If $B$ the inverse matrix of $A^2$ show that the inverse of $A$ is $AB$ How do I continue from $A(AB) = (BA)A = I$ and how can we justify it if we don't know that $AB=BA$?
EDIT: Also, how can we prove that if $AB=I$ then $ BA = I$? This is seperate from the question above but I felt it didn't need a new post.
 A: The thing is that for matrices over a field, $XY=I_n$ implies $YX=I_n$. You've already seen that $A$ is a left inverse for $AB$, and that implies it is also a right inverse (the inverse) for $AB$.
Added:
For a complete explanation of this, check out this question: If $AB = I$ then $BA = I$

Another way to do it would be completely with determinants, if you're willing. Supposing you are happy to conclude a nonzero determinant means a matrix is invertible, then $\det(A^2B)=\det(A)^2\det(B)=1$ should convince you $\det(AB)\neq 0$, so that $AB$ is invertible. Thus the one sided inverse $A$ is twosided.
A: As
$A^2B = I$
we can write
$B = A^{-2}$
whence
$AB = AA^{-2} = A^{-1}$
and
$BA = A^{-2}A = A^{-1}$
A: This does not depend on $AB=I$ and $BA=I$ being equivalent, as is true for finite matrices, but not for linear maps in general. It is a pure categorical fact that if $a,b$ are arrows $X\to X$ for some object $X$ of a category $\mathcal C$ (one cannot have two objects involved if $a\circ a$ is to make sense), and $b$ is the inverse of $a\circ a$ (that is, $b\circ a\circ a=1_X=a\circ a\circ b$), then $a$ is the inverse of $a\circ b$.
As noted in the question the $a\circ a\circ b=1_X$ part is given, so consider $a\circ b\circ a$. All one needs to do is append the given cycle of arrows and regroup:
$$
  a\circ b\circ a = 1_X\circ a\circ b\circ a 
 = b\circ a\circ a\circ a\circ b\circ a 
 = b\circ a\circ 1_X\circ a=b\circ a\circ a=1_X.
$$
A: $A(AB)=(AA)B=A^2 B=I$ and $(BA)A=B(AA)=BA^2=I$
