Question

If A and B are symmetric matrices of the same order, then
(i) AB – BA is a _________.
(ii) BA – 2BA is a _________.

Answer 1

It is given that, A and B are symmetric matrices of the same order.

$\therefore A^T=A$ and $B^T=B$ .....(1)

(i)
${\left(AB-BA\right)}^T$

$={\left(AB\right)}^T-{\left(BA\right)}^T\left[{\left(X+Y\right)}^T=X^T+Y^T\right]$

$=B^T{A}^T-A^T{B}^T\left[{\left(XY\right)}^T=Y^T{X}^T\right]$

$=BA-AB$ [Using (1)]

$=-\left(AB-BA\right)$

Since ${\left(AB-BA\right)}^T$$=-\left(AB-BA\right)$, so the matrix AB – BA is a skew-symmetric matrix.

AB – BA is a __skew-symmetric matrix__.

(ii)
Disclaimer: The solution has been provided for the following question.

BA – 2AB is a _________.

${\left(BA-2AB\right)}^T$

$={\left(BA\right)}^T-{\left(2AB\right)}^T\left[{\left(X+Y\right)}^T=X^T+Y^T\right]$

$={\left(BA\right)}^T-2{\left(AB\right)}^T\left[{\left(kX\right)}^T=k{\left(X\right)}^T\right]$

$=A^T{B}^T-2B^T{A}^T\left[{\left(XY\right)}^T=Y^T{X}^T\right]$

$=AB-2BA$ [Using (1)]

Since ${\left(BA-2AB\right)}^T\ne BA-2AB$ or ${\left(BA-2AB\right)}^T\ne -\left(BA-2AB\right)$, so the matrix BA – 2AB is neither symmetric nor skew-symmetric matrix.

BA – 2AB is a __neither symmetric nor skew-symmetric matrix__.