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线性代数极其应用第五版答案
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I
NSTRUCTOR
’
S
S
OLUTIONS
M
ANUAL
J
UDI
J.
M
C
D
ONALD
Washington State University
L
INEAR
A
LGEBRA
AND
I
TS
A
PPLICATIONS
F
IFTH
E
DITION
David C. Lay
University of Maryland
Steven R. Lay
Lee University
Judi J. McDonald
Washington State University
Copyright © 2016 Pearson Education, Inc.
1-1
1.1 SOLUTIONS
Notes:
The key exercises are 7 (or 11 or 12), 19–22, and 25. 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.
1.
12
12
57
27 5
xx
xx
157
275
Replace R2 by R2 + (2)R1 and obtain:
12
2
57
39
xx
x
157
039
Scale R2 by 1/3:
12
2
57
3
xx
x
157
013
Replace R1 by R1 + (–5)R2:
1
2
8
3
x
x
10 8
01 3
The solution is (x
1
, x
2
) = (–8, 3), or simply (–8, 3).
2.
12
12
24 4
57 11
xx
xx
24 4
5711
Scale R1 by 1/2 and obtain:
12
12
22
57 11
xx
xx
12 2
5711
Replace R2 by R2 + (–5)R1:
12
2
22
321
xx
x
122
0321
Scale R2 by –1/3:
12
2
22
7
xx
x
12 2
01 7
Replace R1 by R1 + (–2)R2:
1
2
12
7
x
x
10 12
01 7
The solution is (x
1
, x
2
) = (12, –7), or simply (12, –7).
1-2 CHAPTER 1 • Linear Equations in Linear Algebra
Copyright © 2016 Pearson Education, Inc.
3. The point of intersection satisfies the system of two linear equations:
12
12
57
22
xx
xx
157
122
Replace R2 by R2 + (–1)R1 and obtain:
12
2
57
79
xx
x
157
079
Scale R2 by –1/7:
12
2
57
9/7
xx
x
15 7
019/7
Replace R1 by R1 + (–5)R2:
1
2
4/7
9/7
x
x
104/7
019/7
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:
12
12
51
37 5
xx
xx
151
375
Replace R2 by R2 + (–3)R1 and obtain:
12
2
51
82
xx
x
151
082
Scale R2 by 1/8:
12
2
51
1/ 4
xx
x
15 1
011/4
Replace R1 by R1 + (5)R2:
1
2
9/4
1/4
x
x
109/4
011/4
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
1640 1
02704
00123
000515
. 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 0
x
1
+ 0
x
2
+ 0 x
3
= 1, 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.
1.1 • Solutions 1-3
Copyright © 2016 Pearson Education, Inc.
8. The standard row operations are:
1 490 1 490 1 400 1000
0 1 7 0~0 1 7 0~0 1 0 0~0 1 0 0
0020 0010 0010 0010
The solution set contains one solution: (0, 0, 0).
9. The system has already been reduced to triangular form. Begin by scaling the fourth row by 1/2 and
then replacing R3 by R3 + (3)R4:
11004 11004 11004
01307 01307 01307
~~
0013 1 00131 00105
00024 00012 00012
Next, replace R2 by R2 + (3)R3. Finally, replace R1 by R1 + R2:
1 100 4 10004
01008 01008
~~
00105 00105
0 001 2 00012
The solution set contains one solution: (4, 8, 5, 2).
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.
120 32 12007 10003
01047 01005 01005
~~
00106 00106 00106
00013 00013 00013
The solution set contains one solution: (–3, –5, 6, –3).
11. First, swap R1 and R2. Then replace R3 by R3 + (–3)R1. Finally, replace R3 by R3 + (2)R2.
0145 1352 1352 1352
1 3 5 2~0 1 4 5~0 1 4 5~0 1 4 5
3776 3776 02812 0002
The system is inconsistent, because the last row would require that 0 = 2 if there were a solution.
The solution set is empty.
12. Replace R2 by R2 + (–3)R1 and replace R3 by R3 + (4)R1. Finally, replace R3 by R3 + (3)R2.
1344 1344 1344
3778~0254~0254
4617 06159 0003
The system is inconsistent, because the last row would require that 0 = 3 if there were a solution.
The solution set is empty.
1-4 CHAPTER 1 • Linear Equations in Linear Algebra
Copyright © 2016 Pearson Education, Inc.
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.
1038 1038 1038 1038
2297~02159~0152~0152
0152 0152 02159 0055
10 3 8
~0 1 5 2
00 1 1
100 5
~0 1 0 3
001 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.
1 305 1 305 1 305 1 305
1152~0257~0 110~0110
0 110 0 110 0 257 0 077
1305
~0 1 1 0
0011
1305 1002
~0 1 0 1~0 1 0 1
0011 0011
. The solution is (2, –1, 1).
15. First, replace R4 by R4 + (–3)R1, then replace R3 by R3 + (2)R2, and finally replace R4 by
R4 + (3)R3.
10302 1030 2
010330103 3
~
023210232 1
30075 009711
10 3 0 2
01 0 3 3
~
00 3 4 7
009711
103 0 2
010 3 3
~
003 4 7
000 510
.
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.
16. 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.
10023 10023
022 0 0 022 0 0
~
001 3 1 001 3 1
232 1 5 032 3 1
10023 10023
02 2 0 0 022 0 0
~~
00 1 3 1 001 3 1
00 1 3 1 000 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.
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