Let A be an mxn matrix. Consider the statement. "For each b in R", the equation Ax = b has a solution." Because of a fundamental theorem about such matrix equations, this statement is equivalent to wha statements? Choose all that apply. OA. The rows of A span R". B. Each b in RM is a linear combination of the columns of A. Oc. The columns of A span Rm. OD. The matrix A has a pivot position in each column. D E. The matrix A has a pivot position in each row. Let A be the nxk matrix [v, v2 ... Vk] whose columns are the vectors of S. Since A has fower | there cannot be a pivot position in each than of A. Therefore, by the above theorem, the | of A do not span Why is the proof completo? O A. The proof is complete because a basis for a subspace must span that subspace. O B. The proof is complete by the Spanning Set Theorem. O C. The proof is complete because the subspace spanned by S is RK. O D. The proof is complete because a basis for a subspace must consist of linearly independent vectors.
Let A be an mxn matrix. Consider the statement. "For each b in R", the equation Ax = b has a solution." Because of a fundamental theorem about such matrix equations, this statement is equivalent to wha statements? Choose all that apply. OA. The rows of A span R". B. Each b in RM is a linear combination of the columns of A. Oc. The columns of A span Rm. OD. The matrix A has a pivot position in each column. D E. The matrix A has a pivot position in each row. Let A be the nxk matrix [v, v2 ... Vk] whose columns are the vectors of S. Since A has fower | there cannot be a pivot position in each than of A. Therefore, by the above theorem, the | of A do not span Why is the proof completo? O A. The proof is complete because a basis for a subspace must span that subspace. O B. The proof is complete by the Spanning Set Theorem. O C. The proof is complete because the subspace spanned by S is RK. O D. The proof is complete because a basis for a subspace must consist of linearly independent vectors.
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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![Let S= (V1,., Vk be a set of k vectors in R", with k<n. Use a theorem about the matrix equation Ax = b to explain why S cannot be a basis for R".
Let A be an mxn matrix. Consider the statement. "For each b in Rm, the equation Ax = b has a solution." Because of a fundamental theorem about such matrix equations, this statement is equivalent to what other
statements? Choose all that apply.
O A. The rows of A span R".
O B. Each b in Rm is a linear combination of the columns of A.
O C. The columns of A span Rm.
O D. The matrix A has a pivot position
each column.
O E. The matrix A has a pivot position in each row.
Let A be the n xk matrix [V V2 ... Vk] whose columns are the vectors of S. Since A has fewer
than
V there cannot be a pivot position in each
V of A.
Therefore, by the above theorem, the
V of A do not span
Why is the proof complete?
O A
The proof is complete because a basis for a subspace must span that subspace.
O B. The proof is complete by the Spanning Set Theorem.
O C. The proof is complete because the subspace spanned by S is RK.
O D. The proof is complete because a basis for a subspace must consist
linearly independent vectors.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffba142dd-1494-4b78-9e41-2122df5fca48%2Fce6cde6d-24da-4fd0-ba8f-19ed390deec1%2Fe0asvmh_processed.png&w=3840&q=75)
Transcribed Image Text:Let S= (V1,., Vk be a set of k vectors in R", with k<n. Use a theorem about the matrix equation Ax = b to explain why S cannot be a basis for R".
Let A be an mxn matrix. Consider the statement. "For each b in Rm, the equation Ax = b has a solution." Because of a fundamental theorem about such matrix equations, this statement is equivalent to what other
statements? Choose all that apply.
O A. The rows of A span R".
O B. Each b in Rm is a linear combination of the columns of A.
O C. The columns of A span Rm.
O D. The matrix A has a pivot position
each column.
O E. The matrix A has a pivot position in each row.
Let A be the n xk matrix [V V2 ... Vk] whose columns are the vectors of S. Since A has fewer
than
V there cannot be a pivot position in each
V of A.
Therefore, by the above theorem, the
V of A do not span
Why is the proof complete?
O A
The proof is complete because a basis for a subspace must span that subspace.
O B. The proof is complete by the Spanning Set Theorem.
O C. The proof is complete because the subspace spanned by S is RK.
O D. The proof is complete because a basis for a subspace must consist
linearly independent vectors.
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