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, 21) + (x2, Y2, 22) = (x1 + x2 1 + y2, Z1 + Z2) c(x, y, z) = (0, cy, cz) 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, 21) + (x2, yz, Zz) = (0, 0, 0) c(x, y, z) = (cx, cy, cz) 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, 21) + (x2, Y2. 22) = (x1 + x2 + 6, y₁ + y2 + 6, Z₁ + Z2 + 6) c(x, y, z) = (cx, cy, cz) O The set is a vector space. The set is not a vector space because the additive identity property is not satisfied. 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.
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, 21) + (x2, Y2, 22) = (x1 + x2 1 + y2, Z1 + Z2) c(x, y, z) = (0, cy, cz) 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, 21) + (x2, yz, Zz) = (0, 0, 0) c(x, y, z) = (cx, cy, cz) 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, 21) + (x2, Y2. 22) = (x1 + x2 + 6, y₁ + y2 + 6, Z₁ + Z2 + 6) c(x, y, z) = (cx, cy, cz) O The set is a vector space. The set is not a vector space because the additive identity property is not satisfied. 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.
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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