## s3a 2 years ago Why does it matter what c is equal in problem #3a to if it disappears thanks to elimination? (Sorry if that's a dumb question.): http://ocw.mit.edu/courses/mathematics/18-06sc-linear-algebra-fall-2011/ax-b-and-the-four-subspaces/exam-1/MIT18_06SCF11_ex1s.pdf Any input would be greatly appreciated!

1. ingenuus

if c=3, you cannot divide a number by (c-3)

2. ingenuus

it is zero.. so they have seperated two cases, one when c=3 and the other when c !=3

3. s3a

Oh. I see; it's all about the pivot being non-zero.

4. ingenuus

no no no it matters because you can't divide a number by 0

5. ingenuus

if c=3 a thing such as -4/(c-3) doesn't exist!!

6. ingenuus

so you can't make the arguments as those cases for c !=3

7. s3a

Ya, that too BUT if we do not divide by (c-3) = 0, there is nothing illegal and we get what would be a pivot to be a 0.

8. ingenuus

what i am saying is the only reason for seperating the cases

9. ingenuus

pivot to be 0 doesn't make any sense because the first appearing 1s are called pivots

10. ingenuus

you can do your elimination process because c!=3(the one you did previously) you can't argue something like 'Why does it matter what c is equal in problem #3a to if it disappears thanks to elimination' since it's not the case! for c=3 you need to make a whole new different arguments

11. ingenuus

i mean look at the basis for C(A) they are different!! it matters!!

12. ingenuus

two different case, two different logics, two different consequences

13. s3a

if c = 3, then it's a simple plugging in of c = 3, and then you have 3 - 3 = 0 and then you have a matrix with only regular numbers. I think we're miscommunicating again, lol.

14. s3a

I know that the bases are different. That's because the pivot from c!=3 is no longer a pivot; it's now a 0.

15. ingenuus

what you just said is right well what is it then that you don't understand?

16. s3a

I now do understand this problem. :)