Question

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!

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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.

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

two different case, two different logics, two different consequences

Answer 13

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.

Answer 14

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.

Answer 15

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

Answer 16

I now do understand this problem. :)