Determinant and matrix representation questions here is a pair of quick questions that I am wondering about, I would appreciate any help:
Is it always the case that any matrix $A$ can be represented as $E_{1}...E_{n}$ * rref($A$) (and in special case when the matrix is invertible, the representation becomes simply $E_{1}...E_{n}$) where $E_{i}$ is an elementary matrix?
And is it always the case that we can write the determinant of any matrix $A$ in the form $(-1)^{s}k_{1}...k_{n}$ * det(rref($A$)) where $s$ signifies the number of row swaps needed to get to the rref form, and $k_{i}$ are the scalars by which we divided rows of $A$ to get to its rref form (and in the special case when the matrix is not invertible, det(rref($A$)) is equal to zero, so the product of the whole expression equals to zero)?
Thank you!
 A: Yes to both.
Multiplying on the left by an elementary matrix corresponds to doing an elementary row operation. Conversely, every elementary row operation can be "performed" by multiplying on the left by an appropriate elementary matrix. 
When you go from $A$ to the reduced row-echelon form of $A$, $\mathrm{rref}A$, you do so by performing a (finite) sequence of elementary row operations; these can be represented instead by performing multiplications on the left by appropriate elementary matrices. So, you can find a sequence of elementary matrices $F_1,F_2,\ldots,F_n$, corresponding to the actions you took, that will make the following true:
$$F_nF_{n-1}\cdots F_1A = \mathrm{rref}A.$$
Now, every elementary matrix is invertible, and its inverse is an elementary matrix (of the same type). So if we let $E_i = F_i^{-1}$, then multiplying on the left both sides of the equation by $E_n$, $E_{n-1},\ldots, E_1$, you get
$$A = E_1E_2E_3\cdots E_n\mathrm{rref}A,$$
with each $E_i$ an elementary matrix.
An elementary matrix that corresponds to "add a multiple of one row to another row" has an inverse of the same type, and in both cases the determinant is just $1$, so you don't need to keep track of them.
An elementary matrix that corresponds to "swap two rows" has an inverse of the same type (in fact, it is its own inverse), and has determinant $-1$, so you only need to keep track of how many of these matrices are around.
But an elementary matrix that corresponds to "divide one row by $k_i$" has as inverse the elementary matrix that corresponds to "multiply one row by $k_i$". The former has determinant $\frac{1}{k_i}$, the latter has determinant $k_i$. So here you need to keep track of exactly what you divded by, and the taking of inverses is why for the determinant computation you multiply by the $k_i$ whereas you originally divided by them. 
Since the determinant of the product is the product of the determinant (when all matrices are square), from
$$A = E_1\cdots E_n\mathrm{rref}A$$
you get the formula you write for the determinant.
