not really weird or odd, but one method I just recently learned for getting a better Ax=B solution for the x vector is to re-use the residual vector as another solution Ac=x(k) where k is the # of iterations and x(0) is the original x vector, and using x(0)+x(k)+x(k+1)+... etc until you reach the number of precision you need.... (or if it's an illconditioned problem, until your set maxiumum k is reached)...