That's a lot of trouble. Why not try a determinant? 2r - 10 = 0?
The question is : Determine, if possible, a value of r for which the given set
{[1,0,-1,1],[0,-1,2,1],[-1,1,1,0],[-1,9,r,2]} is linearly dependent.
I put it into an augmented matrix with the "b" vector being the zero vector to determine linear dependence/independence.
1 0 -1 -1 0
0 -1 1 9 0
-1 2 1 r 0
1 1 0 2 0
and proceded to use elementary row operations as part of gaussian elimination
R3+R1>R3
R4-R1>R4
to get
1 0 -1 -1 0
0 -1 1 9 0
0 2 0 (r-1) 0
0 1 1 3 0
then, R3+2R2>R3
and R4+R2 > R4
for
1 0 -1 -1 0
0 -1 1 9 0
0 0 2 (r-17) 0
0 0 2 12 0
and finally R4- R3>R4
to get
1 0 -1 -1 0
0 -1 1 9 0
0 0 2 (r-17) 0
0 0 0 (29-r) 0
which would lead me to believe that r=29 would make this system have inf solutions because x4 would be a free variable therefore making the set DEPENDENT.
However, when I plug the un-augmented matrix into my calculator and use the RREF function I get the identity matrix... which would mean there is exactly one solution to the system, which of course would be 0,0,0,0 making the set INDEPENDENT.
any help would be muchly appreciated
ah just because we hadn't learned that yet and this was a question on a quiz.
But anyways I know where I messed up, the r-17 should be an r+17 and instead of 29 I get r to be -5 which is correct.
Thank you for your help, though.