The test for linear dependence of 4 vectors is:
k1a+k2b+k3c+k4d=0 when at least one of the k values is non-zero.
Let's apply this to the given vector elements:
A: 9k1+3k2+3k3-3k4=0
B: -22k1-2k2-4k3+6k4=0
C: 7k1+k2+2k3-k4=0
D: 20k1+2k2+3k3-7k4=0
We need to find a non-trivial solution to prove linear dependence.
The equations are labelled for manipulation purposes.
E:B+D (eliminates k2): -2k1-k3-k4=0, that is, 2k1+k3+k4=0
F:B+2C (eliminates k2 and k3): -8k1+4k4=0, so 4k4=8k1, k4=2k1 and E becomes 2k1+k3+2k1=0, so k3=-4k1.
Let's rewrite A, B, C, D substituting for k3 and k4:
A: 9k1+3k2-12k1-6k1=0, -9k1+3k2=0, 3k2=9k1, k2=3k1. We can now also substitute for k2 in the remaining equations:
B: -22k1-6k1+16k1+12k1=0 is satisfied without having to make k1=0.
C: 7k1+3k1-8k1-2k1=0 is similarly satisfied.
D: 20k1+6k1-12k1-14k1=0 is similarly satisfied.
So, since k1 doesn't need to be zero, the other constants can be derived from k1 so that no constant is zero. For example, if k=1, then k2=3, k3=-4, k4=2 and:
a+3b-4c+2d=0 or more generally: ka+3kb-4kc+2kd=0 where k is any scalar, which proves the vectors are linearly dependent.