If A is a real skew-symmetric matrix such that A^2+1=O, then

We have,

$A^2+1=O$

$A^2=-I= |A^2|=|I|⇒ |A|^2 =1⇒ |A|=±1$

Let the order of A be $n×n$.

Since A is skew-symmetric matrix.

$∴A^T=- A$

$⇒|A^T|=|-A|$

$⇒ |A|=(-1)^n |A|$

$⇒|A|=-|A|$, if n is odd

$⇒ |A|=0$, if n is odd.

But, $|A|≠0$. So, A cannot be a skew-symmetric matrix of odd order. Infact, the determinant of a skew-symmetric matrix of even order is a perfect square.

$∴|A|=1$