# Linear Algebra basic problem

1. Jan 15, 2009

### Sheneron

1. The problem statement, all variables and given/known data
Find a number for k in which the system yields no unique solution.

$$x+y+kz = 3$$
$$x+ky+z = 2$$
$$kx+y+z= 1$$

3. The attempt at a solution

I know that I need to find a number so that there is no solution and so that there is an infinite number of solutions, I just am not sure how to go about it. I have tried setting it up in row echelon form and here is what I got.

$$x+y+kz = 3$$
$$(k-1)y - (k-1)z = -1$$
$$kx+y+z=1$$

So I only changed the middle one and I know that if k = 1 then the system will yield no solutions, so there is one number. But I looked in the back and there was another answer. I don't know how to get the other one, and is that even right so far?

2. Jan 15, 2009

### Dick

Do you know that if the determinant of the coefficients matrix vanishes, then you can't get a unique solution? That's the general way to do it. If you want to see an example of how the other root creates an inconsistancy, just add all of the equations.

3. Jan 15, 2009

### Sheneron

I did not know that. All we have been given is the row echelon form. We have had nothing about determinants yet...

4. Jan 15, 2009

### Dick

Then you should actually push through solving the whole system in terms of k and see what values are singular. So far you've only solved it partially and you have only one value.

5. Jan 16, 2009

### Sheneron

Alright, I kept solving it and it worked. Thanks for the help.