aiiaviBackSubstitutionb(n)det(A.)=nx1a(n)aiaiinnbl"-Za'x;j=i+1(=n-1,n-2,..,2,1x:=(i)diiTheory:If all the sequence ofAis not O,then Gausseliminationcanbreaktthrough to the end, the only solution can be woeked out.Warning:Actually,if A-I is existed,with the successiveelimination byswitching element and the line,the eguations for the triangularequationsand the only solutionwill be worked out.上页下页返回
上页 下页 返回 ( ) ( ) n nn n n n a b x = i = n −1,n − 2, ,2,1 ( ) 1 ( ) ( ) i i i n j i j i i j i i i a b a x x = + − = Theory: If all the sequence of A is not 0, then Gauss elimination can break through to the end, the only solution can be woeked out. Warning:Actually,if A−1 is existed,with the successive elimination by switching element and the line, the equations for the triangular equations and the only solution will be worked out. Back Substitution i ii i i a a a a A . . . . . det( ) 1 11 1 =
examplel: Solute this system of linearequations with the gauss eliminationmethodXi+ X+ X =64x2- X, = 5L2x -2x, +X,=1Solute:bytheaugmented matrix:613-2ri(Ab) =51-111n13+1251上页-6下页返圆
上页 下页 返回 2 2 1 4 5 6 1 2 3 2 3 1 2 3 − + = − = + + = x x x x x x x x example 1: Solute this system of linear equations with the gauss elimination method Solute: by the augmented matrix: (A b) − = − 2 2 1 1 0 4 1 5 1 1 1 6 − − − ⎯→ − − 0 4 1 11 0 4 1 5 1 1 1 6 3 2 1 r r − − ⎯→ − + 0 0 2 6 0 4 1 5 1 1 1 6 3 2 r r
The triangular equations:Xi+ X+ X =64x2- Xg = 5- 2x, = -6Back substitution:X3 =3X =2x =10: Xi =1,X2 =2,X, = 3上页下页返圆
上页 下页 返回 2 6 4 5 6 3 2 3 1 2 3 − = − − = + + = x x x x x x Back substitution: The triangular equations: 3 x3 = x2 = 2 x1 = 1 so: x1 = 1, x2 = 2, x3 = 3
2,The Gauss elimination method by chosing the pivots[10-′x + X2 = Example :The single precision solutionsystem ofequations:X+=28bit8bit1-1.00...0100...andx,= 2- x, = 0.99 ... 9899..*/￥*precision solution:x,1-10-Calculated using the Gauss elimination method:10°m21 = a21 / ai18个a22=1-m2,×1=0.0...01×10°-10°=-10gb, =2-m2i×1=-109The small main element may10元leadtothefailureof-10°-10°0computation.上页(0X2 =1,XiL下页返回
上页 下页 返回 Example :The single precision solution system of equations: + = + = − 2 10 1 1 2 1 2 9 x x x x /* precision solution: 1.00.0100. and */ 1 10 1 1 9 = − = − x 8bit 2 0.99 . 9899. 2 1 x = − x = 8bit Calculated using the Gauss elimination method: 9 m21 = a21 / a11 = 10 9 9 9 a2 2 = 1− m2 1 1 = 0.0.0110 −10 = −10 8个 9 b2 = 2− m21 1 = −10 − − − 9 9 9 0 10 10 10 1 1 x2 = 1, x1 = 0 The small main element may lead to the failure of computation. 2,The Gauss elimination method by chosing the pivots
The pivoting elimination methodSelect the element whose absolute value is max in each step ,guarante: I mik /≤ 1Step k: ① choose: IaixixI=max. Iai, I+0;k≤i,j≤nIf ik+k then exchange kth rowandikrth row;If jk+ k then exchange kth colum andjsth colum;Elimination.Note: column exchange order changes the order of Xi and record theexchange order.exchangeagainaftersoluting The colum'spivotingelimination methodSave the steps of exchanging colums, instead of only choosing the largestelement in the column.上页Iaix,k/=maxlaik|±0k<i<n下页返圆
上页 下页 返回 The pivoting elimination method Select the element whose absolute value is max in each step ,guarante: | mik | 1 Step k: ① choose: | | max | | 0; , = ij k i j n ai j a k k ② If ik k then exchange kth row and ikt th row; If jk k then exchange kth colum and jk th colum; ③ Elimination. Note: column exchange order changes the order of Xi and record the exchange order.exchange again after soluting. The colum's pivoting elimination method Save the steps of exchanging colums, instead of only choosing the largest element in the column. | , | = max | | 0 i k k i n ai k a k