# The consistency of linear systems

1. Feb 10, 2013

### MoreDrinks

1. The problem statement, all variables and given/known data
What conditions do x, y and z need to fulfill for the system of equations to be consistent?

2. Relevant equations
2a + 3b + 5c = x
a + 3c = y
a - b + c = z

3. The attempt at a solution
Not quite sure. The notion of consistency was never discussed in class, nor is it addressed in the linear algebra text. I understand that it simply means having some sort of solution, and that inconsistency means not having any sort of solution - but examples almost always give integers on the right sides of the equations so that you can reduce the rows and see if there's a contradiction (such as 0+0+0=1) or not.

Is it just "If a, b and c are zero, then x, y and z must not be other than zero" or something of that nature?

2. Feb 10, 2013

### Karnage1993

Have you learned about matrices yet? If so, row reduce the matrix until you reach row reduced echelon form. Then use your example of "0 + 0 + 0 = 1", which is an inconsistent solution, to find out which values for x,y, and z the system is consistent.

3. Feb 10, 2013

### Dick

a, b and c are the variables you want to solve for. So the rows are [2,3,5], [1,0,3] and [1,-1,1]. Yes, reduce them to see if there might be any requirements on x, y and z to get consistency.

4. Feb 10, 2013

### MoreDrinks

I think I get it. Thanks.

Last edited: Feb 10, 2013
5. Feb 10, 2013

### MoreDrinks

But it would seem that to do this, one would need to use an augmented matrix, right? And in that case, isn't it going to be difficult or impossible to isolate x, y and z in order to see what requirements there might be?

6. Feb 10, 2013

### Dick

You could do that. But if you reduce the matrix of just the coefficients and you get that it has full rank, i.e. no zero rows, then there will be no requirements for consistency on x, y and z. Yes? It wouldn't matter much what happens on the augmented side. I think you will get full rank. If I did it right.

7. Feb 10, 2013

### MoreDrinks

Thank you. There's one other where I do not seem to be getting full rank:

1 0 2
2 1 5
1-1 1

First operations: -2R1+R2, -R1+R3

1 0 2
0 0 1
0-1-1

Actually, would the following work? -R3+R2, R2+R1

1 0 2
0 1 2
0 0 1

then using the final row to eliminate the 2's?

Last edited: Feb 10, 2013
8. Feb 10, 2013

### Dick

You do seem to be getting full rank. But that's because you are doing it wrong. It doesn't have full rank. Do the operations one at a time and specify the row you are substituting for. If you wind up with a row of zeros, then you should do the augmented matrix. The expressions in x, y and z in the rows with all zeros are your consistency constraints.

9. Feb 10, 2013

### MoreDrinks

Ahh. By not screwing up the very first operation, it seems to go like so

1 0 2 x
2 1 5 y
1-1 1 z

1 0 2 x
0 1 1 y-2x
1-1 1 z

1 0 2 x
0 1 1 y-2x
0-1-1 z-x

then finally, add R2 to R3

1 0 2 x
0 1 1 y-2x
0 0 0 z-3x+y

So the constraint would be z-3x+y = 0, or else the system is inconsistent.

10. Feb 10, 2013

### Dick

Yes, but you meant add -R1 to R3 in the second step, yes?

11. Feb 10, 2013

### MoreDrinks

Yes, sorry. I'm exhausted. Thank you!

12. Feb 10, 2013

### Dick

Me too. But you did it right. That's what counts. Very welcome!