\begin{vmatrix} \lambda & 1 & 1\\ 1 & \lambda & 1 \\ 1 & 1 & \lambda \end{vmatrix}
If we want to see that the determinant is non zero we can do modulo 2, and we get if $\lambda$ is even:
$$ \begin{vmatrix} 0 & 1 & 1\\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{vmatrix}$$
The determinant is in this case not zero.
But… if we input $\lambda = -2$, then the determinant is zero. Shouldn't the determinant be non zero for every even number in the diagonal since it's modulo 2 determinant is not zero.
EDIT: We want to see if determinant is non zero in integers and used modulo 2 only as a tool for quicker computation. Like shown here in the first answer Is the following matrix invertible?
Best Answer
$$ \begin{vmatrix} 0 & 1 & 1\\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{vmatrix}=2\equiv0\pmod2$$