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数学代写-Q1
时间：2022-06-20
2022/6/20 18:01 University of New South Wales - 成绩单
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第 1 页，共 1 页 1 - 7 行，共 7 行 - 0
试题
Q1 2/2.0 查看 原始作答 未过滤的答案
Let , , be vectors in .
Suppose that , , , .
For each of the following, select True if the statement is correct and select False if otherwise.
(a) The dimension of span is 2.
您的应答 正确应答
True
自动评分的 评分: 5/5.0
(b) The dimension of span is 3.
您的应答 正确应答
False
自动评分的 评分: 5/5.0
(c) The set is a basis for .
您的应答 正确应答
False
自动评分的 评分: 2/2.0
(d) The set is a basis for span .
您的应答 正确应答
False
自动评分的 评分: 4/4.0
(e) The set is a basis for span .
您的应答 正确应答
True
自动评分的 评分: 2/2.0
查看面板- 格式 数值
学生详细信息- Quiz 1 -- Week 3
分数: 历时: 40 min
7/10.0 已开始: 2022/6/20 AEST 上午12:24:19已结束: 2022/6/20 AEST 上午1:04:41
Enhua Zhang- 考试任务邮件地址: enhua.zhang@student.unsw.edu.au已结束: 0活动的: 0
学号: z5325328 等待人工阅卷: 0及格的: 0
=
v
1
⎛
⎝
⎜
⎜
⎜
5
−1
−4
−5
⎞
⎠
⎟
⎟
⎟
=
v
2
⎛
⎝
⎜
⎜
⎜
−3
−3
−4
5
⎞
⎠
⎟
⎟
⎟
=
v
3
⎛
⎝
⎜
⎜
⎜
−2
4
8
0
⎞
⎠
⎟
⎟
⎟
R
4
=e
1
⎛
⎝
⎜
⎜
⎜
1
0
0
0
⎞
⎠
⎟
⎟
⎟
=e
2
⎛
⎝
⎜
⎜
⎜
0
1
0
0
⎞
⎠
⎟
⎟
⎟
=e
3
⎛
⎝
⎜
⎜
⎜
0
0
1
0
⎞
⎠
⎟
⎟
⎟
=e
4
⎛
⎝
⎜
⎜
⎜
0
0
0
1
⎞
⎠
⎟
⎟
⎟
( , )v
1
v
2
( , , )v
1
v
2
v
3
{ , , }v
1
v
2
v
3
R
4
{ , , }v
1
v
2
v
3
( , , )v
1
v
2
v
3
{ , }v
1
v
2
( , , )v
1
v
2
v
3
2022/6/20 18:01 University of New South Wales - 成绩单
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试题
(f) The set is a basis for .
您的应答 正确应答
False
自动评分的 评分: 2/2.0
总分： 1.0×5/20 + 1.0×5/20 + 1.0×2/20 + 1.0×4/20 + 1.0×2/20 + 1.0×2/20 = 25% + 25% + 10% + 20% + 10% +
10%
反馈信息：
(a) is true. The two non-zero vectors are not scalar multiple of one another. The set
is linearly independent and it spans span .
(b) is false. The vector is a linear combination of . Hence the dimension of
span is the same as the dimension of span .
(c) is false. The dimension of is 4. The set of three vectors cannot be a basis.
(d) is false. The set is linearly dependent because is a linear
combination of . The set cannot be a basis for the span.
(e) is true. span = span , and part (a).
(f) is false. The dimension of is 4. The set of five vectors cannot be a basis.
Q2 1/1.0 查看 原始作答 未过滤的答案
Let be invertible matrices of the same size. Suppose that
is symmetric, i.e. ;
is skew-symmetric, i.e. ; and
is orthogonal, i.e. .
您的应答 正确应答
自动评分的 评分: 1/1.0
总分： 1.0×1/1 = 100%
反馈信息：
Note, in general, .
The following properties are useful.
, and .
