Let A and B be 3 × 3 matrices of real numbers, where A is symmetric, B is skew-symmetric, and $(A + B) (A-B) = (A-B) (A + B)$ if $(AB)^T =(-1)^n AB$, then

n is an odd natural number

We have,

$(A + B) (A-B) = (A - B) (A + B)$

$⇒AB = BA$  ...(i)

Now, $(AB)^T=(-1)^n AB$

$⇒ B^T A^T =(-1)^n AB$

$⇒(-B) A=(-1)^n AB$  [∵ $B^T-B$ and $A^T = A$]

$⇒-BA=(-1)^nBA$  [Using (i)]

$⇒(-1)^n=-1$

⇒ n is an odd natural number