The question is as follows:
Suppose $\textbf{A}$, $\textbf{B}$, and $\textbf{C}$ are square matrices such that $\textbf{ABC}=\textbf{I}$. Which of the following statements are true?
- $\textbf{C}^{-1} = \textbf{AB}$
- $\textbf{C} = \textbf{B}^{-1}\textbf{A}^{-1}$
- $\textbf{A}$ is invertible.
- $\textbf{C} = \textbf{A}^{-1}\textbf{B}^{-1}$
- $\textbf{C}^{-1} = \textbf{A}^{-1}\textbf{B}^{-1}$
So far I've determined that 1, 2, and 3 are the only statements that are true:
$$\begin{align*}
\textbf{ABC} &= \textbf{I}\\
\textbf{ABC}\textbf{C}^{-1} &= \textbf{I}\textbf{C}^{-1}\\
\textbf{AB} &= \textbf{C}^{-1}\\
\therefore \textbf{C}^{-1} &= \textbf{AB}
\end{align*}$$
Thus, statement 1 is true.
2.
$$\begin{align*}
\textbf{ABC} &= \textbf{I}\\
\textbf{B}^{-1}\textbf{A}^{-1}\textbf{ABC} &= \textbf{B}^{-1}\textbf{A}^{-1}\textbf{I}\\
\therefore\textbf{C} &= \textbf{B}^{-1}\textbf{A}^{-1}\\
\end{align*}$$
Thus, statement 2 is true.
3.
$$\begin{align*}
\textbf{ABC} &= \textbf{I}\\
\textbf{ABC}\textbf{C}^{-1}\textbf{B}^{-1} &= \textbf{I}\textbf{C}^{-1}\textbf{B}^{-1}\\
\textbf{A} &= \textbf{C}^{-1}\textbf{B}^{-1}
\end{align*}$$
Assuming $\textbf{A}$ is invertible,
$$\begin{align*}
\textbf{ABC} &= \textbf{I}\\
\textbf{BCA} &= \textbf{I}\\
\textbf{BC}\textbf{C}^{-1}\textbf{B}^{-1} &= \textbf{I}\\
\textbf{I} &= \textbf{I}
\end{align*}$$
Thus, statement 3 is true.
4.
False due to statement 2.
5.
False due to statement 1.
Is there anything I've missed in terms of calculations or matrix properties? Any form of clarification would be greatly appreciated.
Best Answer
Careful: you're assuming that the inverses of the matrices exist. In question 1, for example, you've multiplied on the right by $C^{-1}$. How do you know it exists?
Instead, use the definition of inverse. Given a square matrix $M$, if $MN = I$, then we say that $N$ is the inverse of $M$, and we can call it $M^{-1}$. In question 1, you have $ABC=I$. Writing this as $(AB)C=I$, we see that $C^{-1} = AB$ by definition.
This problem stands out especially in question 3, where your argument is: "Assuming $A$ is invertible, then $A$ is invertible". Do you see why this doesn't work? Much simpler: $A$ is invertible because its inverse, again by definition, is $BC$.
For questions 4 and 5, it is incorrect to assume that $B^{-1}A^{-1} \neq A^{-1}B^{-1}$. It's true that they will be different in general, but your argument is insufficient. (As @openspace points out in a comment above, you might have $A=B=C=I$.) Instead, find a specific counterexample where the two sides are different.