Let $A$ be a real $n \times n$ matrix. Denote by $\operatorname{cof} A$ The cofactor matrix of $A$. By definition, $A (\operatorname{cof} A)^T=\det A \cdot I$.

Thus, it is immediate that $A \in \operatorname{SO}_n$ if and only if $$ (**) \operatorname{cof} A =A,\det A =1$$

However, if $n \neq 2$ the condition on the determinant is superfluous:

$ \operatorname{cof} A =A \Rightarrow AA^T=\det A \cdot I \Rightarrow (\det A)^2=(\det A)^n \Rightarrow \det A \in \{1,-1\}$. However, Since $\det A \cdot I=AA^T$ is positive semidefinite $\det A \ge 0$ so $\det A = 1$ and $A \in \operatorname{SO}_n$.

For the case $n=2$, an easy calculation shows $\operatorname{cof} A=A$ if and only if $A$ is a **scaled rotation**.

**Question:**

While the above derivations are easy to do algebraically, I would like to find a more *geometric* explanation of these results. I think this amounts to obtaining a better geometric interpretation for the cofactor matrix. (I know it measure in some sense the volume of $n-1$ dimensional parallelepiped, see here).

In particular,

**Is there any geometric intution behind the condition $A \in\operatorname{SO}_n \iff (**) \operatorname{cof} A =A,\det A =1$?**

**Is there any explanation for why dimension $2$ is special?**

Note: The condition $(**)$ for characterizing matrices in $\operatorname{SO}_n$ is not a mere game. In some contexts this is the only way to show some transformations are indeed isometries. (For instance in proofs of Reshetnyak’s rigidity theorem).