Linear_Algebra_and_Applications-Lay-solutions.pdf
(
5941 KB
)
Pobierz
584744378 UNPDF
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
.
5 7
27 5
x
1
+=
x
2
157
275
−− =−
x
−−−
1
2
Replace R2 by R2 + (2)R1 and obtain:
x
1
+=
57
39
x
2
157
039
=
2
Scale R2 by 1/3:
x
1
+=
57
3
x
2
157
013
=
2
Replace R1 by R1 + (–5)R2:
x
1
=−
=
8
3
10 8
01 3
−
x
2
The solution is (
x
1
,
x
2
) = (–8, 3), or simply (–8, 3).
2
.
24 4
57 1
1
+ −
x
2
57
−
x
+=
1
2
Scale R1 by 1/2 and obtain:
2 2
57 1
1
+ −
x
2
x
+=
1
2
Replace R2 by R2 + (–5)R1:
x
1
+ −
2
x
2
2
12 2
03 1
−
−=
3
21
−
2
Scale R2 by –1/3:
x
1
+ −
2
x
2
2
7
=−
−
2
Replace R1 by R1 + (–2)R2:
x
1
=
=−
12
7
10 2
01 7
x
−
2
The solution is (
x
1
,
x
2
) = (12, –7), or simply (12, –7).
1
x
24 4
x
12 2
57
−
12 2
01
−
2 CHAPTER 1 • Linear Equations in Linear Algebra
3
. The point of intersection satisfies the system of two linear equations:
xx
x
+=
5
2
7
157
122
1
−−
2
2
2
−−
Replace R2 by R2 + (–1)R1 and obtain:
x
1
+=
5
x
2
7
157
079
−=−
7
9
−−
2
Scale R2 by –1/7:
x
1
+=
5
x
2
7
9/7
15 7
019/7
=
2
Replace R1 by R1 + (–5)R2:
x
1
=
=
4/7
9/7
10 4/7
01/7
x
2
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:
51
37 5
1
−=
2
151
375
−
1
−=
2
−
Replace R2 by R2 + (–3)R1 and obtain:
xx
−=
51
82
2
151
082
=
2
Scale R2 by 1/8:
xx
1
−=
5
2
1
1/ 4
15 1
01/4
=
2
Replace R1 by R1 + (5)R2:
x
1
=
=
9/4
1/ 4
109/4
01/4
x
2
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
16 40 1
02704
00 123
00055
−
−
−
produces
. 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
xx
x
−
1
−
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:
1100 4 11004 1100 4
01307 01307 01307
~ ~
00 13 1 00 13 1 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
−
−
~
~
0010 5 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 028 2 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 06 59 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
1038 1038 1038 1038
2297~0259~0152~0152
0152 0152 0259 0055
−
−
−
−
13
.
−
−
−
−
−
−
−
10 3 8 100 5
~0 1 5 2~0 1 0 3
00 1 1 001 1
−
−
. The solution is (5, 3, –1).
−
−
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
−
−
−
−
14
.
−
−
−
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 10 30 2
010330103 3
~
02321 0232 1
30075 00971
−
−
−
−
−
−
−
10 3 0 2 10 3 0 2
01 0 3 3 010 3 3
−
−
~
~
00 3 4 7 003 4 7
00 9 7 1 000 50
−
−
−
−
−
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
.
~
368 063 4
h
4
1
h
4
Write
c
for 6 – 3
h
. If
c
= 0, that is, if
h
= 2, then the system has no
−
h
−
solution, because 0 cannot equal –4. Otherwise, when
h
≠
2, the system has a solution.
20
.
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
.
1 3 1 3
~ .
24 6 042 0
h
−
h
−
−
+
h
21
.
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
.
13 2 1 3
−
~
−
2
.
−
4
h
8
0
h
+
12 0
22
.
23 23
~ .
695 0053
−
h
−
h
The system is consistent if and only if 5 + 3
h
= 0, that is, if and only
+
h
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).
1
−
Plik z chomika:
Kuya
Inne pliki z tego folderu:
A_First_Course_in_Algebra-Fraleigh-7e-solutions.pdf
(1064 KB)
Introduction_to_Linear_Algebra-Strang-3e-solutions.pdf
(531 KB)
Linear_Algebra-Friedburg-4e-solutions.pdf
(853 KB)
Linear_Algebra-Hefferon-solutions.pdf
(3409 KB)
Linear_Algebra_and_Applications-Lay-solutions.pdf
(5941 KB)
Inne foldery tego chomika:
calculus
differential_equations
discrete
methods
prob_and_stats
Zgłoś jeśli
naruszono regulamin