(X1, Y1 Z1) + (x2; Y2 Z2) = (X1.+ X2+ 1, Y1+ Y2+ 1, 21 + 22+ 1) c(x, y, z) = (cx +c- 1, cy + c - 1, cz + c- 1) (d) O The set is a vector space. O The set is not a vector space because the additive identity property is not satisfied. O The set is not a vector space because it is not closed under scalar multiplication. O The set is not a vector space because the distributive property is not satisfied. O The set is not a vector space because the multiplicative identity property is not satisfied.

Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Is r3 a vector space?

 

Ñ#$

(d) (x1 Y1, Z1)
(X2, Y21 Z2) = (X1.+ X2 + 1, y1 + V2+ 1, 21 + Z2 + 1)
C(x, y, z) = (cx +C - 1, cy +c- 1, cz +C- 1)
%3D
%3D
The set is a vector space.
The set is not a vector space because the additive identity property is not satisfied.
The set is not a vector space because it is not closed under scalar multiplication.
OThe set is not a vector space because the distributive property is not satisfied.
The set is not a vector space because the multiplicative identity property is not satisfied.
Transcribed Image Text:(d) (x1 Y1, Z1) (X2, Y21 Z2) = (X1.+ X2 + 1, y1 + V2+ 1, 21 + Z2 + 1) C(x, y, z) = (cx +C - 1, cy +c- 1, cz +C- 1) %3D %3D The set is a vector space. The set is not a vector space because the additive identity property is not satisfied. The set is not a vector space because it is not closed under scalar multiplication. OThe set is not a vector space because the distributive property is not satisfied. The set is not a vector space because the multiplicative identity property is not satisfied.
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