(X1, Y1, Z1) + (x2, Y2, Z2) = (x1 + X2 + 6, y1 + Y2 + 6, Z1 + Z2 + 6) %3D с(х, у, 2) %3D (сх + бс 6, су + бс - 6, сz + бс б) 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.
(X1, Y1, Z1) + (x2, Y2, Z2) = (x1 + X2 + 6, y1 + Y2 + 6, Z1 + Z2 + 6) %3D с(х, у, 2) %3D (сх + бс 6, су + бс - 6, сz + бс б) 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.
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
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.
Transcribed Image Text:(X1, Y1, Z1) + (x2, Y2, Z2) = (x1 + X2 + 6, y1 + Y2 + 6, Z1 + Z2 + 6)
(p).
c(x, y, z) = (cx + 6c – 6, cy + 6c - 6, cz + 6c – 6)
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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