Coefficients that make Gaussian elimination impossible?

[Mohamed Abdul]

Homework Statement

Given this matrix:

I am asked to find values of the coefficient of the second value of the third row that would make it impossible to proceed and make elimination break down.

Homework Equations

Gaussian elimination methods I used given here:
http://mathworld.wolfram.com/GaussianElimination.html

The Attempt at a Solution

I managed to write the original matrix in its reduced echelon form; however I'm having trouble finding the value that would make elimination impossible, even when interchanging rows. I've tried zero, but that didn't help me at all.

 

[Mark44]

Mentor note: Thread moved from Intro Physics to Precalc Math

Homework Statement

Given this matrix:
View attachment 214753
I am asked to find values of the coefficient of the second value of the third row that would make it impossible to proceed and make elimination break down.

Homework Equations

Gaussian elimination methods I used given here:
http://mathworld.wolfram.com/GaussianElimination.html

The Attempt at a Solution

I managed to write the original matrix in its reduced echelon form; however I'm having trouble finding the value that would make elimination impossible, even when interchanging rows. I've tried zero, but that didn't help me at all.

There are three possible things that could happen with a system of equations:

1. The system has a unique solution.
2. The system has an infinite number of solutions.
3. The system has no solutions at all.

What they are asking, is how could the 2nd entry of the third row be changed so that the system has no solutions? In the image you posted, it looks like you changed the 3rd entry of the second row, not the 2nd entry of the third row.

 

[Mohamed Abdul]

Mentor note: Thread moved from Intro Physics to Precalc Math

There are three possible things that could happen with a system of equations:

1. The system has a unique solution.
2. The system has an infinite number of solutions.
3. The system has no solutions at all.

What they are asking, is how could the 2nd entry of the third row be changed so that the system has no solutions? In the image you posted, it looks like you changed the 3rd entry of the second row, not the 2nd entry of the third row.

I didn't mean to change any values, I just copied it down wrong and had to erase my original number.
As for finding if there are no solutions, is there any method or trick to go about doing that? Or do I have to brute force through the equations picking numbers until I find the one that works?

 

[mfb]

I am asked to find values of the coefficient of the second value of the third row that would make it impossible to proceed and make elimination break down.

The "-1"?

You can replace it by a variable and proceed until you divide by something involving this variable, then you can choose it to make the denominator zero, breaking the process.
Alternatively, have a look at row 1 and 3. The rightmost entry is different. What happens if all three entries on the left are the same?

 

[Mohamed Abdul]

The "-1"?

You can replace it by a variable and proceed until you divide by something involving this variable, then you can choose it to make the denominator zero, breaking the process.
Alternatively, have a look at row 1 and 3. The rightmost entry is different. What happens if all three entries on the left are the same?

Yes, I am trying to replace the -1. Also, regarding what you said, I noticed that if I change the -1 to a 1, that makes two same equations that equal to separate values, -2 and -1. Does this count as inconsistency that would break down the system? It definitely doesn't make sense to have two of the same exact equations to be equal to different numbers, after all.

 

[mfb]

Right.
To show that formally you can also subtract the two equations and then you are left with 0=1.