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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!

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Yes to both. Though the notation looks suspiciously like that of math.stackexchange.com/questions/22610/… –  Arturo Magidin Feb 19 '11 at 22:58
    
Thank you! I am struggling to log in. –  LinAlgStudent Feb 19 '11 at 23:00
    
@LinAlgStudent: Did you post that previous question? –  Arturo Magidin Feb 19 '11 at 23:01
    
@Arturo Magidin: I did indeed. Just checking whether such representation is consistent for both invertible and non-invertible matrices (and their determinants). –  LinAlgStudent Feb 19 '11 at 23:02
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@LinAlgStudent @user5157, considering that you just logged into the user5157 account 20 minutes ago, I take it any difficulty has been resolved. I've thus merged your accounts. Also, please do provide a valid e-mail address in your profile, so the moderators can contact you in situations such as this. –  Willie Wong Feb 19 '11 at 23:26

1 Answer 1

up vote 2 down vote accepted

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.

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