What kind of problem is this and what tools can we use to tackle it?

[LuculentCabal]

Basically I have the following system:

[tex] ac = A [/tex]

[tex] ad + bc = B [/tex]

[tex] bd = C [/tex]


where [tex]A[/tex], [tex]B[/tex], and [tex]C[/tex] are constants.

Solving for [tex]a, b, c,[/tex] and [tex]d[/tex], what kind of problem/system am I encountering and what appropriate tools (vectors and/or numerical methods perhaps[?]) would help to find the set of solutions for [tex]a, b, c,[/tex] and [tex]d[/tex]?

I know that the system factors a trinomial but I don't know what kind of problem is presented by the system itself.

Thanks in advance for any insight.

 

[betel]

Inserting the first and third equation into the second gives you a quadratic equation (altghough for a product of two unknowns).
You will still end up with one unknown variable as you do not have enough equations.

 

[LuculentCabal]

You will still end up with one unknown variable as you do not have enough equations.

Could there still be numerical methods for finding the solutions?

Thanks for your reply.

 

[mikeph]

My messing around on MATLAB. Not sure how useful it is...

EDU>> syms a b c A B C
EDU>> sol = solve('a*c = A','a*d + b*c = B','b*d = C');
EDU>> [a;sol.b(1);sol.c(1);sol.d(1)]
ans =
                                 a
 a*(1/2*B-1/2*(B^2-4*A*C)^(1/2))/A
                               A/a
   (1/2*B+1/2*(B^2-4*A*C)^(1/2))/a
EDU>> [a;sol.b(2);sol.c(2);sol.d(2)]
ans =
                                 a
 a*(1/2*B+1/2*(B^2-4*A*C)^(1/2))/A
                               A/a
   (1/2*B-1/2*(B^2-4*A*C)^(1/2))/a

I guess that's two sets of solutions?

edit - I generally prefer linear algebra to sets of equations where variables are multiplied. It's kind of a nightmare actually - I can see a uniqueness/existence problem if a = 0 right away, or B^2-4*A*C < 0? What a mess...

 

[Gerenuk]

I recommend substituting the top and bottom into the middle.
[tex]Cx+\frac{A}{x}=B[/tex]
(where [itex]x=\frac{a}{b}[/itex]) and solve for x.

The variable b is arbitrary (but not equal 0) and [itex]a=xb[/itex]
Use top and bottom equation to find rest.

EDIT: just as bebel said :D

 

[LuculentCabal]

Thanks for the output MikeyW, your MATLAB solution looks consistent with what I have found. I have found that the system can be simplified to a function of [tex]b, c[/tex], and the constants. Letting one of the variables assume any value, I can solve for the other with the quadratic formula or Newtons method, and then find the other two variables.

After simplifying:

[tex]
{b^2}{c^2} - Bbc + AC = 0
[/tex]

[tex]
a =\frac{A}{c}
[/tex]

[tex]
d = \frac{C}{b}
[/tex]

I don't know if this would necessarily work for higher degree polynomials (especially an odd number of variables), and wonder if there is a solution using vectors/linear algebra. My linear algebra knowledge is still elementary and haven't had to encounter multiple variables before so I don't know if there are tools that can help.

 

[LuculentCabal]

A little more complex would be the following system which factors a quintic when [tex]a, b, c, d, e, f, g, h, i [/tex] and [tex]j[/tex] are solved for with [tex]A, B, C, D, E,[/tex] and [tex]F[/tex] being constants.


[tex] acegi = A[/tex]

[tex] acegj + acehi + acfgi + egadi + egbci = B [/tex]

[tex] acehj + acfgj + egadj + egbcj + acfhi + adehi + adfgi + bcehi + bcfgi + bdegi = C [/tex]

[tex] acfhj + adehj + bcehj + bcfgj + bdegj + adfhi + bcfhi + ehbdi + fgbdi + adfgj = D [/tex]

[tex] adfhj + bcfhj + ehbdj + fgbdj + bdfhi = E [/tex]

[tex] bdfhj = F [/tex]


Is there any way to solve for the above ten variables with a method more streamlined than the "plug and chug" method used previously for the quadratic?

*I am not as interested in any solution as I am in any algorithm ("plug and chug" or other)

Thanks again for any info. It is greatly appreciated.