Elementary Matrices: Solving Homework Problem

roam
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Homework Statement



Here's the problem I don't understand and its solution:

http://img11.imageshack.us/img11/5867/24545624.gif

Homework Equations





The Attempt at a Solution



I want to write the original matrix as a product of elementary matrices but I don't know they got that set of 4 elementary matrices. Can anyone help me to see how to get those elementary matrices?
I can row reduce the original matrix to rref but I don't see how that helps. (Actually the matrix when in reduced row echelon form is the identity, indicating that it is invertible.)
 
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Each time you perform row reduction, record that step. Then find the particular elementary matrix associated with that row reduction. Remember that the elementary matrix corresponding to a particular step is simply the row operation in question applied to the identity matrix.
 
The representation of a matrix as a product of elementary matrices isn't unique -- there are lots of correct answers, including the one you produced when you put your matrix into reduced row echelon form. (Assuming you actually kept track of what row operations you used, did all the arithmetic correctly, and converted that into elementary row operations correctly)

That said, the answer you showed us is wrong; the third term in the product on the right hand side isn't elementary!

P.S. if the rref wasn't the identity, then that matrix cannot possibly be the product of elementary row matrices. (Do you understand why?)
 
Hurkyl said:
The representation of a matrix as a product of elementary matrices isn't unique -- there are lots of correct answers, including the one you produced when you put your matrix into reduced row echelon form. (Assuming you actually kept track of what row operations you used, did all the arithmetic correctly, and converted that into elementary row operations correctly)

For example one row operation would be to add -2 times the first row to the second. We get the matrix:

111
010
112

What will be say, E1, the elementary matrix corresponding to this particular step?

That said, the answer you showed us is wrong; the third term in the product on the right hand side isn't elementary!

Really? Is it because there has to be a 1 instead of 3 in the main diagonal?

P.S. if the rref wasn't the identity, then that matrix cannot possibly be the product of elementary row matrices. (Do you understand why?)

A matrix is invertible if it can be written as a product of elementary matrices. Since the reduced row echelon form of A is assumed to be In then there is a sequence of elementary row operations that reduces A to identity. Each of these operations is done with multipication by an elementary matrix. So there is a sequence of elementary operations E1, E2, ...,Ek such that: Ek...E2E1 = I
 
roam said:
For example one row operation would be to add -2 times the first row to the second. We get the matrix:

111
010
112

What will be say, E1, the elementary matrix corresponding to this particular step?
The elementary matrix would simply be

100
-210
001

Because that is what you get you perform this row operation on the identity matrix.



Really? Is it because there has to be a 1 instead of 3 in the main diagonal?
No, an elementary matrix is the result of applying a single row operation to the identity matrix. You can't get that matrix by using only 1 row operation.



A matrix is invertible if it can be written as a product of elementary matrices. Since the reduced row echelon form of A is assumed to be In then there is a sequence of elementary row operations that reduces A to identity. Each of these operations is done with multipication by an elementary matrix. So there is a sequence of elementary operations E1, E2, ...,Ek such that: Ek...E2E1 = I
Almost correct. The last sentence should read: Ek...E2E1A=I.
 
Ok, thank you, I understand!

It looks different from the matrices in the answer (there's a -2 instead of 2) but since the presentation of a matrix as a product of elementary matrices isn't unique then there shouldn't be a problem. :-p
 
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