Consider matrix, 1-3 2-3 A= 2 0 3 1-2-4 LI --T 3-1 3 9 -9 7 -6 -6 Let us convect A in reduced row echelon form ANTI-3 2-3 333 -15 13 -3-3-9 0 6-3 9 0-10 5 6 AN 10 1/2 3/2 9/27 1/2 อว 18 0100 -000 AN 3/2 -12-12 0 1/2 3/2 9/2 1 -1/2 3/2 1/2 0 -12-12 000-000-0 0-000000 ANTIO 1/20 1-1/20 3 60000-00 -000-000 100 by R2-2R1, R3+2R1, R₂-R, --by R4-R2, R₁+1 R₂ R3+ 10 R2 then 16 R2 by Rs Ry 18 I 1/2 0 -1/2 Thus 1st, 2nd, 4th 18 by R3 then R1-R3, R2-3/2R3 --by to R4 then R1-3R4, R2-R4, R3-R4 & 5th column contains pivet. so only { U, V2, V4, Vs) are linearly independent. So Span(s) Span VN2 N3 N4 N5] 1 = span [V1, V2, V4, US} so basis of span(s) is, {V1, V2, V4, V5]

Consider matrix, 1-3 2-3 A= 2 0 3 1-2-4 LI --T 3-1 3 9 -9 7 -6 -6 Let us convect A in reduced row echelon form ANTI-3 2-3 333 -15 13 -3-3-9 0 6-3 9 0-10 5 6 AN 10 1/2 3/2 9/27 1/2 อว 18 0100 -000 AN 3/2 -12-12 0 1/2 3/2 9/2 1 -1/2 3/2 1/2 0 -12-12 000-000-0 0-000000 ANTIO 1/20 1-1/20 3 60000-00 -000-000 100 by R2-2R1, R3+2R1, R₂-R, --by R4-R2, R₁+1 R₂ R3+ 10 R2 then 16 R2 by Rs Ry 18 I 1/2 0 -1/2 Thus 1st, 2nd, 4th 18 by R3 then R1-R3, R2-3/2R3 --by to R4 then R1-3R4, R2-R4, R3-R4 & 5th column contains pivet. so only { U, V2, V4, Vs) are linearly independent. So Span(s) Span VN2 N3 N4 N5] 1 = span [V1, V2, V4, US} so basis of span(s) is, {V1, V2, V4, V5]

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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