Final answer:
The set W is not a subspace of R³ because it does not contain the zero vector, it is not closed under vector addition, and it is not closed under scalar multiplication.
Step-by-step explanation:
To determine if the set W is a subspace of R³, we must check if W satisfies the three subspace properties: (1) containing the zero vector, (2) being closed under vector addition, and (3) being closed under scalar multiplication. The set W consists of all vectors in R³ whose third component is -4.
The zero vector in R³ is (0, 0, 0), which is not included in W because the third component is not -4. Therefore, W does not satisfy the first property. Next, if we take two arbitrary vectors from W, say (a, b, -4) and (c, d, -4), and add them together, the resulting vector is (a+c, b+d, -4-4) = (a+c, b+d, -8), which has its third component equal to -8 and not -4, meaning it is not in W. This shows that W is not closed under vector addition.
Lastly, for scalar multiplication, if we multiply any vector in W by a scalar α, the result is (αa, αb, -4α). If α is anything other than 1, the third component will not be -4, which means W is not closed under scalar multiplication. Hence, the set W is not a subspace of R³.