Determine if w = -5 is in NulA where A = 2 5 13 21 23 8 14 19 2 O Yes, w is in NulA because Aw is the zero vector. O Yes, w is in NulA because w is in R³ and Nul Ais a subspace of R³. O No, w is not in NulA because w can only be in R³. O No, w is not in NulA because Aw is not the zero vector. 27 1
Determine if w = -5 is in NulA where A = 2 5 13 21 23 8 14 19 2 O Yes, w is in NulA because Aw is the zero vector. O Yes, w is in NulA because w is in R³ and Nul Ais a subspace of R³. O No, w is not in NulA because w can only be in R³. O No, w is not in NulA because Aw is not the zero vector. 27 1
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![Determine if w =
-5 is in NulA where A =
2
5 21 19
23
14
13
8
O Yes, w is in NulA because Aw is the zero vector.
O Yes, w is in NulA because w is in R³ and Nul Ais a subspace of R³.
O No, w is not in NulA because w can only be in R³.
No, w is not in NulA because Aw is not the zero vector.
21
2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F018dde93-f85e-4b68-97c4-7858fe4b7f30%2F496f70e8-eec3-428a-af92-a74ae196c2f1%2F4ahgov7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Determine if w =
-5 is in NulA where A =
2
5 21 19
23
14
13
8
O Yes, w is in NulA because Aw is the zero vector.
O Yes, w is in NulA because w is in R³ and Nul Ais a subspace of R³.
O No, w is not in NulA because w can only be in R³.
No, w is not in NulA because Aw is not the zero vector.
21
2
![Determine if w =
3
-5 is in NulA where A =
2
5
13 23
8 14
21 19
OYes, w is in NulA because Aw is the zero vector.
O Yes, w is in NulA because w is in R³ and Nul Ais a subspace of R³.
O No, w is not in NulA because w can only be in R³.
O No, w is not in NulA because Aw is not the zero vector.
921](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F018dde93-f85e-4b68-97c4-7858fe4b7f30%2F496f70e8-eec3-428a-af92-a74ae196c2f1%2Fwcd3jz_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Determine if w =
3
-5 is in NulA where A =
2
5
13 23
8 14
21 19
OYes, w is in NulA because Aw is the zero vector.
O Yes, w is in NulA because w is in R³ and Nul Ais a subspace of R³.
O No, w is not in NulA because w can only be in R³.
O No, w is not in NulA because Aw is not the zero vector.
921
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