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首页Linear Algebra and its Applications 3rd edition 习题解析
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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).
2 CHAPTER 1 • Linear Equations in Linear Algebra
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 3
8. The standard row operations are:
1 490 1 490 1 400 1000
0170~0170~0100~0100
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
~~
00 13 1 00 131 00 10 5
0002 4 000 12 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.
1 20 3 2 1 200 7 1000 3
01047 01005 01005
~~
00106 00106 00106
0 00 1 3 0 001 3 0001 3
−−− −
−−−
−−−
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.
4 CHAPTER 1 • Linear Equations in Linear Algebra
13.
1038 1038 1038 1038
2297~02159~0152~0152
0152 0152 02159 0055
−−−−
−−−
−−−−
10 3 8 100 5
~0 1 5 2~0 1 0 3
00 1 1 001 1
−
−
−−
. The solution is (5, 3, –1).
14.
1 305 1 305 1 305 1 305
1 1 5 2~0 2 5 7~0 1 1 0~0 1 1 0
0 110 0 110 0 257 0 077
−−−−
−−
−
1305 1305 1002
~0 1 1 0~0 1 0 1~0 1 0 1.
0 011 0 01 1 001 1
−−
−−
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
~
02321 0232 1
30075 009711
−−
−−
−−−
10 3 0 2 103 0 2
01 0 3 3 010 3 3
~~
00 3 4 7 003 4 7
00 9 7 11 000 510
−−
−−
−− −
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.
1.1 • Solutions 5
17. Row reduce the augmented matrix corresponding to the given system of three equations:
14 1 14 1 14 1
213~075~075
134 075 000
−−−
−−−−
−− −
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.
18. Row reduce the augmented matrix corresponding to the given system of three equations:
12 14 12 1 4 12 1 4
01 11~01 1 1~01 1 1
13 00 01 1 4 00 0 5
−− −
−− −
The third equation, 0 = –5, shows that the system is inconsistent, so the three planes have no point in
common.
19.
141 4
~
368 063 4
hh
h
−−
Write
c for 6 – 3h. 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.
20.
131 3
~.
24 6 042 0
hh
h
−−
−+
Write
c for 4 + 2h. Then the second equation cx
2
= 0 has a solution
for every value of c. So the system is consistent for all
h.
21.
13 2 1 3 2
~.
480120
hh
−−
−+
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.
22.
23 23
~.
695 0053
hh
h
−−
−+
The system is consistent if and only if 5 + 3
h = 0, that is, if and only
if
h = –5/3.
23. a. True. See the remarks following the box titled Elementary Row Operations.
b. False. A 5 × 6 matrix has five rows.
c. False. The description given applied 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.
d. True. See the box before Example 2.
24. a. True. See the box preceding the subsection titled Existence and Uniqueness Questions.
b. False. The definition of row equivalent requires that there exist a sequence of row operations that
transforms one matrix into the other.
c. False. By definition, an inconsistent system has no solution.
d. True. This definition of equivalent systems is in the second paragraph after equation (2).
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