Linear Equations?

Question

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


For what value(s) of k does tha system have a unique solution?

For what value(s) of k does tha system have no solution?

For what value(s) of k does tha system have infinitely many solutions?

Answer 1

for k=1 or k=-2 system have infinite numbers of solutions.
for other value system have 1 solution.
and there is something wrong about what zorro said cause if u put 1 in k^3-3k+2 so it turn to 0 and it means if k=1 have no solution and have many solutions (according to what zorro said).
k=1 and k=-2 are roots of k^3-3k+2
Sorry for my English.

Answer 2

If the determinant of the matrix is zero, then there will be no solution, or infinitely many.

Thus if k^3 - 3k + 2 = 0 there will be no solution or infinitely many

If k = 1 then there are infinitely many solutions.

If k=-2 There could be infinitely many or none. Not sure right off hand.

Any other value of k will result in a unique solution.