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If 'A' is skew-symmetric matrix of odd order then prove that it is singular.
1 Answer
| written 4.6 years ago by |
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.