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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 R³ a vector space? Justify your answers. (d) (x1, Y1, Z1) + (x2, Y2, z2) с(х, у, 2) O The set is a vector space. (x1 + X2 + 3, Yı + Y2 + 3, Z1 + Z2 + 3) (сх + 3с — 3, су + Зс — 3, cz + 3с — 3) 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.

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 R³ a vector space? Justify your answers. (d) (x1, Y1, Z1) + (x2, Y2, z2) с(х, у, 2) O The set is a vector space. (x1 + X2 + 3, Yı + Y2 + 3, Z1 + Z2 + 3) (сх + 3с — 3, су + Зс — 3, cz + 3с — 3) 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 R°, suppose these two operations are defined as follows. With these new definitions, is Rº a vector space? Justify your answers. (d) (X1, Y1, Z1) + (x2, Y2, z2) = (x1 + X2 + 3, y1 + y2 + 3, z1 + z2 + 3) (сх + Зс -3, су + Зс — 3, cz + 3c — 3) с(х, 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. 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: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 Rº a vector space? Justify your answers. (d) (X1, Y1, Z1) + (x2, Y2, z2) = (x1 + X2 + 3, y1 + y2 + 3, z1 + z2 + 3) (сх + Зс -3, су + Зс — 3, cz + 3c — 3) с(х, 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. 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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