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# MTH501 assignment no 2 (due date 8 January 2017)

Anyone who has completed this assignment? I am unable to solve question 2 of this.

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### Replies to This Discussion

video is okay for understanding but it is hard to understand

i think if linear combination of one span of vectors produces any of the vector of other span, the spans would be equal.

Comment if u think it's right.

i dont no

yes it is hard to underrstand

The two sets do not span the same subspace of R^3. One approach to this type of problem---where both sets contain the same number of vectors---is to see if one pair is in the span of the other. That is, can you write

v1 = au1 + bu2

and

v2 = cu1 + du2

for some scalars a,b,c, and d? If you can, then the two sets span the same space. (You can flip it and try to find u1 and u2 in terms of v1 and v2.)

Can (1,0,0) be found as a(1,2,3) + b(1,0,1)? If so, then

a + b = 1
2a = 0
3a + b = 0

but there is no solution because 2a = 0 implies a = 0 and the first and last equation give b = 1 and b = 0. No go!

So v1 is not in the space span{u1,u2}.

If we know anything at all it is that v1 is in span{v1,v2}. That should be obvious. Since v1 is not in span{u1, u2}, we can conclude that span{u1,u2} and span{v1, v2} are different spaces.

Just an FYI: span{u1,u2} is just the set of all linear combinations of the vectors u1, u2. That is, it is the set of all vectors

au1 + bu2

where you let a and b range over all of the real numbers.

just found this at yahoo. hope it helps,,,

solution share kro yr yaha itna mtlbi na bna kro :P

plzzzzzzzzzz share the solution

KL[;

Attachments:

is this full solution?bro

no it is not the full solution plzz share the solution

if any one have a solution file of Assignment please share it

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