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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
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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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.
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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