Question

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
(A) \(\mathrm{A}^{2}\)
(B) \(-\mathrm{B}^{2}\)
(C) \(-\mathrm{A}^{2}\)
(D) \(\mathrm{AB}\)

Answer 1

Ans: (C)
Hint: \(A B=B A, A^{\top}=-A, B^{\top}=-B\)
\(\because\) matrices are non singular
\(\therefore\) order is even
\(A^{2} B^{2}\left(A^{\top} B\right)^{-1}\left(A B^{-1}\right)^{\top}\)
\(=A^{2} B^{2}(-A B)^{-1}\left(A B^{-1}\right)^{T}\)
\(=-\mathrm{A}^{2} \mathrm{~B}^{2} \mathrm{~B}^{-1} \mathrm{~A}^{-1}\left(\mathrm{~B}^{\mathrm{T}}\right)^{-1} \mathrm{~A}^{\top}\)
\(=-\mathrm{A}^{2} \mathrm{~B} \mathrm{~A}^{-1} \mathrm{~B}^{-1} \mathrm{~A}\)
\(=-\mathrm{A}^{2} \mathrm{~B}(\mathrm{BA})^{-1} \mathrm{~A}\)
\(=-\mathrm{A}^{2} \mathrm{~B}(\mathrm{AB})^{-1} \mathrm{~A}=-\mathrm{A}^{2} \mathrm{~B} \mathrm{~B}^{-1} \mathrm{~A}^{-1} \mathrm{~A}=-\mathrm{A}^{2}\)

Answer 2

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