Answer:
1/6
Step-by-step explanation:
For a set of vectors a, b and c to be linearly dependent, the linear combination of the set of vectors must be zero.
C1a + C2b + C3c = 0
From the question, we are given the set S={u−3v,v−2w,w−ku}, the corresponding vectors are a(0, -3, 1), b(-2,1,0) and c(1,0,-k)
The values in parenthesis are the ccoefficients of w, v and u respectively.
On writing this vector as a linear combination, we will have;
C1(0, -3, 1) + C2(-2,1,0) + C3(1,0,-k) = (0,0,0)
0-2C2+C3 = 0........ 1
-3C1+C2 = 0 ........... 2
C1-kC3 = 0 ….......... 3
From equation 2, 3C1 = C2
Substituting into 1, -2(3C1)+C3 = 0
-6C1+C3 = 0
-6C1 = -C3
6C1 = C3.…..4
Substitute 4 into 3 to have
C1-k(6C1) = 0
C1 = k6C1
6k = 1
k = 1/6
Hence the value of k for the set of vectors to be linearly dependent is 1/6
= 1.28