Finding an Elementary Matrix E for A and B

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Homework Statement


A= 1 2 -3 B= -1 2 0
0 1 2 0 1 2
-1 2 0 1 2 -3

Find an elementary matrix E such that EA=B

Homework Equations



None

The Attempt at a Solution


I don't know how to start this problem. Please help. Thank you.
 
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the elementary matrix that you want is gotten by taking the identity matrix and interchanging the first and third column. this emulates the interchanging of the corresponding rows in A. As we would expect elementary matrices are invertible so we can make operations on the matrix in both directions retracing our steps or continuing until we reach a certain form. in numerical linear algebra these matrices (along with certain conditions which ensure a minimum of error in the computations) are used to solve systems of linear equations. if you want to check it out look up the LU factorization.
 
ephemeral1 said:

Homework Statement



A=\begin{pmatrix}1 & 2 & -3 \\ 0 & 1 & 2 \\ -1 & 2 & 0\end{pmatrix}B=\begin{pmatrix}-1 & 2 & 0 \\ 0 & 1 & 2 \\ 1 & 2 & -3\end{pmatrix}

Find an elementary matrix E such that EA=B.

I don't know how to start this problem.
First, what are the elementary row operations? Second, how are they represented by matrices?
 
since this question is basically trivial the answer is given by (0 0 1;0 1 0; 1 0 0). plug this into the equation to verify the result. try solving a system of linear equations. the steps that you take in this process are called elementary operations. giving your problem a context i think is important.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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