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- 9 - 6 4 Let v, = 0 , V2 6 V3 = - 18 and p= Determine if p is in Nul A, where A = V1 V2 V3 18 9 9 8 Is p in Nul A? A. No, because Ap is not equal to the zero vector. В. Yes, because the augmented matrix A p is consistent. O C. Yes, because Ap is equal to the zero vector. D. No, because the augmented matrix [* А р is not consistent.

- 9 - 6 4 Let v, = 0 , V2 6 V3 = - 18 and p= Determine if p is in Nul A, where A = V1 V2 V3 18 9 9 8 Is p in Nul A? A. No, because Ap is not equal to the zero vector. В. Yes, because the augmented matrix A p is consistent. O C. Yes, because Ap is equal to the zero vector. D. No, because the augmented matrix [* А р is not consistent.

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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- 9 - 6 4 Let v1 = V2 = V3 18 and p = Determine if p is in Nul A, where A = v, V2 V3 18 9. 9 - 8 Is p in Nul A? A. No, because Ap is not equal to the zero vector. В. Yes, because the augmented matrix A is consistent. C. Yes, because Ap is equal to the zero vector. O D. No, because the augmented matrix A p is not consistent.
Transcribed Image Text:- 9 - 6 4 Let v1 = V2 = V3 18 and p = Determine if p is in Nul A, where A = v, V2 V3 18 9. 9 - 8 Is p in Nul A? A. No, because Ap is not equal to the zero vector. В. Yes, because the augmented matrix A is consistent. C. Yes, because Ap is equal to the zero vector. O D. No, because the augmented matrix A p is not consistent.
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