If $A$ is an invertible $3\times 3$ real matrix, then $\det A^\star = \det^2A$ 
How do I show that if a matrix $A \in \mathcal{M_{3}}(\mathbb{R})$ has a inverse, then $\det A^{\star}=\det^2A$?

I was trying to use that $A^{\star}= \det(A)\cdot A^{-1}$. Then what and why? If I go to det in that equality why should I get $(\det(A))^3\cdot \det(A^{-1})$?
 A: So, you have $\mathrm{adj}(A)=\det(A)\cdot A^{-1}$. Thus, $$\det(\mathrm{adj}(A))=\det(\det(A)\cdot A^{-1})=\det(A)^3\det(A^{-1})$$
as in general $\det(\alpha X)=\alpha^n\det(X)$ for $X\in\mathbb{R}^{(n,n)}$.
Now, $1=\det(E_3)=\det(AA^{-1})=\det(A)\det(A^{-1})$, i.e. $\det(A^{-1})=\det(A)^{-1}$ where $E_3$ is the $3\times 3$-unit matrix. Thus, 
$$\mathrm{adj}(A)=\det(A)^3\det(A)^{-1}=\det(A)^2$$

Note that, since $\det(\alpha X)=\alpha^n\det(X)$ for $X\in\mathbb{R}^{(n,n)}$ generally holds, you may generalize this to $n\times n$-matrices, i.e. showing for invertible $A\in\mathbb{R}^{(n,n)}$, that $$\det(\mathrm{adj}(A))=\det(\det(A)A^{-1})=\det(A)^n\det(A^{-1})=\det(A)^{n-1}$$

On how to derive $\det(\alpha X)=\alpha^n\det(X)$ for $n$-dimensional square matrix $X$, you can observe that $\alpha X=\alpha E_nX$ and followingly $\det(\alpha X)=\det(\alpha E_nX)=\det(\alpha E_n)\det(X)$. Now, the result follows by considering $\det(\alpha E_n)$.
A: We know that $AA^\star = A^\star A = (\det A) I$. Taking the determinant gives
$$(\det A)(\det A^\star) = \det (AA^\star) = \det \Big[(\det A) I\Big] = (\det A)^3 (\det I) = (\det A)^3$$
because for any  $n\times n$ matrix $B$ we have $\det (\lambda B) = \lambda^n \det (B)$. 
$A$ is invertible so $\det A \ne 0$. Hence we can divide the above relation by $\det A$ to obtain $\det A^\star = (\det A)^2$.
