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.

  • $\begingroup$ @Henry you are doing it by substituting values of n $\endgroup$ – J.Doe Dec 5 '16 at 11:46

$A^{4n-1}=A^{4(n-1)}A^3=A^{4(n-1)}A^2A = I(-I)A= -A=A^T $


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.