Proving that a matrix is invertible without using determinants 
Prove if $A$, $B$, and $C$ are square matrices and $ABC = I$, then $B$ is invertible and  $B^{-1}= CA$.

I know that this proof can be done by taking the determinant of $ABC=I$ and showing that $A$, $B$, and $C$ are invertible and then finding the inverse of $B$. However, in this chapter of the book, we have not yet learned determinants so I would like to solve the problem without determinants. My proof method involves using a contradiction and is as follows:
Assume ${C^{-1}}$ does not exist, then $\exists$ $x$ $\neq$ $0$ such that $Cx = 0$.
$ABCx =Ix$
$AB0 = x$
$0=x$, which is a contradiction since we know that $x$ $\neq$ $0$, and therefore ${C^{-1}}$ exists.
$AB$${C^{-1}}$ $=I$${C^{-1}}$
$AB=$${C^{-1}}$
WLOG, B is invertible
$CAB =C$${C^{-1}}$
$CAB = I$
$CAB$${B^{-1}}$ $=I$${B^{-1}}$
$CA=$${B^{-1}}$
My question is if it is correct to assume ${C^{-1}}$ does not exist since the proof does not mention anything about ${C^{-1}}$ existing or not.
 A: A square matrix is invertible if and only if its rank is $n$.
Also, we know that $\operatorname{rank}(AB) \le \min(\operatorname{rank}(A),\operatorname{rank}(B) )$
In this question
$$ABC=I$$
Hence $\operatorname{rank}(ABC)=n$
$$n \le \min(\operatorname{rank}(A),\operatorname{rank}(B), \operatorname{rank}(C) )$$
Hence $\operatorname{rank}(A)=\operatorname{rank}(B) =\operatorname{rank}(C)=n$ and they are all invertible.
Hence $B=A^{-1}C^{-1}$ and $B^{-1}=(A^{-1}C^{-1})^{-1}=CA$
A: Since, $ABC = I$, then $A$ is invertible (see below). Same with $C$ since $(AB)C=I$. The inverse of $C$ is $AB$. So, $B = A^{-1}C^{-1}$ and $BCA= I$. So, the inverse of $B$ is $CA$.
Proof-sketch of AB=I $\implies$ A and B are invertible and they are inverses of each other
Consider the linear transformations induced by matrices $A$ and $B$ on $\mathbb R^n$. For, convenience I will represent the corresponding linear transformations by $A$ and $B$ as well. Since $AB=I$, it implies the composition of linear transformations $A$ and $B$ is the identity linear transformation. So, $B$ is an injective linear transformation and $A$ is a surjective linear transformation. Since, we are in a finite-dimensional vector space this implies both $A$ and $B$ are invertible linear transformation (by rank-nullity theorem). So, $A$ and $B$ are invertible as matrices. Also not that since $AB=I$, we get $BABA=BA$. Now by multiplying inverse of $A$ and $B$ from right on both sides, we get $BA=I$ as well. Now, from the definition inverse of matrices, we get $A$ and $B$ are inverses of each other.
A: It can be shown, via elementary means, that if $M$ and $N$ are square matrices such that $MN = I$, then $NM= I$.
Thus, if $ABC = A(BC) = I$, then $(BC)A = B(CA) = I$, i.e., the matrix $B$ is invertible and $B^{-1}=CA$.
A: A square matrix $A$ is invertible if and only if there is another matrix $A^{-1}$ such that $A^{-1}A=I$. In the express$ABC=I$, chose $X=AB$ and we have $XC=I$. Thus $C^{-1}=X$. Similarly show $A$ is invertible. Now,
$$ABC=I$$
$$AB=C^{-1}$$
$$CAB=I$$
$$CA=B^{-1}$$
