Transcribed Image Text: 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)

с(x, у, 2) 3D (сх, 0, сz2)

The set is a vector space.

The set is not a vector space because the associative property of addition 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 associative property of multiplication is not satisfied.

The set is not a vector space because the multiplicative identity property is not satisfied.

(6) (х1, У1, Z1) + (x2, У2, Zz)

(0, 0, 0)

(сх, су,

с(x, у, z)

The set is a vector space.

cz)

The set is not a vector space because the commutative property of addition is not satisfied.

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 multiplicative identity property is not satisfied.

(c)

(x1, У1, Z1) + (x2, У2, Z2)

(X1 + x2 + 3, Yı + Y2 + 3, z1 + z2 + 3)

с (х, у, 2)

The set is a vector space.

(сх, су, сz)

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.

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.

(d) (X1, Y1, Z1) + (x2, Y2, Z2)

с(x, у, 2)

(X1 + X2 + 7, Yi + Y2 + 7, z1 + Z2 + 7)

(сх + 7с – 7, су + 7с — 7, cz+ 7с — 7)

%D

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 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.

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