Please solve all below. -7 49 3 Let u = -3 and A= 0 1 -1 Is u in the subset of R³ spanned by the columns of A? Why or why not? 11 31 8 Select the correct choice below and fill in the answer box to complete your choice. (Type an integer or decimal for each matrix element.) O A. Yes, multiplying A by the vector writes u as a linear combination of the columns of A. OB. No, the reduced echelon form of the augmented matrix is , which is an inconsistent system. This is what I get when I try to solve. My example below but was unable to get the correct answer. Please help. (Do not solve below this statement. Solve the problem above.)
Please solve all below. -7 49 3 Let u = -3 and A= 0 1 -1 Is u in the subset of R³ spanned by the columns of A? Why or why not? 11 31 8 Select the correct choice below and fill in the answer box to complete your choice. (Type an integer or decimal for each matrix element.) O A. Yes, multiplying A by the vector writes u as a linear combination of the columns of A. OB. No, the reduced echelon form of the augmented matrix is , which is an inconsistent system. This is what I get when I try to solve. My example below but was unable to get the correct answer. Please help. (Do not solve below this statement. Solve the problem above.)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
Please solve and show all work.
![Please solve all below.
-7
49 3
3
Let u = -3 and A= 0 1 -1. Is u in the subset of R³ spanned by the columns of A? Why or why not?
OB
11
31 8
Select the correct choice below and fill in the answer box to complete your choice.
(Type an integer or decimal for each matrix element.)
OA. Yes, multiplying A by the vector writes u as a linear combination of the columns of A.
OB. No, the reduced echelon form of the augmented matrix is , which is an inconsistent system.
This is what I get when I try to solve. My example below but was unable to get the correct answer.
Please help. (Do not solve below this statement. Solve the problem above.)
The matrix in reduced echelon form is 01-60
01
00
A system is consistent if an echelon form of the corresponding matrix does not have a row of the form
10 120
[o 0 b] with b nonzero. The matrix 0 1 -6 0 has such a row; the third row is [0 0 0 1]
00 0 1
To find the vector of weights, use the reduced echelon form. The value of the first weight is the value found in
the rightmost column of the first row. The value of the second weight is the value found in the rightmost column
of the second row, and so on. If a row is all zeros, then there is no weight for that row. Because the system
is inconsistent, the vector x does not exist.
The vector u is not in the subset of R³ spanned by the columns of A because u cannot be written as a linear
combination of the columns of A.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc3fa098a-800d-440b-b28a-d62ad7aee48d%2Ff233cfa5-1b0b-4f70-9288-e452e4589be5%2Ftjrdoow_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Please solve all below.
-7
49 3
3
Let u = -3 and A= 0 1 -1. Is u in the subset of R³ spanned by the columns of A? Why or why not?
OB
11
31 8
Select the correct choice below and fill in the answer box to complete your choice.
(Type an integer or decimal for each matrix element.)
OA. Yes, multiplying A by the vector writes u as a linear combination of the columns of A.
OB. No, the reduced echelon form of the augmented matrix is , which is an inconsistent system.
This is what I get when I try to solve. My example below but was unable to get the correct answer.
Please help. (Do not solve below this statement. Solve the problem above.)
The matrix in reduced echelon form is 01-60
01
00
A system is consistent if an echelon form of the corresponding matrix does not have a row of the form
10 120
[o 0 b] with b nonzero. The matrix 0 1 -6 0 has such a row; the third row is [0 0 0 1]
00 0 1
To find the vector of weights, use the reduced echelon form. The value of the first weight is the value found in
the rightmost column of the first row. The value of the second weight is the value found in the rightmost column
of the second row, and so on. If a row is all zeros, then there is no weight for that row. Because the system
is inconsistent, the vector x does not exist.
The vector u is not in the subset of R³ spanned by the columns of A because u cannot be written as a linear
combination of the columns of A.
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