連立一次方程式　式(１)に①，②の操作（前進消去）をして上三角行列化し，次に③の代入計算（後退代入）により未知数を下から順に求める方法．
①　第１式をa₁₁で割り，この式にa_i1を掛けた式を第i式から減じ（i=2,n），a₁₁の下の係数を消去する．
②　①と同様の操作をa₂₂からa_n-1n-1について繰返し式(２)の上三角行列を作る．\[\begin{array}{c} \left. {\begin{array}{*{20}{c}} {{a_{11}}{x_1} + {a_{12}}{x_2} + \cdots + {a_{1n}}{x_n} = {b_1}}\\ {{a_{21}}{x_1} + {a_{22}}{x_2} + \cdots + {a_{2n}}{x_n} = {b_2}}\\ { \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots }\\ {{a_{n1}}{x_1} + {a_{n2}}{x_2} + \cdots + {a_{nn}}{x_n} = {b_n}} \end{array}} \right\}(1)\\ \left. {\begin{array}{*{20}{r}} {{a_{11}}{x_1} + {a_{12}}{x_2} + \cdots + {a_{1n}}{x_n} = {b_1}}\\ {{a_{22}}{x_2} + \cdots + {a_{2n}}{x_n} = {b_2}}\\ { \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots }\\ {{a_{nn}}{x_n} = {b_n}} \end{array}} \right\}(2) \end{array}\]③　\({x_k} = ({b_k} - \sum\limits_{j = k + 1}^n {{a_{kj}}{x_j}} )/{a_{kk}}(k = n,1)\)よりx_kを求める．