Final answer:
The set S={(2,0,1),(2,0,−1),(6,0,1),(4,0,2)} does not span R3, but instead spans a plane in R3.
Step-by-step explanation:
To determine whether the set S={(2,0,1),(2,0,−1),(6,0,1),(4,0,2)} spans R3, we need to check whether the vectors in S can generate every possible vector in R3. To do this, we can set up a system of equations and solve for the coefficients of the vectors in S that would produce a given vector in R3.
If we consider a general vector (a,b,c) in R3, we can set up the following system of equations:
a(2,0,1) + b(2,0,−1) + c(6,0,1) + d(4,0,2) = (a(2)+b(2)+6c+4d,0, a+c+2d) = (a(2)+b(2)+6c+4d,0, a+c+2d) = (a(2)+b(2)+6c+4d,0, a+c+2d).
To determine the coefficients a, b, c, and d, we have to find the solution of the system of equations that would work for any (a,b,c) in R3. If we cannot find such a solution, then the set S does not span R3. If we can find a solution, then the set S spans R3.
In this particular case, the system of equations has no solution other than the trivial solution (a,b,c,d) = (0,0,0,0), which means that the set S does not span R3. Instead, it spans a two-dimensional subspace, which is a plane in R3.