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**Pinkk** My professor states in one instance that if $\displaystyle A$ is invertible, then $\displaystyle B=A^{-1}$ when $\displaystyle AB=BA=I$, leading to $\displaystyle AA^{-1}=A^{-1}A=I$. Yet later on in class, during a proof involving the multiplicity of invertible matrices, he makes the claim that $\displaystyle B=A^{-1}$ implies that $\displaystyle B$ is invertible when $\displaystyle BA=AB=I$, leading to $\displaystyle A^{-1}A=AA^{-1}=I$. I understand this theorem fairly easily, but my professor made a big deal about whether the order of the equation is $\displaystyle AB=BA=I$ or $\displaystyle BA=AB=I$, yet he clearly contradicts himself if order is so important. A similar is in once instance, $\displaystyle B=A^{-1}$ leads to $\displaystyle AB=BA=I$, and then later on states that it is $\displaystyle A=B^{-1}$ that leads to $\displaystyle AB=BA=I$. Again, I understand fairly easily that is a matrix is an inverse of another matrix, then the inverse of that first matrix is the other matrix, but this professor is pedantic.

So I need clarification on the theorem. If $\displaystyle AB=BA=I$ then $\displaystyle A$ is invertible and $\displaystyle B=A^{-1}$, or does $\displaystyle B=A^{-1}$ imply $\displaystyle B$ is invertible and therefore $\displaystyle BA=AB=I$. My professor is a real hardass about following the proofs EXACTLY as they are stated, but like I said, the two claims that he made seem to contradict themselves if order is so important.