{ , , , , }v
1
v
2
e
2
e
3
e
4
R
4
{ , }v
1
v
2
( , )v
1
v
2
v
3
{ , }v
1
v
2
( , , )v
1
v
2
v
3
( , )v
1
v
2
R
4
{ , , }v
1
v
2
v
3
v
3
{ , }v
1
v
2
( , )v
1
v
2
( , , )v
1
v
2
v
3
R
4
D, E, F
D = DD
T
E = −EE
T
F =F
T
F
−1
(D (ED F =E
−1
F
T
)
−1
)
T
−F F
E
2
AB ≠ BA
(AB =)
−1
B
−1
A
−1
(AB =)
T
B
T
A
T
2022/6/20 18:01 University of New South Wales - 成绩单
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试题
Q3 1/1.0 查看 原始作答 未过滤的答案
Given that is a matrix and are vectors in such that
where the matrix is invertible. Find the matrix such that
.
Enter the matrix in the box below.
您的应答 正确应答
自动评分的 评分: 1/1.0
Note: Do not enter M = in the answer box.
总分： 1.0×1/1 = 100%
Q4 1/1.0 查看 原始作答 未过滤的答案
Select all equations with infinitely many solutions.
选项 选中的
, where ,
No [答案保留]
, where ,
No [答案保留]
, where ,
No [答案保留]
, where ,
Yes [答案保留]
, where ,
Yes [答案保留]
, where ,
No [答案保留]
, where ,
Yes [答案保留]
, where ,
No [答案保留]
反馈
信
息：
For each of the equations, the augmented matrix is already in row
echelon form.
The equation has infinitely many solutions when the right hand
column of the row-echelon form is non-leading and there is at least one column
on the left is non-leading.
A
, ,
v
1
v
2
v
3
R
3
⎧
⎩
⎨
⎪
⎪
A = 2v
1
v
1
A = −7 + 5v
2
v
1
v
2
A = 6 − 7 − 5v
3
v
1
v
2
v
3
P = ( | | )
v
1
v
2
v
3
M
AP = PM
M
⎡
⎣
⎢
2
0
0
−7
5
0
6
−7
−5
⎤
⎦
⎥
Ax = b A =
⎛
⎝
⎜
4
0
0
5
0
0
5
0
0
⎞
⎠
⎟
b =
⎛
⎝
⎜
2
−1
0
⎞
⎠
⎟
Ax = b A =
⎛
⎝
⎜
5
0
0
5
4
0
3
1
−1
⎞
⎠
⎟
b =
⎛
⎝
⎜
2
2
5
⎞
⎠
⎟
Ax = b A =
⎛
⎝
⎜
4
0
0
1
5
0
−1
0
3
⎞
⎠
⎟
b =
⎛
⎝
⎜
2
5
0
⎞
⎠
⎟
Ax = b A =( )
4
0
5
0
b =( )
−1
0
Ax = b A =
⎛
⎝
⎜
4
0
0
2
5
0
2
5
0
2
2
5
5
0
−1
⎞
⎠
⎟
b =
⎛
⎝
⎜
2
5
2
⎞
⎠
⎟
Ax = b A =
⎛
⎝
⎜
⎜
⎜
4
0
0
0
1
5
0
0
−1
0
3
0
⎞
⎠
⎟
⎟
⎟
b =
⎛
⎝
⎜
⎜
⎜
2
5
2
0
⎞
⎠
⎟
⎟
⎟
Ax = b A =
⎛
⎝
⎜
5
0
0
1
−1
0
5
2
0
4
2
4
⎞
⎠
⎟
b =
⎛
⎝
⎜
−1
0
5
⎞
⎠
⎟
Ax = b A =
⎛
⎝
⎜
3
0
0
5
−1
0
5
2
0
⎞
⎠
⎟
b =
⎛
⎝
⎜
2
2
4
⎞
⎠
⎟
(A|b)
Ax = b
2022/6/20 18:01 University of New South Wales - 成绩单
https://unsw.mobius.cloud/gradebook/Details.do?userId=16183&trId=3082776 4/6
试题
Q5 1/2.0 查看 原始作答 未过滤的答案
(a) Let be the vector space of all the polynomials of degree 7 or less.
The dimension of is
您的应答 正确应答
8 8
自动评分的 评分: 5/5.0
.
(b) Let and be a set of 4 vectors in .
Suppose that is the subspace of spanned by .
Let be the matrix with the vectors in as columns,
and .
The system is inconsistent. There is no solution.
The system has infinitely many solutions.
Fill in the following answer fields using drop down menus.
