The Ax = b system is "solvable" only if b is a (a) Explain why the linear system Ax=b where A = 4 4 -1 and b = 0 [Hint: Examine the dot product of the vector b and the column vectors of A.] of A. is not solvable. (b) The vector bESCR [b being the vector given in part (a), and S a subspace]. Can we identify S as one of the four subspaces of A in part(a)? Is so, which one? If not, explain why not.
The Ax = b system is "solvable" only if b is a (a) Explain why the linear system Ax=b where A = 4 4 -1 and b = 0 [Hint: Examine the dot product of the vector b and the column vectors of A.] of A. is not solvable. (b) The vector bESCR [b being the vector given in part (a), and S a subspace]. Can we identify S as one of the four subspaces of A in part(a)? Is so, which one? If not, explain why not.
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
![The Ax=b system is "solvable" only if b is a
2
-3
(a) Explain why the linear system Ax=b where A =
1
0
-1
[Hint: Examine the dot product of the vector b and the column vectors of A.]
4
4
and b =
1
of A.
is not solvable.
(b) The vector beSCR [b being the vector given in part (a), and S a subspace]. Can we
identify S as one of the four subspaces of A in part(a)? Is so, which one? If not, explain
why not.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa2e9a44e-bc50-46ba-94ac-b8037341f466%2F7f7b8dec-c1e1-4d43-9b17-04820c4bd779%2F18ag2g_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The Ax=b system is "solvable" only if b is a
2
-3
(a) Explain why the linear system Ax=b where A =
1
0
-1
[Hint: Examine the dot product of the vector b and the column vectors of A.]
4
4
and b =
1
of A.
is not solvable.
(b) The vector beSCR [b being the vector given in part (a), and S a subspace]. Can we
identify S as one of the four subspaces of A in part(a)? Is so, which one? If not, explain
why not.
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