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.
