Let $A, B, C$ be $n\times n$ invertible matrices. When you simplify the expression
\[C^{-1}(AB^{-1})^{-1}(CA^{-1})^{-1}C^2,\]
which matrix do you get?
(a) $A$
(b) $C^{-1}A^{-1}BC^{-1}AC^2$
(c) $B$
(d) $C^2$
(e) $C^{-1}BC$
(f) $C$

In this problem, we use the following facts about inverse matrices: if $P, Q$ are invertible matrices, then we have
\[(PQ)^{-1}=Q^{-1}P^{-1} \text{ and } (P^{-1})^{-1}=P.\]

Using these, we simplify the given expression as follows:
\begin{align*}
&C^{-1}(AB^{-1})^{-1}(CA^{-1})^{-1}C^2\\
&=C^{-1}(B^{-1})^{-1}A^{-1}(A^{-1})^{-1}C^{-1}C^2\\
&=C^{-1}BA^{-1}AC^{-1}C^2\\
&=C^{-1}BIC\\
&=C^{-1}BC,
\end{align*}
where we used the identity $A^{-1}A=I$, identity matrix, in the third step.
Thus, the answer is (e).

Common Mistake

This is a midterm exam problem of Lienar Algebra at the Ohio State University.

Here are two common mistakes.

The first one is to use a wrong identity $(PQ)^{-1}=P^{-1}Q^{-1}$. This is WRONG.
If you use this, then most likely you chose (b).

The seocnd one is to think matrix multiplication is commutative. Even though $PQ\neq QP$ for matrices $P, Q$, some students freely cancel terms.
In this case, (c) was chosen.

Be careful, if you have $A^{-1}BA$, then in general it is not equal to $B$. You cannot cancel $A$ becuase $A$ and $A^{-1}$ are not adjacent each other.

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