Value of Adjugate of Determinant

Theorem
Let $$D$$ be the determinant of order $$n$$.

Let $$D^*$$ be the adjugate of $$D$$.

Then $$D^* = D^{n-1}$$.

Proof
Let $$\mathbf{A} = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn}\end{bmatrix}$$ and $$\mathbf{A}^* = \begin{bmatrix} A_{11} & A_{12} & \cdots & A_{1n} \\ A_{21} & A_{22} & \cdots & A_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ A_{n1} & A_{n2} & \cdots & A_{nn}\end{bmatrix}$$.

Thus $$\left({\mathbf{A}^*}\right)^t = \begin{bmatrix} A_{11} & A_{21} & \cdots & A_{n1} \\ A_{12} & A_{22} & \cdots & A_{n2} \\ \vdots & \vdots & \ddots & \vdots \\ A_{1n} & A_{2n} & \cdots & A_{nn}\end{bmatrix}$$ is the transpose of $$\mathbf{A}^*$$.

Let $$c_{ij}$$ be the typical element of $$\mathbf{A} \left({\mathbf{A}^*}\right)^t$$.

Then $$c_{ij} = \sum_{k=1}^n a_{ik} A_{jk}$$ by definition of matrix product.

Thus by the corollary of the Expansion Theorem for Determinants, $$c_{ij} = \delta_{ij} D$$.

So $$\det \left({\mathbf{A} \left({\mathbf{A}^*}\right)^t}\right) = \begin{vmatrix} D & 0 & \cdots & 0 \\ 0 & D & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & D\end{vmatrix} = D^n$$ by Determinant of a Diagonal Matrix.

From Determinant of Matrix Product, $$\det \left({\mathbf{A}}\right) \det \left({\left({\mathbf{A}^*}\right)^t}\right) = \det \left({\mathbf{A} \left({\mathbf{A}^*}\right)^t}\right)$$

From Determinant of Transpose, $$\det \left({\left({\mathbf{A}^*}\right)^t}\right) = \det \left({\mathbf{A}^*}\right)$$.

Thus as $$D = \det \left({\mathbf{A}}\right)$$ and $$D^* = \det \left({\mathbf{A}^*}\right)$$ it follows that $$DD^* = D^n$$.

Now if $$D \ne 0$$, the result follows.

However, if $$D = 0$$ we need to show that $$D^* = 0$$.

Let $$D^* = \begin{vmatrix} A_{11} & A_{12} & \cdots & A_{1n} \\ A_{21} & A_{22} & \cdots & A_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ A_{n1} & A_{n2} & \cdots & A_{nn}\end{vmatrix}$$.

Suppose that at least one element of $$\mathbf{A}$$, say $$a_{rs}$$, is non-zero (otherwise the result follows immediately).

By the Expansion Theorem for Determinants and its corollary, we can expand $$D$$ by row $$r$$, and get:

$$D = 0 = \sum_{j=1}^n A_{ij} t_j, \forall i = 1, 2, \ldots, n$$

for all $$t_1 = a_{r1}, t_2 = a_{r2}, \ldots, t_n = a_{rn}$$.

But $$t_s = a_{rs} \ne 0$$.

So, by (work in progress), $$D^* = \begin{vmatrix} A_{11} & A_{12} & \cdots & A_{1n} \\ A_{21} & A_{22} & \cdots & A_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ A_{n1} & A_{n2} & \cdots & A_{nn}\end{vmatrix} = 0$$.