If $A$ and $B$ are invertible matrices, then $A^2B^{-1}$ is invertible. Let $A$ be an $n \times n$ matrix and $B$ be an $n \times n$ matrix. I know:
$$
\begin{align*}
AA^{-1}=A^{-1}A=I_n \tag{1}\\
BB^{-1}=B^{-1}B=I_n \tag{2}
\end{align*}
$$
Starting from $B^{-1}B=I_n$, I have the following series of equalities:
$$
\begin{align*}
A^2B^{-1}B=A^2I_n \tag{Left multiplying by $A^2$}\\
A^2BB^{-1}=A^2(BB^{-1}) \tag{Replacing with (2)}\\
\vdots
\end{align*}
$$

*

*Is this proof heading in the right direction ?

*Do I need to assume:

*

*$A$ and $B$ are square matrices, and

*They have the same number of rows and columns ?



 A: Perhaps you are overcomplicating things. It suffices to find a multiplicative inverse, and $BA^{-2}$ will do, as you can easily check.
To answer your other questions, the matrices must be square because else the notion of invertibility does not even make sense.
If you know about determinants, you can just note that $A\in M_{n\times n}(k)$ is invertible if and only $\det(A)\ne 0$. As a consequence, $\det(A)\ne 0$ and $\det(B)\ne 0$. So,
$$ \det(A^2B^{-1})=\frac{\det(A)^2}{\det B}\ne 0$$
and you are done.
A: An invertible matrix is a square matrix by definition. They also have to be of the same order, otherwise you wouldn't be able to multiply them. Now, what do you need to do with $A^2B^{-1}$ to get the identity matrix? First multiply by the inverse of $B^{-1}$ from the right side to cancel $B$, then multiply the result by $A^{-2}$ from the right side to cancel $A^2$. In other words:
$A^2B^{-1}(BA^{-2})=I$
And I believe you know that if a matrix has a right inverse then this inverse works from the left side as well. (if you don't know it, just multiply by $BA^{-2}$ from the left side and check you again get the identity matrix). So $A^2B^{-1}$ is invertible, $BA^{-2}$ is the inverse.
In general, there is a simple rule: if $A_1,...,A_k$ are invertible matrices of order $n$ then their product is invertible and $(A_1...A_k)^{-1}=A_k^{-1}...A_1^{-1}$, the product of inverses in the opposite order. This follows from direct computation.
A: Yes you have to assume that otherwise  the product makes no sense. Then $A^2B^{-1}\cdot BA^{-2}$ is $I$ so your product has an inverse.
