I'm looking at Apostol's Calculus and I found something that might be trivial but I can't see it directly.
$det A_{12}'=\begin{vmatrix} 0 & 1 & 0 & ...& 0\\ a_{21} & 0 & a_{23} & ...& a_{2n}\\ . & . & . & ...& .\\ . & . & . & ...&.\\ .&.&.&.&.&\\ a_{n1}&0&a_{n3}&...&a_{nn} \end{vmatrix}=detA_{12}\begin{vmatrix} 0 & 1 & 0 & ...& 0\\ 1 & 0 & 0 & ...& 0\\ 0 & 0 & 1 & ...& 0\\ 0 & 0 & 0 & 1...& 0\\ .&.&.&.&.&\\ 0&0&0&...&1 \end{vmatrix} $
$detA_{12}'$ here is the cofactor and $A_{12}$ is the minor. It said the equation above is from uniqueness theorem. I can see this function satisfies Axiom 1-3, but I don't know why the $d(A_1,...,A_n)=detA_{12}$ and why $f(I)$ would be the right-hand side determinant of this non-identity matrix. Can anyone explain this? I'm not sure if I misunderstood some basic concepts here...
The uniqueness theorem I have is: Let $d$ and $f$ be functions satisfying AXIOMS 1-4 and 1-3 respectively. Then, $f(A_1, . . . , A_n) = d(A_1, · · · , A_n)f(I_1, . . . , I_n).$ In particular, if $f$ satisfies, in addition, AXIOM 4, then $f = d$.
Besides, Apostol explained $J$ is the identity matrix of order $n-1$ but I don't know why $f(J)$ would look like the matrix below. I thought if it's identity matrix then $f(J)$ would be 1. Thank you for helping me clarify on this! I'm really confused.
$f(J)$ is by definition of the detriment of the matrix: \begin{matrix} 0 & 1 & 0 & ...& 0\\ 1 & 0 & 0 & ...& 0\\ 0 & 0 & 1 & ...& 0\\ 0 & 0 & 0 & 1...& 0\\ .&.&.&.&.&\\ 0&0&0&...&1 \end{matrix}
I haven't the book at hand and I don't know which pages or section you are referring to, but the idea is to view $$ \det A_{12}'=\begin{vmatrix} 0 & 1 & 0 &\cdots& 0\\ a_{21} & 0 & a_{23} &\cdots& a_{2n}\\ \vdots&\vdots&\vdots&\cdots&\vdots\\ \vdots&\vdots&\vdots&\cdots&\vdots\\ a_{n1}&0&a_{n3}&\cdots&a_{nn} \end{vmatrix}\tag{1} $$ as a function $f(B_1,B_3,\ldots,B_n)$ of the rows/columns (depending on the convention adopted by Apostol) of $(n-1)\times(n-1)$ submatrix $$ B=\,\begin{bmatrix} a_{21} & a_{23} &\cdots& a_{2n}\\ \vdots&\vdots&\cdots&\vdots\\ \vdots&\vdots&\cdots&\vdots\\ a_{n1}&a_{n3}&\cdots&a_{nn} \end{bmatrix}. $$ It should be straightforward to verify that $f$ is an alternating multilinear function (i.e. a function satisfying the axioms you mentioned) of $B_1,B_3,\ldots,B_n$. It follows from the uniqueness theorem that $$ f(B_1,B_3,\ldots,B_n)=d(B_1,B_3,\ldots,B_n)f(J_1,J_2,\ldots,J_{n-1}), $$ where $J_1,J_2,\ldots,J_{n-1}$ are the rows/columns of the $(n-1)\times(n-1)$ identity matrix $J$. However, $d(B_1,B_3,\ldots,B_n)$ is precisely the minor $\det A_{12}$ and $f(J_1,J_2,\ldots,J_{n-1})$ is precisely the RHS of $(1)$ when the $a_{ij}$s are replaced by the entries of $J$. Hence $$ f(J_1,J_2,\ldots,J_{n-1})= \begin{vmatrix} 0&1&0&0&\cdots&0\\ 1&0&0&\ddots&\ddots&\vdots\\ 0&0&1&0&\ddots&\vdots\\ 0&\ddots&0&1&\ddots&0\\ \ddots&\ddots&\ddots&\ddots&\ddots&0\\ 0&\cdots&0&0&0&1 \end{vmatrix} $$ and the result follows.