Solving 5 eqns with 6 unknowns in a 2x3 contingency matrix, is there a unique solution? [closed]

Background

I have the following equations:

$$a+b+c=6$$ $$d+e+f=15$$ $$a+d=5$$ $$b+e=7$$ $$c+f=9$$

This is a 2x3 matrix $[a b c, d e f]$ where the marginal totals are fixed. In addition, all of the unknowns are positive.

I incorrectly told a student that he could do this analytically, and he returned with the answer that, no it could not be solved analytically, but provided a solution that uses the method of iterative proportional fitting.

Question

Is it possible to find a unique solution (and know it is unique) for the set of six unknowns using iterative proportional fitting or some similar approach?

Note

I posted a related question on math.stackexchange that was limited to the analytical solution, where the initial response was that there was no unique solution; a subsequent response led me to the exact trivial solution and now I see why this was closed.

-

closed as too localized by Will Jagy, Ben Webster♦Jan 25 '11 at 1:41

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

MathOverflow is for questions which are of interest to research mathematicians; this is essentially an undergraduate linear algebra exercise (even if it's not HW, it could be given as an HW in any linear algebra class). If you're not satisfied with the answers you have here, I would suggest asking on math.SE (it's not clear to me why you switched sites). – Ben Webster Jan 25 '11 at 1:44
@Ben Thanks I did not realize how simple it was. I moved it here after my student returned with the iterative proportional fitting solution, which did not appear to provide a unique solution, but now I realize that the analytical solution is trivial and exact. – David LeBauer Jan 27 '11 at 18:12

Since you have 5 linear equations but 6 unknowns so there cannot be an unique solution and there is at least one degree of freedom for the solution, for instance 5 of the unknowns depend on the 6'th.
After that you introduce one restriction: all unknowns should be positive and integer (because you define, the equations stem from a contingency-table) - such restriction may fix some specific solution, but don't see this here (because Will's comment shows multiple positive solutions). One uniqueness criterion is the "expected frequencies" in a contingency tables. A contingency table with given margins like your example (sorry, don't know how to separate the margins in the matrix)

$\begin{matrix} a & b & c & 6 & \\\ d & e & f & 15 \\\ 5 & 7 & 9 & 21 \end{matrix}$

has expected frequencies:

$\begin{matrix} 6*5/21 & 6*7/21 & 6*9/21 & 6 & \\\ 15*5/21 & 15*7/21 & 15*9/21 & 15 \\\ 5 & 7 & 9 & 21 \end{matrix}$

and this is

$\begin{matrix} 10/7 & 2 & 18/7 & 6 & \\\ 25/7 & 5 & 45/7 & 15 \\\ 5 & 7 & 9 & 21 \end{matrix}$

so a=10/7, b=2 , c=18/7, d=25/7, e=5, f=45/7 is one unique solution, if you put it into the framework of the chi-square-rationale

[update] if the values are required to be integer (which is not explicite from the question, but may be meaningful) then there I'd assign "uniqueness" to that solution which requires least modification, in this case uses the next integer by subtracting / adding 3/7:

$\begin{matrix} 1 & 2 & 3 & 6 & \\\ 4 & 5 & 6 & 15 \\\ 5 & 7 & 9 & 21 \end{matrix}$

[end update]

-
@Gottfried Thank you for your answer - now I understand @Will's point better. Perhaps I misused the term contingency table: the values do not have to be integers. Are you saying that if I assume that the variable has a chi-square distribution? – David LeBauer Jan 25 '11 at 0:40
@David: I don't understand your last question. A distribution is a property of a random-variable, that means something which can have many realizations. But here you have 6 variables in an arrangement of a contingency-table with given margins, you have 2 degrees of freedom: 2 variables (not in the same column) can vary continuously - (with some range-restriction when negative values are forbidden). If -for the varying a,b,c,..., - the resp. chi-square-value $\Chi$ is computed you'll compare it with a "chi-square-distribution for 2 degrees of freedom" – Gottfried Helms Jan 25 '11 at 1:11
@Will: "contingency table" is just a bivariate table of frequencies. And yes, a 2x3-contingency-table has 2 "degrees of freedom" , so 2 parameters (2 frequencies) are free and 6-4=2 frequencies are then determined. (But the question is closed, so this comment will likely not appear...; the question should have occured in stat.exchange instead) – Gottfried Helms Jan 25 '11 at 2:16
Thanks, Gottfried. Comments can continue on closed questions. – Will Jagy Jan 25 '11 at 3:25