a non trivial linear relation

Question

?Linearly IndependentLinearly Dependent 1. Determine whether or not the four vectors listed above are linearly independent or linearly dependent.

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 = 9, -22, 7, 20

b = 3, -2, 1, 2

c = 3, -4, 2, 3

d = -3, 6, -1, -7

Answer 1

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.