Chris Rorres
University of Pennsyivania
WILEY
JOHN WILEY & SONS,I NC.
Cover art: Norm Christiansen
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ISBN 978-0-470-45822-8
10 9 8 76 5 4 3
Printed and bound by Bind-Rite/Robbinsville
Contents
Chapter 1 1
Chapter 2 44
Chapter 3 61
Chapter 4 81
Chapter 5 120
Chapter 6 138
Chapter 7 170
Chapter 8 185
Chapter 9 199
Chapter 10 211

Chapter 1
Systems of Linear Equations and Matrices
Section 1.1
Exercise Set 1.1
1. (a), (c), and (f) are linear equations in X], a:2, and X3.
(b) is not linear because of the term X1X3.
(d) is not linear because of the term x~'^.
(e) is not linear because of the term
3. (a) and (d) are linear systems,
(b) is not a linear system because the first and second equations are not linear,
(c) is not a linear system because the second equation is not linear.
5. By inspection,( a) and (d) are both consistent; x^ =3, X2 = 2, X3 = -2, X4 = 1 is a solution of( a) and
X| = 1, X2 = 3, X3 = 2, X4 = 2 is a solution of( d). Note that both systems have infinitely many solutions.
7. (a),( d), and (e) are solutions,
(b) and (c) do not satisfy any of the equations.
9. (a) 7x-5>- = 3
5 3
X =-y +-
7 7
Let y = t. The solution is
5 3
X =-y +-
7 7
y = t
(b) -8x1 + 2x2“ 5 x3 + 6x4 = 1
1 5 3 1
X, =-X2 X3 +-X4
' 4 ^ 8 4 ^ 8
Let X2 = r, X3 = and X4 = t. The solution is
1 5 3 1
Xi = — r — ^ + —f —
' 4 8 4 8
X2=r
X3 =i
X4 =t
2 0 0
11. (a) 3 -4 0 corresponds to
0 1
2x = 0
1
3xi -4x2 = 0-
X2 =1
1
Chapter 1: Systems of Linear Equations and Matrices SSM: Elementary Linear Algebra
15. If( a, b, c) is a solution of the system, then
3 0 -2 5
aaf +bX]+c = y,, ax\ +bx2+c = y2, and
(b) 7 4 -3 corresponds to
0 -2 7 ax^ +bx2, + c = y3 which simply means that the
3a- -2x3 =5 points are on the curve.
1
7x| +X2 +4x3 =“3-
17. The solutions of x^+kx2-c are X| -c-kt,
-2x2 + -'^3 = ^
X2 = t where t is any real number. If these
satisfy x^+1x2 =d, c- kt + It = d or
7 2 1 -3 5
(c) corresponds to
J 2 4 0 1 c-d ={k- 1) t for all real numbers t. In
particular, if r = 0, then c = d, and if r = 1, then
7xi + 2x2 + -^3“ ■ ^■^4 “ 5
k=l.
X| + 2x2 + 4-^3 = •
True/False 1.1
1 0 0 0 7
(a) True; Xj = X2 = ■ • ■ = x„ =0 will be a solution.
0 1 0 0
(d) corresponds to
0 0 1 0
(b) False; only multiplication by nonzero constants
.0 0 0 1 4
is acceptable.
^1 7
^2 -2 (c) True; if k = 6 the system has infinitely many
^3 3 solutions, while if k^6, the system has no
solution.
X4 4
(d) True; the equation can be solved for one variable
13. (a) The augmented matrix for -2x| = 6
in terms of the other(s), yielding parametric
3x, =8
1 equations that give infinitely many solutions.
9x, =-3
(e) False; the system 3x-5y = -7 has the
-2 6
2x + 9y = 20
IS 3
6x-10y = -14
9 -3
solution X = 1, y = 2.
(b) The augmented matrix for (f) False; multiplying an equation by a nonzero
6x1 -X7 +3x3 = 4 is ^ * 3 4 constant c does not change the solutions of the
,0 5 1 system.
5X2 “-^3=1
(g) True; subtracting one equation from another is
(c) The augmented matrix for the same as multiplying an equation by -1 and
2x2 - 3x4 + X5 = 0 adding it to another.
-3x| - X2 + X3
(b) False; the second row corresponds to the
bX] + 2x2 - -'^3 + 2x4 - 3x5 = 6
equation 0x| +0x2 = or 0 = -1 which is false.
0 2 0 -3 1 0
IS -3 1 0 0 Section 1.2
6 2 -1 2 -3 6
Exercise Set 1.2
(d) The augmented matrix for X| -Xj =7 is 1. (a) The matrix is in both row echelon and
reduced row echelon form.
