How to prove the cofactor formula for determinants, using a different definition of the determinant? So, I am very interested on this theorem (Laplace expansion), but I am still a high school student. I have four books about matrices, but only one of them have proo and that doesn't start with my definition of determinant. Also I don't understand the proof of Laplace expansion in wikipedia, because there is too much symbol and terms I didn't learn and that proof dosn't start with my definition. So may someone give me the outline of the proof start with my definition?
Here's my definition of the $n \times n$ determinant: 
The value of the determinant of a matrix of order $n$ is defined as the sum of $n!$ terms of the form $(-1)^k a_{1 i_1} a_{2 i_2} \cdots a_{n i_n}$. Each term contains one and only one element from each row and one and only element from each column; i.e., the second subscripts $i_1, i_2 , \ldots, i_n$ are equal to $1,2, \ldots, n$ taken in some order. The exponent $k$ represents the number of interchanges of two elements necessary for the second subscripts to be placed in the order $1,2, \ldots, n$. For example, consider the term containing $a_{13} a_{21} a_{34} a_{42}$ in the evaluation of the determinant of a matrix of order four. The value of $k$ is $3$ since three interchanges of two elements are necessary for the second subscripts to be placed in the order $(1,2,3,4)$.
 A: The proof in Wikipedia which you linked to in your previous attempt at asking this question is in fact using the very same definition of "determinant" that you are using, only with different terminology.
Wikipedia uses the definition of determinant as the sum of all terms of the form
$$\mathrm{sgn}(\tau)a_{1\tau(1)}\cdots a_{n\tau(n)}$$
where $a_{ij}$ is the $(i,j)$th entry of the matrix, $\tau$ is a permutation of $\{1,2,\ldots,n\}$ (that is, a bijection $\tau\colon\{1,2,\ldots,n\}\to\{1,2,\ldots,n\}$), and $\mathrm{sgn}(\tau)$ is the sign of $\tau$.
In your description, the numbers $i_1,\ldots,i_n$ must be precisely the numbers $\{1,2,\ldots,n\}$ in some order; given a summand
$$(-1)^ka_{1i_1}\cdots a_{ni_n}$$
as in your description, we define $\tau\colon\{1,2,\ldots,n\}\to\{1,2,\ldots,n\}$ by $\tau(j) = i_j$. Then $\tau$ is a permutation of $\{1,2,\ldots,n\}$, and we can express the summand as
$$(-1)^ka_{1\tau(1)}\cdots a_{n\tau(n)}.$$
So the only thing left is to show that your factor $(-1)^k$ is precisely equal to $\mathrm{sgn}(\tau)$. 
The sign of a permutation $\tau$ is equal to $1$ if $\tau$ can be written as a product of an even number of transpositions, and is equal to $-1$ if it can be written as a product of an odd number of transpositions. A "transposition" is precisely an "exchange of two elements".
For each "exchange" you perform to the list $i_1,i_2,\ldots,i_n$, consider the transposition that corresponds to that exchange. Performing the exchange is equivalent to composing $\tau$ with that transposition. The fact that after performing $k$ of these exchanges you get the identity tells you that $\tau$ can be written as a product of $k$ transpositions, and therefore that $\mathrm{sgn}(\tau)=(-1)^k$... exactly the same as the factor you have.
For instance, in your example $a_{13}a_{21}a_{34}a_{42}$, the permutation $\tau$ associated to this factor is, in disjoint cycle notation $\tau=(1,3,4,2)$, and in two-line notation
$$\tau = \left(\begin{array}{cccc}
1&2&3&4\\
3&1&4&2
\end{array}\right).$$
Exchanging the $1$ and $3$ is equivalent to composing $\tau$ with the transposition that exchanges $1$ and $3$ (which is $(1,3)$ in disjoint cycle notation); then exchanging $2$ and $3$; and finally exchanging $3$ and $4$, so you get $(-1)^3 = -1$. But this is the same as saying that $\tau$ is the composition of the three permutations I just mentioned: that is, that $(1,3,4,2) = (1,3)(2,3)(3,4)$, which is easy to verify. So the transposition has odd parity, hence $\mathrm{sgn}(\tau)=-1$, yielding the exact same result as yours.
Since the two definitions of determinant are in fact identical, except for the terminology used to describe the summands, the proof in Wikipedia is in fact "the proof [...] with [your] definition." 
A: Arturo Magidin, thanks for your answer. I now understand the steps, namely:
1) All determinants of a matrix of size n will be calculated by a sum of n! terms (easy to prove by induction)
2) these n! terms will each be a product of n numbers (evident)
3) By the definition of how a determinant is defined, these n numbers will be taken from different rows and columns, for a total of n! possible terms, that is: from looking at the matrix, we can know exactly which terms will appear in the end sum of the determinant. All we need is to find out the sign of each term.
I believe that a crucial element is missing: how can we be sure that the sign calculus will be the same, no matter the order that we choose to calculate it? In order to prove that we can begin with any column/row and reach the same determinant, this seems necessary. For example, in a n=4 matrix
x x b x
a x x x
x c x x
x x x d
let's ignore the x's, which can be any number, and say we are calculating the sign of the abcd term that will appear in the determinant sum. How can we be sure that we will end up with the same sign if we choose any order of abcd? that is, take b and calculate its cofactor's sign, then to calculate b's cofactor's sign, take a, and then c and d, namely b-a-c-d, or do the same for c-b-a-d. 
Because as the matrices for the new calculations become smaller, the sign of the cofactors of a, b, c, d can also change. Is there a proof in this regard, that shows that multiplying the resulting sign here is the same for any sequence that we choose to use? This is evident only for the main diagonal. This seems to me to transcend the scope of Linear Algebra and an answer might come from a different field of Mathematics.
