#### Answer

the system has non trivial solutions if and only if $\lambda=3$ or $\lambda=1$.

#### Work Step by Step

The system has non trivial solution if the determinant of the coefficient matrix is zero, that is
$$\left| \begin{array} {cc} \lambda -2 &1\\1&\lambda -2 \end{array} \right|=0,$$
so we have
$$(\lambda -2)^2-1=0 \Longrightarrow \lambda^2-4\lambda +3=0. $$
By factorization the above equation lead to
$$(\lambda-3)(\lambda -1)=0$$
and we have the solution $\lambda=3$ or $\lambda=1$.
Hence, the system has non trivial solutions if and only if $\lambda=3$ or $\lambda=1$.