1001 01 10 1-1Find a basis of the row space of A=by reducing A to reduced echelon form.1012 20 200-2]O2B= ((100 10].[0 13 0 1])Ob.B= {[100 1 0).[1101-1).[10 1 2 2]}OcB={[10110),[0 10 0 -1).[00 10 2 )}Od. none of theseOe.B= ([100 10),[0 100 -1).[0 0 1 1 2 ]}

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1001 0 110 1-1 Find a basis of the row space of A= by reducing A to reduced echelon form. 1012 2 0200-2J O2B= ((100 10].[0 130 1]) Ob.B= {(10 01 0).[110 1-1).[101 2 2]} OcB-{[10110),[0 100 -1).[00 10 2 )} Od. none of these Oe.B= ([100 10],[0 100 -1).[0 0 1 1 2 ]}

1001 0 110 1-1 Find a basis of the row space of A= by reducing A to reduced echelon form. 1012 2 0200-2J O2B= ((100 10].[0 130 1]) Ob.B= {(10 01 0).[110 1-1).[101 2 2]} OcB-{[10110),[0 100 -1).[00 10 2 )} Od. none of these Oe.B= ([100 10],[0 100 -1).[0 0 1 1 2 ]}

Linear Algebra: A Modern Introduction

4th Edition

ISBN:9781285463247

Author:David Poole

Publisher:David Poole

Chapter5: Orthogonality

Section: Chapter Questions

Problem 20RQ

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1001 0
1 10 1-1
Find a basis of the row space of A=
by reducing A to reduced echelon form.
1012 2
0 200-2]
O2B= ((100 10].[0 13 0 1])
Ob.B= {[100 1 0).[1101-1).[10 1 2 2]}
OcB={[10110),[0 10 0 -1).[00 10 2 )}
Od. none of these
Oe.B= ([100 10),[0 100 -1).[0 0 1 1 2 ]}

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