Whats the answer for these Spanning sets and linear independence questions :

Question

Q1:determine whether each vector can be written as a linear combination of the vectors in S.

1.S={(2,-1,3),(5,0,4)}

a) u=(1,1,-1). b) v=(8,-1/4,27/4).

c) w=(1,-8,12). d) z=(-1,-2,2)

Q2:determine whether the set S spans R².if the set does not span R²;give geometric description of the subspace that it does span.

2.S={(1,-1),(2,1)}

3.S={(-3,5)}

4.S={(1,3),(-2,-6),(4,12)}

Q3:determine whether the set S spans R³.if the set does not span R³.give a geometric description of the subspace that it does span.

5.S={(1,-2,0),(0,0,1),(-1,2,0)}

Q4:determine whether the set S is linearly independent or linearly dependent.

6.S={(-2,4),(1,-2)}

7.S={(1,0),(1,1),(2,-1)}

Q5:show that the set is linearly dependent by finding a nontrivial linear combination (of vectors in the set) whose sum is the zero vector.Then express one of the vectors in the set as a linear combination of the other vectors in the set.

8.S={(3,4),(-1,1),(2,0)}

Answer 1

(1,1,-1) = k(2,-1,3) + C(5,0,4)

(1,1,-1) = (2k,-k,3k)+(5c,0,4c)

= (2k+5c, -k, 3k+4c)

Then

2k+5c = 1

-k = 1

3k + 4c = -1

So k=-1

the first equation forces c=3/5

the third equation forces c=1/2

so no, it cannot

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(1, -8, 12) = k(2,-1,3)+ C(5,0,4)

(1, -8, 12) = (2k, -k , 3k) + (5c, 0, 4c)

(1,-8,12) = (2k + 5c, -k, 3k+4c) <--- you can use this right

half for parts b and d

2k + 5c = 1

-k = -8

3k + 4c = 12

so k=8

first equation forces c=-3

third equation agrees. So yes