7 8 11 Let A = -5 1 and v = Compute Au and Av, and compare them with b. Could u possibly be a least-squares 7 8 3 solution of Ax = b? (Answer this without computing a least-squares solution.) Au = |(Simplify your answer.) Ay = |(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. O 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. O D. Au and Av are equally close to b. Thus, neither u nor v can be a least-squares solution of Ax = b.
7 8 11 Let A = -5 1 and v = Compute Au and Av, and compare them with b. Could u possibly be a least-squares 7 8 3 solution of Ax = b? (Answer this without computing a least-squares solution.) Au = |(Simplify your answer.) Ay = |(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. O 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. O 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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Transcribed Image Text:7 8
11
Let A =
- 5 1
b =
- 8
u =
and v =
Compute Au and Av, and compare them with b. Could u possibly be a least-squares
-8
7 8
3
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
O B. Au is closer to b than Av is. Thus, u could possibly be a least-squares solution of Ax = b.
O C. Au and Av are equally close to b. Thus, both u and v can be a least-squares solution of Ax = b.
OD.
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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