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.

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

- "runner b is moving at a slower rate."

- "the graphs shown are graphs of distance(km) against time(min).

the slopes of this graphs are are given by:

slope = (change in y)/(change in x)

= (change in distance)/(change in time)

this is the formula for finding the speed. so, the slope of these graphs represent the speed of the runner.

taking each division in the graph to represent 1 on both axis:

runner a ⇒ speed = (2-0)/(1-0)

= 2/1

= 2 km/min

runner b ⇒ speed = (3-0)/(3-0)

= 3/3

= 1 km/min

this shows that   runner b is moving at a slower rate."

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step-by-step explanation:

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