Let, A be a skew-symmetric square matrix of $ n \times n $ , where n is odd, By general properties of determinants,

$ det(A) \; = \; det(A^T) \; \; \; \ldots (i)$

However, since A is a skew-symmetric matrix where

$ a_{ij} \; = \; -a_{ij} \; \; $ (i,j are rows and column numbers ),

$\therefore$ In case of skew-symmetric matrix,

$ A^T \; = \; -A \\ \; \\ Now \; det(-A) \; = \; det(A^T) \\ \; \\ But, \; det(-A) \; = \; (-1)^n \; det(A) $

where n is no. of rows/columns in a square Matrix.

$ \therefore det(A^T) \; = \; (-1)^n \; det(A) \\ \; \\ \because n \; is \; odd, \; (-1)^n \; = \; -1 \\ \; \\ \therefore det(A^T) \; = \; - \; det(A) \; \; \; \ldots(ii) \\ \; \\ \; \\ Subtracting \; equation \; (ii) \; from \; (i), \\ \; \\ \therefore 2 det(A) \; = \; det(A^T)-det(A^T) \; = \; 0 \\ \; \\ \therefore det(A) \; = \; 0 $

Hence matrix A is singular.... By definition of singular matrix

Hence proved.