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This question already has an answer here:

$A$ is a square anti symmetric matrix with dimension $n\times n$.

It is known that $n$ is an odd number. Prove that $A$ is not invertible.

How do I prove this? any hints please?

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marked as duplicate by user26857, user228113, Jeremy Rickard, Takumi Murayama, pjs36 Apr 24 '16 at 0:33

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$$\det(A)=\det(A^T)=\det(-A)=(-1)^n\det(A)=-\det(A)$$ since $n$ is odd. hence $$\det(A)=0$$

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  • $\begingroup$ I did not get it, why $$\det(A)=\det(A^T)$$ and why $$\det(A^T)=\det(-A)$$ $\endgroup$ – LiziPizi Nov 29 '15 at 17:10
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    $\begingroup$ because determinant of a matrix and determinant of its transpose are equal always. $\endgroup$ – Kushal Bhuyan Nov 29 '15 at 17:12
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    $\begingroup$ antisymmetric means $A^T=-A$, by definition $\endgroup$ – Kushal Bhuyan Nov 29 '15 at 17:13

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