[1 0 0 0 -1 7].
(b) The matrix is in both row echelon and
reduced row echelon form.
2
SSM: Elementary Linear Algebra Section 1.2
(c) The matrix is in both row echelon and (d) The last line of the matrix corresponds to the
reduced row echelon form. equation OX] +0x2 +0-^3 “ which is not
satisfied by any values of X\, X2, and X3,
(d) The matrix is in both row echelon and
reduced row echelon form. so the system is inconsistent.
(e) The matrix is in both row echelon and I I 2
reduced row echelon form. 5. The augmented matrix is -I -2 3
3 -7 4 lO
(f) The matrix is in both row echelon and
reduced row echelon form. Add the first row to the second row and add -3
times the first row to the third row.
(g) The matrix is in row echelon form.
I I 2 8
3. (a) The matrix corresponds to the system 0 - I 5 9
X] -3 x2 X] = 7+ 3x2“ 4 x3 0 -lO -2 -14
X2 + 2x3 =2 or X2 = 2- 2x3
Multiply the second row by -1.
X3 = 5 X3 = 5
2 8
Thus X3 = 5, X2 = 2- 2(5)= -8, and 0 1 -5 -9
X| =7+ 3(-8)-4(5)= -37. The solution is 0 -10 -2 -14
X| =-37, X2 =-8, X3 =5. Add 10 times the second row to the third row.
1 2 8
(b) The matrix corresponds to the system
0 1 -5 -9
X, + 8x3 -5x4 =6
0 0 -52 -104
X2 + 4x3 -9 x4 = 3 or
X3 +X4=2 1
Multiply the third row by .
X| = 6- 8 x3 +5 x4 52
X2 = 3- 4 x3 +9 x4 . 1 1 2
X3 = 2- X4 0 1 -5 -9
Let X4 = t, then X3 =2-t, 0 0 1 2
X2 =3-4(2-0+ 9r = 13/-5, and Add 5 times the third row to the second row and
-2 times the third row to the first row.
X] = 6- 8 (2- r )+ 5f = 13r -10. The solution
isx|=13f-10, X2=13?-5, X3=-r + 2, 1 1 0 4
0 1 0 1
X4 =t.
0 0 1 2
(c) The matrix corresponds to the system Add -1 times the second row to the first row.
X] +7x2 “2x3 -8x5 =-3
1 0 0 3
X3 + X4 + 6x5 =5 or
0 1 0 1
X4 +3x5 =9
0 0 1 2
X] --3-1x2+ ^^3 ^ As
The solution is X| = 3, X2 = 1, X3 = 2.
X3 = 5- X4 -6 x5
X4 =9-3x5
7. The augmented matrix is
Let X2=s and X5 = t, then X4 = 9- 3 t,
1 -1 2 -1 -1
X3 = 5- ( 9- 3 0- 6 f = -3f - 4, and 2 1 -2-2 -2
X, =-3-7^+ 2(-3r-4)+ 8f = -75 + 2r-IL -1 2 -4 1 1 ‘
The solution is X] =-75 +2f-l 1, X2=s, , 3 0 0 -3 -3.
Add -2 times the first row to the second row, 1
X3 = -3t- 4 , X4 = -3t + 9, X5 = t.
times the first row to the third row, and -3 times
the first row to the fourth row.
3
Chapter 1: Systems of Linear Equations and Matrices SSM: Elementary Linear Algebra
13. Since the system has more unknowns( 4) than
1 -1 2 -1 -1
equations (3), it has nontrivial solutions.
0 3 -6 0 0
0 1 -2 0 0
15. Since the system has more unknowns( 3) than
,0 3 -6 0 0. equations (2), it has nontrivial solutions.
1
Multiply the second row by - and then add -1
2 1 3 0
times the new second row to the third row and 17. The augmented matrix is 1 2 0 0
add -3 times the new second row to the fourth 0 1 1 0
row.
Interchange the first and second rows, then add
1 -1 2 -1 -1 -2 times the new first row to the second row.
0 1 -2 0 0
1 2 0 0
0 0 0 0 0
0-3 3 0
.0 0 0 0 0.
0 1 1 0
Add the second row to the first row.
"l 0 0-1 -l’
Multiply the second row by —, then add -1
0 1 -2 0 0 3
0 0 0 0 0 times the new second row to the third row.
