YAMathT: Can someone double check my linear algebra answer?

[Ricemarine]

***Instead of x1, x2, x3, I'll use x, y, z since there aren't really subscripts here***

x + y = 3
x + (k^2-8)y = k

So, by reducing it, I got it to

x + y = 3
x + (k^2-9)y = k-3

in which equation 2 is able to be reduced to

(k+3)y = 1

So now I am trying to find out when the system has no solution, a unique solution, and infinitely many...

Right now, I've figured that for no solution, k would have to be -3.
As for the unique solution, I believe that the answer should be all reals except -3.
For infinitely many, I'm trying to think logically, but I can't seem to find an answer to when that's the case... since this is in R2...


Thanks for helping.

 

[Tweak155]

I'm more amazed by your upside down and backwards sig.

 

[RichieZ]

i don't remember how to do any of this anymore... sad

 

[Ricemarine]

Originally posted by: Tweak155
I'm more amazed by your upside down and backwards sig.

lol, I wonder who else does it.

 

[dighn]

if k is 3, then you have infinite solutions

 

[Ricemarine]

Originally posted by: dighn
if k is 3, then you have infinite solutions

Because then they'd be parallel...

x+y =3
x+y = 3...


ohhhh....

Thanks!