LU分解（不考虑行交换）

$A=LU$

对矩阵$A$做$LU$分解（不考虑行交换）
$$A=\left[\begin{array}{cccc}{a}_{11}& a_{12}& a_{13}& a_{14}\\ a_{21}& a_{22}& a_{23}& a_{24}\\ a_{31}& a_{32}& a_{33}& a_{34}\\ a_{41}& a_{42}& a_{43}& a_{44}\end{array}\right]$$
第一步，构造矩阵$M_1$
$$M_1=\left[\begin{array}{cccc}1& 0& 0& 0\\ -l_{21}& 1& 0& 0\\ -l_{31}& 0& 1& 0\\ -l_{41}& 0& 0& 1\end{array}\right]$$
其中$l_{i1}=\frac{a_{i1}}{a_{11}},i=2,3,4.$ 用矩阵$M_1$左乘$A$
$$M_{1}A=\left[\begin{array}{cccc}{a}_{11}& a_{12}& a_{13}& a_{14}\\ 0& a_{22}^{\prime }& a_{23}^{\prime }& a_{24}^{\prime }\\ 0& a_{32}^{\prime }& a_{33}^{\prime }& a_{34}^{\prime }\\ 0& a_{42}^{\prime }& a_{43}^{\prime }& a_{44}^{\prime }\end{array}\right]$$
第二步，构造矩阵$M_2$
$$M_2=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& -l_{32}& 1& 0\\ 0& -l_{42}& 0& 1\end{array}\right]$$
其中$l_{i2}=\frac{a_{i2}^{\prime }}{a_{22}^{\prime }},i=3,4.$ 用矩阵$M_2$左乘$M_{1}A$
$$M_2{M}_{1}A=\left[\begin{array}{cccc}{a}_{11}& a_{12}& a_{13}& a_{14}\\ 0& a_{22}^{\prime }& a_{23}^{\prime }& a_{24}^{\prime }\\ 0& 0& a_{33}^{\prime \prime }& a_{34}^{\prime \prime }\\ 0& 0& a_{43}^{\prime \prime }& a_{44}^{\prime \prime }\end{array}\right]$$
第三步，构造矩阵$M_3$
$$M_3=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& -l_{43}& 1\end{array}\right]$$
其中$l_{43}=\frac{a_{43}^{\prime \prime }}{a_{33}^{\prime \prime }}.$ 用矩阵$M_3$左乘矩阵$M_2{M}_{1}A$
$$\begin{array}{rl}U& =M_3{M}_2{M}_{1}A\\ & =\left[\begin{array}{cccc}{a}_{11}& a_{12}& a_{13}& a_{14}\\ 0& a_{22}^{\prime }& a_{23}^{\prime }& a_{24}^{\prime }\\ 0& 0& a_{33}^{\prime \prime }& a_{34}^{\prime \prime }\\ 0& 0& 0& a_{44}^{\prime \prime \prime }\end{array}\right]\end{array}$$
用矩阵$(M_3{M}_2{M}_1)$的逆矩阵$(M_3{M}_2{M}_1{)}^{-1}$左乘上式
$$\begin{array}{rl}A& =(M_3{M}_2{M}_1{)}^{-1}U\\ & =M_1^{-1}{M}_2^{-1}{M}_3^{-1}U\\ & =LU\end{array}$$
其中
$$M_1^{-1}=\left[\begin{array}{cccc}1& 0& 0& 0\\ l_{21}& 1& 0& 0\\ l_{31}& 0& 1& 0\\ l_{41}& 0& 0& 1\end{array}\right]$$
$$M_2^{-1}=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& l_{32}& 1& 0\\ 0& l_{42}& 0& 1\end{array}\right]$$
$$M_3^{-1}=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& l_{43}& 1\end{array}\right]$$
$$\begin{array}{rl}L& =M_1^{-1}{M}_2^{-1}{M}_3^{-1}\\ & =\left[\begin{array}{cccc}1& 0& 0& 0\\ l_{21}& 1& 0& 0\\ l_{31}& l_{32}& 1& 0\\ l_{41}& l_{42}& l_{43}& 1\end{array}\right]\end{array}$$

算例

$$A=\left[\begin{array}{ccc}10& -7& 0\\ 5& -1& 5\\ -3& 2& 6\end{array}\right]$$
按上述步骤对矩阵$A$做$LU$分解。第一步，构造矩阵$M_1$
$$M_1=\left[\begin{array}{ccc}1& 0& 0\\ -0.5& 1& 0\\ 0.3& 0& 1\end{array}\right]$$
$$M_1^{-1}=\left[\begin{array}{ccc}1& 0& 0\\ 0.5& 1& 0\\ -0.3& 0& 1\end{array}\right]$$
用矩阵$M_1$左乘$A$
$$M_{1}A=\left[\begin{array}{ccc}10& -7& 0\\ 0& 2.5& 5\\ 0& -0.1& 6\end{array}\right]$$
第二步，构造矩阵$M_2$
$$M_2=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0.04& 1\end{array}\right]$$
$$M_2^{-1}=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& -0.04& 1\end{array}\right]$$
用矩阵$M_2$左乘$M_{1}A$
$$\begin{array}{rl}U& =M_2{M}_{1}A\\ & =\left[\begin{array}{ccc}10& -7& 0\\ 0& 2.5& 5\\ 0& 0& 6.2\end{array}\right]\end{array}$$
$$\begin{array}{rl}L& =M_1^{-1}{M}_2^{-1}\\ & =\left[\begin{array}{ccc}1& 0& 0\\ 0.5& 1& 0\\ -0.3& -0.04& 1\end{array}\right]\end{array}$$