Solving Linear Dependence of a Matrix

[mr_coffee]

Hello everyone, I'm finishing up some matrices review and im' confused on this question i have the matrix:
-1 -3 -1 2
5 13 3 -8
3 10 9 -8
1 4 7 -4

I row reduced got this:
1 0 0 3/5
0 1 0 -4/5
0 0 1 -1/5
0 0 0 0

So you can see that this isn't a basis due to column 5 not being 0 0 0 1, but what does this mean the questions says:
If they are linearly dependent, determine a non-trivial linear relation - (a non-trivial relation is three numbers which are not all three zero.) Otherwise, if the vectors are linearly independent, enter 0's for the coefficients, since that relationship always holds.
?A + ?B + ?C + ?D = 0.
I tried 1 1 1 3/5
1 1 1 0, i tried actually 14 times, all of them are wrong hah, any help?>

 

[matt grime]

Did you record the row operations that you used? Because they tell you the relationship, the reduced row of zeroes is a linear combination of the 4 rows, just recall what the combination is.

 

[mr_coffee]

I used a Ti-83 calculator to find the row reduction, he said to use them for these problems!

 

[shmoe]

You know how to solve a homogeneous system? If A is your matrix of column vectors, a non-trivial solution to the homogeneous system AX=0 will give you a non-trivial linear relation between your vectors.

AX is just a linear combination of the columns of A after all.

edit-are you concerned with a linear combination of the rows or the columns of your matrix? matt and i answered the different interpretations (in that order).