33. Given A = -2 -6 7 21 -3 -9 oll bow120 DE find one nontrivial solution of Ax = 0 by inspection. [Hint: Think of the equation Ax = 0 written as a vector equation.] iw d. nontoo? I dsinsy 002 ni anoi "Jugovo-jugai" tait
33. Given A = -2 -6 7 21 -3 -9 oll bow120 DE find one nontrivial solution of Ax = 0 by inspection. [Hint: Think of the equation Ax = 0 written as a vector equation.] iw d. nontoo? I dsinsy 002 ni anoi "Jugovo-jugai" tait
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
Related questions
Question
33
![e. The solution set of Ax = b is obtained by translating thei ne
solution set of Ax = 0.
34.
25. Prove the second part of Theorem 6: Let w be any solution of
Ax = b, and define v₁ = w - p. Show that V₁, is a solution
of Ax = 0. This shows that every solution of Ax = b has the
form w = p + Vh, with p a particular solution of Ax = b and
Vh a solution of Ax = 0.
27. Suppose A is the 3 x 3 zero matrix (with all zero entries).
Describe the solution set of the equation Ax = 0.
28. If b0, can the solution set of Ax = b be a plane through
the origin? Explain.
In Exercises 29-32, (a) does the equation Ax = 0 have a nontriv-
ial solution and (b) does the equation Ax = b have at least one
solution for every possible b?
29. A is a 3 x 3 matrix with three pivot positions.
30. A is a 3 x 3 matrix with two pivot positions.
31. A is a 3 x 2 matrix with two pivot positions.
32. A is a 2 x 4 matrix with two pivot positions.
-2 -6
7
21
-3
-9
33. Given A =
9
Chow
find one nontrivial solution of
26. Suppose Ax = b has a solution. Explain why the solution is
unique precisely when Ax = 0 has only the trivial solution. 36. Construct a 3 x 3 nonzero matrix A such that the vector
odi to alugiuo Istos odos bougi
all soonsled yliosx9 101558 does to
20
Ax = 0 by inspection. [Hint: Think of the equation Ax = 0
written as a vector equation.]
15 21.qq.1801 190
1912 bm.(15wog) sinfol Iso
ols
vode 28 2101052 200
101 21045928 to 2hsq fanitos o insesiqen n
constant
to suquo Istol odi isdi axna
20 minuus' sdi brs loo12 of 902 leo lor
ali oqo of 15bio ni awoni ti se
a
endoglA
bid:
voll
boteskus 4
--
-6
12
6 -9
Ax = 0 by inspection.
Given A =
201 aids
35. Construct a 3 x 3 nonzero matrix A such that the vector
H
1.5 Solution Sets of Linear Systems 49
noitur-2
1
2 moi 2-teit
37. Construct a 2 x 2 matrix A such that the solution set of the
equation Ax = 0 is the line in R2 through (4, 1) and the
origin. Then, find a vector b in R2 such that the solution set
lozof Ax = b is not a line in R2 parallel to the solution set of
Ax = 0. Why does this not contradict Theorem 6?
zoldsinsy 002 ni anoiteups 002
bubong suquo-juqni" toitnos s as mo
Crood sjom nadw.ljeto arom mi Ich
Syrien isiqmie s is dool
bebivib 21 monoso a noiian se
'noitens seoqque
sirdi onlinexs
to T
1]ang
ivisa
SOLUTIONS TO PRACTICE PROBLEMS
Jesmond
opzione 101 16/1 2200
bobivib ei tugiuo eidi
To sulav sellob isto odi od 1. Row reduce the augmented matrix:
aluzon gniwollol od bovorq leht
laitnos tuo tadi to ty si bollo ad
4 -5
0 3
3
2-18 9]-[ 9]-[624]
1 -2 -1
ing-
-8
is a solution of Ax = 0.
Mate
X1
x2 =
X3
on oitos
2M3
38. Suppose A is a 3 x 3 matrix and y is a vector in R³ such that
the equation Ax = y does not have a solution. Does there
exist a vector z in R3 such that the equation Ax = z has a
unique solution? Discuss.
Jed
tuloe
39. Let A be an m x n matrix and let u be a vector in R" that satis-
fies the equation Ax = 0. Show that for any scalar c, the vec-
tor cu also satisfies Ax = 0. [That is, show that A (cu) = 0.]
40. Let A be an m x n matrix, and let u and v be vectors in R"
with the property that Au = 0 and Av = 0. Explain why
A(u + v) must be the zero vector. Then explain why
A(cu + dv) = 0 for each pair of scalars c and d.
3
is a solution of Ax = 0.
