用mathematica求六元一次方程组且方程个数比变量个数少一个

问题详见知乎:https://www.zhihu.com/question/68000713

用mathematica求六元一次方程组且方程个数比变量个数少一个

我的问题:有5个方程,6个变量,其实我是想求出来de1=(系数)*dS1的形式,系数有Cij组成,Cij为常数。怎样求?具体方程下面代码可以看出(sys1里时方程,var1里时变量,b是一个大于等于0小于等于1常数)

我是用mathematica求的,参考http://reference.wolfram.com/language/tutorial/SolvingEquations.html

Eliminate[{de1-C11*dS1-C12*dS2-C13*dS3==0,de2-C21*dS1-C22*dS2-C23*dS3==0,de3-C31*dS1-C32*dS2-C33*dS3==0,de1+de2+de3==0,(dS1-dS3)*b-(dS2-dS3)==0},{de2,de3,dS2,dS3}]

Eliminate[{de1 - C11*dS1 - C12*dS2 - C13*dS3 == 0,
de2 - C21*dS1 - C22*dS2 - C23*dS3 == 0,
de3 - C31*dS1 - C32*dS2 - C33*dS3 == 0, (de1 + de2 + de3)/de1 ==
0.5, (dS1 - dS3)*b - (dS2 - dS3) == 0}, {de2, de3, dS2, dS3}] Solve[C11 (2.` C22 - 2.` b C22 + 2.` C23 + 2.` C32 - 2.` b C32 +
2.` C33) dS1 ==
C12 de1 - 1.` b C12 de1 + C13 de1 + 2.` C22 de1 - 2.` b C22 de1 +
2.` C23 de1 + 2.` C32 de1 - 2.` b C32 de1 + 2.` C33 de1 +
2.` C12 C21 dS1 - 2.` b C12 C21 dS1 + 2.` C13 C21 dS1 +
2.` b C13 C22 dS1 - 2.` b C12 C23 dS1 + 2.` C12 C31 dS1 -
2.` b C12 C31 dS1 + 2.` C13 C31 dS1 + 2.` b C13 C32 dS1 -
2.` b C12 C33 dS1, {de1, dS1}]

{{dS1 -> 0. - (0.5 (-1. C12 + 1. b C12 - 1. C13 - 2. C22 + 2. b C22 - 2. C23 - 2. C32 + 2. b C32 - 2. C33) de1)/(-1. C12 C21 + 1. b C12 C21 - 1. C13 C21 + 1. C11 C22 - 1. b C11 C22 - 1. b C13 C22 + 1. C11 C23 + 1. b C12 C23 - 1. C12 C31 + 1. b C12 C31 - 1. C13 C31 + 1. C11 C32 - 1. b C11 C32 - 1. b C13 C32 + 1. C11 C33 + 1. b C12 C33)}}

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