What can we do to matrices that we can do to the regular good ol' numbers? I am a high school student and my formal Linear Algebra education consisted merely of the definition of matrices as a list of numbers and then some random properties. While reading the BetterExplained article on Linear Algebra and some of 3Blue1Brown's videos on the same did help, I still often face difficulties while solving problems and when I look up the solutions, all I'm thinking is "Wait. You're allowed to do that to matrices too?"
So my question is - what can we do to these list of numbers that we can do to individual real and complex numbers?
To clarify, I do know that we can add matrices by adding their individual elements together and multiply matrices in a row-column order and stuff like that. My doubts are along the lines of the ones listed below:

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*Does $AB = CD$ imply that $B^{-1}A^{-1}=D^{-1}C^{-1}$ ? (where A, B, C and D are 4 non-singular matrices of appropriate orders)

*Is $A \times A^n = A^n \times A$ valid? (where A is a matrix and n is a natural number)

*Is multiplication of matrices commutative only when a matrix is being multiplied by a null matrix or unit matrix of appropriate order?

*Is the inverse of the matrix $A^n$ the same as the inverse of $A$ multiplied $n$ times to itself?

I am not asking for proofs of the problems listed above - instead, I would greatly appreciate it if someone who has noticed some "patterns" in the doubts listed above would point me to some resource that I can study to clear all such doubts in my conceptual understanding. Alternatively, how should I approach matrix multiplication and their inverses in general that would solve these and other similar problems that I may be having?
Thank you.

Edit: These issues have been solved adequately and then some by Dave L. Renfro's comments on this question.
 A: Everything that follows is under that assumption that we work over square matrices only. For non-square matrices some things are no longer well defined, e.g. there are no two-sided inverses (and one-sided are never unique), you can't multiply a matrix by itself, and so on.


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*Does $AB = CD$ imply that $B^{-1}A^{-1}=D^{-1}C^{-1}$ ? (where A, B, C and D are 4 non-singular matrices of appropriate orders)


Yes. This follows from two general rules (taken from group theory): (1) that $(gh)^{-1}=h^{-1}g^{-1}$ and (2) that inverses are unique.



*Is $A \times A^n = A^n \times A$ valid? (where A is a matrix and n is a natural number)


I assume that by "$\times$" you mean matrix multiplication. The answer is yes, these are the same. This follows from the fact that matrix multiplication is associative. By applying induction on $n$. Also to solve that you need a proper definition of $A^n$ which formally is defined recursively as: $A^0:=I$ is the identity matrix (diagonal matrix with $1$s on the diagonal) and $A^n:=AA^{n-1}$ for $n\geq 1$. With that the induction step is quite simple:
$$AA^n=A(AA^{n-1})=A(A^{n-1}A)=(AA^{n-1})A=A^nA$$
The initial step for $n=0$ is the trivial $AI=IA$ equation.



*Is multiplication of matrices commutative only when a matrix is being multiplied by a null matrix or unit matrix of appropriate order?


Clearly $AB=BA$ whenever $A=B$, regardless of what matrix you pick.
This is however true under some stronger assumption, namely: if $A$ is a square matrix such that $AB=BA$ for any other square matrix $B$ then $A=\lambda I$ where $I$ the identity matrix and $\lambda$ is a scalar (at least over fields). Or in other words when $A$ is a diagonal matrix with a single fixed value on the diagonal. This is not a trivial result though, requires some work.



*Is the inverse of the matrix $A^n$ the same as the inverse of $A$ multiplied $n$ times to itself?


Yes. By induction on $n$ and the fact that inverses are unique. Indeed, write down that $B:=(A^n)^{-1}$. Then $I=A^nB=A^{n-1}AB$ and thus $AB$ is the inverse for $A^{n-1}$. By the induction step and uniqueness of inverses this means that $AB=(A^{-1})^{n-1}$ and thus by multiplying by $A^{-1}$ on the left (note that $A$ is invertible) we get $B=(A^{-1})^{n}$.
The case $n=0$ is trivial since $I^{-1}=I$.
