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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, 22) = (X1 + X2, Y1 + Y2, Z1 + 22) c(x, y, z) = (cx, 0, 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, Z1) + (X2, Y2, Z2) = (0, o, 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.

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, 22) = (X1 + X2, Y1 + Y2, Z1 + 22) c(x, y, z) = (cx, 0, 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, Z1) + (X2, Y2, Z2) = (0, o, 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.

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 2EQ
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(c) (x1, Y1, Z1) + (x2, Y2, Z2) = (X1 + X2 + 6, Y1 + Y2 + 6, z1 + Z2 + 6) с(x, у, 2) %3 (сх, су, сг) 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. (d) (X1, Y1, Z1) + (X2, Y2, Z2) = (X1 + X2 + 8, Y1 + Y2 + 8, z1 + Z2 + 8) c(x, у, 2) 3 (сх + 8с — 8, су + 8с - 8, сz + 8с - 8) %3D The set is a vector space. 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. 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) = (X1 + X2 + 6, Y1 + Y2 + 6, z1 + Z2 + 6) с(x, у, 2) %3 (сх, су, сг) 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. (d) (X1, Y1, Z1) + (X2, Y2, Z2) = (X1 + X2 + 8, Y1 + Y2 + 8, z1 + Z2 + 8) c(x, у, 2) 3 (сх + 8с — 8, су + 8с - 8, сz + 8с - 8) %3D The set is a vector space. 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. 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 R?, 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) = (X1 + X2, Y1 + Y2, Z1 + Z2) C(x, y, z) = (cx, 0, 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, Z1) + (X2, Y2, 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.
Transcribed Image Text: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 R3 a vector space? Justify your answers. (a) (x1, Y1, Z1) + (X2, Y2, Z2) = (X1 + X2, Y1 + Y2, Z1 + Z2) C(x, y, z) = (cx, 0, 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, Z1) + (X2, Y2, 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.
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