(c) (X1, Y1, Z1) + (x2, Y2, Z2) = (xX1 + X2 + 7, y1 + y2 + 7, z1 + Z2 + 7) с(x, у, 2) %3D (сх, су, с2) !! 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 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.

(c) (X1, Y1, Z1) + (x2, Y2, Z2) = (xX1 + X2 + 7, y1 + y2 + 7, z1 + Z2 + 7) с(x, у, 2) %3D (сх, су, с2) !! 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 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.

Elementary Linear Algebra (MindTap Course List)

8th Edition

ISBN:9781305658004

Author:Ron Larson

Publisher:Ron Larson

Chapter4: Vector Spaces

Section4.2: Vector Spaces

Problem 37E: Let V be the set of all positive real numbers. Determine whether V is a vector space with the...

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Question

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.

(c)
(X1, Y1, Z1) + (x2, Y2, Z2) = (x1 + X2 + 7, y1 + y2 + 7, Z1 + Z2 + 7)
c(x, y, z) = (cx, cy, cz)
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 the additive inverse 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 distributive property is not satisfied.
O O

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