Adjugate(classical adjoint) of a matrix https://en.m.wikipedia.org/wiki/Adjugate_matrix is the inverse rotation without scaling. When a matrix is multipled by its inverse, off diagonal entries are zero and diagonal entries have the value scaled by matrix.
But according to the Matt's answer in Geometric interpretation of $\det(A^T) = \det(A)$ , transpose too is stated as inverse rotation without scaling. So adjugate is equal to transpose? Which is clearly wrong. Where am I going wrong?
Let $A^*$ be the adjoint of $A$. The defining property of the adjoint is that it satisfies $\langle Ax,y\rangle = \langle x,A^*y\rangle$ for some inner product. For a real vector space this will be the transpose however for a complex vector space it will be the conjugate transpose.
To see this let’s restrict ourselves to real vector spaces with $x$ and $y$ column vectors so that we can represent the Euclidean inner product with matrix multiplication as $\langle x,y \rangle = x^Ty$. This means we have $$\langle Ax,y\rangle = (Ax)^Ty = x^TA^Ty=\langle x , A^Ty\rangle $$
so the adjoint is the transpose. The complex case can be proved in the same way although the geometric meaning is less clear. Note that there is a nice relationship between the inner product, its norm, and the metric it induces. This allows the geometric interpretation given to the adjoint since the inner product is directly related to distances and thus scaling.
