Question

Let u,v,wu,v,w be three linearly independent vectors in R7R7. Determine a value of kk, k=k= , so that the set S={u−3v,v−2w,w−ku}S={u−3v,v−2w,w−ku} is linearly dependent.

Answer 1

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