The vector
您的应答 正确应答
is not is not
自动评分的 评分: 5/5.0
in the span of .
The set is a linearly
您的应答 正确应答
independent dependent
自动评分的 评分: 0/5.0
set in .
The dimension of
您的应答 正确应答
is 4 is less than 4
自动评分的 评分: 0/5.0
.
总分： 1.0×5/20 + 1.0×5/20 + 0.0×5/20 + 0.0×5/20 = 25% + 25% + 0% + 0%
反馈信息：
The dimension of is .
The dimension of is .
For a vector space , if is a finite spanning set for and is a linearly
independent set in , then
| | dim( ) | |,
where | | and | | are the number of elements of and , respectively.
A basis for a vector space is a linearly independent spanning set for .
The dimension of a vector space , with a finite spanning set, is the number of
vectors in a basis for .
P
7
P
7
V =
R
6
S = { ,…, }
v
1
v
4
V
W = span(S) V S
A = ( |⋯| )v
1
v
4
S
b ∈ R
6
Ax = b
Ax = 0
b
W
S
V
W
M
m,n
m× n
P
n
n+ 1
V S V T
V
T ≤ V ≤ S
T S T S
V V
V
V
2022/6/20 18:01 University of New South Wales - 成绩单
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试题
Q6 0/2.0 查看 原始作答 未过滤的答案
Select ALL functions which are linear transformations.
您的应答 正确应答
选项 3:
Suppose that is a matrix and
such that
, for all .
选项 4:
The function defined by
, for all
matrix. .
Here is the set of all real matrices
and is the identity matrix.
反馈信息：
选项 3: The function satisfies both the addition and
scalar multiplication conditions.
选项 4: The function does not map zero of the domain
to zero of the codomain.
The function defined
by
, for all
.
Suppose that is a matrix
and such that
, for all .
The function defined
by
,
for all .
自动评分的 评分: 0/1.0
总分： 0.0×1/1 = 0%
A 3 × 2
T : →R
2
R
3
T (v) = Av v ∈ R
2
F : →M
22
M
22
F (A) = 3A+ 4I 2 × 2
A
M
22
2 × 2
I 2 × 2
T : →R
3
R
2
T
⎛
⎝
⎜
⎜
x
1
x
2
x
3
⎞
⎠
⎟
⎟
=
( )
−4 − 3 −x
1
x
2
x
3
−3 +x
2
x
3
⎛
⎝
⎜
⎜
x
1
x
2
x
3
⎞
⎠
⎟
⎟
∈ R
3
A 3 × 2
T : →R
2
R
3
T (v) = Av v ∈ R
2
T : →R
3
P
2
T
⎛
⎝
⎜
⎜
x
1
x
2
x
3
⎞
⎠
⎟
⎟
=
+ (2 + ) t+ ( + 3 ) x
1
x
1
x
3
x
2
x
3
t
2
⎛
⎝
⎜
⎜
x
1
x
2
x
3
⎞
⎠
⎟
⎟
∈ R
3
2022/6/20 18:01 University of New South Wales - 成绩单
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试题
Q7 1/1.0 查看 原始作答 未过滤的答案
Suppose that is a basis for , where
, , and .
Find the vector such that its coordinate vector is given by
.
Enter the vector in the box below, in Maple syntax.
您的应答 正确应答
<10,-7,16>
自动评分的 评分: 1/1.0
Note: The vector , in Maple syntax, should be entered as
总分： 1.0×1/1 = 100%
反馈信息：
If the coordinate vector of with respect to the basis is
, by definition,
B = { , , }v
1
v
2
v
3
R
3
=
v
1
⎛
⎝
⎜
1
4
−5
⎞
⎠
⎟
=
v
2
⎛
⎝
⎜
−6
4
−5
⎞
⎠
⎟
=
v
3
⎛
⎝
⎜
−1
5
1
⎞
⎠
⎟
v
[v =]
B
⎛
⎝
⎜
−1
−2
1
⎞
⎠
⎟
v
v =
⎛
⎝
⎜
a
b
c
⎞
⎠
⎟
v B = { , , }v
1
v
2
v
3
⎛
⎝
⎜
α
β
γ
⎞
⎠
⎟
v = α + β + γ .v
1
v
2
v
3
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第 1 页，共 1 页 1 - 7 行，共 7 行 - 0
试题
Q1 2/2.0 查看 原始作答 未过滤的答案
Let , , be vectors in .
