I have a matrix A and wish to solve the equation Ax=b such that b is a vector filled with the same constant B. The value of B is unimportant, thus for any solution x and any constant C, Cx is also a solution. For my purposes I am interested in the solution normed to 1 but it is sufficient to find any one nonzero solution.
Of course inverting A and multiplying the inverse with a constant vector b produces such a solution, but in my case A is large and I was wondering if there was a more suitable way of solving the equation since b has such a unique structure.
Can this be solved without inverting A?
I think I understand the question now. It is asking if there is a "nice" way to find all $\mathbf{x}$ for some matrix $A$ such that $A \mathbf{x}$ results in a vector whose entries are all the same.
Unfortunately, the answer is no. There are two cases here:
The resulting vector is all zeroes. Then we must solve $A \mathbf{x} = \mathbf{0}$.
If the resulting vector $\mathbf{b}$ is filled with non-zero values $B$, then the question reduces to finding the solution $A \mathbf{x} = \mathbf{1}$, because $A \mathbf{x} = \mathbf{b} \iff A (\mathbf{x}/B) = \mathbf{1}$. So the only solutions to the equation in this case are vectors of the form $B\mathbf{u}$, where $A \mathbf{u} = \mathbf{1}$.
In general, there is no "special" way to solve $A \mathbf{x} = \mathbf{b}$ for any value of $\mathbf{b}$. As outlined above, what you want is equivalent to solving this equation for either $\mathbf{b} = \mathbf{1}$ or $\mathbf{b} = \mathbf{0}$.