aaltaoa-ao-aa(1)a...ad0a2.k+1...........(k-1),(k-1)at!假设第k-1步消元后A(k-1)=00ak,n+1kk(k-1)(k-1)altl.(k-1)00ak+1,kak+1,k+1ak+1,n+1......a(k-1)α(k-1)a(k-1)a(k-1)00n,k+1nknnn,n+1= α(k-1) / α(k-1)Stepk：若ak-10，计算lk(i= k +1, ..., n)kkα(k-1)a(t-1)以及a(k)10(i= k+1.., n; j=k+1...,n+l)5iikkjaaaaloa a.+Oadadaad0......ik'k1(k-1)g(k-1).(k-1).(k-1)00kkak,n+1ak,k+1aknAA(k)A(k-1)...(k)(k)(k)000i=k+l,.,nak+1,k+1ak+1,n+1ak+1,n.........o..(k)(k)(k)000an,k+1annan,n+1-6-
-6- 假设第k 1步消元后 k k n n k k n n k k k k k kk k k kn k n k k k k k k k a a a a a a a a a a a A a a a a a a a (0) (0) (0) (0) (0) (0) 11 12 1 1, 1 1 1, 1 (1) (1) (1) (1) (1) 22 2 2, 1 2 2, 1 ( 1) ( 1) ( 1) ( 1) ( 1) , 1 , 1 ( 1) ( 1) 1, 1, 1 . . . . . . . . . . 0 0 0 0 0 . . . . . . . . . k k n k n k k k k nk n k nn n n a a a a a ( 1) ( 1) 1, 1, 1 ( 1) ( 1) ( 1) ( 1) , 1 , 1 . . . 0 0 . . . . . . . . Step k： i k n 1, ., i ik k r l r k kk a ( 1) 若 0， k k ik ik kk l a a ( 1) ( 1) 计算 / ( 1, ., ) i k n k k k ij ij ik kj a a l a 以及 ( ) ( 1) ( 1) ( 1 . ; 1,., 1) i k n j k n , , k A ( 1) ( ) k A k k n n k k n n k k k k kk k k kn k n k k n k k k n k k a a a a a a a a a a a a a a a a a a (0) (0) (0) (0) (0) (0) 11 12 1 1, 1 1 1, 1 (1) (1) (1) (1) (1) 22 2 2, 1 2 2, 1 ( 1) ( 1) ( 1) ( 1) , 1 , 1 1, 1 1, 1, 1 ( ) ( ) ( . . . . . . . . . . . 0 0 0 0 0 0 . . . . . n k nn n n k k k a a a , 1 ) ( ) ( ) ( ) , 1 . . . . . . . . . 0 0 . 0 .
消元结束，A化为上三角矩阵Step n-l:aaaalf+aaaada..aad0att!atl!at-l)A(n-a(k)att.at.010ak+1,k+1....a(ni)a(n-1)0nn,n+1k =1,2,..,n-l, a 0i= k+1,..,n在实际编程中为了节省内存a不引入新变量j = k + 1,..., n+1消元过程记为ay-aikak=a-7-
-7- Step n-1： k k n n k k n n n k k k k kk k k kn k n k k k k k n k n a a a a a a a a a a a A a a a a a a a (0) (0) (0) (0) (0) (0) 11 12 1 1, 1 1 1, 1 (1) (1) (1) (1) (1) 22 2 2, 1 2 2, 1 ( 1) ( 1) ( 1) ( 1) ( 1) , 1 , 1 ( ) ( ) 1, 1 1, 1, . . . . . . . . . . . . 0 0 . 0 . . 0 0 . . . 0 k n n nn n n a a ( ) 1 ( 1) ( 1) , 1 . . . . . . . 0 . . . 0 0 . 0 消元结束，A化为上三角矩阵： 1 2 1 0 kk k n a , ,., ， i k n 1,., ik ik ik kk a l a a j k n 1,., 1 ij ik kj ij a a a a 在实际编程中， 为了节省内存， 不引入新变量， 消元过程记为：
