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с (x, у, 2) %3D (сх, 0, с2) 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, 71) + (x2, Y2, z2) = (0, 0, 0) с(x, у, 2) %3D (сх, су, с2) O 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, Z1) + (x2, Y21 Z2) = (x1 + x2 + 2, Y1 + Y2 + 2, z1 + z2 + 2) с(х, у, 2) %3D (сх, су, сг) 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. 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. (d) (X1, Y1, Zq) + (X2, Y21 Z2) = (x1 + ×2 + 5, Y1 * Y2 * 5, z1 + 22 + 5) c(x, y, z) = (cx + 5c - 5, cy + 5c - 5, cz + Sc - 5) 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.

с (x, у, 2) %3D (сх, 0, с2) 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, 71) + (x2, Y2, z2) = (0, 0, 0) с(x, у, 2) %3D (сх, су, с2) O 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, Z1) + (x2, Y21 Z2) = (x1 + x2 + 2, Y1 + Y2 + 2, z1 + z2 + 2) с(х, у, 2) %3D (сх, су, сг) 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. 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. (d) (X1, Y1, Zq) + (X2, Y21 Z2) = (x1 + ×2 + 5, Y1 * Y2 * 5, z1 + 22 + 5) c(x, y, z) = (cx + 5c - 5, cy + 5c - 5, cz + Sc - 5) 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. (a) (x1, Y1, Z1) + (x2, Y2, z2) = (X1 + X2, Y1 + Y2, Zı + z2) с (x, у, 2) %3D (сх, о, с2) 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, Z1) + (x2, Y2, z2) = (0, 0, 0) с(x, у, 2) %3D (сх, су, с2) O 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, Z1) + (x2, Y2, Z2) = (x1 + X2 + 2, y1 + Y2 + 2, z1 + Z2 + 2) с(x, у, 2) %3D (сх, су, с2) 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. 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. (d) (X1, Y1, Z1) + (x2, Y2, Z2) = (x1 + x2 + 5, y1 + Y2 + 5, z1 + z2 + 5) с (x, у, 2) %3D (сх + 5c - 5, су + 5c — 5, сz + 5с — 5) 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 R3, suppose these two operations are defined as follows. With these new definitions, is R3 a vector space? Justify your answers. (a) (x1, Y1, Z1) + (x2, Y2, z2) = (X1 + X2, Y1 + Y2, Zı + z2) с (x, у, 2) %3D (сх, о, с2) 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, Z1) + (x2, Y2, z2) = (0, 0, 0) с(x, у, 2) %3D (сх, су, с2) O 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, Z1) + (x2, Y2, Z2) = (x1 + X2 + 2, y1 + Y2 + 2, z1 + Z2 + 2) с(x, у, 2) %3D (сх, су, с2) 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. 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. (d) (X1, Y1, Z1) + (x2, Y2, Z2) = (x1 + x2 + 5, y1 + Y2 + 5, z1 + z2 + 5) с (x, у, 2) %3D (сх + 5c - 5, су + 5c — 5, сz + 5с — 5) 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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