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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:

ANo solutions
BOne solution
CA lot of solutions ...?

Verified Answer :

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.

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