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# Linear Algebra Foundations #8 - Systems of Equations

# Linear Algebra Foundations #8 - Systems of Equations

JLiersch + 2 comments I suggest adding "b not equal to 0" to the problem statement.

kiner_shah + 1 comment Is the ans 2?

JLiersch + 0 comments 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.

himanshu_123j + 1 comment 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".

JLiersch + 0 comments 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.

agu3rra + 0 comments 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!

Quantris + 1 comment 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.tordjarv + 1 comment 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

Quantris + 1 comment 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.

tordjarv + 0 comments 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

tilper + 2 comments Hint: Write system in matrix form. What property of a matrix tells you whether such a system has a solution or not?

kiner_shah + 1 comment no. of independent variables < rank of matrix?

tilper + 0 comments 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.

PsychicPhoenix + 0 comments the determinant property works just fine. use it to get multiple values for "a" then your answer will be the least

amigo7 + 0 comments 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.

himanshu_123j + 1 comment 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 ??

wthered + 0 comments Yeap!!! -1 is the solution. Just type it and dont ask why

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