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If matrix $A$ is invertible, then the inverse is unique. Perhaps the simplest proof is the following: $$B = BI = B(AC) = (BA)C = IC = C,$$

if $B,C$ are any pair of matrices that satisfies $AB=BA=AC=CA=I.$

When I was grading an exam for introductory linear algebra class, I saw many "proofs" that goes like this: Assuming $AB=AC=I$, for some matrices $B,C$, we get $$A^{-1}AB = A^{-1}AC\Rightarrow B=C.$$

Now I have a problem with this proof because when the student is multiplying both sides by $A^{-1}$, the equality might not be preserved since we are not assuming its uniqueness. However, this approach is easily salvaged by choosing $A^{-1}$ to be either $B$ or $C$.

Therefore, I deducted points from students who did not specify what $A^{-1}$ exactly is. Hence, I am wondering if it is right to consider the above as an incomplete proof and I want to hear opinions of more experienced professors/teachers.

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    Actually the equality is preserved, but the notation is ambiguous, as no canonical $A^{-1}$ exists a priori. So I would subtract points, but for this reason (i.e that it is not well-defined) – 2017-02-23
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    The proof is ok if formulated "Assume $AB=AC=I$ and $A^{-1}A=I$" (where $A^{-1}$ may be one of $B,c$ or a completely different thing. In other words, we prove: *If there exists at least one left-inverse, then there exists at most one right inverse* – 2017-02-23
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    Yes, I subtracted points only because they had not said anything about $A^{-1}$. I was just not completely sure because I am fairly new to teaching math to non-math major students in the US. – 2017-02-23

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