2 1 7 4 and v = 6 Compute Au and Av, and compare them with b. Is it possible that at least one of u or v could be a Let A = -9 - 3 - 36 u= - 3 4 33 least-squares solution of Ax = b? (Answer this without computing a least-squares solution.) Au = (Simplify your answer.) Av = (Simplify your answer.) Compare Au and Av with b. Is it possible that at least one of u or v could be a least-squares solution of Ax = b? O A. Av is closer to b than Au is. Thus, u cannot be a least-squares solution of Ax = b, but v can be. O B. Au is closer to b than Av is. Thus, v cannot be a least-squares solution of Ax = b, but u can be. C. Au and Av are equally close to b. Thus, both can be the least-squares solution of Ax = b. O D. Au and Av are equally close to b. Thus, neither can be the least-squares solution of Ax = b.

2 1 7 4 and v = 6 Compute Au and Av, and compare them with b. Is it possible that at least one of u or v could be a Let A = -9 - 3 - 36 u= - 3 4 33 least-squares solution of Ax = b? (Answer this without computing a least-squares solution.) Au = (Simplify your answer.) Av = (Simplify your answer.) Compare Au and Av with b. Is it possible that at least one of u or v could be a least-squares solution of Ax = b? O A. Av is closer to b than Au is. Thus, u cannot be a least-squares solution of Ax = b, but v can be. O B. Au is closer to b than Av is. Thus, v cannot be a least-squares solution of Ax = b, but u can be. C. Au and Av are equally close to b. Thus, both can be the least-squares solution of Ax = b. O D. Au and Av are equally close to b. Thus, neither can be the least-squares solution of Ax = b.

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

2
1
7
4
and v =
6
Compute Au and Av, and compare them with b. Is it possible that at least one of u or v could be a
Let A =
-9 - 3
- 36
u=
- 3
4
33
least-squares solution of Ax = b?
(Answer this without computing a least-squares solution.)
Au =
(Simplify your answer.)
Av =
(Simplify your answer.)
Compare Au and Av with b. Is it possible that at least one of u or v could be a least-squares solution of Ax = b?
O A. Av is closer to b than Au is. Thus, u cannot be a least-squares solution of Ax = b, but v can be.
O B. Au is closer to b than Av is. Thus, v cannot be a least-squares solution of Ax = b, but u can be.
C. Au and Av are equally close to b. Thus, both can be the least-squares solution of Ax = b.
O D. Au and Av are equally close to b. Thus, neither can be the least-squares solution of Ax = b.

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