Finding Non-Trivial Solution(s) For 3x3 Matrix

[Ascendant0]

Homework Statement

-

Relevant Equations

-

I didn't have any good way to put this in the homework statement, but this is what the question is asking:

For what c-value(s) will there be a non-trivial solution:

$x_1 - x_2 + x_3 = 0$
$2x_1 + x_2 + x_3 = 0$
$-x_1 + (c)x_2 + 2x_3 = 0$

I have spent a good couple hours looking at various ways to find non-trivial solutions, but I couldn't find one with a variable like "c" in it. I figured out how to solve without a variable like "c" in them, but not sure what to do with this one.

I tried following the methods in other non-trivial solution videos. I tried to get each line to have the "1s" diagonally down to the right, and put c in the center, but this is the best I've gotten to as far as that:

$[ M= \left[ {\begin{array}{cc} 1 & -1 & 1 \\ -1 & c & 2 \\ 0 & 3 & 1 \\ \end{array} } \right] ]$

From here, I wasn't seeing any way to clear anything else out without either using fractions, or by including "c" with one of the other equations, which I feel is going to get real sloppy. I know it has to be something simpler than that considering the answer to the problem is $c = -8$. I just can't figure out how to get there. Help would be greatly appreciated.

 

[docnet]

The system has a non-trivial solution when the coefficient matrix has a non-zero determinant. Do you know how to find the determinant of a $3\times 3$ matrix?

 

[Ascendant0]

The system has a non-trivial solution when the augmented matrix has a non-zero determinant. Do you know how to find the determinant of a $3\times 3$ matrix?

Wow, it was SO much easier than I was making it out to be. Thanks for pointing me in the right direction.

I see how to do a problem like this now, which is great, but it's still not making complete sense to me. When I calculate the determinant, I end up with the $c + 8$, and then I'm assuming from the answer in the solutions, you just treat it like $c + 8 = 0$ and solve for c. So, isn't solving for c in that way making it so that you *are* setting it to 0? That's throwing me off a bit here.

 

[docnet]

Sorry, I just found an error with my comment. the right term was coefficient matrix, not augmented matrix, it was my mistake.

I'm getting a different result for the determinant.

yes, you can find values of c that make the determinant equal to 0. The answer will be the complement of those values.

 

[PeroK]

The system has a non-trivial solution when the coefficient matrix has a non-zero determinant. Do you know how to find the determinant of a $3\times 3$ matrix?

It's the other way round. If M has non-zero determinant then it is invertible and multiplying both sides by the inverse of M shows that we have only the trivial solution.

If, however, M has zero determinant, then there exist non-trivial solutions.

 

[Ascendant0]

Sorry, I just found an error with my comment. the right term was coefficient matrix, not augmented matrix, it was my mistake.

I'm getting a different result for the determinant.

yes, you can find values of c that make the determinant equal to 0. The answer will be the complement of those values.

Try it with the original problem statement. I made a mistake in the matrix I made. I tried moving things around and multiplying, subtracting and adding, etc. It's the right answer to the problem before I messed with it, lol. I went through the original problem and got exactly what the solution had. When I did it with the matrix I was working on, I got c = -16. So, I definitely messed up somewhere.

But anyway, where I'm getting confused is in your first statement (and from what I've seen online about this type of problem) - it's non-trivial when the determinant is non-zero, but then in this statement quoted, you're saying that we are looking for the value(s) of c that make the determinant 0. That's what's throwing me off - everything tells me the non-trivial solution is when the determinant is "non-zero", but then we're setting the determinant to 0 in order to find what value of c would make it zero. From how I'm viewing it, it seems like c = -8 is the answer that *makes* it a trivial solution. I know I'm looking at this wrong, but the explanation seems contradictory to me?

 

[PeroK]

Try it with the original problem statement. I made a mistake in the matrix I made. I tried moving things around and multiplying, subtracting and adding, etc. It's the right answer to the problem before I messed with it, lol. I went through the original problem and got exactly what the solution had. When I did it with the matrix I was working on, I got c = -16. So, I definitely messed up somewhere.

But anyway, where I'm getting confused is in your first statement (and from what I've seen online about this type of problem) - it's non-trivial when the determinant is non-zero, but then in this statement quoted, you're saying that we are looking for the value(s) of c that make the determinant 0. That's what's throwing me off - everything tells me the non-trivial solution is when the determinant is "non-zero", but then we're setting the determinant to 0 in order to find what value of c would make it zero. From how I'm viewing it, it seems like c = -8 is the answer that *makes* it a trivial solution. I know I'm looking at this wrong, but the explanation seems contradictory to me?

See my post above, which explains why the determinant must be zero for non-trivial solutions.

 

[Ascendant0]

It's the other way round. If M has non-zero determinant then it is invertible and multiplying both sides by the inverse of M shows that we have only the trivial solution.

If, however, M has zero determinant, then there exist non-trivial solutions.

Thanks, didn't see this prior to my post. So, does this mean that if there is a determinant, there is only one solution, but if the determinant is 0, then there are multiple solutions?

 

[Orodruin]

Thanks, didn't see this prior to my post. So, does this mean that if there is a determinant, there is only one solution, but if the determinant is 0, then there are multiple solutions?

There is always a determinant. The question is if that determinant is zero or not.

If it is non-zero the matrix is invertible and there exists a single unique solution. If it is zero there are multiple or no solutions to Mx = a (and multiple solutions to Mx=0)

 

[docnet]

yikes! it seems I already forgot my linear algebra. it's probably because I use it so rarely, if at all. hopefully my mistake didn't cause too much confusion, and sorry