Proving that a matrix is invertible without using determinants

It can be shown, via elementary means, that if $M$ and $N$ are square matrices such that $MN = I$, then $NM= I$.

Thus, if $ABC = A(BC) = I$, then $(BC)A = B(CA) = I$, which shows that $B$ is invertible and $B^{-1}=CA$.

A square matrix $A$ is invertible if and only if there is another matrix $A^{-1}$ such that $A^{-1}A=I$. In the express$ABC=I$, chose $X=AB$ and we have $XC=I$. Thus $C^{-1}=X$. Similarly show $A$ is invertible. Now, $$ABC=I$$ $$AB=C^{-1}$$ $$CAB=I$$ $$CA=B^{-1}$$

A square matrix is invertible if and only if its rank is $n$.

Also, we know that $\operatorname{rank}(AB) \le \min(\operatorname{rank}(A),\operatorname{rank}(B) )$

In this question

$$ABC=I$$

Hence $\operatorname{rank}(ABC)=n$

$$n \le \min(\operatorname{rank}(A),\operatorname{rank}(B), \operatorname{rank}(C) )$$

Hence $\operatorname{rank}(A)=\operatorname{rank}(B) =\operatorname{rank}(C)=n$ and they are all invertible.

Hence $B=A^{-1}C^{-1}$ and $B^{-1}=(A^{-1}C^{-1})^{-1}=CA$