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(iii) {(x, y, z) = R³: (x, y, z). (1, 1, 1) = 0} = {(x, y, z) = R³ : x+y+z = 0} = {(1,-1,0), (1, 0, -1)} and these two vectors give a basis.

(iii) {(x, y, z) = R³: (x, y, z). (1, 1, 1) = 0} = {(x, y, z) = R³ : x+y+z = 0} = {(1,-1,0), (1, 0, -1)} and these two vectors give a basis.

Calculus: Early Transcendentals
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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Can you explain how we got the L part
(iii) {(x, y, z) = R³: (x, y, z). (1, 1, 1) = 0} = {(x, y, z) = R³ : x+y+z = 0} = {(1,-1,0), (1, 0, -1)} and these two vectors give a basis. (ixz)
Transcribed Image Text:(iii) {(x, y, z) = R³: (x, y, z). (1, 1, 1) = 0} = {(x, y, z) = R³ : x+y+z = 0} = {(1,-1,0), (1, 0, -1)} and these two vectors give a basis. (ixz)
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