则有1(41.b)=(4,4) k=1,2,3,…,n-1 因此 L2·L1(AD,b①)=(4(0,b0) 从而Ln1…:L2·L1A=A0 故A=1L2…Ln140)=LU L=L1L2…Ln1为单位下三角矩阵 U=A)为上三角矩阵则有 ( , ) (k ) (k ) Lk × A b ( , ) ( +1) ( +1) = k k A b k = 1,2,3,L,n - 1 ( , ) (1) (1) × L × L2 × L1 × A b Ln-1 ( , ) (n) (n) 因此 = A b 从而 Ln-1 × L × L2 × L1 × A 1 ( ) 1 1 2 1 1 n L L Ln A - - - - 故 A = L U = A (n)为上三角矩阵 L = L - 1 1 L - 2 1LL - n 1 -1为单位下三角矩阵 = LU (n) = A