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I like the way there a physical meaning tied to the determinant as being related to the geometric volume. Since the determinant can be calculated through Laplace's formula where the cofactor matrix is involved, I'm wondering what would be the physical/geometric meaning of the cofactor matrix? What about the adjugate matrix?

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  • $\begingroup$ I think this might require a long line of reasoning, but it's based on even simpler logic than the Jacobian matrix. I personally find that it's adequate to start with the understanding that the determinant is just the volume of the hyper-parallelipiped. The reason transposition is involved in the adjugate matrix is because its product with the original matrix would represent dot products. $\endgroup$ – Tunococ Feb 7 '16 at 2:31
  • $\begingroup$ @Tunococ Based on your comment, I'm going to edit my post. I agree, I should have asked about determinants in general $\endgroup$ – ThatsRightJack Feb 7 '16 at 2:37
  • $\begingroup$ I'm not sure if this would help. Then using the same geometric understanding, Cramer's rule follows easily. You can then think of the adjugate matrix as an application of Cramer's rule. $\endgroup$ – Tunococ Feb 7 '16 at 3:23
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As you may know, an interpretation could be as vectors and volumes in the dual space.

But it is not very physically meaningful...

I will give here a geometrical explanation in $\mathbb{R^3}$.

Let $u,v,w \in \mathbb{R^3}$. Call $A$ the matrix with these columns $u,v,w$.

Imagine the parallelotope $P$ generated by $u,v,w \in \mathbb{R}^3$.

The volume of $P$ is precisely $det(A)$.

Consider the cross products $a = u \times v$, $b = v \times w$ and $c = w \times u$.

Let us call $B$ the matrix with these columns $a,b,c$.

Geometrical view: Vectors $a,b,c$ are normal to the faces $(u,v)$, $(v,w)$ and $(w,u)$.

They generate in turn a parallelotope $P'$.

We are going to see that this parallelope $P'$ is a way to "visualize" the inverse of $A$

(I prefer to use the inverse matrix formulation instead of the cofactor matrix or adjugate matrix).

Indeed, by considering the products of lines of $A^T$ with columns of $B$, one gets

$A^TB=k I_3 \ \ \ (1)$

where k is the triple product $[u,v,w]$ which is nothing else than $det(A)$.

Taking determinants on both sides of (1), we obtain $det(A) det(B) = det(A)^3$, from which

$det(B)=det(A)^2$.

In words, the volume of the parallelotope $P'$ generated by $a = u \times v, b=v \times w, c=w \times u$ is the square of the volume of the initial parallelotope $P$ generated by $u,v,w$.

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Let $x_1,...x_n$ be the column of your matrix $M$. Consider the simplex they generate with the vector $0$, (a tetraedron in dimension 3). In the dual of your space (ie the set of line vectors)look at the dual simplex, namely the set of (line) vectors such that $0\leq <x_1,y> \leq 1$. This dual simplex is the convex hull of $0,y_1,...y_n$, with $<y_i,x_j>= \delta _i^j$. So the $y_i$ are exactly the lines of the inverse of $M$, and their coordinates (in the dual base) are the cofactors divided by the determinant.

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