AB=AC=I $\rightarrow$ B=C Let there be $A \in F^{n*n}$

If $AB=AC=I \rightarrow B=C$ 
Because we know that every matrix have one rref (Reduced row echelon form) that mean that B and C are group of elementary actions, and due to the uniqueness of the rref of $A$ they must be the same elementary actions, therefore $B=C$
am I right?  
 A: $AB=AC=I$ means $B$ and $C$ are the inverse of $A$. By the uniqueness of inverse, $B=C$.
A: The main non-trivial question here seems to be whether $A$ is invertible. It actually is and here is an easy proof of that: $$1=\det I=\det(AB)=(\det A)(\det B).$$ In particular, $\det A\neq 0$, which is equivalent to the existence of $A^{-1}$.
Now, pre-multiply $AB=AC\,(\,=I)$ by $A^{-1}$ to conclude that $B=C\,(\,=A^{-1})$.
A: In general, thinking in terms of rref is a useful strategy, but I'm skeptical of attempting to do this here. A prior, a sequence of row operations putting $B$ into rref may not also put $C$ into rref. I'm also not sure what you mean by "the same elementary actions". 
Hint: Recall that the ring of $n \times n$ matrices is isomorphic to the ring of linear maps from $F^n$ to $F^n$. If we let $\mathcal A$, $\mathcal B$, $\mathcal C$, and $\mathcal I$ be the linear maps represented by $A$, $B$,$C$, and $I$, respectively, it's enough to show that $\mathcal A \circ \mathcal B = \mathcal A \circ \mathcal C \implies \mathcal B= \mathcal C$.
First, try to prove that $\mathcal A$ is a linear isomorphism, and therefore has two-sided inverse, $\mathcal D$. For this, note that $\mathcal A$ is surjective under our hypotheses, since $\mathcal B$ is a right inverse. Now apply the rank-nullity theorem to show $\mathcal A$ is surjective as well. It follows that there exists some $\mathcal D$ so that $\mathcal D \mathcal \circ A = \mathcal I$. Do you see how to show that $\mathcal B= \mathcal C$ now? 
