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, Z1 + Z2) c(x, у, г) %3 (0, су, 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, Yı, Z1) + (x2, Y2, z2) = (0, 0, 0) c(х, у, 2) %3D (сх, су, сz) 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 + 5, y1 + y2 + 5, z1 + Z2 + 5) c(x, 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, Z1) + (x2, Y2, z2) = (x1 + x2 + 3, y1 + Y2 + 3, z1 + z2 + 3) с(x, у, 2) %3D (сх + 3с — 3, су + Зс — 3, сz + 3с —- 3) 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.

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, Z1 + Z2) c(x, у, г) %3 (0, су, 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, Yı, Z1) + (x2, Y2, z2) = (0, 0, 0) c(х, у, 2) %3D (сх, су, сz) 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 + 5, y1 + y2 + 5, z1 + Z2 + 5) c(x, 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, Z1) + (x2, Y2, z2) = (x1 + x2 + 3, y1 + Y2 + 3, z1 + z2 + 3) с(x, у, 2) %3D (сх + 3с — 3, су + Зс — 3, сz + 3с —- 3) 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.

Linear Algebra: A Modern Introduction

4th Edition

ISBN:9781285463247

Author:David Poole

Publisher:David Poole

Chapter6: Vector Spaces

Section6.1: Vector Spaces And Subspaces

Problem 2EQ

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