Is it possible to find the Eigenvalues of a square matrix by reducing it into echelon form then finding what values lead to a 0 in the diagonal?

Question

I was taught that the determinant of a triangular matrix is just the product of all the entries in its diagonal. Yet when I reduce the matrix:

\begin{bmatrix} 2 & 0 & 3\\ 0 & 3 & 0\\ 4 & 0 & 1 \end{bmatrix} to: \begin{bmatrix} 2 & 0 & 3\\ 0 & 3 & 0\\ 0 & 0 & -5 \end{bmatrix} The Eigenvalues I find with the equation $\det(A- \lambda I) = 0$ are 2, 3, and -5. The correct values are 3, 5, and -2. What am I doing wrong?