Question

If $A$ is a non-singular skew-symmetric matrix and $B$ is a square matrix such that $B={\left(\right(A^T{B}^T\left)A^{-1}\right)}^T,$ then $(A+B{)}^2$ is equal to

• A. $O$
• B. $A+B$
• C. $A^2+B^2$
• D. $A^2+B^2+2AB$

Answer 1

The correct option is C $A^2+B^2$

$B={\left(\left(A^T{B}^T\right)A^{-1}\right)}^T$
$={\left({\left(BA\right)}^T{A}^{-1}\right)}^T$
$={\left(A^{-1}\right)}^T{\left(\right(BA{)}^T)}^T$
Given that $A$ is skew-symmetric matrix, then $A^{-1}$ is also skew-symmetric matrix and hence ${\left(A^{-1}\right)}^T=-A^{-1}$
We get, $B=-A^{-1}BA$

Now, $(A+B{)}^2=(A+B\left)\right(A+B)$
$=A^2+AB+BA+B^2=A^2+A(-A^{-1}BA)+BA+B^2[\because B=-A^{-1}BA]$
$=A^2-\left(A{A}^{-1}\right)\left(BA\right)+BA+B^2$ [By Associative law]
$=A^2-BA+BA+B^2[\because A{A}^{-1}=I]=A^2+B^2$