中国矿亚大整CHINA UNIVERSITYOFMININGANDTECHNOLOGYCH3线性方程组解法/*Direct Method forSolving Linear Systems */$1消去法$2矩阵分解法$3方程组的性态和条件数$4迭代法
CHINA UNIVERSITY OF MINING AND TECHNOLOGY CH3 线性方程组解法 /* Direct Method for Solving Linear Systems */ §1 消去法 §2 矩阵分解法 §3 方程组的性态和条件数 §4 迭代法
中国矿亚大医CHINAUNIVERSITY OFMININGANDTECHNOLOGY81消去法一、高斯顺序消去法/*GaussianElimination*2x,+x2 +x,=7口例题 求解34x+5x,-x,=11-2x2 +x,=0X解 Stepl：消元23+A=(A,b)=-2r3x25-3r3-0.5r0-2710X, =-6/(-2)= 3X2=(-3+3×3)/3=2Step2:回代人X,=(7-1×3-1×2)/2=1
CHINA UNIVERSITY OF MINING AND TECHNOLOGY §1 消去法 解 1 23 1 23 1 23 7 4 5 11 2 0 x xx x xx x xx ⎧ 2 + + = ⎪⎨ + − = ⎪ − + = ⎩ Step1:消元 21 1 7 ( , ) 4 5 1 11 1 21 0 A Ab ⎡ ⎤ ⎢ ⎥ == − ⎢ ⎥ ⎢ − ⎥ ⎣ ⎦ 21 1 7 03 3 3 00 2 6 ⎡ ⎤ ⎢ ⎥ − − ⎢ ⎥ ⎢ − − ⎥ ⎣ ⎦ 3 2 53 2 r r + ⎯⎯⎯⎯× → Step2:回代 ⎧⎪⎨⎪⎩ 5 1 7 22 2 21 1 7 03 3 3 0 ⎡ ⎤ ⎢ ⎥ − − ⎢ ⎥ ⎢ − − ⎥ ⎣ ⎦ 2 1 3 1 20.5 r r r r − ⎯⎯⎯→ − 例题 求解 x3 = − −= 6/( 2) 3 x2 = ( 3 3 3)/ 3 2 −+ × = x1 = ( )/ 71312 2 1 −×−× = 一、高斯顺序消去法 /* Gaussian Elimination */
中国矿亚大医CHINA UNIVERSITY OFMININGAND TECHNOLOGY1.计算机上所用的公式ax, +ax, +...+ainx,=ain+ia2x, +a22x, +...+a2n, =a2,n+1求解方程：am,+an,+...+amX,=ann+1下面研究它的计算规律：
CHINA UNIVERSITY OF MINING AND TECHNOLOGY nn n nn n n n nn n n n ax ax ax a ax ax ax a ax ax ax a + + + ⎧ + ++ = ⎪ ⎪ + ++ = ⎨ ⎪ ⎪ + ++ = ⎩ 11 1 12 2 1 1, 1 21 1 22 2 2 2, 1 11 22 , 1 " " """""""""""" " 求解方程： 1.计算机上所用的公式 下面研究它的计算规律：
中国矿亚大鉴CHINAUNIVERSITYOF MININGANDTECHNOLOGYal+aa.faalo...doaoao(0)a0a.0a2,n+l..............aafaaC1)消元记4(=at0)aa.(0)(0)(0)(0)(0)ak+1,2ak+1,kak+1,+1ak+1,#ak+l,n+1...............alk+a(0)aaatoan,n+1假设0,令=/0Stepl:(i=2, ... n)aaao(0)(0)ai,k+1ai,n+1.....agaadlada0....lin'A(0).............allaaalktadalhat1i=2,...n...(1)(1)(1)(1)(1)ak+1,2ak+1nak+1,kak+1,k+1ak+1,n+1.............a0aaaltaaal)0n2n,n+i
CHINA UNIVERSITY OF MINING AND TECHNOLOGY Step 1： 0 0 1 1 11 2 i i laa i n = = () () 令 / ( , ., ) i n = 2, ., i i r lr − 1 1 a ≠ (0) 11 假设 ，0 (0) (0) (0) (0) (0) (0) 11 12 1 1, 1 1 1, 1 (0) (0) (0) (0) (0) (0) 21 22 2 2, 1 2 2, 1 (0) (0) (0) (0) (0) (0) (0) 1 2 ,1 ,1 (0) (0) (0) 1,1 1,2 1, . . . . . . . . . . . . . kk nn kk nn k k kk k k kn k n k k kk k aa aa aa aa aa aa A aa aa aa aa aa + + + + + + ++ + + = (0) (0) (0) 1, 1 1, 1, 1 (0) (0) (0) (0) (0) (0) 1 2 ,1 ,1 . . . . . . . . . k kn kn n n nk n k nn n n a a aa aa aa + + ++ + + ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ kk nn kk nn k kk k k kn k n k kk kk kn k aa a a a a a aa aa a aa aa a aa aa + + + + + + + + ++ + + (0) (0) (0) (0) (0) (0) 11 12 1 1, (1) (1) (1) (1) (1) (1) (1) (1) (1) (1) (1) ( 1 1 1, 1 22 2 2, 1 2 2, 1 2 , 1) (1) (1) 1 ,1 1,2 1, 1, 1 1, . . . . . . . . . . . . . 0 0 . 0 n n nk n k nn n n a aa aa + + + ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ (1) (1) (1) ( 1, 1 2 ,1 ,1 1) (1) (1) . . . . . 0 . . . A(0) 1)消元 记
a0aloal+ala中国矿亚大鉴CHINA UNIVEaadaadat0.....R-1(k-1)(k1)(k-1)A(k-l)假设第k-1步消元后1ak,n+1ak.k+11kn(k-l)(k-1)(k-1)ak+1,kak+1,k+1ak+1,nak+l,n+1.........(k-1)(k1)(k-1)(k-1)1aan,n+1Ian,k+1nknn(k-1(k-1)Step k： 若ak-" 0, 计算lx =ak(i=k+l,...,n)akk以及a(k)=a(t-1) -liak(i=k+l.., n; j=k+l,....n+1)(0)(0)1o(0)(0)aodikai.k+1dinai.n+1(agha.ada..(1)ad122............(k-1)at!ikK(k-1)(k-1)ak.k+lkkdknA(K)(k)(k)(k)-i=k+l,.nak+1,n+1ak+1,k+1ak+1.n...(k)(k)(k)a11n,k+1nnn,n+1
CHINA UNIVERSITY OF MINING AND TECHNOLOGY 假设第 步消元后 k − 1 kk nn kk nn k kk kk kk k k kn k n k k kk kk k aa a a a a a aa aa A aa aa aa 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 kn kk kk nk n k nn n n a aa aa − − + + −− −− + + ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ( 1) ( 1) 1, 1, 1 ( 1) ( 1) ( 1) ( 1) ,1 ,1 . . . 0 0 . . . . . . . . Step k ： ik n = + 1, ., i ik k r lr − k kk a − ≠ ( 1) 若 ，0 k k ik ik kk la a − − = ( 1) ( 1) 计算 / ( 1, ., ) ik n = + kk k ij ij ik kj a a la − − = − 以及 ( ) ( 1) ( 1) ( 1 . ; 1,., 1) ik n = + =+ + , , j k n k A( 1) − ( ) k A kk nn kk nn kk kk kk k k kn k n k kn k k k n k k aa a a a a a aa aa aa aa a aa + + + + −− −− + + ++ + ++ (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 00 0 . . . . . n k nn n n k kk a aa + + ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ , 1 ) () () () , 1 . . . . . . . . . 0 0 . 0 . Δ