Q:

Why there is a $(-1)$ factor in the cofactor - determinant relation?

In the book of Linear Algebra by Werner Greub, at page $116$, section $4.14$, it says that

Given an $n\times n$ matrix, $A = (\alpha_j^i)$, denote for each pair
  $(i,j)$ by $S_i^j$ the $(n-1) \times (n-1)$ matrix obtained from A by
  deleting the i-th row and the j-th column.We are going to show that 
  $$cof (\alpha_i^j) = (-1)^{i+j} det S_i^j.$$
In fact, by $(i-1)$ interchanges of rows and $(j-1)$ interchanges of
  columns we can transform $C_i^j$ into the matrix 

, where $C_i^j$ is 

So here is my question: Why do we need $(i-1)$ interchanges of rows or $(j-1)$ interchanges of columns. I mean can't we directly interchange the j-th column with the first column, and the i-th row with the first row ?
In this way, determinant would be changes its sign twice, which would give us the same determinant.

A:

The goal is to preserve $S_i^j$. By interchanging the $i$th row with the $(i-1)$th row, then the $(i-2)$th row, et cetera, you extract the $i$th row without changing the sign of $\det(S_i^j)$. The way of thinking about this that you have suggested accomplishes essentially the same thing in terms of extracting the $i$th row and $j$th column, but the determinant of the matrix you get is related to $\det(S_i^j)$ by the parity of $i$ and $j$ anyway.

