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.

Set square i think so