If $A$ is $n \times n$ skew-symmetric matrix where $n$ is an odd integer number, then $A$ is singular.

Question

In order to answer, I must first state whether the statement is true or false, followed by an explanation

If $A$ is $n \times n$ skew-symmetric matrix where $n$ is an odd integer number, then $A$ is singular.

The answer is True:
\begin{align}
A^{\mathsf{T}} &= -A \\
|A^{\mathsf{T}}| &= |-A| \\[1pt]
|A| &= (-1)^n |A|
\end{align}

That is the step I know, but I need assistance with what follows how to prove that it is True.

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Tanks to @Sammy Black
I have got the right answer to the question – the answer is:

$𝐴ᵀ = -𝐴$

$|𝐴ᵀ| = |-𝐴|$

$|𝐴| = (-1)^𝑛 |𝐴|$

$|𝐴| = -|𝐴|$

$2|A| = 0 $

$2|A|/2 = 0 $

$|A| = 0$

therefore the answer is true A is singular since $|A| = 0$

Answer 1

You have correctly deduced that the scalar $d = |A|$ satisfies the equation $$ d = (-1)^n d $$ for any natural $n$. Using the fact that $n$ is odd, $(-1)^n = -1$, so $d = -d$ or $$ 2d = 0. $$

Here's where we need to assume that the characteristic of the ring of scalars that the matrices are defined over is not $2$. For instance, if the matrices are over the field $\mathbb{R}$ or $\mathbb{C}$, where characteristic is $0$, we're good. We may conclude that $d = 0$, hence the matrix is singular.

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However, over a field of characteristic $2$, where $-1 = 1$ and $2 = 0$, we cannot reach this conclusion, since $2d = 0$ is true for any $d$. In fact, in such a situation, being skew-symmetric and being symmetric are the same thing! As an explicit, example, consider the $3 \times 3$ identity matrix, which is skew-symmetric over, say, $\mathbb{F}_2$, and is certainly not singular.