.0 0 0 0 0. 1 2 0 0
The corresponding system of equations is 0 1 -1 0
X -H’= -l x= w-l 0 0 2 0
y-2z = 0 or y = 2z n
Let z = s and w = t. The solution is x = f - 1, Multiply the third row by —, then add the new
y-2s,z-s,w = t.
third row to the second row.
9. In Exercise 5, the following row echelon matrix 1 2 0 0
occurred. 0 1 0 0
1 1 2 8 0 0 1 0
0 1 -5 -9
Add -2 times the second row to the first row.
0 0 1 2
1 0 0 0
The corresponding system of equations is
0 1 0 0
Xj + X2 + 2x^ — 8 Xi =-X2 -2x3 +8
0 0 1 0
X2 -5 x3 = -9 or X2 = 6x3 -9
The solution, which can be read from the matrix,
X3=2 X3 -2
is X] = 0, X2 = 0, X3 = 0.
Since X3 = 2, X2 = 5(2)- 9 = 1, and
X] =-1-2(2)+ 8 = 3. The solution is X| =3,
3 1 1 1 0
19. The augmented matrix is
X2=l, X3=2. ,5 -1 1 -1 Oj
1
11. From Exercise 7, one row echelon form of the Multiply the first row by —, then add -5 times
3
1 -1 2 -1 -1 the new first row to the second row.
0 1 -2 0 0
augmented matrix is 1 I I j »■
0 0 0 0 0 3 3
.0 0 0 0 0. 0 8 _2 f “J
3 3
The corresponding system is
3
x-y + 2z-w = -l x= y-22 + vv-l Multiply the second row by —.
or 8
y-2z =0 y = 2z
fl -L -L i Ol
Let z = s and w = t. Then y = 2s and 3 3 3
x=2s-2s + t - 1 = r- 1. The solution is 0 1 T 1 0
4
x = t- \,y = 2s, z = s,w = t.
4
SSM: Elementary Linear Algebra Section 1.2
1 w -y =0 w= y
Add — times the second row to the first row.
3 x+y =0 or x = -y .
z=0 z=0
1 0^0 0
4 Let y = t. The solution is w = t, x = -t, y = t,
0 1 ^ 1 0 z = 0.
4
This corresponds to the system
2 -1 3 4 9
1
x. +7 ^3 = 0 ^1 =-7-^3 1 0 -2 7 11
4 4 23. The augmented matrix is ^
or . - 3 1 5
1
X2 + — Xj + X4 — 0 ^2 =-7^3--3^4 2 1 4 4 10
4
Interchange the first and second rows,
Let X3 = 4i and X4 = t. Then Xj = -s and
'l 0 -2 7 If
X2 = -s-t. The solution is x^ -s, X2 =-s-t,
2 -1 3 4 9
X3 = 45, X4 = t.
3 -3 1 5 8
2 1 4 4 10
0 2 2 4 0
Add -2 times the first row to the second and
1 0-1 -3 0
21. The augmented matrix is fourth rows, and add -3 times the first row to the
2 3 1 0
third row.
-2 1 3 -2 0
1 0-2 7 11
Interchange the first and second rows. 0 -1 7 -10 -13
'1 0-1 -3 0' 0-3 7 -16 -25
0 2 2 4 0 0 1 8 -10 -12
2 3 1 1 0
Multiply the second row by -1, then add 3 times
-2 1 3 -2 0
the new second row to the third row and -1
Add -2 times the first row to the third row and 2 times the new second row to the fourth row.
times the first row to the fourth row. '1 0 -2 7 if
'1 0 -1 -3 0' 0 1 -7 10 13
0 2 2 4 0 0 0 -14 14 14
0 3 3 7 0 0 0 15 -20 -25
0 1 1 -8 0
1
Multiply the third row by , then add -15
14
Multiply the second row by —, then add -3
2 times the new third row to the fourth row.
times the new second row to the third row and 1 0-2 7 11
-1 times the new second row to the fourth row. 0 1 -7 10 13
1 0-1 -3 0 0 0 1 -1
o i l 2 0 0 0 0 -5 -10
0 0 0 1 0
1
0 0 0 -10 0 Multiply the fourth row by —, then add the
5
Add 10 times the third row to the fourth row, -2
new fourth row to the third row, add -10 times
times the third row to the second row, and 3
the new fourth row to the second row, and add
times the third row to the first row.
-7 times the new fourth row to the first row.
'1 0-1 0 0'
'1 0-2 0 -3'
0 1 1 0 0
0 1 -7 0-7
0 0 0 1 0
0 0 1 0 1
0 0 0 0 0
0 0 0 1 2
The corresponding system is
Add 7 times the third row to the second row and
2 times the third row to the first row.
5