c) (X1, Y1, Z1) + (x2, Y2, z2) = (x1 + X2 + 3, yı + y2 + 3, z1 + z2 + 3) (сх, су, с2) с(х, у, 2) %3D O The set is a vector space. O The set is not a vector space because the additive identity property is not satisfied. The set is not a vector space because the additive inverse 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. d) (X1, Y1, Z1) + (x2, Y2, z2) с (х, у, z) O The set is a vector space. (X1 + X2 + 7, Y1 + Y2 + 7, z1 + z2 + 7) (cx + 7c – 7, cy + 7c – 7, cz + 7c – 7) 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.
c) (X1, Y1, Z1) + (x2, Y2, z2) = (x1 + X2 + 3, yı + y2 + 3, z1 + z2 + 3) (сх, су, с2) с(х, у, 2) %3D O The set is a vector space. O The set is not a vector space because the additive identity property is not satisfied. The set is not a vector space because the additive inverse 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. d) (X1, Y1, Z1) + (x2, Y2, z2) с (х, у, z) O The set is a vector space. (X1 + X2 + 7, Y1 + Y2 + 7, z1 + z2 + 7) (cx + 7c – 7, cy + 7c – 7, cz + 7c – 7) 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.
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,...
Related questions
Question
answer c,d

Transcribed Image Text:(c) (X1, Y1, Z1) + (×2, Y2, z2)
с (х, у, z)
O The set is a vector space.
(X1 + X2 + 3, Y1 + Y2 + 3, z1 + z2 + 3)
(сх, су, сz)
The set is not a vector space because the additive identity property is not satisfied.
The set is not a vector space because the additive inverse 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.
(d)
(X1, Y1, Z1) + (x2, Y2, z2) = (x1 + X2 + 7, y1 + y2 + 7, z1 + z2 + 7)
c(x, у, 2)
O The set is a vector space.
(сх + 7с - 7, су + 7с — 7, сz + 7с - 7)
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.

Transcribed Image Text:Rather than use the standard definitions of addition and scalar multiplication in R³, suppose these two operations are defined as follows. With these new definitions, is R3 a vector
space? Justify your answers.
(a)
(X1, Y1, Z1) + (×2, Y2, Z2)
(x1 + X2, Y1 + Y2, z1 + z2)
с(х, у, 2) %3D (сх, су, 0)
O The set is a vector space.
O The set is not a vector space because the associative property of addition 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 associative property of multiplication is not satisfied.
O The set is not a vector space because the multiplicative identity property is not satisfied.
(Б)
(X1, У1, Z1) + (X2, У2, Z2)
(0, 0, 0)
c(x, у, 2) %3D (сх, су, с2)
O The set is a vector space.
O The set is not a vector space because the commutative property of addition is not satisfied.
O 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.
O The set is not a vector space because the multiplicative identity property is not satisfied.
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