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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