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If $A$ is a skew symmetric matrix such that $A^{T}A=I$ , then $A^{4n-1} (n \in \mathbb N)$ is equal to:
$(A)$ $-A^{T}$ $(B)$ $I$ $(C)$ $-I$ $(D)$ $A^{T}$.

In this I tried as $A = - A^T$ and then $-A^2=I$, but I don't know how to proceed after this. Any help is appreciated.

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  • $\begingroup$ @Henry you are doing it by substituting values of n $\endgroup$ – J.Doe Dec 5 '16 at 11:46
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$A^{4n-1}=A^{4(n-1)}A^3=A^{4(n-1)}A^2A = I(-I)A= -A=A^T $

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