Leibniz formula and determinants Let $A \in M_n(\mathbb C)$, then
$$\det(A) = \sum_{\sigma \in S_n} \mbox{sign}(\sigma) a_{1\sigma(1)} a_{2\sigma(2)} \cdots a_{n\sigma(n)} = \sum_{\sigma \in S_n}\prod_{i=1}^n a_{1\sigma(i)}$$
I looked at different resources to try and understand how the formula works, but it hasn't really clicked with me yet. I even read through Understanding the Leibniz formula for determinants and it helped a bit, but I still need some clarification. What throws me off the most are the subscripts for the elements in $A$, and I don't understand how you know which elements to multiply together.
If anyone can break it down into simpler terms, that would be greatly appreciated. Thanks!
 A: *

*Consider $A$ as on the $n\times n$ chessboard. Then each product would correspond to a setting of $n$ rooks which cannot hit each other (each row and each column will contain exactly one rook, the column index of the rook in terms of the row index is the permutation $\sigma$). Then, if the permutation is odd (contains an odd number of cycles of length even (e.g. transpositions)) the product of the entities gets a minus sign.


*Probably a recursive definition would get you closer (expanding by the first row):



*

*for a $1\times 1$ matrix $\det\pmatrix{a}:=a$.

*for $A\in M_{n}(\Bbb C)$, let $A_{i}$ denote the matrix where we delete the $i$th column and first row from $A$, and let the first row be $(a_1,\dots,a_n)$, then define
$$\det\pmatrix{a_1&\dots&a_n\\ \vdots & \ddots & \vdots \\
&\dots &}:=a_1\det(A_1)-a_2\det(A_2)+a_3\det(A_3)-\dots$$
By induction it is possible to prove that we get back the same expression (each permutation will be counted, exactly once, with the appropriate sign).



*The geometrical meaning! 
Let $v_1,..,v_n$ be vectors in $\Bbb R^n$, written -say- as column vectors in the matrix $A$. Then $\det(A)$ is just the (signed) $\,n$ dimensional volume of the $n$ dimensional parallelepipedone spanned by $v_1,...,v_n$. The sign is positive iff the vectors $v_1,...,v_n$ in this order are in the 'same direction' as the standard basis $e_1,..,e_n$.


*And to show that the quantity described at 3. indeed gives the determinant defined at 2, observe the following:



*

*Let $V$ denote the signed volume described in 3., then it is multilinear, in particular
$$\lambda_i\, V( e_i,v_2,..,v_n)+\lambda_j\, V( e_j,v_2,..,v_n)=
V(\lambda_i e_i+\lambda_j e_j,\,v_2,..,v_n)$$

*If we consider a parallelepippedon spanned by $v_1,..,v_n$, placed on the origin, and we move its 'top face' along a vector that is on the 'base face', then the height doesn't change, so neither changes $V$, meaning algebraically:
$$V(v_1,v_2,v_3,...,v_n)=V(v_1,\,v_2-\lambda v_1,v_3,...,v_n)$$
Applying these steps is called the Gauss elimination of the matrix (actually, one more step is included: to multiply a vector by a nonzero scalar), and it can always lead either to a matrix with a $0$ column or to the identity matrix $I$.

Knowing that, all you have to check is that the determinant and the signed volume coincides for these final cases, and that the steps of Gauss elimination do exactly the same effect on the determinant as on the signed volume.
A: It's not as bad as it seems! It means pick one element from each row of A such that no two elements are in the same column, then multiply them together. Add all such possible products together, with some plus and some minus. You put a plus if you chose the diagonal or something which can be placed on the diagonal by interchanging an even number of columns, and a minus if you have to interchange an odd number of columns to place your set on the diagonal.
