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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.

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
Section4.2: Vector Spaces
Problem 42E: Rather than use the standard definitions of addition and scalar multiplication in R3, let these two...
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(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:(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.
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.
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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