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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, 21) + (x2, Y2, 22) = (x1 + x2 1 + y2, Z1 + Z2) c(x, y, z) = (0, cy, cz) 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, 21) + (x2, yz, Zz) = (0, 0, 0) c(x, y, z) = (cx, cy, cz) 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, 21) + (x2, Y2. 22) = (x1 + x2 + 6, y₁ + y2 + 6, Z₁ + Z2 + 6) c(x, y, z) = (cx, cy, cz) O 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 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.

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, 21) + (x2, Y2, 22) = (x1 + x2 1 + y2, Z1 + Z2) c(x, y, z) = (0, cy, cz) 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, 21) + (x2, yz, Zz) = (0, 0, 0) c(x, y, z) = (cx, cy, cz) 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, 21) + (x2, Y2. 22) = (x1 + x2 + 6, y₁ + y2 + 6, Z₁ + Z2 + 6) c(x, y, z) = (cx, cy, cz) O 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 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.

Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter6: Vector Spaces
Section6.1: Vector Spaces And Subspaces
Problem 49EQ
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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, 1, 2₁) + (x2, y/2, Z2) = (x₁ + x2, Y₁ + y2, Z1 + Z2) c(x, y, z) = (0, cy, cz) O The set is a vector space. O The set is not a vector space because the associative property of addition 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 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, 2₁) + (x2, yz, Z2) = (0, 0, 0) c(x, y, z)= (cx, cy, cz) 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, 21) + (x2, y2, 22) = (x1 + x2 + 6, y₁ + y2 + 6, Z₁ + Z2 + 6) 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. 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, 2₁) + (x2, Y2, 22) = (x1 + x2 + 3, y₁ + y2 + 3, Z₁ + Z2 + 3) c(x, y, z) = (cx + 3c-3, cy+ 3c-3, cz + 3c-3) 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: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, 1, 2₁) + (x2, y/2, Z2) = (x₁ + x2, Y₁ + y2, Z1 + Z2) c(x, y, z) = (0, cy, cz) O The set is a vector space. O The set is not a vector space because the associative property of addition 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 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, 2₁) + (x2, yz, Z2) = (0, 0, 0) c(x, y, z)= (cx, cy, cz) 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, 21) + (x2, y2, 22) = (x1 + x2 + 6, y₁ + y2 + 6, Z₁ + Z2 + 6) 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. 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, 2₁) + (x2, Y2, 22) = (x1 + x2 + 3, y₁ + y2 + 3, Z₁ + Z2 + 3) c(x, y, z) = (cx + 3c-3, cy+ 3c-3, cz + 3c-3) 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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