Answer:
The given system can be represented as the augmented matrix:
[ 1 1 1 1 ]
[10 1 0 ]
[ 0 1 0 1 ]
To determine whether the set {u, v} spans the solution space, we need to check if all the vectors in the solution space can be written as a linear combination of u and v.
Let's consider option a: u = (1, 0, -1, 0) and v = (0, 1, 0, -1).
To check if {u, v} spans the solution space, we need to row reduce the augmented matrix to its row echelon form (REF).
Performing row operations, we can reduce the given matrix to the following REF:
[ 1 1 1 1 ]
[ 0 9 -1 -10]
[ 0 0 -1 0 ]
Notice that the third row of the REF has a pivot element of -1. This means that the third variable is leading (non-free), which implies that the system has a unique solution.
Since the system only has one solution (x = 0), it means that the solution space is the trivial solution {0}.
Therefore, the set {u, v} does not span the solution space because no linear combination of u and v can produce the trivial solution.
Now, let's consider option b: u = (1, 0, -1, 0) and v = (1, 1, -1, -1).
Performing row operations on the augmented matrix using u and v as row multipliers, we get:
[ 1 1 1 1 ]
[11 1 -1 11 ]
[ 0 0 -1 0 ]
Notice that the third row of the REF still has a pivot element of -1, indicating that the system still has a unique solution.
Since the system only has one solution (x = 0), it means that the solution space is still the trivial solution {0}.
Therefore, the set {u, v} also does not span the solution space because no linear combination of u and v can produce the trivial solution.
In conclusion, neither option a nor option b spans the solution space of the given system because the solution space only consists of the trivial solution {0}.