Finding the eigenvalues of a $3 \times 3$ matrix without the determinant I am trying to find the eigenvalues of 
$$A = \begin{pmatrix} 0 & -2 & -3 \\ -1 & 1 & -1 \\ 2 & 2 & 5\end{pmatrix}$$
Without using the determinant. I first tried doing something like I did here Finding the eigenvalues of $\begin{pmatrix} a & b \\ b & a \end{pmatrix}$ matrix without the determinant
And from the system 
$\begin{cases}
-\lambda x_1 & -2x_2 & -3x_3 &= 0 \\
-x_1  &+ \ (1-\lambda)x_2 &- x_3 &= 0 \\
2x_1 &+ \ 2x_2 &+ \ (5 - \lambda)x_3 &= 0
\end{cases}$
Tried to solve for
$\cfrac{-\lambda x_1 - 2x_2}{3} = -x_1 + (1 - \lambda)x_2 = \cfrac{2x_1 + 2x_2}{\lambda - 5}$
But nothings seems to come of that. I then proceeded to try to find $\lambda$ such that
$$\begin{pmatrix} -\lambda & -2 & -3 \\ -1 & 1-\lambda & -1 \\ 2 & 2 & 5-\lambda\end{pmatrix}$$ is non-invertible, by looking at the reduced row echelon form and see for what $\lambda$ the rank is less than 3.
$\begin{equation}
\begin{split}
\begin{pmatrix} 1 & \lambda - 1 & 1 \\ 0 & \lambda^2 - \lambda - 2 & \lambda - 3 \\ 0 & 2(2-\lambda) & 3 - \lambda \end{pmatrix}
\rightarrow
\begin{pmatrix} 1 & \lambda - 1 & 1 \\ 0 & 1 & \cfrac{\lambda - 3}{\lambda^2 - \lambda - 2} \\ 0 & 2(2-\lambda) & 3 - \lambda \end{pmatrix}
\end{split}
\end{equation}$
Where $\lambda \neq 2$. Another operation yields: 
$\begin{equation}
\begin{split}
\rightarrow 
\begin{pmatrix} 1 & \lambda - 1 & 1 \\ 0 & 1 & \cfrac{\lambda - 3}{\lambda^2 - \lambda - 2} \\ 0 & 0 & 3 - \lambda \Big (1 + \cfrac{2(2-\lambda}{\lambda ^2 - \lambda - 2} \Big) \end{pmatrix} \rightarrow  
\begin{pmatrix} 1 & \lambda - 1 & 1 \\ 0 & 1 & \cfrac{\lambda - 3}{\lambda^2 - \lambda - 2} \\ 0 & 0 & 1 \end{pmatrix}
\end{split}
\end{equation}$
Where $\lambda \neq 1, 3$ (because these are roots of the polynomial).
Now it happens that the eigenvalues are indeed $\lambda_1 = 1, \lambda_2 = 2, \lambda_3 = 3$. But I'm not sure how what I've done makes sense as a mathematical process for finding eigenvalues. Moreover it doesn't even seem to fully work in this other example
$B = \begin{pmatrix} 1 & 5 \\ 3 & 3 \end{pmatrix}$
$\begin{equation}
\begin{split}
\begin{pmatrix} 1 - \lambda & 5 \\ 3 & 3 - \lambda \end{pmatrix} \rightarrow \begin{pmatrix} 1 & \cfrac{5}{1 - \lambda} \\ 3 & 3 \end{pmatrix}
\end{split}
\end{equation}$
Where $\lambda \neq 5$. 
$\begin{equation}
\begin{split}
\begin{pmatrix} 1 & \cfrac{5}{1 - \lambda} \\ 0 & \cfrac{-15}{1 - \lambda} + 3 - \lambda \end{pmatrix} \rightarrow \begin{pmatrix} 1 & \cfrac{5}{1 - \lambda} \\ 0 & 1  \end{pmatrix}
\end{split}
\end{equation}$
Where $\lambda \neq -2, 6$. 
Here it happens that $B$ has eigenvalues $\lambda_1 = -2, \lambda_2 = 6$, but there is no eigenvalue for $1$. So what's going on here?

You have to look at the factors you're dividing the rows by to make diagonal elements $1$: their product turns out to be $(+/-)$ the characteristic polynomial, so its roots will be the eigenvalues.  You say you want to do this "without using the determinant", but your row reduction procedure (keeping track of those factors) is exactly what you use to compute the determinant.
Thus in your first example, you 
interchanged two rows (which multiplies the determinant by $-1$), then divided by $-1$ to get the $1$ in position $(1,1)$, then added multiples of the first row to get $0$'s in the rest of the first column.  Then you divided the second row by
$\lambda^2 - \lambda - 2$, and added a multiple of the second row to the third to get a $0$ in position $(3,2)$.  Finally you divided the third row by
$$3-\lambda+{\frac { \left( -4+2\,\lambda \right)  \left( -3+\lambda
 \right) }{{\lambda}^{2}-\lambda-2}}=-{\frac { \left( -1+\lambda
 \right)  \left( -3+\lambda \right) }{\lambda+1}}
$$
So the product of the factors is 
$$ - (\lambda^2 - \lambda - 2) {\frac { \left( -1+\lambda
 \right)  \left( -3+\lambda \right) }{\lambda+1}} =- (\lambda - 1)(\lambda - 2)(\lambda - 3) $$
and the characteristic polynomial is $(\lambda - 1)(\lambda - 2)(\lambda-3)$
so the eigenvalues are $1, 2, 3$.
