Unique Solution for Linear Algebra System with Nonsingular Matrix

[emergentecon]

Homework Statement

Consider the system:
(1) 6x + ky = 0
(2) 4x + 6y = 0

The system will have a unique solution when k is:
(a) equal to 9
(b) any real number

Which statements are true.

Homework Equations

(1) 6x + ky = 0
(2) 4x + 6y = 0

The Attempt at a Solution

If m=n (number of equations is equal to number of unknowns) and the matrix is nonsingular (the determinant does not equal zero), then the system has a unique solution in the n variables.

Here, the determinant (ad-bc) equals zero when k = 9.
9 is also a real number.

Therefore, neither (a) or (b) can be true.

 

[BruceW]

and the matrix is nonsingular (the determinant does not equal zero), then the system has a unique solution in the n variables.

I think you've got this mixed up. If the matrix is nonsingular, then it has an inverse, right? so then would it make sense that there is a unique solution if the matrix has an inverse?

p.s. welcome to physicsforums :)

edit: hmm. actually, see what happens if you assume the matrix has an inverse. Is there a unique solution in that case? And then consider the other case, where k is any number. So, think of the two cases separately.

second edit: sorry If I'm being confusing. I'm going back on what I originally said. you did not have it mixed up.

 

[emergentecon]

Hold on, because I lost something here.
To have a unique solution, two conditions must be satisfied:
(1) m=n
which is true in my example

(2) system must be nonsingular (determinant must not equal zero / have an inverse).
this is only true when k does not equal 9 (when k=9, then ad-bc = 0)
9 is also a real number.

So technically the answer would be any real number, excluding 9.
Which means options A and B are both wrong.
IMO.
No?

Hehe, thank for the welcome, and the swift response!

 

[BruceW]

hehe, sorry about my first post, I rushed to give an answer without properly thinking, which is usually a bad idea, as I'm sure you know.

OK, yes there needs to be m=n and determinant not equal zero for a unique solution. And in this case, for k=9, the determinant is zero. So can't have k=9, if you want unique solution. so a) is definitely incorrect. and b) ... well, you're right, b) looks to be incorrect too, since 9 is a real number.

edit: although, the phrase for b) is: "The system will have a unique solution when k is any real number" It feels like they are trying to say "the system will have a unique solution for some k, which is a real number". The second phrase would be correct. But since they have actually written the first phrase, I think you must take it literally, and so the phrase is incorrect.

edit again: I should probably say 'statement' instead of 'phrase', since I think 'statement' is the proper way to talk about a logical yes or no.

 

[HallsofIvy]

I wonder if you read or copied the problem correctly (or if it had been originally written correctly). If it had said "The system will not have a unique solution when" then the two options would make sense.

 

[emergentecon]

Yeah, question is written exactly as it is in my notes ;)