Let W = {(a, b, c, d)| 5a – c = 0 } be a subspace of R. Then a basis for W is O {(5,1,0,0),(0,0,1,0),(0,0,0,1)} O {(1,0,5,0)} O ((1,0,5,0).(0,1,0,0),(0,0,0,1)) O (1,0,5,0).(0,0,0.1)}
Let W = {(a, b, c, d)| 5a – c = 0 } be a subspace of R. Then a basis for W is O {(5,1,0,0),(0,0,1,0),(0,0,0,1)} O {(1,0,5,0)} O ((1,0,5,0).(0,1,0,0),(0,0,0,1)) O (1,0,5,0).(0,0,0.1)}
Elementary Linear Algebra (MindTap Course List)
8th Edition
ISBN:9781305658004
Author:Ron Larson
Publisher:Ron Larson
Chapter4: Vector Spaces
Section4.3: Subspaces Of Vector Spaces
Problem 52E
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![Let W = {(a, b, c, d)| 5a – c = 0 } be a subspace of R. Then a basis for W is
O {(5,1,0,0).(0,0,1,0),(0,0,0,1)}
O {(1,0,5,0)}
O (1,0,5,0).(0,1,0,0).(0,0,0,1)}
O (1,0,5,0).(0,0,0.1)}
Letk be a real numberand A=
Then determinant of A is](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F38e89155-bbae-48a0-ab72-049822bea051%2F9f7a6ec1-82ad-47ba-b68b-35d67ef1afe3%2Fs8s68ir_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Let W = {(a, b, c, d)| 5a – c = 0 } be a subspace of R. Then a basis for W is
O {(5,1,0,0).(0,0,1,0),(0,0,0,1)}
O {(1,0,5,0)}
O (1,0,5,0).(0,1,0,0).(0,0,0,1)}
O (1,0,5,0).(0,0,0.1)}
Letk be a real numberand A=
Then determinant of A is
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