Rather than use the standard definitions of addition and scalar multiplication in R3, let these two operations be defined as shown below.With each of these new definitions, is R3 a vector space? Justify your answers. (a) (x1, Y1, z,) + (x,, V2, Z,)= (x, + x,, y1 + y2, Z1 + Z2)с(х, у, г) %3D (сх, су, 0)(b) (х,. У, 2,) + (х, у, 2,) %3D (0, 0, 0)с(х, у, г) — (сх, су, с2)(c) (x1, y1, 71) + (x2, Y2, Z2)=(x+ x, + 1, y, + y2 + 1, z, + Z2 + 1)с(х, у, г) — (сх, су, сә)(d) (x1, y1, Z1) + (x2, Y2, Z2)= (x, + x, + 1, y1 + y2 + 1, z1 + Z2 + 1)с(х, у, г) — (сх +с — 1, су + с — 1, сz + с — 1)-

(a) (x1, Y1, z,) + (x,, V2, Z,) = (x, + x,, y1 + y2, Z1 + Z2) с(х, у, г) %3D (сх, су, 0) (b) (х,. У, 2,) + (х, у, 2,) %3D (0, 0, 0) с(х, у, г) — (сх, су, с2) (c) (x1, y1, 71) + (x2, Y2, Z2) =(x+ x, + 1, y, + y2 + 1, z, + Z2 + 1) с(х, у, г) — (сх, су, сә) (d) (x1, y1, Z1) + (x2, Y2, Z2) = (x, + x, + 1, y1 + y2 + 1, z1 + Z2 + 1) с(х, у, г) — (сх +с — 1, су + с — 1, сz + с — 1) -

(a) (x1, Y1, z,) + (x,, V2, Z,) = (x, + x,, y1 + y2, Z1 + Z2) с(х, у, г) %3D (сх, су, 0) (b) (х,. У, 2,) + (х, у, 2,) %3D (0, 0, 0) с(х, у, г) — (сх, су, с2) (c) (x1, y1, 71) + (x2, Y2, Z2) =(x+ x, + 1, y, + y2 + 1, z, + Z2 + 1) с(х, у, г) — (сх, су, сә) (d) (x1, y1, Z1) + (x2, Y2, Z2) = (x, + x, + 1, y1 + y2 + 1, z1 + Z2 + 1) с(х, у, г) — (сх +с — 1, су + с — 1, сz + с — 1) -

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, z1 + Z2) C(x, 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) c(x, y, z) = (cx, cy, cz) 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…