Suppose that , , , .
For each of the following, select True if the statement is correct and select False if otherwise.
(a) The dimension of span is 2.
您的应答 正确应答
True
自动评分的 评分: 5/5.0
(b) The dimension of span is 3.
您的应答 正确应答
False
自动评分的 评分: 5/5.0
(c) The set is a basis for .
您的应答 正确应答
False
自动评分的 评分: 2/2.0
(d) The set is a basis for span .
您的应答 正确应答
False
自动评分的 评分: 4/4.0
(e) The set is a basis for span .
您的应答 正确应答
True
自动评分的 评分: 2/2.0
查看面板- 格式 数值
学生详细信息- Quiz 1 -- Week 3
分数: 历时: 40 min
7/10.0 已开始: 2022/6/20 AEST 上午12:24:19已结束: 2022/6/20 AEST 上午1:04:41
Enhua Zhang- 考试任务邮件地址: enhua.zhang@student.unsw.edu.au已结束: 0活动的: 0
学号: z5325328 等待人工阅卷: 0及格的: 0
=
v
1
⎛
⎝
⎜
⎜
⎜
5
−1
−4
−5
⎞
⎠
⎟
⎟
⎟
=
v
2
⎛
⎝
⎜
⎜
⎜
−3
−3
−4
5
⎞
⎠
⎟
⎟
⎟
=
v
3
⎛
⎝
⎜
⎜
⎜
−2
4
8
0
⎞
⎠
⎟
⎟
⎟
R
4
=e
1
⎛
⎝
⎜
⎜
⎜
1
0
0
0
⎞
⎠
⎟
⎟
⎟
=e
2
⎛
⎝
⎜
⎜
⎜
0
1
0
0
⎞
⎠
⎟
⎟
⎟
=e
3
⎛
⎝
⎜
⎜
⎜
0
0
1
0
⎞
⎠
⎟
⎟
⎟
=e
4
⎛
⎝
⎜
⎜
⎜
0
0
0
1
⎞
⎠
⎟
⎟
⎟
( , )v
1
v
2
( , , )v
1
v
2
v
3
{ , , }v
1
v
2
v
3
R
4
{ , , }v
1
v
2
v
3
( , , )v
1
v
2
v
3
{ , }v
1
v
2
( , , )v
1
v
2
v
3
2022/6/20 18:01 University of New South Wales - 成绩单
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试题
(f) The set is a basis for .
您的应答 正确应答
False
自动评分的 评分: 2/2.0
总分： 1.0×5/20 + 1.0×5/20 + 1.0×2/20 + 1.0×4/20 + 1.0×2/20 + 1.0×2/20 = 25% + 25% + 10% + 20% + 10% +
10%
反馈信息：
(a) is true. The two non-zero vectors are not scalar multiple of one another. The set
is linearly independent and it spans span .
(b) is false. The vector is a linear combination of . Hence the dimension of
span is the same as the dimension of span .
(c) is false. The dimension of is 4. The set of three vectors cannot be a basis.
(d) is false. The set is linearly dependent because is a linear
combination of . The set cannot be a basis for the span.
(e) is true. span = span , and part (a).
(f) is false. The dimension of is 4. The set of five vectors cannot be a basis.
Q2 1/1.0 查看 原始作答 未过滤的答案
Let be invertible matrices of the same size. Suppose that
is symmetric, i.e. ;
is skew-symmetric, i.e. ; and
is orthogonal, i.e. .
您的应答 正确应答
自动评分的 评分: 1/1.0
总分： 1.0×1/1 = 100%
反馈信息：
Note, in general, .
The following properties are useful.
, and .
{ , , , , }v
1
v
2
e
2
e
3
e
4
R
4
{ , }v
1
v
2
( , )v
1
v
2
v
3
{ , }v
1
v
2
( , , )v
1
v
2
v
3
( , )v
1
v
2
R
4
{ , , }v
1
v
2
v
3
v
3
{ , }v
1
v
2
( , )v
1
v
2
( , , )v
1
v
2
v
3
R
4
D, E, F
D = DD
T
E = −EE
T
F =F
T
F
−1
(D (ED F =E
−1
F
T
)
−1
)
T
−F F
E
2
AB ≠ BA
(AB =)
−1
B
−1
A
−1
(AB =)
T
B
T
A
T
2022/6/20 18:01 University of New South Wales - 成绩单
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试题
Q3 1/1.0 查看 原始作答 未过滤的答案
Given that is a matrix and are vectors in such that
where the matrix is invertible. Find the matrix such that
.
