Why does a determinant always give the same value for expanding about any row or column? The expansion of determinant by different row and column always gives same value.
why does it true?
i need proof for general case.
 A: The determinant is a highly symmetrical $n^{\rm th}$ degree polynomial function of the matrix elements. It consists of $n!$ terms, each of which is a product of $n$ matrix elements – exactly one  from each row and each column – and a sign factor. "Expanding" an $n$-determinant with respect to a row is part of a recursive scheme, like computing a multiple integral first with respect to the "innermost" variable. Due to the overwhelming symmetry it does not play a rôle which row you use for the reduction.
A: Because the definition of the determinant of an $n\times n$ matrix $A=\begin{bmatrix}a_{ij}\end{bmatrix}$ is
$$\det A=\sum_{\sigma\in S_n}(-1)^{\varepsilon(\sigma)}a_{1\mkern2mu\sigma(1)}a_{2\mkern2mu\sigma(2)}\dots a_{n\mkern2mu\sigma(n)},$$
where $\varepsilon(\sigma)$ is the signature of the permutation $\sigma$, and this definition does not depend on the row or the column along which the determinant is expanded.
A: Determinant is a linear function, so, for any $n\times n$ determinant, we have the following formula(take the first row as an example , in fact we can do same operation to any other row or column):
$$\begin{bmatrix} { a }_{ 11 } & { a }_{ 12 } & \cdots  & { a }_{ 1n } \\ { a }_{ 21 } & { a }_{ 22 } & \cdots  & { a }_{ 2n } \\ \vdots  & \vdots  & \ddots  & \vdots  \\ { a }_{ n1 } & { a }_{ n2 } & \cdots  & { a }_{ nn } \end{bmatrix}={ a }_{ 11 }\begin{bmatrix} 1 & 0 & \cdots  & 0 \\ { a }_{ 21 } & { a }_{ 22 } & \cdots  & { a }_{ 2n } \\ \vdots  & \vdots  & \ddots  & \vdots  \\ { a }_{ n1 } & { a }_{ n2 } & \cdots  & { a }_{ nn } \end{bmatrix}+{ a }_{ 12 }\begin{bmatrix} 0 & 1 & \cdots  & 0 \\ { a }_{ 21 } & { a }_{ 22 } & \cdots  & { a }_{ 2n } \\ \vdots  & \vdots  & \ddots  & \vdots  \\ { a }_{ n1 } & { a }_{ n2 } & \cdots  & { a }_{ nn } \end{bmatrix}+\cdots +{ a }_{ 12 }\begin{bmatrix} 0 & 0 & \cdots  & 1 \\ { a }_{ 21 } & { a }_{ 22 } & \cdots  & { a }_{ 2n } \\ \vdots  & \vdots  & \ddots  & \vdots  \\ { a }_{ n1 } & { a }_{ n2 } & \cdots  & { a }_{ nn } \end{bmatrix}$$
If you want to expand this determinant by other row or column(for example, the $i$th row), there is no difference because: because you can exchange this row with the first row, and this only changes its symbol. After finish your calculation, you need to change the symbol back. 
So the next thing you need to do is to proof that
$$\begin{bmatrix} 0 & 0 & \cdots  & 1 & \cdots  & 0 \\ { a }_{ 21 } & { a }_{ 22 } & \cdots  & { a }_{ 2i } & \cdots  & { a }_{ 2n } \\ \vdots  & \vdots  &  & \vdots  &  & \vdots  \\ { a }_{ n1 } & { a }_{ n2 } & \cdots  & { a }_{ ni } & \cdots  & { a }_{ nn } \end{bmatrix}=\begin{bmatrix} { a }_{ 21 } & { a }_{ 22 } & \cdots  & { a }_{ 2(i-1) } & { a }_{ 2(i+1) } & \cdots  & { a }_{ 2n } \\ \vdots  & \vdots  &  & \vdots  & \vdots  &  & \vdots  \\ { a }_{ n1 } & { a }_{ n2 } & \cdots  & { a }_{ n(i-1) } & { a }_{ n(i+1) } & \cdots  & { a }_{ nn } \end{bmatrix} $$
This is the expansion by the first row which column $i$ is $1$.
In fact, this can be easily deduced from the basic definition of determinant which has been given in the first answer to this question. This requires no special skill but carefulness. You may have a try.
One more thing to remember, the value of a determinant is given by its basic definition, and the Laplace expansion is just a simpler way to calculate the value. As you can see, the definition is too complex to use, especially while calculating a “big” determinant.
