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I am trying to write down a formal proof.

Attempt: Firstly, we settle on the notations and considerations:

$A$, $B$, $C$ are all $m\times n$ matrices where $m$ is fixed and $n$ is arbitrary over the ground field $F$.

$e_1^r(A): \text{ multiplication of the $r^{th}$ row of A by any $0\neq c \in F$}$

$e_2^r(A): \text{ $r^{th} $ row+$c.$$s^{th}$ row, any $c\in F, r\neq s$}$

$e_3^r(A): \text{ Interchange of $r^{th}$ and $s^{th}$ rows}$.

Let the inverses of these elementary row operation be denoted by $ f_1^r(A),f_2^r(A),f_3^r(A)$ respectively, all of which are themselves elementary row operations.

We wish to prove the symmetry of the relation by the method of Induction.

$P(n):$ If $B$ is obtained from $A$ after any $n$ row operations, then $A$ can be obtained from $B$ by $n$ elementary row operations.

$P(1):$ For any $e_u^r(A)=B\implies f_u^r(e^r_u(A))=f_u^r(B)\implies A=f_u^r(B) $, $u \in \{1,2,3\}$

Let $P(m)$ hold.

To prove $P(m+1)$ holds: Let $e_{u_{m+1}}^{r_{m+1}}(...e_{u_2}^{r_2}(e_{u_1}^{r_1}(A))...)=B\implies f_{u_{m+1}}^{r_{m+1}}( e_{u_{m+1}}^{r_{m+1}}(...e_{u_2}^{r_2}(e_{u_1}^{r_1}(A))...))= f_{u_{m+1}}^{r_{m+1}}(B)= e_{u_{m}}^{r_{m}}(...e_{u_2}^{r_2}(e_{u_1}^{r_1}(A))...) $.

Now, $P(m)$ being true, $P(m+1)$ is true as well.

The reflex and transitive properties are evident.[ For transitive, $B=$ $s$ row operations on $A$, $C=$ $t $ row operations on $A$. Composition of mappings is associative, so $C= t$ row operations $((s$ row operations on $A))$.] For reflexive, we take $e_1^1(A)$, set $c=1$.

Is the proof correct? Please verify.

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