第k步:设≠叶计算因子l1=-a)1a(i=k+1,,m) 且计算{4" (k) (k+1) (k) b (k) kk (ij=k+1,…,n) 共进行n-1步,得到 (1) (1) n b1) 11 1 (2) 2 2 (n)0 ( )  k 第k步：设 akk ，计算因子 且计算 ( ) ( ) / ( 1, ..., ) k k ik ik kk l a a i k n = − = + ( 1) ( ) ( ) ( 1) ( ) ( ) ( , 1, ..., ) k k k ij ij ik kj k k k i i ik k a a l a b b l b i j k n + +  = +   = + = + 共进行 n − 1步，得到               =                           ( ) (2) 2 (1) 1 2 1 ( ) (2) 2 (2) 2 2 (1) 1 (1) 1 2 (1) 1 1 . . . . . . . . . ... ... ... n n n n nn n n b b b x x x a a a a a a