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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

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Try turning that into eschelon form.
I am thinking a bit differently...

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Other answers:

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...
I'm not sure myself either...
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 \(\det A\neq0\)
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 \(A\)....
a, b, and c are any real number such that each condition is met
right, any matrix \(\vec b\)
now, do I look for conditions in which the three cases can be met?
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
I'll start with that then. these kinds of questions happen to be the even numbered problems. thanks
sure, wish I could help more

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