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Suppose that $A$ is invertible $n \times n$ matrices. Then using the property $A~\mathrm{adj}(A) = \det(A)I_n$, prove that $\det(\mathrm{adj}(A^3)) = (\det(A))^{3n-3}$

I started with.

$$ A~\mathrm{adj}(A) = \det(A)I_n $$

I took $\det()$ of both sides, then raised both side to power of $3$, then I multiplied both sides by $1/\det(A^3)$, then I use the rule $(\mathrm{adj}(A))^n = \mathrm{adj}(A^n)$ on the left side, and combine exponents on the right side to complete the proof that

$$ \det(\mathrm{adj}(A^3)) = (\det(A))^{3n-3} $$

the only thing I wanted to confirm if i did this right is if $(\mathrm{adj}(A))^n = \mathrm{adj}(A^n)$?

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Yes (link). Also, as $A^3$ is also invertible, we can start with:$$A^3{\rm adj}(A^3)={\rm det}(A^3)I_n.$$ Then, taking determinant on both sides: $$ {\rm det}(A^3){\rm det}({\rm adj}(A^3))= ({\rm det}(A^3))^n.$$

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