<s>[INST] 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?
 [/INST] \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.
</s>