Enter the matrix in the box below.
您的应答 正确应答
自动评分的 评分: 1/1.0
Note: Do not enter M = in the answer box.
总分： 1.0×1/1 = 100%
Q4 1/1.0 查看 原始作答 未过滤的答案
Select all equations with infinitely many solutions.
选项 选中的
, where ,
No [答案保留]
, where ,
No [答案保留]
, where ,
No [答案保留]
, where ,
Yes [答案保留]
, where ,
Yes [答案保留]
, where ,
No [答案保留]
, where ,
Yes [答案保留]
, where ,
No [答案保留]
反馈
信
息：
For each of the equations, the augmented matrix is already in row
echelon form.
The equation has infinitely many solutions when the right hand
column of the row-echelon form is non-leading and there is at least one column
on the left is non-leading.
A
, ,
v
1
v
2
v
3
R
3
⎧
⎩
⎨
⎪
⎪
A = 2v
1
v
1
A = −7 + 5v
2
v
1
v
2
A = 6 − 7 − 5v
3
v
1
v
2
v
3
P = ( | | )
v
1
v
2
v
3
M
AP = PM
M
⎡
⎣
⎢
2
0
0
−7
5
0
6
−7
−5
⎤
⎦
⎥
Ax = b A =
⎛
⎝
⎜
4
0
0
5
0
0
5
0
0
⎞
⎠
⎟
b =
⎛
⎝
⎜
2
−1
0
⎞
⎠
⎟
Ax = b A =
⎛
⎝
⎜
5
0
0
5
4
0
3
1
−1
⎞
⎠
⎟
b =
⎛
⎝
⎜
2
2
5
⎞
⎠
⎟
Ax = b A =
⎛
⎝
⎜
4
0
0
1
5
0
−1
0
3
⎞
⎠
⎟
b =
⎛
⎝
⎜
2
5
0
⎞
⎠
⎟
Ax = b A =( )
4
0
5
0
b =( )
−1
0
Ax = b A =
⎛
⎝
⎜
4
0
0
2
5
0
2
5
0
2
2
5
5
0
−1
⎞
⎠
⎟
b =
⎛
⎝
⎜
2
5
2
⎞
⎠
⎟
Ax = b A =
⎛
⎝
⎜
⎜
⎜
4
0
0
0
1
5
0
0
−1
0
3
0
⎞
⎠
⎟
⎟
⎟
b =
⎛
⎝
⎜
⎜
⎜
2
5
2
0
⎞
⎠
⎟
⎟
⎟
Ax = b A =
⎛
⎝
⎜
5
0
0
1
−1
0
5
2
0
4
2
4
⎞
⎠
⎟
b =
⎛
⎝
⎜
−1
0
5
⎞
⎠
⎟
Ax = b A =
⎛
⎝
⎜
3
0
0
5
−1
0
5
2
0
⎞
⎠
⎟
b =
⎛
⎝
⎜
2
2
4
⎞
⎠
⎟
(A|b)
Ax = b
2022/6/20 18:01 University of New South Wales - 成绩单
https://unsw.mobius.cloud/gradebook/Details.do?userId=16183&trId=3082776 4/6
试题
Q5 1/2.0 查看 原始作答 未过滤的答案
(a) Let be the vector space of all the polynomials of degree 7 or less.
The dimension of is
您的应答 正确应答
8 8
自动评分的 评分: 5/5.0
.
(b) Let and be a set of 4 vectors in .
Suppose that is the subspace of spanned by .
Let be the matrix with the vectors in as columns,
and .
The system is inconsistent. There is no solution.
The system has infinitely many solutions.
Fill in the following answer fields using drop down menus.
The vector
您的应答 正确应答
is not is not
自动评分的 评分: 5/5.0
in the span of .
The set is a linearly
您的应答 正确应答
independent dependent
自动评分的 评分: 0/5.0
set in .
