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(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
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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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.
(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.
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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