Independence Test Homework Problem Solution

[EvLer]

Hello,
I have a homework problem I need some help with.
Given 3 matrices:
A = [1,2,1,2]^t
B = [2,3,-1,0]^t
C = [1,0,1,0]^t
I need to use test of independence to find out whether they are independent.
So, the matrix I ended up with is this
1 2 1 0
0 1 2 0
0 0 0 0
0 0 0 0
The answer says it's independent and I can see that from the original matrices. But I do not know how to proceed with the indep. test to prove that; for one thing, I have more equations than unknowns (I know they are all zeros, so I just disregarded the last row) and for another, if I make it a row-echelon form it looks like column 3 (not being a pivot matrix) is actually a lin. comb. of col.1 and 2 which means that the set is lin.dep.
Where did I go wrong? :yuck:
Thanks.

 

[HallsofIvy]

How did you end with that matrix?

Using the given vectors as rows of a matrix, you start with
1 2 1 2
2 3 -1 0
1 0 1 0

Now "row reduce":
subtract twice the first row from the second row and subtract the first row from the third row to get
1 2 1 2
0 -1 -3 -4
0 -2 0 -2

Now subtract twice the second row from the third row:
1 2 1 2
0 -1 -3 -4
0 0 -6 6

Look independent to me!

 

[EvLer]

well, since they are transposes (I put '^t' there), following the dependence equation I got initially this:
xA+yB+zD = 0 (vector, i.e. zero vector 4 by 1)
which is
1x + 2y + z = 0
2x + 3y + 0 = 0
1x - y + z = 0
2x + 0 + 0 = 0
and then I got the augmented matrix which reduced to what I posted above and I cannot 'make' it independent.
Thank you for reply.

 

[HallsofIvy]

Take a look at your last equation: 2x= 0. Doesn't that say that x= 0? Now put that back into the second equation: 3y= 0. y must be 0! finally, either of the remaining equations and you get z= 0. Looks independent to me!

 

[EvLer]

That obvious...
Thank you very much.