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Rather than use the standard definitions of addition and scalar multiplication in R, suppose these two operations are defined as follows. With these new definitions, is R- a vector space? Justify your answers. (a) (x1, Y1, Z1) + (x2, Y2, z2) = (X1 + ×2, Y1 + Y2, z, + Z2) с(x, 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. (b) (x1, Y1, Z1) + (x2, Y2, z2) = (0, 0, 0) c(x, у, 2) 3D (сх, су, сг) 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. O The set is not a vector space because it is not closed under scalar multiplication. O The cot horauco thr ltiplicative idontity pro porty ic pot sat+icfiod

Rather than use the standard definitions of addition and scalar multiplication in R, suppose these two operations are defined as follows. With these new definitions, is R- a vector space? Justify your answers. (a) (x1, Y1, Z1) + (x2, Y2, z2) = (X1 + ×2, Y1 + Y2, z, + Z2) с(x, 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. (b) (x1, Y1, Z1) + (x2, Y2, z2) = (0, 0, 0) c(x, у, 2) 3D (сх, су, сг) 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. O The set is not a vector space because it is not closed under scalar multiplication. O The cot horauco thr ltiplicative idontity pro porty ic pot sat+icfiod

Elementary Linear Algebra (MindTap Course List)
Elementary Linear Algebra (MindTap Course List)
8th Edition
ISBN:9781305658004
Author:Ron Larson
Publisher:Ron Larson
Chapter4: Vector Spaces
Section4.1: Vector In R^n
Problem 61E: Illustrate properties 110 of Theorem 4.2 for u=(2,1,3,6), v=(1,4,0,1), w=(3,0,2,0), c=5, and d=2....
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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) + (x2, Y2, z2) с(х, у, 2) O The set is a vector space. (x1 + X2, Y1 + Y2, Z1 + z2) (сх, су, 0) 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. : (0,0, 0) с (х, у, 2) %3D (сх, су, сz) (Б) (X1, Y1, Z1) + (x2, Y2, z2) 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. 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 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) + (x2, Y2, z2) с(х, у, 2) O The set is a vector space. (x1 + X2, Y1 + Y2, Z1 + z2) (сх, су, 0) 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. : (0,0, 0) с (х, у, 2) %3D (сх, су, сz) (Б) (X1, Y1, Z1) + (x2, Y2, z2) 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. 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 multiplicative identity property is not satisfied.
(c) (X1, Y1, Z1) + (x2, Y2, z2) с(x, у, 2) O The set is a vector space. (X1 + X2 + 3, y1 + Y2 + 3, z1 + z2 + 3) (сх, су, сz) 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. (X1 + X2 + 7, Y1 + Y2 + 7, z1 + z2 + 7) c(x, у, 2) %3D (сх + 7с - 7, су + 7с - 7, сz + 7с -7) (d) (x1, Y1, Z1) + (x2, Y2, z2) 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.
Transcribed Image Text:(c) (X1, Y1, Z1) + (x2, Y2, z2) с(x, у, 2) O The set is a vector space. (X1 + X2 + 3, y1 + Y2 + 3, z1 + z2 + 3) (сх, су, сz) 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. (X1 + X2 + 7, Y1 + Y2 + 7, z1 + z2 + 7) c(x, у, 2) %3D (сх + 7с - 7, су + 7с - 7, сz + 7с -7) (d) (x1, Y1, Z1) + (x2, Y2, z2) 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.
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