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Determine whether w= 1 is in the column space of A = 7 -5 9 19 O The product of A and w is something that indicates that w is not in COLA. O Yes, because when solving for Ax = 0, we realize that w is actually a solution. O The augmented matrix [A O The augmented matrix [A 6-4 1 -1 0-2 -11 7-3 7 1 -9 31 w] helps solves Ax= w. So finding RREF of [Aw], we realize that [Aw] is consistent, hence yes, w E Col4. w] helps solves Ax= w. So finding RREF of A w], we realize that A w is inconsistent, hence no, w is not in ColA

Determine whether w= 1 is in the column space of A = 7 -5 9 19 O The product of A and w is something that indicates that w is not in COLA. O Yes, because when solving for Ax = 0, we realize that w is actually a solution. O The augmented matrix [A O The augmented matrix [A 6-4 1 -1 0-2 -11 7-3 7 1 -9 31 w] helps solves Ax= w. So finding RREF of [Aw], we realize that [Aw] is consistent, hence yes, w E Col4. w] helps solves Ax= w. So finding RREF of A w], we realize that A w is inconsistent, hence no, w is not in ColA

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 whether w= -3 is in the column space of A = 7 -5 6-4 1 0-2 7-3 -1 9 -11 19 -9 7 1 O The product of A and w is something that indicates that w is not in COLA. OYes, because when solving for Ax = 0, we realize that w is actually a solution. O The augmented matrix [Aw] helps solves Ax= w. So finding RREF of [Aw], we realize that [Aw] is consistent, hence yes, w E COLA. The augmented matrix [Aw] helps solves Ax=w. So finding RREF of [Aw], we realize that [Aw] is inconsistent, hence no, w is not in COLA.
Transcribed Image Text:Determine whether w= -3 is in the column space of A = 7 -5 6-4 1 0-2 7-3 -1 9 -11 19 -9 7 1 O The product of A and w is something that indicates that w is not in COLA. OYes, because when solving for Ax = 0, we realize that w is actually a solution. O The augmented matrix [Aw] helps solves Ax= w. So finding RREF of [Aw], we realize that [Aw] is consistent, hence yes, w E COLA. The augmented matrix [Aw] helps solves Ax=w. So finding RREF of [Aw], we realize that [Aw] is inconsistent, hence no, w is not in COLA.
Determine whether u is in the column space of A. If it is, write u as a linear combination of the column vectors of A. A = O 5 4 U= U= u= -17 -25 u= -171 -25 17 25 -17 -25 is in the column space of A and is in the column space of A and is not in the column space of A - O None of these choices. -17 -25 -17 -25 5 -5 [3] + ] = -5 [5] + +2 +2 5 [1]
Transcribed Image Text:Determine whether u is in the column space of A. If it is, write u as a linear combination of the column vectors of A. A = O 5 4 U= U= u= -17 -25 u= -171 -25 17 25 -17 -25 is in the column space of A and is in the column space of A and is not in the column space of A - O None of these choices. -17 -25 -17 -25 5 -5 [3] + ] = -5 [5] + +2 +2 5 [1]
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