文档计算机科学数值分析线性方程组的直接解法线性方程组的直接解法消元法通用过程规定线性方程组为 Ax=bA\boldsymbol x = \boldsymbol bAx=b,有 nnn 个 nnn 元方程。消元法的通用过程就是将 AAA 分解为下三角矩阵 LLL 和上三角矩阵 UUU 相乘,再回代求解。对第 iii 行:选定 liil_{ii}lii。令 uij=aij−∑k=1i−1likukjlii (j≥i)u_{ij} = \dfrac{a_{ij} - \sum_{k = 1}^{i-1} l_{ik}u_{kj}}{l_{ii}}\ (j\ge i)uij=liiaij−∑k=1i−1likukj (j≥i),ziz_izi 类似。令 lji=aji−∑k=1i−1ljkukiuii (j<i)l_{ji} = \dfrac{a_{ji} - \sum_{k = 1}^{i-1} l_{jk}u_{ki}}{u_{ii}}\ (j < i)lji=uiiaji−∑k=1i−1ljkuki (j<i)对于一个三元线性方程组按照如下方式列表格:l11,u11=a11l11l_{11},u_{11} = \dfrac{a_{11}}{l_{11}}l11,u11=l11a11u12=a12l11u_{12} = \dfrac{a_{12}}{l_{11}}u12=l11a12u13=a13l11u_{13} = \dfrac{a_{13}}{l_{11}}u13=l11a13z1=b1l11z_1 = \dfrac{b_1}{l_{11}}z1=l11b1l21=a21u11l_{21} = \dfrac{a_{21}}{u_{11}}l21=u11a21l22,u22=a22−l21u12l22l_{22},u_{22} = \dfrac{a_{22} - l_{21}u_{12}}{l_{22}}l22,u22=l22a22−l21u12u23=a23−l21u13l22u_{23} = \dfrac{a_{23} - l_{21}u_{13}}{l_{22}}u23=l22a23−l21u13z2=b2−l21z1l22z_2 = \dfrac{b_2 - l_{21}z_1}{l_{22}}z2=l22b2−l21z1l31=a31u11l_{31} = \dfrac{a_{31}}{u_{11}}l31=u11a31l32=a32−l31u12u22l_{32} = \dfrac{a_{32} - l_{31}u_{12}}{u_{22}}l32=u22a32−l31u12l33,u33=a33−l31u13−l32u23l33l_{33},u_{33} = \dfrac{a_{33} - l_{31}u_{13} - l_{32}u_{23}}{l_{33}}l33,u33=l33a33−l31u13−l32u23z3=b3−l31z1−l32z2l33z_3 = \dfrac{b_3 - l_{31}z_1 - l_{32}z_2}{l_{33}}z3=l33b3−l31z1−l32z2高斯消元法对于所有的 liil_{ii}lii,均选定 111。克劳特消元法对于所有的 liil_{ii}lii,均选定使 uii=1u_{ii} = 1uii=1,即 lii=aii−∑k=1i−1likukil_{ii} = a_{ii} - \displaystyle\sum_{k = 1}^{i-1} l_{ik}u_{ki}lii=aii−k=1∑i−1likuki。平方根法对于所有的 liil_{ii}lii,均选定使 uii=liiu_{ii} = l_{ii}uii=lii,即 lii=aii−∑k=1i−1likukil_{ii} = \sqrt{a_{ii} - \displaystyle\sum_{k = 1}^{i-1} l_{ik}u_{ki}}lii=aii−k=1∑i−1likuki。主元法列主元法为减小误差,消元进行到第 iii 行时,选定第 iii 列系数绝对值最大的一个为主元,交换主元行与当前行。交换后,选择 lii=1l_{ii} = 1lii=1。全主元法在 a(i∼n)(i∼n)a_{(i\sim n)(i \sim n)}a(i∼n)(i∼n) 子矩阵中选择主元,需要交换行和未知数的顺序。方程组的迭代解法线性方程组的迭代解法