Question

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) (*1, Y₁, Z₁) + (x₂, ₂, ₂) = (x₁ + X₂, Y₁ + y₂, Z₁ + Z₂) c(x, y, z) = (cx, 0, cz) 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) OThe set is a vector space. O The set is not a vector space because the commutative property of addition is not satisfied. OThe 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. OThe set is not a vector space because the multiplicative identity property is not satisfied. (C) (X₁, Y₁, Z₁) + (x₂, ₂, Z₂) = (x₁ + x₂ + 7, Y₁ + y₂ + 7, Z₁ + Z₂ + 7) c(x, y, z) = (cx, cy, cz) OThe set is a vector space. OThe set is not a vector space because the additive identity property is not satisfied. OThe set is not a vector space because the additive inverse property is not satisfied. OThe set is not a vector space because it is not closed under scalar multiplication. OThe set is not a vector space because the distributive property is not satisfied. (d) (X₁, Y₁, Z₁) + (x₂, Y₂, Z₂) = (x₁ + x₂ + 9, Y₁ + y₂ + 9, Z₁ + Z₂ + 9) c(x, y, z) = (cx + 90 - 9, cy + 90 - 9, cz + 90 - 9) OThe set is a vector space. O The set is not a vector space because the additive identity property is not satisfied. OThe set is not a vector space because it is not closed under scalar multiplication. OThe set is not a vector space because the distributive property is not satisfied. OThe set is not a vector space because the multiplicative identity property is not satisfied.

Answer 1

R³ with the given operations does not constitute a vector space because it fails to satisfy some of the fundamental properties required for vector spaces, such as closure under scalar multiplication, the existence of an additive identity, and the multiplicative identity property.

Step-by-step explanation:

To determine whether R³ with the given operations is a vector space, we need to check if the set satisfies all vector space axioms such as closure under addition and scalar multiplication, the existence of an additive identity and additive inverses, and the distributive properties, among others.

(a) The set is not a vector space because it is not closed under scalar multiplication. Indeed, scalar multiplication as defined c(x, y, z) = (cx, 0, cz) does not satisfy the property that the second component should be cy, not 0, to maintain closure within the space.

(b) The set is not a vector space because the additive identity property is not satisfied. The operation (X₁, Y₁, Z₁) + (x₂, Y₂, Z₂) = (0, 0, 0) does not allow for an additive identity (0, 0, 0) to exist such that v + 0 = v for any vector v in the set.

(c) The set is not a vector space because the additive identity property is not satisfied. Addition defined as (X₁, Y₁, Z₁) + (x₂, y₂, Z₂) = (x₁ + x₂ + 7, Y₁ + y₂ + 7, Z₁ + Z₂ + 7) means there is no vector that can act as an additive identity, which is a requirement for a set to be considered a vector space.

(d) The set is not a vector space because the multiplicative identity property is not satisfied. Scalar multiplication as defined c(x, y, z) = (cx + 90 - 9, cy + 90 - 9, cz + 90 - 9) does not satisfy the multiplicative identity property, which requires that 1v = v for every vector v in the set.