(X1, Y1, Z1) + (x2, Y2, z2) = (x1 + x2 + 3, Y1 + Y2 + 3, z1 + z2 + 3) с (x, у, z) %3D (сх, су, с2) (c) 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. 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, Z1) + (x2, Y2, z2) = (x1 + x2 + 3, Y1 + Y2 + 3, z1 + z2 + 3) с (x, у, z) %3D (сх, су, с2) (c) 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. 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.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:(c)
(X1, Y1, Z1) + (x2, Y2, z2)
(X1 + X2 + 3, Y1 + y2 + 3, z1 + z2 + 3)
с(х, у, 2) %3D (сх, су, сz)
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.
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)
с(х, у, 2) %3
O The set is a vector space.
(сх + 7с - 7, су + 7с — 7, cz + 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.
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