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
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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
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
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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