Originally Posted by
Wass
Treat the matrix in a normal way, try to ease it up so it becomes easier, with several numbers being zero etc. This will probably lead to some numbers being a-x or x-a (x being a random number), and when you can't simplify it more try to put in logical numbers that will give NO solution to the matrix. Find out if there are any numbers that will give you a solution, and that'd be the answer.
I had a similar task with a matrix like this:
a 1 2 = 4
1 1 1 = 1
1 a 2 = 0
After a couple modifications I got it to:
1 1 1 = 1
0 a-1 1 = 1
0 0 3-a = 3-a
From there I could derive that it didn't have any solutions for a = 1, a = 3 had one solution, and any a != (1 or 3) would give infinite solutions.