Proof $\det(AB)=\det(A)\det(B)$ I have read the following proof , in here:

Why can we go from the first line to the second one? 
why $Det(E^k)\cdot Det(E^m)=Det(E^k\cdot E^m)$? 
is it because $\det(E^k)\in \mathbb{F}$ for all $k$?
 A: The property $\det(E^iE^j) = \det(E^i)\det(E^j)$ doesn't just hold for two elementary matrices $E^i,E^j$, it holds for any elementary matrix $E$ and any matrix $A$:
$$\det(EA) = \det(E)\det(A)$$

Proof:
There are three types of elementary matrices:


*

*row-scaling

*row-swapping

*row-adding


If I have some matrix $A$ with determinant $\det(A)$, and I multiply $A$ by some elementary matrix $E$ to get $EA$, then...


*

*if $E$ is a row-scaling matrix which scales a row of $A$ by a constant $c$, we can easily see that $\det(E) =c$ since $E$ is the identity matrix, but with one diagonal element changed to $c$. We can also show that $\det(EA) = c\det(A) = \det(E)\det(A)$ [proof]

*if $E$ is a row-swapping matrix, then $det(E) = -1$ since it is simply the identity matrix with two rows swapped. We can show that $\det(EA) = -\det(A) = \det(E)\det(A)$ [proof]

*if $E$ is a row-adding matrix which adds two rows of $A$ together, with one of the rows scaled by a constant $k$, then $\det(E) = 1$ since it is simply the identity matrix with one non-zero off-diagonal element $k$. We can also show that $\det(EA) = \det(A) = \det(E)\det(A)$ [proof]
A: The determinant is a multiplicative function on $M_n(R)$, where $R$ is the base (commutative) ring:
$$\det(AB)=\det A\cdot\det B$$
for any $n\times n$ matrices with coefficients in $R$, and not only when one is an elementary matrix.
