Originally Posted by
alexandrabel90 sorry if i was not being clear.
i was given 3 simultaneous equations and asked to find the number of vectors that form the basis for them..what i did was i did gaussian elimination..i was left with 2 leading ones and one free parameter. so from there, i concluded that the rank is 2 and the dimension is 1.
Well, at least this time it's a little clearer...but it seems to be that you were asked to find the dimension of the subspace spanned by those three equations (in the space of all equations with three unknowns over some given field), so that subspace's dimension = rank of the coefficients' matrix of the system is 2. Why do you write "dimension is 1"? Unless you mean, and I'm guessing since you said nothing, that 1 is the dimension of the kernel of that matrix = the matrix's null space.
Then, the two non-zero lin. independent rows you get after applying Gauss method are a basis for that subspace=the span of those three equations).
Tonio
now im wondering if the no of vectors that form B={ } is 2 vectors(the rank) or that one vector( that has the free parameter)