Here's the question you clicked on:
s3a
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!
if c=3, you cannot divide a number by (c-3)
it is zero.. so they have seperated two cases, one when c=3 and the other when c !=3
Oh. I see; it's all about the pivot being non-zero.
no no no it matters because you can't divide a number by 0
if c=3 a thing such as -4/(c-3) doesn't exist!!
so you can't make the arguments as those cases for c !=3
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.
what i am saying is the only reason for seperating the cases
pivot to be 0 doesn't make any sense because the first appearing 1s are called pivots
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
i mean look at the basis for C(A) they are different!! it matters!!
two different case, two different logics, two different consequences
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.
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.
what you just said is right well what is it then that you don't understand?
I now do understand this problem. :)