anonymous
  • anonymous
Linear Algebra: Find (if possible) values a, b, and c such that the system of linear equations has: a) no solution b) exactly one solution c) infinitely many solutions 2x - y + z = a x +y +2z = b 3y + 3z = c
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
|dw:1348174076973:dw|
inkyvoyd
  • inkyvoyd
Try turning that into eschelon form.
TuringTest
  • TuringTest
I am thinking a bit differently...

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inkyvoyd
  • inkyvoyd
Yeah, I skimmed through a textbok about 30 mins ago, I'm haven't exactly even taken this ;). I know a solution though, it's probably not the most efficient way...
TuringTest
  • TuringTest
I'm not sure myself either...
TuringTest
  • TuringTest
For any \(n\times n\) matrix \(A\) the system\[A\vec x=\vec b\]has exactly one solution for each \(n\times1\) matrix \(\vec b\) if and only if \(\det A\neq0\)
TuringTest
  • TuringTest
so what does a,b, and c have to do with the answer to your question I'm not even sure, the answer seems to be dependent on the matrix \(A\)....
anonymous
  • anonymous
a, b, and c are any real number such that each condition is met
TuringTest
  • TuringTest
right, any matrix \(\vec b\)
anonymous
  • anonymous
|dw:1348174993843:dw|
anonymous
  • anonymous
now, do I look for conditions in which the three cases can be met?
TuringTest
  • TuringTest
if this is always true then there are infinite solutions, so that would suggest that a=-3, c=3, and b is whatever (easy to figure out) how to test the other two conditions I'm not so sure
anonymous
  • anonymous
I'll start with that then. these kinds of questions happen to be the even numbered problems. thanks
TuringTest
  • TuringTest
sure, wish I could help more

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