Determinant of a 3x3 matrix vanishes?

Question

I have to find the eigenvalues of the following Jacobian matrix. I usually just proceed by computing the determinant $det(J - \lambda \mathbb I)$, however in this case I can't do this as my formula vanishes due to the 1st column of zeros.

In this case, what can I do?

$$ J = \begin{pmatrix} 0 & a & b\\ 0 & c & 0 \\ 0 & d & e \end{pmatrix} $$

Answer 1

\begin{eqnarray*} \det \begin{bmatrix} -\lambda & a & b \\ 0 & c-\lambda & 0 \\ 0 & d & e-\lambda \\ \end{bmatrix} = -\lambda(c -\lambda)(e-\lambda). \end{eqnarray*}