Finding a Unique Solution to a System of Equations

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The Head
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Homework Statement
Find when the system of equations is unique:

x-y-2z-2w= 3
y+z+w= 4a+3
z+3w= -4a-4
(-a+2)w= b+4a^2-4a-7
Relevant Equations
Full Rank = Unique
It makes sense that a=2 would cause problems because then we wouldn't have a matrix of full rank and we'd be unable to determine a value for w. But the key also says that when b+4a^2-4a-7≠0. Why is that an issue? For example, if a=1, that just says implies that w=0. Through back-subsitution, we get z=-8, y=15, x=2. And the solution: (2, 15,-8, 0) is unique still because it's the only possible solution, right?

What am I missing here?
 
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Hmm, yeah, maybe it was just the wording that is a little ambiguous. It says for a=2 AND b≠-1, the solution is inconsistent (makes sense, because then you get something like 0=1), for a≠2 AND b+4a^2-4a-7≠0, it is unique, and for a=2 AND b=-1 there are infinitely many solutions (also makes sense, since then you get 0=0). As long as I'm not missing something fundamental and the basic conditions I'm looking for are correct, I'm not too worried about it.

Thanks for helping out with this. I appreciate it.
 
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The Head said:
Hmm, yeah, maybe it was just the wording that is a little ambiguous. It says for a=2 AND b≠-1, the solution is inconsistent (makes sense, because then you get something like 0=1), for a≠2 AND b+4a^2-4a-7≠0, it is unique, and for a=2 AND b=-1 there are infinitely many solutions (also makes sense, since then you get 0=0). As long as I'm not missing something fundamental and the basic conditions I'm looking for are correct, I'm not too worried about it.

Thanks for helping out with this. I appreciate it.
But that list leaves out a≠2 AND b+4a^2-4a-7=0, and as you say that case does yield a unique solution. So you are right, the "AND b+4a^2-4a-7≠0" is entirely redundant.
 
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