Let $A$ be a real $n \times n$ matrix. Denote by $\operatorname{cof} A$ The [cofactor matrix](https://en.wikipedia.org/wiki/Adjugate_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](https://math.stackexchange.com/a/594513/104576)).


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).



What is special in dimension $2$ (When characterizing isometries using the cofactor matrix)?
This is somewhat informal. My argument is in the spirit of dimensional analysis in physics.


The essential point is that the cofactor matrix encapsulates $n-1$ dimensional volume, while the dimension of the matrix $A$ in this sense is $1$. (Think the of operator norm, for instance. The "size" of the elements of $A$ roughly corresponds to dilation of length, i.e to *one-dimensional* volume).


Thus, if $n-1 \neq 1$ it is very unlikely that equality like $A=\operatorname{Cof} A$ will hold, since it compares volumes of different "scales". Indeed, it turns out the only case where this happens is when $A$ is an isometry, i.e *there is no difference* between different scales\dimensions (think on the toy example $1^d=1$).


Of course, for $n=2$ the incompatibility of scales disappears, so it's reasonable to expect for more instances of $A=\operatorname{Cof} A$, other than the rigid ones. 


Naturally, this heuristic argument does not explain why the scaled rotations are the only ones which stand out, but this is probably related to some specific details regarding *exactly how* the cofactor encodes what it encodes...


