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? Justif (a) (x1, Y1, Z1) + (x2, Y2, Z2) = (X1 + X2, Y1 + Y2, Zı + Z2) c(x, y, z) = (cx, cy, 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, 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, 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. 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 R, suppose these two operations are defined as follows. With these new definitions, is R a vector space? Justif (a) (x1, Y1, Z1) + (x2, Y2, Z2) = (X1 + X2, Y1 + Y2, Zı + Z2) c(x, y, z) = (cx, cy, 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, 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, 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. 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 42E: Rather than use the standard definitions of addition and scalar multiplication in R3, let these two...
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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
(a) (x1, Y1, Z1) + (X2, Y2, Z2) = (X1 + X2, Y1 + Y2, Z1 + Z2)
c(x, y, z) = (cx, cy, 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, 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, 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.
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.
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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.
(X1, Y1, Z1) + (X2, Y2, Z2)
c(x, y, z)
The set is a vector space.
(c)
|(X1 + X2 + 7, Y1 + Y2 + 7, zı +Z
(сх, су, с2)
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.
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, Y1, 21) + (X2, Y2, Z2) = (X1 + x2 + 5, y1 + Y2 + 5, z1 + Z2 + 5)
(cx + 5c - 5, cy + 5c - 5, cz + 5c – 5)
(d)
c(x, у, 2)
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.
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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