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Question

A is a skew symmetric matrix such that $$A^TA = I$$, then $$A^{4n-1} (n \in N)$$ is equal to


A
AT
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B
I
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C
I
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D
AT
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Solution

The correct option is D $$A^T$$
$$A^T = - A $$ & $$ A^TA = I$$
$$\Rightarrow  A^2 = -I \Rightarrow  A^{4n} = I$$ 
$$\therefore A^{4n-1}  = A^{-1}  = A^T$$

Mathematics

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