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

Prove that in a given vector space V, the additive inverse of a vector is unique.

Find a basis for R2 that includes the vector (2,2).

Rather than use the standard definitions of addition and scalar multiplication in R3, let these two operations be defined as shown below. (a) (x1,y1,z1)+(x2,y2,z2)=(x1+x2,y1+y2,z1+z2) c(x,y,z)=(cx,cy,0) (b) (x1,y1,z1)+(x2,y2,z2)=(0,0,0) c(x,y,z)=(cx,cy,cz) (c) (x1,y1,z1)+(x2,y2,z2)=(x1+x2+1,y1+y2+1,z1+z2+1) c(x,y,z)=(cx,cy,cz) (d) (x1,y1,z1)+(x2,y2,z2)=(x1+x2+1,y1+y2+1,z1+z2+1) c(x,y,z)=(cx+c1,cy+c1,cz+c1) With each of these new definitions, is R3 a vector space? Justify your answers.