Prove that $\operatorname{adj}A^t = \operatorname{adj} A$ Let $A$ be an anti-symmetric ($A^t = -A$), squared matrix ($n \times n$, while $n$ is uneven).
Prove that ${\rm adj}\;A^t = {\rm adj}\;A$.
 A: Let $X := (x_{ij})$ by an $n$-by-$n$ matrix. The adjugate matrix, $A := (a_{ij})$, is the transpose of the cofactor matrix. The cofactor matrix is defined in the following way. 
For each $1 \le i \le j \le n$, define the sub-matrix $[X]_{ij}$ to be the matrix derived from $X$ by deleting the $i^{\text{th}}$ row and the $j^{\text{th}}$ column. 
Each sub-matrix will, of course, be an $(n-1)$-by-$(n-1)$ matrix. Let $K := (\kappa_{ij})$ denote the cofactor matrix. The entries are given, by definition, by:
$$\kappa_{ij} := (-1)^{i+j}\det [X]_{ij}$$
The adjugate matrix is give, by definition, by $A = K^{\top}$, meaning that $a_{ij} = \kappa_{ji}$. Hence:
$$a_{ij} = (-1)^{j+i}\det[X]_{ji}$$
We aim to show that, if $X$ is skew-symmetric, then the adjugate of $X$ is identical to the adjugate of $X^{\top}$. Let $B := (b_{ij})$ be the adjugate matrix of $X^{\top}$. By definition:
$$b_{ij} = (-1)^{j+i}\det\left[X^{\top}\right]_{ji}$$
We need only show that $a_{ij} = b_{ij}$ for all $1 \le i \le j \le n$. Since $X$ is skew-symmetric $X^{\top} = -X$. Hence $\det[X^{\top}]_{ji} = \det[-X]_{ji}$. Since $n$ is odd, $n-1$ is even and for any even dimensional square matrix we have $\det(Y) = \det(-Y)$. Recall that the sub-matrices $[-X]_{ji}$ are $(n-1)$-by-$(n-1)$ matrices and so $\det[-X]_{ji} = \det(-[-X]_{ji}) = \det[X]_{ji}$. Since $\det[-X]_{ji} = \det[X]_{ji}$, $a_{ij} = b_{ij}$.
A: Anti-symmetry of $A$ gives you $\def\adj{\operatorname{adj}}\adj A^t=\adj(-A)$. And for any odd-size square (not squared) matrix $B$ one has $\adj(-B)=\adj B$, because each coefficient of $\adj(-B)$ is given by an even-size $k\times k$ determinant whose entries are all negated with respect to the entries of the corresponding coefficient of $\adj B$, which signs by multilinearity of the determinant affect the result by a factor $(-1)^k=1$ since $k$ is even. Then
$$
  \adj A^t=\adj(-A)=\adj A.
$$
While not in the question you may equally easily conclude that this is also $(\adj A)^t$, so $\adj A$ is in fact a symmetric matrix.
A: $A^t=-A$ therefore $|A^{t}|=|-A|$
$\Rightarrow |A| = (-1)^n \cdot |A|$
This equation can only take place for $|A| = 0$ since $n \in \mathbb{N}_{odd}$ ($x = -x$).
From a theorem we get that
$$A \cdot  {\rm adj}\;A = I |A|$$
$$\Rightarrow A \cdot  {\rm adj}\;A = 0 \space\space {\color{Red} \bigstar}$$
We will look at the transposes of both sides of the equation and get that:
$$A^t \cdot {\rm adj}\;A^t = 0$$
And we know that $A^t = -A$ thus,
$$-A \cdot {\rm adj}\;A^t = 0$$
We will now look at ${\color{Red} \bigstar}$ and both sides of the equation by $(-1)$ and get,
$$-A \cdot {\rm adj}\;A = 0$$
By looking at both equation we understand that ${\rm adj}\;A = {\rm adj}\;A^t$,
Q.E.D.
