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CompuChip
#2
Mar18-08, 04:20 AM
Sci Advisor
HW Helper
P: 4,300
There are the following possibilities:
  • All columns are independent
  • The first and the second are dependent, but independent of the third
  • The first and the third are dependent, but independent of the second
  • The second and the third are dependent, but independent of the first
  • All of them are dependent
Or, if you prefer, you can do this with the rows.
Now for each case, you can write down an equation and solve it for a. For example,let me do the third case (first and third are dependent, but independent of the second).
If the third column is a multiple n of the first one, you must have
1 = n a
-2 = 2 n
a = n 1

From the second equation you see that there is just one possibility for n. Then you get a solution for a from one of the others. Finally, use the remaining equation to see if this value indeed satisfies all of them. Then plug this value into the 4a in the second column, and check that it is indeed independent of the first (and/or third)