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WBJEE · Maths · Matrices

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

  1. A \(\mathrm{A}^{2}\)
  2. B \(-\mathrm{B}^{2}\)
  3. C \(-\mathrm{A}^{2}\)
  4. D \(\mathrm{AB}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(-\mathrm{A}^{2}\)

Step-by-step Solution

Detailed explanation

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