**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](http://en.wikipedia.org/wiki/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](https://math.stackexchange.com/q/18780/3733) 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.



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


