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.
Best Answer
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 $B^{-1}=CA$.