Math Help - what does it mean about the solutions to a system if the det is 0

1. what does it mean about the solutions to a system if the det is 0

question:find all real numbers λ such that the homogenous system
In-A)x=0 has a nontrivial solution
$A=\left[\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 1 & -1 \\ 1 & 0 & 0 \end{array} \right]$

work: $(\Lambda I_n - A)=\left[\begin{array}{ccc} \Lambda & 0 & 0 \\ 0 & \Lambda-1 & -1 \\ 1 & 0 & \Lambda \end{array} \right]$. Then the answer key says that (λIn-A)x=0 has a nontrivial solution if and only if its determinant=0. So then to solve the problem the determinant is taken and the values for λ are used such that the determinant is 0. I'm sceptical about setting the determinant to 0 because if a system has a determinant that is 0 then it means the system only has the trivial solution or infinetly many solutions. So couldn't this mean that the system only has trivial solutions? or am I lossing sight of the question being asked

2. Yes if the determinant is zero it is not invertible.

If it were invertible then we would simply end up with the trivial solution.

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From Wikipedia

When a transformation is represented by a square matrix A, the eigenvalue equation can be expressed as

where I is the identity matrix. This can be rearranged to

If there exists an inverse

then both sides can be left multiplied by the inverse to obtain the trivial solution: x = 0. Thus we require there to be no inverse by assuming from linear algebra that the determinant equals zero:

3. Originally Posted by superdude
question:find all real numbers λ such that the homogenous system
In-A)x=0 has a nontrivial solution
$A=\left[\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 1 & -1 \\ 1 & 0 & 0 \end{array} \right]$

work: $(\Lambda I_n - A)=\left[\begin{array}{ccc} \Lambda & 0 & 0 \\ 0 & \Lambda-1 & -1 \\ 1 & 0 & \Lambda \end{array} \right]$. Then the answer key says that (λIn-A)x=0 has a nontrivial solution if and only if its determinant=0. So then to solve the problem the determinant is taken and the values for λ are used such that the determinant is 0. I'm sceptical about setting the determinant to 0 because if a system has a determinant that is 0 then it means the system only has the trivial solution or infinetly many solutions (emphasis added).
No. This is wrong. A system of equations has a single solution if and only if the determinant of its coefficient matrix is non zero. If the determinant is 0 then it either has no solution or an infinite number of solutions. If the system is homogeneous then it always has the trivial solution so the "no solution" case is impossible. If a homogeneous system has determinant 0, there must be an infinite number of other, non-trivial, solutions.

So couldn't this mean that the system only has trivial solutions? or am I lossing sight of the question being asked

4. so in this situation if λ is 0 or 1 then the system will have infinetly many solutions. Shouldn't the question ask find the values of λ that make the system have infinetly many solutions because it doesn't make sense to ask for a value of λ that makes the system of only one solution?

5. Originally Posted by superdude
so in this situation if λ is 0 or 1 then the system will have infinetly many solutions. Shouldn't the question ask find the values of λ that make the system have infinetly many solutions because it doesn't make sense to ask for a value of λ that makes the system of only one solution?
Non-trivial solution does not mean unique solution.

6. Originally Posted by mr fantastic
Non-trivial solution does not mean unique solution.
I still don't see how this answers my question.
According to HallsoIvy "If a homogeneous system has determinant 0, there must be an infinite number of other, non-trivial, solutions." When the problems says "find all real numbers λ such that the homogenous system (λIn-A)x=0 has a nontrivial solution" by setting the determinant to 0 that would meen there are infinetly many trivial solutions. Am I correct in that understanding?

7. Originally Posted by superdude
I still don't see how this answers my question.
According to HallsoIvy "If a homogeneous system has determinant 0, there must be an infinite number of other, non-trivial, solutions." When the problems says "find all real numbers λ such that the homogenous system (λIn-A)x=0 has a nontrivial solution" by setting the determinant to 0 that would me there are infinetly many trivial solutions. Am I correct in that understanding?
No. You are not correctly reading the replies you have got. Setting the determinant equal to zero gives you the non-trivial solution. This has been said clearly already. The non-trivial solution contains a parameter, which is why you get an inifinite number of solutions from it.

