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Let $A$ and $B$ be two non-singular skew symmetric matrices such that $A B=B A$, then $A^{2} B^{2}\left(A^{\top} B\right)^{-1}\left(A B^{-1}\right)^{\top}$ is equal to
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The correct answer is:
$-\mathrm{A}^{2}$
$\mathrm{AB}=\mathrm{BA}, \mathrm{A}^{\top}=-\mathrm{A}, \mathrm{B}^{\top}=-\mathrm{B}$
$\because$ matrices are non singular
$\therefore$ order is even
$\mathrm{A}^{2} \mathrm{~B}^{2}\left(\mathrm{~A}^{\top} \mathrm{B}\right)^{-1}\left(\mathrm{AB}^{-1}\right)^{\top}$
$\therefore$ order is even
$A^{2} B^{2}\left(A^{\top} B\right)^{-1}\left(A B^{-1}\right)^{\top}$
$=A^{2} B^{2}(-A B)^{-1}\left(A B^{-1}\right)^{\top}$
$=-A^{2} B^{2} B^{-1} A^{-1}\left(B^{\top}\right)^{-1} A^{\top}$
$=-A^{2} B A^{-1} B^{-1} A$
$=-A^{2} B(B A)^{-1} A$
$=-A^{2} B(A B)^{-1} A=-A^{2} B B^{-1} A^{-1} A=-A^{2}$
$\because$ matrices are non singular
$\therefore$ order is even
$\mathrm{A}^{2} \mathrm{~B}^{2}\left(\mathrm{~A}^{\top} \mathrm{B}\right)^{-1}\left(\mathrm{AB}^{-1}\right)^{\top}$
$\therefore$ order is even
$A^{2} B^{2}\left(A^{\top} B\right)^{-1}\left(A B^{-1}\right)^{\top}$
$=A^{2} B^{2}(-A B)^{-1}\left(A B^{-1}\right)^{\top}$
$=-A^{2} B^{2} B^{-1} A^{-1}\left(B^{\top}\right)^{-1} A^{\top}$
$=-A^{2} B A^{-1} B^{-1} A$
$=-A^{2} B(B A)^{-1} A$
$=-A^{2} B(A B)^{-1} A=-A^{2} B B^{-1} A^{-1} A=-A^{2}$
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