Question:

The system is homogeneous. For a non-trivial solution, the determinant of the coefficient matrix must be zero: |1 \lambda -1; \lambda -1 -1; 1 1 -\lambda| = 0. Expanding the determinant: 1(\lambda - (-1)) - \lambda(-\lambda^2 - (-1)) - 1(\lambda - (-1)) = 0. (\lambda + 1) - \lambda(1 - \lambda^2) - (\lambda + 1) = 0. \lambda + 1 - \lambda + \lambda^3 - \lambda - 1 = 0. \lambda^3 - \lambda = 0. \lambda(\lambda^2 - 1) = 0. \lambda(\lambda - 1)(\lambda + 1) = 0. Thus, \lambda = 0, 1, -1. There are exactly three values of \lambda.