Can anyone prove this:
$\det(M) =\begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} = a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{31}a_{22}a_{13}-a_{32}a_{23}a_{11}-a_{33}a_{21}a_{12}$
You can just calculate :
$$\det(M)=\begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}=a_{11}\times\begin{vmatrix}a_{22}&a_{23}\\a_{32}&a_{33}\end{vmatrix}-a_{12}\times\begin{vmatrix}a_{21}&a_{23}\\a_{31}&a_{33}\end{vmatrix}+a_{13}\times\begin{vmatrix}a_{21}&a_{22}\\a_{31}&a_{32}\end{vmatrix}$$ and as $\begin{vmatrix}a&b\\c&d\end{vmatrix}=ad-bc$ you get your result.
Sarrus Rule is the graphical method, extending the matrix and then using the diagonals.
More useful is to learn the Laplace development of a determinant:
$$ \text{det}(A) = \sum_j (-1)^{i+j} a_{ij} \, \text{det}(S_{ij}) = \sum_i (-1)^{i+j} a_{ij} \, \text{det}(S_{ij}) $$ where $S_{ij}$ is the matrix with row $i$ and column $j$ removed, because it reduces the dimension of the determinants involved by one each time one uses it, until one hits the easy to calculate 2-determinants (or 1-determinants).