The dimension of
您的应答 正确应答
is 4 is less than 4
自动评分的 评分: 0/5.0
.
总分： 1.0×5/20 + 1.0×5/20 + 0.0×5/20 + 0.0×5/20 = 25% + 25% + 0% + 0%
反馈信息：
The dimension of is .
The dimension of is .
For a vector space , if is a finite spanning set for and is a linearly
independent set in , then
| | dim( ) | |,
where | | and | | are the number of elements of and , respectively.
A basis for a vector space is a linearly independent spanning set for .
The dimension of a vector space , with a finite spanning set, is the number of
vectors in a basis for .
P
7
P
7
V =
R
6
S = { ,…, }
v
1
v
4
V
W = span(S) V S
A = ( |⋯| )v
1
v
4
S
b ∈ R
6
Ax = b
Ax = 0
b
W
S
V
W
M
m,n
m× n
P
n
n+ 1
V S V T
V
T ≤ V ≤ S
T S T S
V V
V
V
2022/6/20 18:01 University of New South Wales - 成绩单
https://unsw.mobius.cloud/gradebook/Details.do?userId=16183&trId=3082776 5/6
试题
Q6 0/2.0 查看 原始作答 未过滤的答案
Select ALL functions which are linear transformations.
您的应答 正确应答
选项 3:
Suppose that is a matrix and
such that
, for all .
选项 4:
The function defined by
, for all
matrix. .
Here is the set of all real matrices
and is the identity matrix.
反馈信息：
选项 3: The function satisfies both the addition and
scalar multiplication conditions.
选项 4: The function does not map zero of the domain
to zero of the codomain.
The function defined
by
, for all
.
Suppose that is a matrix
and such that
, for all .
The function defined
by
,
for all .
自动评分的 评分: 0/1.0
总分： 0.0×1/1 = 0%
A 3 × 2
T : →R
2
R
3
T (v) = Av v ∈ R
2
F : →M
22
M
22
F (A) = 3A+ 4I 2 × 2
A
M
22
2 × 2
I 2 × 2
T : →R
3
R
2
T
⎛
⎝
⎜
⎜
x
1
x
2
x
3
⎞
⎠
⎟
⎟
=
( )
−4 − 3 −x
1
x
2
x
3
−3 +x
2
x
3
⎛
⎝
⎜
⎜
x
1
x
2
x
3
⎞
⎠
⎟
⎟
∈ R
3
A 3 × 2
T : →R
2
R
3
T (v) = Av v ∈ R
2
T : →R
3
P
2
T
⎛
⎝
⎜
⎜
x
1
x
2
x
3
⎞
⎠
⎟
⎟
=
+ (2 + ) t+ ( + 3 ) x
1
x
1
x
3
x
2
x
3
t
2
⎛
⎝
⎜
⎜
x
1
x
2
x
3
⎞
⎠
⎟
⎟
∈ R
3
2022/6/20 18:01 University of New South Wales - 成绩单
https://unsw.mobius.cloud/gradebook/Details.do?userId=16183&trId=3082776 6/6
试题
Q7 1/1.0 查看 原始作答 未过滤的答案
Suppose that is a basis for , where
, , and .
Find the vector such that its coordinate vector is given by
.
Enter the vector in the box below, in Maple syntax.
您的应答 正确应答
<10,-7,16>
自动评分的 评分: 1/1.0
Note: The vector , in Maple syntax, should be entered as
总分： 1.0×1/1 = 100%
反馈信息：
If the coordinate vector of with respect to the basis is
, by definition,
B = { , , }v
1
v
2
v
3
R
3
=
v
1
⎛
⎝
⎜
1
4
−5
⎞
⎠
⎟
=
v
2
⎛
⎝
⎜
−6
4
−5
⎞
⎠
⎟
=
v
3
⎛
⎝
⎜
−1
5
1
⎞
⎠
⎟
v
[v =]
B
⎛
⎝
⎜
−1
−2
1
⎞
⎠
⎟
v
v =
⎛
⎝
⎜
a
b
c
⎞
⎠
⎟
v B = { , , }v
1
v
2
v
3
⎛
⎝
⎜
α
β
γ
⎞
⎠
⎟
v = α + β + γ .v
1
v
2
v
3
