Which of the following are the subspaces of R3: (a) {(x, y, z)|x ≥ 0}. (b) {(x, y, z)|x + y = z}. (c) {(x, y, z)|x = y2}

To determine whether a set is a subspace of R3, it must satisfy three conditions:

1. The zero vector is in the set.
2. The set is closed under vector addition.
3. The set is closed under scalar multiplication.

Let's examine each set:

(a) {(x, y, z)|x ≥ 0}:

1. The zero vector (0,0,0) is in the set.
2. However, the set is not closed under vector addition. For example, if we add the vectors (1,0,0) and (-1,0,0), we get (0,0,0), which is not in the set because x is not greater than or equal to 0.
3. The set is not closed under scalar multiplication. For example, if we multiply the vector (1,0,0) by -1, we get (-1,0,0), which is not in the set because x is not greater than or equal to 0. So, (a) is not a subspace of R3.

(b) {(x, y, z)|x + y = z}:

1. The zero vector (0,0,0) is in the set because 0+0=0.
2. The set is closed under vector addition. If we add two vectors (x1,y1,z1) and (x2,y2,z2) in the set, we get (x1+x2, y1+y2, z1+z2). Since x1+y1=z1 and x2+y2=z2, it follows that (x1+x2) + (y1+y2) = z1+z2, so the resulting vector is in the set.
3. The set is closed under scalar multiplication. If we multiply a vector (x,y,z) in the set by a scalar c, we get (cx,cy,cz). Since x+y=z, it follows that cx+cy = cz, so the resulting vector is in the set. So, (b) is a subspace of R3.

(c) {(x, y, z)|x = y2}:

1. The zero vector (0,0,0) is in the set because 0=0^2.
2. However, the set is not closed under vector addition. For example, if we add the vectors (1,1,0) and (4,-2,0), we get (5,-1,0), which is not in the set because 5 is not equal to (-1)^2.
3. The set is not closed under scalar multiplication. For example, if we multiply the vector (1,1,0) by -1, we get (-1,-1,0), which is not in the set because -1 is not equal to (-1)^2. So, (c) is not a subspace of R3.

In conclusion, only (b) is a subspace of R3.