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If 'A' is skew-symmetric matrix of odd order then prove that it is singular.
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For a skew - symmetric matrix,

$A^T= -A$

Taking determinant on both sides,

$|A^T|= |-A| \ \cdots \ \cdots (X)$

$Now, \ |A^T|=|A| \ \cdots \ \cdots (Y)$

$|A|=(-1)^n|A|$

Where n is the order of the matix

For matrix with odd order, n is odd, (-1)odd =-1

$|-A|=-|A| \ \cdots \ \cdots (Z)$

From (X), (Y) and (Z),

$|A| = -|A|$

$\to |A|=0$

∴, A is a singular matrix.

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