Y2, 22) = c(x, y, z) = (cx, cy, 0) a vector space. X2, Y1 + Y2, Z1 + z2) %3! not a vector space because the associative property a not a vector space because it is not closed under scal not a vector space because the associative property c not a vector space because the multiplicative identity E) + (x2, Y2, z2) c(x, у, 2) - a vector space. (0, 0, 0) (cx, cy, cz) !3! = not a vector space because the commutative property s not a vector space because the additive identity prop s not a vector space because it is not closed under scal s not a vector space because the multiplicative identity (x1 + X2 + 5, Y1 + y2 + 5, z1 +2 z1) + (x2, Y2, z2) c(x, у, 2) sa vector space. (cx, cy, cz) is not a vector space because the additive identity prop is not a vector space because the additive inverse prop- is not a vector space because it is not closed under scal is not a vector space because the distributive property , z1) + (x2, Y2, Z2) = (x1 + X2 + 3, Y1 + Y2 + 3, z1 + (cx + 3c - 3, cy + 3c - 3, cz + %3! C(x, y, z) is a vector space. is not a vector space because the additive identity prop
Y2, 22) = c(x, y, z) = (cx, cy, 0) a vector space. X2, Y1 + Y2, Z1 + z2) %3! not a vector space because the associative property a not a vector space because it is not closed under scal not a vector space because the associative property c not a vector space because the multiplicative identity E) + (x2, Y2, z2) c(x, у, 2) - a vector space. (0, 0, 0) (cx, cy, cz) !3! = not a vector space because the commutative property s not a vector space because the additive identity prop s not a vector space because it is not closed under scal s not a vector space because the multiplicative identity (x1 + X2 + 5, Y1 + y2 + 5, z1 +2 z1) + (x2, Y2, z2) c(x, у, 2) sa vector space. (cx, cy, cz) is not a vector space because the additive identity prop is not a vector space because the additive inverse prop- is not a vector space because it is not closed under scal is not a vector space because the distributive property , z1) + (x2, Y2, Z2) = (x1 + X2 + 3, Y1 + Y2 + 3, z1 + (cx + 3c - 3, cy + 3c - 3, cz + %3! C(x, y, z) is a vector space. is not a vector space because the additive identity prop
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
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) = (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 + 5, Y1 + Y2 + 5, z1 + z2 + 5)
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, Zz1) + (x2, Y2, Z2) = (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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