$A^2$ means to multiple the matrix by itself, and $A^{-1}$ refers to the matrix's inverse. Would $A^{-2}$ be the square of the inverse or the inverse of the square?
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$\begingroup$It means $(A^{-1})^2$, so the square of the inverse, which also happens to be the inverse of the square.
$\endgroup$ 3 $\begingroup$$$\left(A^{-1}\right)^2A^2=\left(A^{-1}A^{-1}\right)(AA) = A^{-1}\left(A^{-1}A\right)A=A^{-1}IA=A^{-1}A=I $$which says that $$\left(A^{-1}\right)^2=\left(A^2\right)^{-1} $$This is essentially the same as the proof that $$\left({1\over x}\right)^2={1\over x^2}$$
$\endgroup$ $\begingroup$It can be shown, via induction, that $(A^{-1})^n=(A^n)^{-1}$, $\forall n \in \mathbb{N}$. Thus, whenever $A$ is invertible, $A^{-n}:= (A^{-1})^n=(A^n)^{-1}$.
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