2 4 11 7 and v = 7 Compute Au and Av, and compare them with b. Could u possibly be a least-squares Let A = - 9 1 b = - 8 u= - 2 - 3 2 4 9 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. Could u possibly be a least-squares solution of Ax = b? O A. Av is closer to b than Au is. Thus, u cannot possibly be a least-squares solution of Ax = b. B. Au is closer to b than Av is. Thus, u could possibly be a least-squares solution of Ax = b. OC. Au and Av are equally close to b. Thus, both u and v can be a least-squares solution of Ax = b. D. Au and Av are equally close to b. Thus, neither u nor v can be a 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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2 4
11
7
and v =
7
Compute Au and Av, and compare them with b. Could u possibly be a least-squares
Let A =
- 9 1
b =
- 8
u=
- 2
- 3
2 4
9
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. Could u possibly be a least-squares solution of Ax = b?
O A. Av is closer to b than Au is. Thus, u cannot possibly be a least-squares solution of Ax = b.
B. Au is closer to b than Av is. Thus, u could possibly be a least-squares solution of Ax = b.
OC. Au and Av are equally close to b. Thus, both u and v can be a least-squares solution of Ax = b.
D. Au and Av are equally close to b. Thus, neither u nor v can be a least-squares solution of Ax =b.
Transcribed Image Text:2 4 11 7 and v = 7 Compute Au and Av, and compare them with b. Could u possibly be a least-squares Let A = - 9 1 b = - 8 u= - 2 - 3 2 4 9 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. Could u possibly be a least-squares solution of Ax = b? O A. Av is closer to b than Au is. Thus, u cannot possibly be a least-squares solution of Ax = b. B. Au is closer to b than Av is. Thus, u could possibly be a least-squares solution of Ax = b. OC. Au and Av are equally close to b. Thus, both u and v can be a least-squares solution of Ax = b. D. Au and Av are equally close to b. Thus, neither u nor v can be a least-squares solution of Ax =b.
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