Matrix Inverse and transpose

Question

I got this question:

Let $A$ be a square matrix $n\times n$. Assume $A + BA = I_n$ for some matrix $B$ $n\times n$.

1. What is the inverse of $A^T$?

2. Prove that $AB = BA$.

We did not learn about $\det$ yet so if the solution should not be using it. I'm confused as to the steps needed to be made for $1$. and how to actually get the $A^T$ of a generic matrix. Also, what does the fact that $A + BA = I$ gives me?

I tried playing around with $2$ as well but I'm assuming it involves something I learned for $1$ and so I was not successful as well.

Any help is appreciated.

Answer 1

$$A+BA=I \Rightarrow A(I+B)=I$$

$1)$ The above expression means that both $A$ and $I+B$ are invertible. So,

$$A(I+B)(I+B)^{-1}=I(I+B)^{-1} \Rightarrow A=(I+B)^{-1}$$

But we also have that $(X^T)^{-1}=(X^{-1})^T$ and $(X+Y)^T=X^T+Y^T$

$$A^T=\left[(I+B)^{-1}\right]^T=\left[(I+B)^{T}\right]^{-1}=\left[I+B^T\right]^{-1} \Rightarrow \left(A^T\right)^{-1}=I+B^T$$

$2)$ We already know that $A$ is invertible, so:

$$A+AB=I \Rightarrow AB=I-A \quad(1)$$

but,

$$A^{-1}AB=A^{-1}(I-A) \Rightarrow B=A^{-1}-I \Rightarrow BA=(A^{-1}-I)A=I-A \quad(2)$$

comparing $(1)$ and $(2)$ we get $AB=BA$.

Answer 2

\begin{eqnarray*} I=A+BA=(I+B)A \end{eqnarray*} then $A^{-1}=(I+B)$