Let A and B be two non-singular skew symmetric matrices such that A B=B A, then A2 B2(A T B)-1(A B-1) T is equal to

Question

Let A and B be two non-singular skew symmetric matrices such that A B=B A, then A2 B2(A T B)-1(A B-1) T is equal to (A) A2 (B) -B2 (C) -A2 (D) A B

Tardigrade Admin · Accepted Answer

A B=B A, A T =-A, B T =-B ∵ matrices are non singular ∴ order is even A2 B2(A T B)-1(A B-1) T =A2 B2(-A B)-1(A B-1) T =-A2 B2 B-1 A-1(B T )-1 A T =-A2 B A-1 B-1 A =-A2 B(B A)-1 A =-A2 B(A B)-1 A =-A2 B B-1 A-1 A=-A2

Q. Let $A$ and $B$ be two non-singular skew symmetric matrices such that $A B = B A$ , then $A^{2} B^{2} (A^{T} B)^{- 1} (A B^{- 1})^{T}$ is equal to

$A^{2}$

$- B^{2}$

$- A^{2}$

$A B$

Q. Let A and B be two non-singular skew symmetric matrices such that AB=BA, then A2B2(ATB)−1(AB−1)T is equal to

A2

−B2

−A2

AB

AB=BA,AT=−A,BT=−B ∵ matrices are non singular ∴ order is even A2B2(ATB)−1(AB−1)T =A2B2(−AB)−1(AB−1)T =−A2B2B−1A−1(BT)−1AT =−A2BA−1B−1A =−A2B(BA)−1A =−A2B(AB)−1A =−A2BB−1A−1A=−A2

Q. Let $A$ and $B$ be two non-singular skew symmetric matrices such that $A B = B A$ , then $A^{2} B^{2} (A^{T} B)^{- 1} (A B^{- 1})^{T}$ is equal to

$A^{2}$

$- B^{2}$

$- A^{2}$

$A B$