Will an element always appear in every term? In Shilov's Linear Algebra p.22 about Laplace's theorem
it said
"Finally, let the rows of the determinant $D$ with indices $i_1,i_2,\ldots,i_k$ be fixed; some elements from these rows appear in every term of D."
Why the sentence after ; is true?
In the text, Shilov formed a minor $M^{i_1,i_2,\ldots,i_k}_{j_1,j_2,\ldots,j_k}$ with those k rows and k of the n columns
and a cofactor $\overline{A}^{i_1,i_2,\ldots,i_k}_{j_1,j_2,\ldots.j_k}$ of the minor.
And the terms are now divided into groups $M^{i_1,i_2,\ldots,i_k}_{j_1,j_2,\ldots,j_k} \overline{A}^{i_1,i_2,\ldots,i_k}_{j_1,j_2,\ldots.j_k}$
Note that,
$a_{\alpha_1,1} a_{\alpha_2,2} \cdots a_{\alpha_n,n}$ is a term of $D$, $a_{\alpha_1,1}$ is an element on the first column of the matrix of $D$, $\alpha_1,\alpha_2,\ldots\alpha_n$ are unique.
$a_{i,j}$ is an element of the matrix of $D$.
 A: In fact his follows from the definition of determinant of a square matrix $\,n\times n\,$, which is the sum of $\,n!\,$ products, each one containing exactly one unique element from each row and one unique element from each column of the matrix...
A: Each term of the determinant in fact contains one entry from each row and one entry from each column of the matrix. Think about the way you compute $2\times2$ or $3\times3$ determinants, or look at the general formulas. 
A: Consider the minor (part of a term) mentioned above.
The k rows are fixed and k columns $j_1,j_2,\ldots,j_k$ are chosen from n.
So, e.g. in the first row of the minor, would be consist of some element $a_{i_1,j_1},a_{i_11,j_2},\ldots,a_{i_1,j_k}$ from the matrix of $D$.
It is true for every other rows of the minor and other minors.
Because $D$ is consist of the sum of the product of minors and its corresponding cofactor.
So, every term of $D$ is made of of some elements of those k rows.
I think this is what Shilvo trying to said.
