Finding k for a System with No Unique Solution

[Sheneron]

Homework Statement

Find a number for k in which the system yields no unique solution.

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

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?

 

[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.

 

[Sheneron]

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

 

[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.

 

[Sheneron]

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