Your question has been answered. I suggest you go and review your class notes or textbook to clarify your understanding of what has been said.

8. Originally Posted by superdude
I still don't see how this answers my question.
According to HallsoIvy "If a homogeneous system has determinant 0, there must be an infinite number of other, non-trivial, solutions." When the problems says "find all real numbers λ such that the homogenous system (λIn-A)x=0 has a nontrivial solution" by setting the determinant to 0 that would me there are infinetly many trivial solutions. Am I correct in that understanding?
You are incorrect in your understanding of the phrase "trivial solution".
You seem to be thinking that when you have an infinite number of solutions, they are all then "trivial solutions"- that is wrong. A "trivial solution" is simply the 0 solution. You can't have "infinitely many trivial solutions"- there is only one trivial solution!

If in any system of equations, the determinant of the coefficient matrix is non-zero, there is a unique solution. Any homogoneous system of equations has a "trivial solution"- the 0 solution- so if it is also true that the determinant is non-zero, there is only the trivial solution. If a homogeneous system of equations has zero determinant, then there exist infinitely many other, non-trivial, solutions.

9. So there is only one non-trivial solution, which contains a parameter and thus gives infinetly many solutions to the sysetm. Is that correct?
And this is a why it makes sense for the question to ask for a non-trivial solution, and given that non-trivial solution the system has infinetly many solutions?

or should I consider the prervious posts further?

10. Originally Posted by superdude
So there is only one non-trivial solution, which contains a parameter and thus gives infinetly many solutions to the sysetm. Is that correct?
And this is a why it makes sense for the question to ask for a non-trivial solution, and given that non-trivial solution the system has infinetly many solutions?

or should I consider the prervious posts further?
You have the right idea.

11. Finally. Now that I've been thinking about this question for the past few days I'm behind on my studies. I do not wish to learn any new material because if it builds ontop of my missunderstandings then my missconceptions grow. Would you advise I not get hung up over details such as this?

12. Originally Posted by superdude
Finally. Now that I've been thinking about this question for the past few days I'm behind on my studies. I do not wish to learn any new material because if it builds ontop of my missunderstandings then my missconceptions grow. Would you advise I not get hung up over details such as this?
Having recently completed Linear Algebra II (just a semester ago really) I can honestly say that every little grain of theory matters in this course as it is very theoretical. I passed it by doing a lot more theory than actual problems and it is a course most people fail.

I studied only theory for my first semester test and got 70+ % whereas the majority of the class barely scraped through or failed.
Not that I'm bragging, I just wish to indicate how important it was to study the theoretical aspect of it.

Personally, I found that it is very important to understand even the simplest concepts thoroughly.

But that's just my opinion. Your course might be different to the one I had.

13. Really? My opinion of Linear Algebra is that while it does include a lot of theory, it is also has very concrete examples and so make a perfect "bridge" from courses such as Calculus to the really theoretical course like Analysis and Modern Algebra.

And it is certainly NOT my experience that "most people fail".

I will say this- learn the definitions! I mean that you should learn the precise words of the definitions, not just the general idea. In mathematics, definitions are "working definitions"- you use the precise words of the definitions in proofs.

14. Originally Posted by HallsofIvy
Really? My opinion of Linear Algebra is that while it does include a lot of theory, it is also has very concrete examples and so make a perfect "bridge" from courses such as Calculus to the really theoretical course like Analysis and Modern Algebra.

And it is certainly NOT my experience that "most people fail".

I will say this- learn the definitions! I mean that you should learn the precise words of the definitions, not just the general idea. In mathematics, definitions are "working definitions"- you use the precise words of the definitions in proofs.

Well I suppose it differs between curriculae and the lecturers. My lecturer focused heavily on theory, and even warned us that the course has a very high failure rate.