[代數] 三元二次方程組 解法問題
各位好! 小弟我在研究中出現了一個三元二次方程組 :
{f(x,y,z) = [Aj]*x^2 + [Bj]*y^2 + [Cj]*z^2 + [Dj]*x*y*z + [Ej] = 0, j=1,2,3 }
似乎有兩種解法, 但我用MATLAB算出來的答案好像不太一樣,
我想請問這兩種解法有何差別? 還是有什麼地方我沒注意到的?
<<< 解法一 >>>
<Step1-1>
{f(x,y,z) = [Aj]*x^2 + [Bj]*y^2 + [Cj]*z^2 + [Dj]*x*y*z + [Ej] = 0, j=1,2,3 }
=> { [Aj/Dj]*x^2 + [Bj/Dj]*y^2 + [Cj/Dj]*z^2 + [Ej/Dj] = -x*y*z, j=1,2,3 }
<Step1-2>
{ {f(x,y,z), j=n} - {f(x,y,z),j=m} , n =\= m, n=1,2,3 ,m=1,2,3 } :
=> { g(x,y,z) = [Aj]*x^2 + [Bj]*y^2 + [Cj]*z^2 + [Ej] = 0, j=4,5,6 }
<Step2-1>
{ g(x,y,z), j=1 } * (A5/A4) - { g(x,y,z), j=2 } :
K11*y^2 + K12*z^2 + K13 = 0 ... (2-1)
{ g(x,y,z), j=1 } * (A6/A4) - { g(x,y,z), j=3 } :
K21*y^2 + K22*z^2 + K23 = 0 ... (2-2)
<Step2-2>
(2-1)*(K21/K11)-(2-2) :
K31*z^2 + K32 = 0
=> z^2 = -(K32/K31) ... (2-3)
<Step2-3>
(2-3) into (2-1) => y^2 = -(1/K11)*(K13-K12*(K32/K31)) ... (2-4)
<Step2-4>
(2-3) & (2-4) into g(x,y,z) :
=> x can be found !
Finally [x y z] is solved !
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<<< 解法二 >>> 由板友wope大大提供
A1 * x^2 + B1 * y^2 + C1 * z^2 + D1 *xyz = E1 (1-1)
A2 * x^2 + B2 * y^2 + C2 * z^2 + D2 *xyz = E2 (1-2)
A3 * x^2 + B3 * y^2 + C3 * z^2 + D3 *xyz = E3 (1-3)
(1-1)*A2/A1 :
A2 * x^2 + A2*B1 * y^2/A1 + A2*C1 * z^2/A1 + A2*D1 *xyz/A1 = A2*E1/A1 (1-4)
(1-1)*A3/A1 :
A3 * x^2 + A3*B1 * y^2/A1 + A3*C1 * z^2/A1 + A3*D1 *xyz/A1 = A3*E1/A1 (1-5)
(1-4)-(1-2):
(A2*B1/A1-B2)*y^2 + (A2*C1/A1-C2)*z^2 + (A2*D1/A1-D2)*xyz = (A2*E1/A1-E2)
(1-6)
取
A11=(A2*B1/A1-B2)
A12=(A2*C1/A1-C2)
A13=(A2*D1/A1-D2)
A14=(A2*E1/A1-E2)
A11*y^2 + A12*z^2 + A13*xyz = A14 (2-1)
(1-5)-(1-3):
(A3*B1/A1-B3)*y^2 + (A3*C1/A1-C3)*z^2 + (A3*D1/A1-D3)*xyz = (A3*E1/A1-E3)
(1-7)
取
A21=(A3*B1/A1-B3)
A22=(A3*C1/A1-C3)
A23=(A3*D1/A1-D3)
A24=(A3*E1/A1-E3)
A21*y^2 + A22*z^2 + A23*xyz = A24 (2-2)
(2-2)*A11/A21:
A11*y^2 + A11*A22*z^2/A21 + A11*A23*xyz/A21 = A11*A24/A21 (2-3)
(2-3)-(2-1):
(A11*A22/A21-A12)*z^2 + (A11*A23/A21-A13)*xyz = (A11*A24/A21-A14) (2-4)
取
A31=(A11*A22/A21-A12)
A32=(A11*A23/A21-A13)
A33=(A11*A24/A21-A14)
A31*z^2 + A32*xyz = A33 (2-5)
取
A41=A32/A31
A42=A33/A31
z^2 + A41*xyz = A42 (2-6)
-->z^2=A42-A41*xyz (2-7)
代回(1-1)(1-2)(1-3)可得
A1 * x^2 + B1 * y^2 + (D1-c1*A41) *xyz = E1-A42*C1 (3-1)
A2 * x^2 + B2 * y^2 + (D2-c2*A41) *xyz = E2-A42*C2 (3-2)
A3 * x^2 + B3 * y^2 + (D3-C3*A41) *xyz = E3-A42*C3 (3-3)
取
Y1=x^2 (3-4)
Y2=y^2 (3-5)
Y3=xyz (3-6)
F1=(D1-c1*A41)
F2=(D2-c2*A41)
F3=(D3-C3*A41)
G1=E1-A42*C1
G2=E2-A42*C2
G3=E3-A42*C3
A1 * Y1 + B1 * Y2 + F1 *Y3 = G1 (4-1)
A2 * Y1 + B2 * Y2 + F2 *Y3 = G2 (4-2)
A3 * Y1 + B3 * Y2 + F3 *Y3 = G3 (4-3)
這裡Y1,Y2,Y3該就可以解了
再代回(3-4)(3-5)(3-6)
Y1=x^2
Y2=y^2
Y3=xyz
就可以得到x,y,z的值
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以上,感謝耐心閱讀
※ 編輯: joe7078 來自: 140.110.200.15 (08/21 19:11)