X1
+ 3x3 = 4
250ng mundiliups of brit of wod ewoda alqmxe x2 - 2x3 = -1
1
4
0 -9
Thus x₁ = 43x3, x2 = -1 + 2x3, with x3 free. The general solution in parametric
vector form is
9
find one nontrivial solution of
4 - 3x3
-1 + 2x3
X3
10
230195
-5
18
H
=
4
P
-3
+ x3 2
The intersection of the two planes is the line through p in the direction of v.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6c34d2b4-e6b2-47b1-ad1e-68f13eeb5b48%2Fee29302e-71a5-4e5a-9bc2-aecfa999765a%2Fq1aoyzb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:e. The solution set of Ax = b is obtained by translating thei ne
solution set of Ax = 0.
34.
25. Prove the second part of Theorem 6: Let w be any solution of
Ax = b, and define v₁ = w - p. Show that V₁, is a solution
of Ax = 0. This shows that every solution of Ax = b has the
form w = p + Vh, with p a particular solution of Ax = b and
Vh a solution of Ax = 0.
27. Suppose A is the 3 x 3 zero matrix (with all zero entries).
Describe the solution set of the equation Ax = 0.
28. If b0, can the solution set of Ax = b be a plane through
the origin? Explain.
In Exercises 29-32, (a) does the equation Ax = 0 have a nontriv-
ial solution and (b) does the equation Ax = b have at least one
solution for every possible b?
29. A is a 3 x 3 matrix with three pivot positions.
30. A is a 3 x 3 matrix with two pivot positions.
31. A is a 3 x 2 matrix with two pivot positions.
32. A is a 2 x 4 matrix with two pivot positions.
-2 -6
7
21
-3
-9
33. Given A =
9
Chow
find one nontrivial solution of
26. Suppose Ax = b has a solution. Explain why the solution is
unique precisely when Ax = 0 has only the trivial solution. 36. Construct a 3 x 3 nonzero matrix A such that the vector
odi to alugiuo Istos odos bougi
all soonsled yliosx9 101558 does to
20
Ax = 0 by inspection. [Hint: Think of the equation Ax = 0
written as a vector equation.]
15 21.qq.1801 190
1912 bm.(15wog) sinfol Iso
ols
vode 28 2101052 200
101 21045928 to 2hsq fanitos o insesiqen n
constant
to suquo Istol odi isdi axna
20 minuus' sdi brs loo12 of 902 leo lor
ali oqo of 15bio ni awoni ti se
a
endoglA
bid:
voll
boteskus 4
--
-6
12
6 -9
Ax = 0 by inspection.
Given A =
201 aids
35. Construct a 3 x 3 nonzero matrix A such that the vector
H
1.5 Solution Sets of Linear Systems 49
noitur-2
1
2 moi 2-teit
37. Construct a 2 x 2 matrix A such that the solution set of the
equation Ax = 0 is the line in R2 through (4, 1) and the
origin. Then, find a vector b in R2 such that the solution set
lozof Ax = b is not a line in R2 parallel to the solution set of
Ax = 0. Why does this not contradict Theorem 6?
zoldsinsy 002 ni anoiteups 002
bubong suquo-juqni" toitnos s as mo
Crood sjom nadw.ljeto arom mi Ich
Syrien isiqmie s is dool
bebivib 21 monoso a noiian se
'noitens seoqque
sirdi onlinexs
to T
1]ang
ivisa
SOLUTIONS TO PRACTICE PROBLEMS
Jesmond
opzione 101 16/1 2200
bobivib ei tugiuo eidi
To sulav sellob isto odi od 1. Row reduce the augmented matrix:
aluzon gniwollol od bovorq leht
laitnos tuo tadi to ty si bollo ad
4 -5
0 3
3
2-18 9]-[ 9]-[624]
1 -2 -1
ing-
-8
is a solution of Ax = 0.
Mate
X1
x2 =
X3
on oitos
2M3
38. Suppose A is a 3 x 3 matrix and y is a vector in R³ such that
the equation Ax = y does not have a solution. Does there
exist a vector z in R3 such that the equation Ax = z has a
unique solution? Discuss.
Jed
tuloe
39. Let A be an m x n matrix and let u be a vector in R" that satis-
fies the equation Ax = 0. Show that for any scalar c, the vec-
tor cu also satisfies Ax = 0. [That is, show that A (cu) = 0.]
40. Let A be an m x n matrix, and let u and v be vectors in R"
with the property that Au = 0 and Av = 0. Explain why
A(u + v) must be the zero vector. Then explain why
A(cu + dv) = 0 for each pair of scalars c and d.
3
is a solution of Ax = 0.
X1
+ 3x3 = 4
250ng mundiliups of brit of wod ewoda alqmxe x2 - 2x3 = -1
1
4
0 -9
Thus x₁ = 43x3, x2 = -1 + 2x3, with x3 free. The general solution in parametric
vector form is
9
find one nontrivial solution of
4 - 3x3
-1 + 2x3
X3
10
230195
-5
18
H
=
4
P
-3
+ x3 2
The intersection of the two planes is the line through p in the direction of v.
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