for a $3\times 3$ matrix (or any $n\times n$), one way to work out $A^{-1}$ is using $$A^{-1} = \frac{1}{\det(A)}C^{T}$$ where $C^T$ is the cofactor matrix of A.
This can be extended to powers of A: $$(A^2)^{-1} = \frac{1}{\det(A^2)}C^{T}$$ where $C^T$ here is the cofactor matrix of $A^2$.
I am wondering how, if $A$, $\det(A)$ and $A^{-1}$ are already known, is there a faster way to work out the following: $(A^2)^{-1}$, $\frac{1}{\det(A^2)}$ and $C^{T}$?
ie without doing $AA$, $\det(AA)$ etc
1 $$ C^T = \det(A) \cdot A^{-1} $$ 2 $$ \det(A^n) = \det(A)^n $$ 3 $$ A^{-n} = (A^{n})^{-1} = (A^{-1})^{n} $$