Back substitution ($x_i=(b_i-{\sum }_{j>i}{a}_{ij}{x}_j)/a_{ii}$)

An upper triangular matrix system with nonzero diagonal is solved from the last row upward. Once $x_{i+1},\dots ,x_n$ are known, use $x_i=(b_i-{\sum }_{j>i}{a}_{ij}{x}_j)/a_{ii}$. The nonzero diagonal guarantees a unique solution; this is the final solve after a QR factorization.