# Homework Help: Linear algebra question

1. Aug 3, 2009

### tgt

1. The problem statement, all variables and given/known data
If a linear system of n equations in n unkowns has n leading 1's in the reduced row-echelon form of its augmented matrix, then the system has exactly one solution.

3. The attempt at a solution
True to me but is it?

2. Aug 3, 2009

### kof9595995

Yes it is, but is this all your attempt to the solution?

3. Aug 3, 2009

### tgt

The book said False. Must be a typo?

If there are n equstions and n leading ones, all variables must equal to a specific number. And that is the only solution in the system.

4. Aug 3, 2009

### kof9595995

Em.....unless 1's in the augmented part can be called "leading 1",then the matrix can be like this:
1 0 2
0 0 1
But I have a very vague memory that the 1 in augmented part(1 in second row) can't be called "leading 1",but you should check it out.

5. Aug 5, 2009

### tgt

But there is n leading 1's with n variables. Your example have 2 leading 1's in 3 variables.

The question does say n leading 1's in the reduced echelon matrix. The fact that its reduced echelon suggest that all columns have 0 elsewhere.

6. Aug 5, 2009

### roam

Kof is right. A leading 1 is a pivot position.

EDIT:

Here:
1 0 2
0 0 1
The 1 in the 2nd row isn't a leading 1, but here it is:
1 0 2
0 1 0

7. Aug 5, 2009

### kof9595995

No, my example is an augmented matrix, the 3rd column is the right hand side of the equations.

8. Aug 5, 2009

### tgt

So you are saying that if a linear system of n equations in n unkowns has n leading 1's in the reduced row-echelon form of its augmented matrix then it is inconsistent. hence the book was right.

9. Aug 5, 2009

### kof9595995

Personally I don't think 1 in augmented part can not be called leading one, and with roam's confirmation, probably it's just a typo in the book.

10. Aug 6, 2009

### tgt

The book I'm using does not draw a vertical line before the last column even for an augmented matrix. So in that case the book was right, no typo.