Given $B$ and $A^mBA^n=I$ determine whether $A$ is invertible and find $A^{m+n}$.

Question

Let $m,n\in\Bbb N$ and let $A$ be a matrix such that $$A^mBA^n=I$$ where $$B=\begin{pmatrix}1&-1&3&1\\ 1&1&2&1\\ 2&-1&3&2\\ -1&-2&1&2 \end{pmatrix}$$

$1)$ Is $A$ invertible?

$2)$ Determine $A^{m+n}$.

I'm completely lost...Okay, I know that I need to see whether $\det A=0$ for the first question, and for the second one I know $A^{m+n}=A^m\cdot A^n$. But how do I even start here? I would really appreciate some hint, thanks!

Answer 1

$A$ is invertible for a bunch of reasons, for example because the rank of a product of matrices is bounded by the rank of the factors, so $4\geq \text{rank}(A)\geq 4$. Another way to see it is that the determinant of a product of matrices is the determinant of the products, we conclude $\det(A)\neq 0$ and so $A$ is invertible.

Using this:

$A^mBA^n=I\implies A^mB=A^{-n}\implies B=A^{-m}A^{-n} =A^{-m-n}\implies A^{m+n}=B^{-1}$

Answer 2

1)It is, and its inverse is $A^m\dot{}B\dot{}A^{n-1}$

2)well explained by Jorge Fernández