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Determine if w= 3 5 is in NulA where A = 2 -5 21 -19 2 13 -23 8 14 -1 OYes, w is in NulA because Aw is the zero vector. O Yes, w is in NulA because w is in R³ and NulAis 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.

Determine if w= 3 5 is in NulA where A = 2 -5 21 -19 2 13 -23 8 14 -1 OYes, w is in NulA because Aw is the zero vector. O Yes, w is in NulA because w is in R³ and NulAis 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.

Advanced Engineering Mathematics
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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Determine if w = 3 5 is in NulA where A = 2 -5 21 -19 13 -23 2 8 14 -1 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.
Transcribed Image Text:Determine if w = 3 5 is in NulA where A = 2 -5 21 -19 13 -23 2 8 14 -1 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.
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