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# Which of the following sets of vectors span R^3 (R=real numbers)? watch

1. Ive done the questions and i know some of them are wrong (we have to answer them online and get immediate feedback) but I dont know why or where ive gone wrong as everything looks fine, can anyone help?
A: (-4,4,4), (2,4,4), (-1,3,2) i said they span.
B: (-3,-1,-1), (5,-1,-4), i said they span.
C: (2,0,0), (0,3,0), (0,0,1) i said they span.
D: (4,-4,-3),(1,4,2),(4,7,7) i said they span.
E: (-3,2,3), (4,-1,4),(-4,-4,5) i said they span.
F: (4,6,-2),(2,3,-1),(-3,-1,-1), (-7,-8,8) i said they didn't span.
Thanks!
2. B is obviously wrong since you need atleast 3 vectors to span R^3.
3. oops! thank you.
can you help with any of the others please?
4. the others are correct - except F.
5. I have a feeling that you're trying to guess these answers. If not, post the method for one of them and I'll be glad to help.
6. i just put in
A span
B dont span
C span
D span
E span
F dont span

and its still only partially correct! i dont know what to do!
7. ive been putting each vector in the formula:

a*vector1 + b*vector2 + c*vector3 = (x,y,z)

then using gaussian elimination to see if a,b,c can be written in terms of x,y,z and if they can they the vectors span and if they cant then they dont span. ive done this for each one and i cant see where ive gone wrong!
8. That list is correct - you must have typed them wrong on here.
9. (Original post by **B**)
i just put in
A span
B dont span
C span
D span
E span
F dont span

and its still only partially correct! i dont know what to do!
F spans!
10. i havent ive check with my question sheet and the webpage question!
11. thanks thats correct!
can you explain why though?
i thought that because when i was using gaussian elimination and had 4 variables and 3 rows that there couldnt be a solution?
12. (Original post by **B**)
thanks thats correct!
can you explain why though?
i thought that because when i was using gaussian elimination and had 4 variables and 3 rows that there couldnt be a solution?
A spanning set for R^3 requires atleast 3 linearly independent vectors; extra vectors are ok.
13. (Original post by angelafleming)
A spanning set for R^3 requires 3 linearly independent vectors; you can ignore extra vectors.
oops
14. ok so aslong as you cant write any one of the vectors given as a linear combination of the others then they span?
15. (Original post by Hancock orbital)
oops
Sorry, but I haven't made a mistake.
16. (Original post by **B**)
ok so aslong as you cant write any one of the vectors given as a linear combination of the others then they span?
Yes, but only if you're talking of a 3-vector set.
17. thank you
18. (Original post by angelafleming)
Sorry, but I haven't made a mistake.
I was talking about my mistake.
19. (Original post by **B**)
thank you
You're welcome!

btw which stage of education are you in?
20. (Original post by Hancock orbital)
I was talking about my mistake.
Sorry, I didn't understand that. My apologies.

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