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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 $D$.

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    By the way, your question on the heading has the answer: no, but that question isn't the same as in the body of your question...2012-08-21

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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...

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    I think you guys are correct in some sense. But I finally figure it out, It isn't what we where talking about. The fixed row is for constructing the minors. And the latter sentence is for grouping of the terms of $D$.2012-08-23
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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.

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    In Robert's example, there are 6 terms in the determinant. An element from row 1 appears in every one of those 6 terms; an element from row 2 appears in every one of those 6 terms; an element from row 3 appears in every one of those 6 terms. That's what Shilov is saying.2012-08-21
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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.

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    Why don't you thin$k$ Shilov is trying to say what Don$A$$n$tonio and Robert Israel and I agree Shilov is trying to say?2012-08-21