# Thread: Elimination

1. ## Elimination

I've been struggling with this problem for a while.

find values of k so that the system in unknowns x y and z has a unique solution, no solution, infinite solution

kx + y + z = 1
x + ky + z = 1
x + y + kz = 1

I know that the forms:
0 = 0 is an infinite solution
0 = 1 is no solution
nonzero number = other nonzero number is a unique solution

2. Take the determinant of your coefficient matrix and set it equal to zero. You should get a cubic polynomial in k, and get up to three solutions.

For all values of k other than those (up to) three solutions, the determinant is non-zero and the matrix is non-singular. If this is the case, then there is always a unique solution, no matter what the right hand side is.

For k which make the determinant equal to zero, the matrix is singular. This means that there is always either no solution or infinitely many solutions. Since there are at most three such values of k, you can test each one by hand to see which is the case.

For example, k = 1 gives a zero determinant. For k = 1, all three equations are identical and you get infinitely many solutions.

3. I appreciate the answer but I have to solve this using elimination. Also, I only know what a determinant is for a 2x2 matrix.

4. Then get started eliminating. I think I would be inclined to divide the last equation by k (assuming k is not 0) to get
kx+ y+ z= 1
x+ ky+ z= 1
x/k+ y/k+ z= 1/k

Now, just subtract the second equation from the first and subtract the third equation from the first to eliminate x. What do you need to do to eliminate, say, y from the resulting two equations? You will get something like Ax= B where A and B both depend on k. In order to solve that last equation for x, you will need to divide A. There will be no solution if A is 0 but B is not. There will be an infinite number of solutions if A and B are both 0. Don't forget to check what happens if k= 0.