If A and B are square symmetric matrices of the same order, then $AB'-B'A$ is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Skew symmetric
As A and B are symmetric matrix.
$A^T=A$ and $B^T=B$
$(AB'-B'A)^T=(AB')^T-(B'A)^T$
$=BA^T-A^TB$
$=BA'-A'B$
$=B'A-AB'$