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I suggest adding "b not equal to 0" to the problem statement.
Is the ans 2?
It was not as I solved the problem. Since then the "b" that made the problem ambiguous somehow removed / set to a fixed value. So I guess the solution to the problem is still not 2.
If b = 0 , then system will definitely have a solution(0,0,0).
So, for no solution case , it shud be obvious that "b != 0".
Yes, you can deduce, that b has to be unequal to 0. But that is not the point. The point is, that the problem statement can be interpreted such that there are no solutions for ALL b in IR. This ambiguity is unnecessarily misleading and potentially frustrating.
Tip: If the coefficients matrix of a given system of equations has a zero determinant, then one or more vectors represented by it are linearly dependent (one is a scaled version of the other). Which in turn tells you one of the equations shown is a duplicate of the other. If b=0, then (0,0,0) will always be a solution, which is why I believe they placed b in the problem statement. Good luck, lads!
The problem statement lies, there is not more than one value of a for which the system has no solution (and even to get one, we have to assume b cannot be zero since otherwise there is always a trivial solution). It seems like bad question design to have written a system that admits the trivial solution and not mention that it is disallowed, since one could have just provided a system that actually doesn't allow the trivial solution by putting a non-zero on the RHS.
Assuming that b != 0 then I find two values of a for which are -1 and 2. So the only problem with the problem formulation is that it is missing the requirement that b!=0
Like I said, "we have to assume b cannot be zero" for there to be any "bad" value of a. It could just be an issue of ambiguous wording (as written, it sounds like it is stating a conclusion about the system for arbitrary b, whereas maybe it is meant to be taken as an extra property that allows us to deduce that b is nonzero).
But even with b != 0, the only possible bad value of a is -1 (and this is bad for all nonzero b). I'm not sure how you arrived at a = 2 but in that case there's actually infinitely many solutions for any given value of b (IOW a = 2 is arguably the most wrong answer possible here); for example (x, y, z) = (1, -4, 1) when b = -6. More specifically, if a = 2 and given any value of b, we are free to pick any value of x, and then take z = -x-b/3, and y = 2b/3.
So the condition stated by the problem (that there's more than one value of a which leads to a system with no solutions) is actually an impossible condition. It's a bad statement because it probably leads people to believe wrong things about the relationship between determininant and whether system has no solutions.
Sorry my bad, you are obviously right. I just looked at when the determinant is 0 and thus makes the corresponding matrix not invertible. I should have been more careful and considered the case of infinite solutions
Hint: Write system in matrix form. What property of a matrix tells you whether such a system has a solution or not?
no. of independent variables < rank of matrix?
That should work. I was thinking about determinants. But the invertible matrix theorem gives a bunch of equivalent conditions and I'm pretty sure that's one of them.
the determinant property works just fine. use it to get multiple values for "a" then your answer will be the least
If b is equal to zero the system always has a solution x=y=z=0. If you assume that b is not equal to 0 the sentence "There are more than one integer values of a for which (...)" still remains false for the given system.
This question says , system has no solution for more than one integer values of a.
I got two values of a = -1,2
for a = -1 (no solution)
for a = 2 (many solutions)
so, only one integer(-1) for which we r having no solution ??
Yeap!!! -1 is the solution. Just type it and dont ask why
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