For the following system.

kx + y + z = 1

x + ky + z = 1

x + y + kz = 1

Determine for what values of k the system has:

a) No solutions

b) One solution

c) A lot of solutions ... ?

This problem can be converted into a linear algebra problem. The condition is that if the derminant below is not zero, then the system has one solution.

| k 1 1 |

| 1 k 1 | = k^3 - 3k + 2 = 0

| 1 1 k |

Solving for the roots, k = - 2, and k = 1.

When k = 1, the three equations are the same so there are infinite solutions.

When k = - 2, there are no solutions.

When k / = - 2 and k / = 1, there is one solution.