Answer
the system has non trivial solutions if and only if $\lambda=2$ or $\lambda=-1$.
Work Step by Step
Given the matrix
$$\left[ \begin{array} {cc} \lambda -1&2\\1&\lambda \end{array} \right].$$
Adding the first row to $(1-\lambda)$ times the second row, we get
$$\left[ \begin{array} {cc} \lambda -1&2\\0&\lambda (1-\lambda)+2 \end{array} \right]. $$
The system has non trivial solutions if and only if
$$\lambda (1-\lambda)+2=0.$$
By factorization the above equation lead to
$$(\lambda-2)(\lambda +1)=0$$
and we have the solution $\lambda=2$ or $\lambda=-1$.
Hence, the system has non trivial solutions if and only if $\lambda=2$ or $\lambda=-1$.