消元束后，增广矩阵化为如下“形式”：al.aaa0aala...a..adaad..........alt!(k-1)os,(k-1)Iki1akkak,n+1(k)a(k).(k)Ik+iak+1,nak+1,k+1ak+1,n+1'k+1,kk+1.2...*.*e..,(n-1)a(n-1)InVnkIn.k+11ma中电业nnn,n+12）回代aaa0al.axi....adall(1)ad)X2a2,k+1......-/ax, =a(n-)a.............2a(k-1)a(k-1),(k-1)IIh2(k-1)(k-1)(k-)aknXk-1akkak,k+1X=(a)akjakk.n+1allat.j=k+1Ik+1.!1Xk..*k+1.2"k+1,k(k=n-1,.,1)....a(n-1)41In.k+1x....nn-8-
-8- 消元束后，增广矩阵化为如下“形式” ： k k n n k k n n k k k k kk k k kn k n k k k k k k k k k k a a a a a a l a a l l a a a a a a l a l l a a (0) (0) (0) (0) (0) (0) 11 12 1 1, 1 1 1, 1 (1) (1) (1) (1) (1) 22 2 2, 1 2 2, 1 ( 1) ( 1) ( 1) ( 1) , 1 , 1 ( ) 1 , 21 1 2 1,1 1.2 1, 1 1 . . . . . . . . . . . . . . . n n nk n k k n k n n n k nn n n l l l l a a a ( ) ( ) , 1, 1 ( 1) ( 1) 1 2 1 , , 1 . . . . . . . . . . . n n n n n nn n k k k k k n kj j kk j k x a a x a a x a k n ( 1) 1 , 1 ( 1) ( 1) ( 1) , 1 1 / ( ) / ( 1, ,1) 1 (0) (0) (0) (0) (0) 11 12 1 1, 1 1 (1) (1) (1) (1) 22 2 2, 1 2 ( 1) ( 1) ( 1) , 1 ( ) 21 1 2 1,1 1.2 1, ( ) 1, 1 , 2 1 1 . . . . . . . . . . . . . . . . . . . . . . . . k k n k k n k k k kk k k kn k k k k k n k k k k k k k k l l l a a a a a a a a a a a x x l a l a a x l x 1 ) 1 ( , 1 2 . . . . . . . . n n nk n k n nn n l l l l a x 2）回代
0口一个模拟计算机求解的例子11511X+X++=521-1-24X+2x2 -x + 4x4 =-2解A=求解23-3-2-5-2x -3x2 +2x - 5x4 =31231103x +x,+2x,+x=10111511151-23111-2271/1A→26-1/11342-311-2/1-19-543-2/13/1-2-12-511111511113-23-22223006-1-1-22334-2-5/24-2-5/2-9-
-9- 一个模拟计算机求解的例子 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 5 2 4 2 2 3 2 5 3 3 2 10 x x x x x x x x x x x x x x x x 求解 解 A 1 1/ 2 1/ 3 1/ 2 3 1 1/ 2 1/ 1 –2 3 –7 1 1 1 1 5 –1 4 – 3 13 –2 –1 –2 –5 2 0 6 –5 4 –19 1 1 1 1 5 1 1 –2 3 –7 –5/2 4 –4 1 1 1 1 5 –2 –1 2 0 6 1 1 –2 3 –7 3 –2 A 1 1 1 1 5 1 2 1 4 2 2 3 2 5 3 3 1 2 1 10 1 1 1 1 1 –2 –1 2 0 3 –5/2 4 –1 1 1 –2 3 2 3 –2
4消去法成立的条件：(k-1)+0, (k=1,2...,n-1)akk动态变化！2nn5计算量次乘法+n33-10-
-10- ( 1) 0, ( 1,2 , 1) k kk a k n 4消去法成立的条件： 5计算量： 3 2 3 3 n n n 次乘法 动态变化!