<s>[INST] Easier way to find eigenvalues of Matrices? I am trying to find eigenvalues for this matrix,
A = 
$\begin{bmatrix}
3 & 2 & -3 \\
-3 & -4 & 9 \\
-1 & -2 & 5 \\
\end{bmatrix}$ 
I find the characteristic equation here:
$(\lambda I - A)
=
\begin{bmatrix}
\lambda - 3 & -2 & 3 \\
3 & \lambda + 4 & -9 \\
1 & 2 & \lambda - 5 \\
\end{bmatrix}$ 
The difficult part I am having is finding the determinant of the characteristic equation, mainly that it becomes insanely difficult for me to keep track of the factoring to get the eigenvalues...here I use rule of Sarrus to try to calculate the eigenvalues.
$\begin{vmatrix}
\lambda I - A \\
\end{vmatrix}
= 
\begin{bmatrix}
\lambda - 3 & -2 & 3 & \lambda -3 & -2 \\
3 & \lambda + 4 & -9 & 3 & \lambda + 4 \\
1 & 2 & \lambda - 5 & 1 & 2 \\
\end{bmatrix}$
= $(1)(\lambda + 4)(3)$
$+ (2)(-9)(\lambda - 3)$
$+ (\lambda - 5)(3)(-2)$
$- (\lambda - 3)(\lambda + 4)(\lambda - 5)$
$- (-2)(-9)(1)$
$- (3)(3)(2)$ 
= $(3)(\lambda + 4) + (-18)(\lambda - 3) + (-6)(\lambda -5) - (\lambda - 3)(\lambda + 4)(\lambda - 5) - 18 - 18$
= $(3)(\lambda + 4) + (-18)(\lambda - 3) + (-6)(\lambda -5) - (\lambda - 3)(\lambda + 4)(\lambda - 5) - 36$
= $((\lambda + 4)(3) -(\lambda -3)(\lambda -5)) + (-18)(\lambda -3) + (-6)(\lambda - 5) - 36$
= $(\lambda - 3)((\lambda + 4)(3) -(\lambda - 5)) -18 + (-6)(\lambda - 5) - 36$
= $(\lambda - 5)(\lambda -3)((\lambda + 4)(3) - 1)) - 18 + (-6)(1) - 36$
= $(\lambda - 5)(\lambda -3)((\lambda + 4)(3) - 1)) - 60$
= $(\lambda - 5)(\lambda - 3)(3\lambda + 12 -1 - 60)$
= $(\lambda - 5)(\lambda - 3)(3\lambda - 49)$ 
I end up with 5, 3, and 16.3 as the eigenvalues (16.3 seems off). Obviously that was a ridiculous amount of simplification I had to do just to get eigenvalues and when I feel like I messed up (like here), it is pretty impossible to check my work in an effective manner. The amount of time it takes for me to calculate eigenvalues is unacceptable for my upcoming final exam. Do you guys have any tips or tricks that makes this process easier?
 [/INST] Here is a simple computation:
\begin{align}
\det(\lambda I - A)& = \begin{vmatrix}
\lambda - 3 & -2 & 3 \\
3 & \lambda + 4 & -9 \\
1 & 2 & \lambda - 5 \\
\end{vmatrix}=\begin{vmatrix}
\lambda - 2  & 0 & \lambda - 2  \\
3 & \lambda + 4 & -9 \\
1 & 2 & \lambda - 5 \\
\end{vmatrix}\\& =(\lambda - 2 )\begin{vmatrix}
1 & 0 & 1 \\
3 & \lambda + 4 & -9 \\
1 & 2 & \lambda - 5 \\
\end{vmatrix}=(\lambda - 2 )\biggl(\begin{vmatrix}
\lambda + 4 & -9 \\
2 & \lambda - 5 \\
\end{vmatrix}+\begin{vmatrix}
3 & \lambda + 4  \\
1 & 2  \\
\end{vmatrix}\biggl)\\
&=(\lambda - 2 )\Bigl(\bigl(
(\lambda + 4)(\lambda - 5)+18\bigr) 
+ (6 -\lambda - 4) \Bigl)=(\lambda - 2 )(\lambda^2-2\lambda)\\
&=\lambda(\lambda - 2)^2.
\end{align}
</s>