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 answer (a) (x1, Y1, Z1) + (x2, Y2, z) = (x1 + X2, Y1 + Y2, Z1 + Z2) cx, Y, z) = (cx, cy, 0) 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) сх, у, г) 3 (сх, су, сг) 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, 22) = (x1 + x2 + 7, yı + y2 + 7, 21 + z2 + 7) C(x, y, z) = (cx, cy, cz) 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. (x1, Y1, Z1) + (x2, Y2, z2) = (x1 + x2 + 3, y1 + Y2 + 3, z1 + 22 + 3) C(x, y, z) - (cx + 3c - 3, cy + 3c – 3, cz + 3c – 3) (d) O The set is a vector space. O The set is not a vector space because the additive identity property is not satisfied. The set is not a vector space because it is not closed under scalar multiplication. The set is not a vector space because the distributive property is not satisfied. The set is not a vector space because the multiplicative identity property is not satisfied. 0 0 0 0

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 answer (a) (x1, Y1, Z1) + (x2, Y2, z) = (x1 + X2, Y1 + Y2, Z1 + Z2) cx, Y, z) = (cx, cy, 0) 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) сх, у, г) 3 (сх, су, сг) 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, 22) = (x1 + x2 + 7, yı + y2 + 7, 21 + z2 + 7) C(x, y, z) = (cx, cy, cz) 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. (x1, Y1, Z1) + (x2, Y2, z2) = (x1 + x2 + 3, y1 + Y2 + 3, z1 + 22 + 3) C(x, y, z) - (cx + 3c - 3, cy + 3c – 3, cz + 3c – 3) (d) O The set is a vector space. O The set is not a vector space because the additive identity property is not satisfied. The set is not a vector space because it is not closed under scalar multiplication. The set is not a vector space because the distributive property is not satisfied. The set is not a vector space because the multiplicative identity property is not satisfied. 0 0 0 0

Elementary Linear Algebra (MindTap Course List)

8th Edition

ISBN:9781305658004

Author:Ron Larson

Publisher:Ron Larson

Chapter4: Vector Spaces

Section4.2: Vector Spaces

Problem 42E: Rather than use the standard definitions of addition and scalar multiplication in R3, let these two...

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