If $A=\begin{bmatrix}0&1&-3\\-1&0&5\\3&-5&0\end{bmatrix}$ then the value of $|A^{2025}|$ is

0

The correct answer is Option (4) → 0

Given

$A=\begin{pmatrix}0 & 1 & -3\\[4pt]-1 & 0 & 5\\[4pt]3 & -5 & 0\end{pmatrix}$

Check skew–symmetry:

$A'=\begin{pmatrix}0 & -1 & 3\\[4pt]1 & 0 & -5\\[4pt]-3 & 5 & 0\end{pmatrix}$

$-A=\begin{pmatrix}0 & -1 & 3\\[4pt]1 & 0 & -5\\[4pt]-3 & 5 & 0\end{pmatrix}$

Thus $A'=-A$, so $A$ is a skew–symmetric matrix.

A skew–symmetric matrix of odd order has determinant $0$.

$|A|=0$

$\Rightarrow |A^{2025}| = |A|^{2025} = 0^{2025} = 0$

$|A^{2025}| = 0$