If A and B are symmetric matrices of the same order,
Question:

If $A$ and $B$ are symmetric matrices of the same order, then $A B^{T}-B A^{T}$ is a

(a) skew-symmetric matrix

(b) null matrix

(c) symmetric matrix

(d) none of these

Solution:

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

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

Now,

$\left(A B^{T}-B A^{T}\right)^{T}$

$=(A B-B A)^{T} \quad[U \operatorname{sing}(1)]$

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

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

$=B A-A B$                   [Using (1)]

$=-\left(A B^{T}-B A^{T}\right) \quad[$ Using (1) $]$

We know that, a matrix $X$ is skew-symmetric if $X^{T}=-X$.

Since $\left(A B^{T}-B A^{T}\right)^{T}=-\left(A B^{T}-B A^{T}\right)$, therefore, $A B^{T}-B A^{T}$ is a skew-symmetric matrix.

Hence, the correct answer is option (a).