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

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

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

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

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

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

From Determinant of Transpose:
 * $\det \left({\left({\mathbf A^*}\right)^\intercal}\right) = \det \left({\mathbf A^*}\right)$

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

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$