How should we proof this? Given two row equivalent matrices A and B, then there exists a matrix C such that A=CB.
The focus should be the general existence of a matrix C.
Row Equivalence is defined as:
Definition 1.6. Two matrices are row equivalent if one can obtain the other matrix by any sequence of row operations.
Elementary Matrix is defined as:
An elementary matrix is a square matrix which can be de- scribed as a single row operation on the identity matrix $I_n$
Best Answer
If two matrices are row equivalent then by definition one matrix can be converted to the other through a series of elementary row operations. I don't know how far you are in your linear algebra studies, but every elementary row operation can be represented by an "elementary matrix".
If $E$ is the elementary matrix representing an elementary row operation, then the product $EA$ is equal to that elementary row operation applied to the matrix $A$. In order to find the elementary matrix for a given elementary row operation, all you need to do is apply the elementary row operation to the $n\times n$ identity matrix, where $n$ is the number of rows of matrix $A$.
For example, if A is a matrix with 3 rows, the elementary matrix representing the row operation of multiplying the second row of the matrix by 3 is the matrix $\pmatrix{1 & 0 & 0 \\ 0 & 3& 0 \\ 0 & 0 & 1}$, and the elementary matrix representing the interchanging of rows 1 and 3 is $\pmatrix{0 & 0 & 1 \\ 0 & 1& 0 \\ 1 & 0 & 0}$.
If $A$ and $B$ are row equivalent matrices, then by definition $A=E_k E_{k-1}E_{k-2}... E_3 E_2 E_1 B$, where all $E_i$ are various elementary matrices representing the elementary row operations used to get from $B$ to $A$. By associativity of matrix multiplication, this means $A=(E_k E_{k-1}E_{k-2}... E_3 E_2 E_1)B$, so $C=E_k E_{k-1}E_{k-2}... E_3 E_2 E_1$.