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?
 A: As Ted Schifrin rightly noted, you discarded a lot of information by combining three equations into two the way you did. Instead, you could proceed by back-substitution: from the second equation, we have $x_3=-x_1+(1-\lambda)x_3$, which upon substitution into the other equations produces $$\begin{align} (3-\lambda)x_1+(3\lambda-5)x_2 &= 0 \\ (\lambda-3)x_1+(\lambda^2-6\lambda+7)x_2 &= 0. \end{align}$$ Adding these two equations leaves $(\lambda^2-3\lambda+2)x_2=0$, which gives us two of our eigenvalues, $1$ and $2$. You can get the last eigenvalue, $3$, “for free” by examining the trace of $A$, or grind it out by setting $x_2=0$ in your original equations, which eventually leads to $(\lambda-3)x_1=0$.  
As far as your row-reduction goes, the smaller example suggests that you’re using an invalid inference to decide what the eigenvalues of the matrix are. All that you can really say about the smaller example is that the elementary row operation that you performed is invalid when $\lambda=1$. That, however, doesn’t imply that $\lambda$ is an eigenvalue of $B$. Instead, begin by swapping rows so that you don’t have to divide by a term that might vanish and go from there: $$\begin{bmatrix}1-\lambda&5\\3&3-\lambda\end{bmatrix} \to \begin{bmatrix}3&3-\lambda\\1-\lambda&5\end{bmatrix} \to \begin{bmatrix}3&3-\lambda\\0&\frac13(\lambda-6)(\lambda+2)\end{bmatrix}.$$ This is rank-deficient when $\lambda=6$ or $\lambda=-2$, which are indeed the eigenvalues of $B$.  
Similarly for $A$, all that you can really conclude from examining the denominators of entries in the reduced matrix is that you have to try something else for the values at which they vanish. You can’t immediately conclude that those values are eigenvalues of the matrix. Avoiding dividing by expressions that involve $\lambda$ by judicious row swaps or simply by not normalizing the pivot to $1$, I end up with $$\begin{bmatrix}1&1&\frac12(5-\lambda)\\0&2-\lambda&\frac12(3-\lambda)\\0&0&-\frac12(\lambda-3)(\lambda-1)\end{bmatrix}.$$ We immediately get the eigenvalues $1$ and $3$ from this, and compute $2$ from the trace of $A$. Alternatively, since we have $2-\lambda$ in a pivot position, set $\lambda=2$ and examine the resulting matrix. It has two identical rows, so is also rank-deficient.  
All that aside, I prefer to try Hagen von Eitzen’s answer to the question that you linked before doing anything else: namely, see if I can guess any eigenvectors by trying simple linear combinations of rows and columns. In this case, subtracting the last column from the first produces $(3,0,-3)^T$, so $(1,0,-1)^T$ is an eigenvector with eigenvalue $3$. Subtracting the second column from the first gives $(2,-2,0)^T$ for another eigenvalue. Another way to find this one is to add the first and last rows to each other (left and right eigenvalues of a matrix are identical). The last eigenvector can always be computed from the trace: $6-3-2=1$. 
A: \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}
What you are doing above is finding the null space of the matrix $A-\lambda I$. This consists of all vectors $v$ such that $Av=\lambda v$. This gives you equations that will allow you to find the eigenvalues of the matrix.
A: 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 $1, 2, 3$.
