Matrix problem with natural numbers There are numbers from 3 to 11 and a square 3x3:
  0  1  2
0 __|__|__
1 __|__|__
2   |  |

Fill this square with aforementioned numbers so that the result of multiplication of numbers in rows and columns with the same index be equal. If it is not possible explain why?
answer:
7 6 5 
3 9 8 
10 4 11

but how to solve it?
 A: First we decide what goes on the diagonal. The prime numbers 7 and 11 must be on the diagonal. As there are no other numbers from 3 to 11 that has 7 or 11 as factors, if say 7 is on off diagonal position, then we need another number with 7 as a factor and place it on the mirror position across the diagonal, which is impossible.
Now 9 must be also on the diagonal. If not, then say 9 is in an off diagonal position of row i, then column i must have the two off diagonal entries 3 and 6, since 9 has two factors of 3. But this makes column i having product 3*6*d and row i having product 9*?*d, where d is the corresponding diagonal. This requires ? to be 2. Since we do not have 2 in our list, we must have 9 also on the diagonal.
So far, the diagonal entries must be 7,9,11, to which you can place them any order you want.
Now for the remaining numbers 3,4,5,6,8,10, say start with 3 and put it in any remaining off diagonal position you want. The mirror position of 3 across the diagonal must be a 6, since we need a factor of 3 in those row/column. Now in the row or column that contains 3 we cannot put 4 or 5 as the third off diagonal entry, since we have the mirror position of 3 filled with 6. 
So put either 8 or 10 in the row/column that contains 3. You will find that there will be able to fill the remaining squares with 4 and 5.
Example of a construction:

first put 7,9,11 goes on diagonal:
   11 ? ?
    ? 7 ?
    ? ? 9
next place 3 anywhere remaining spot you want:
   11 ? ?
    ? 7 3
    ? ? 9
we must have 6 to be in mirror position (across diagonal) from 3:
   11 ? ?
    ? 7 3
    ? 6 9
now in the row and column that contains 3, place 10 and 8 however you want:
   11 ? 8
   10 7 3
    ? 6 9
we will be left with one way to finish filling 4 and 5:
   11 5 8
   10 7 3
    4 6 9

(there should be 3!*6*2 = 72 possible solutions.)
