Reduced Row Echelon Forms are used to solve any system for all . We learn that the rank of a matrix reveals the existence and amount of solutions to the system. We augment the column to to get . is only solvable, when is in . If a combination of A gives the zero row, the same combination of b’s entries must give 0 as well.
We determine . One solution, the particular solution is obtained by setting all free variables 0 and solving by back substitution. For multiple solutions to exist, i.e. for some such that
Ax\_{n} = 0 \\ A(x\_{p}+x\_{p}) = b \\ \implies Ax\_{p} = b \end{matrix}$$ Conditions must be met. For the case that $r = n$ there can only exist the solution $x\_{p}$ if it exists at all.