# Product of Elementary Matrices

1. Oct 14, 2008

### ae4jm

1. The problem statement, all variables and given/known data
Write the given permutation matrix as a product of elementary (row interchange) matrices.

\begin{array}{ccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array}

3. The attempt at a solution

I found the row echelon form to be the identity matrix \begin{array}{ccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} by R3 to R1, R2 to R3, and R3 to R4.

I have not been able to get the answer that the book gets, which is
\begin{array}{ccc} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \begin{array}{ccc} 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{array} \begin{array}{ccc} 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{array}
Any help is greatly appreciated.

2. Oct 14, 2008

### jacobrhcp

Your book probably has a theorem like this in it:

'if A is a nxm matrix, and E is an mxm elementary matrix (a matrix that is obtained from the identity matrix by only one operation), then multiplication on the left by E is the same operation on A as the operation that was used on the identity matrix to obtain E.'

So if you can get A to the identity by reducing, you should be able now to write it as a product of elementary matrices, which can be found by letting each of your operation work on the identity to get all different E's. Put them behind eachother and let them work on the identity, and you should be done (pay attention to the order of your product of E's now, though... matrices in general do not commute)

this is probably almost directly the answer in your book, but I have a hard time reading it. If it's not, then they probably just got A to the identity in a different order of reducing. And if you paid attention closely, chances are both your answers are right.

3. Oct 14, 2008

### ae4jm

The given matrix is
0100
0001
1000
0010
then R3<->R1 1000
0001
0100
0010
then R2<->R3 1000
0100
0001
0010
then R3<->R4 1000
0100
0010
0001 And now in REF
The answer in the book is the three matrices

0100 0001 0010
1000 0100 0100
0010 0010 1000
0001 1000 0001

I get the three matrices below because I first got the given to a row echelon form and then I reversed the row interchanges from above to get.

R3<->R4 R2<->R3 R3<->R1
1000 1000 0010
0100 0010 0100
0001 0100 1000
0010 0001 0001

Is this still correct? Thanks a lot for your help and sorry for the confusion from above. I wasn't using the LATEX properly.

4. Oct 14, 2008

### jacobrhcp

you certainly have the right idea, and I'm pretty sure it's right. But if you want to be extremely sure, you can of course just multiply the matrices!

if En...E2E1(I) = A, then its not hard to see and find out!

And if you're really persistent, you can even check if the operations the book gave also add up to the identity.

5. Oct 14, 2008

### ae4jm

The book's answer is correct, but mine is not. I need to reverse the order of my three matrices. After rearranging, I get the correct answer.