I am interested in the following exercise, and I have tried to solve it with the following way. Firstly, could you please check the correctness of the given answer. Secondly, can you give an alternative answer?
$\textbf{Exercise}$: If the product $\mathbf{M} = \mathbf{ABC}$ of three square matrices is invertible, then $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$ are invertible.
$\textbf{Answer}$:
Part 1: If $\mathbf{C}$ is singular, there is $\mathbf{x} \neq 0$ such that $\mathbf{Cx} = 0 \iff \mathbf{ABCx} = 0 \iff \mathbf{Mx} = 0$. This comes in contradiction with the fact the $\mathbf{M}$ is non-singlular by the exercises definition. Thus $\mathbf{C}^{-1}$ exists. Using the last statement, we may write $\mathbf{MC}^{-1} = \mathbf{AB}$. Knowing that $\mathbf{M}$ and $\mathbf{C}^{-1}$ are invertible, we are interested to prove the invertibily of $\mathbf{MC}^{-1}$, so as to continue with similar way with the prof of matrix $\mathbf{B}$ invertibility. We may have:
$$\mathbf{J} = \mathbf{MC}^{-1} \iff \mathbf{M}^{-1}\mathbf{J} = \mathbf{C}^{-1} \iff \mathbf{CM}^{-1}\mathbf{J} = \mathbf{I}$$
This means that matrix $\mathbf{MC}^{-1}$ has a left inverse given by $\mathbf{J}^{-1} = \mathbf{CM}^{-1}$.
Part 2: Based on the last statement, and similarly with the invertibility prof we followed for $\mathbf{C}$, If $\mathbf{B}$ is singular, there is $\mathbf{x} \neq 0$ such that $\mathbf{Bx} = 0 \implies \mathbf{ABx} = 0 \implies \mathbf{MC}^{-1}\mathbf{x} = 0 \implies \mathbf{J}\mathbf{x} = 0$. This comes in contradiction with the fact the $\mathbf{J}$ is non-singlular. Thus $\mathbf{B}^{-1}$ is invertible.
Part 3: Finally, we may write $\mathbf{A} = \mathbf{MC}^{-1}\mathbf{B}^{-1} = \mathbf{J}\mathbf{B}^{-1}$. Matrix $\mathbf{A}$ is non-singular, because:
$$(\mathbf{JB}^{-1})^{-1}\mathbf{JB}^{-1} = \mathbf{I}~~\text{and}~~\mathbf{JB}^{-1}(\mathbf{JB}^{-1})^{-1} = \mathbf{I}$$
Thank you!
PS: Changes have been made taking into account users comments. I hope the post is improved.
Best Answer
Your proof is unclear when you say that $J$ is invertible because $JJ^{-1}=J^{-1}J=I$. Thats the definition of being invertible, not the proof!
As an alternative proof, note that $0\neq\det(ABC)=\det(A)\det(B)\det(C)$.
EDIT
Your proof can be fixed to be valid: $J$ is indeed invertible, and you can prove it (e.g., by writing the explicit form of $J^{-1}$ in terms of $M$ and $C$)