# If a square matrices $A$ and $B$ are such that $AA^{\theta}=A^{\theta}A$, $BB^{\theta}=B^{\theta}B$, $AB^{\theta}=B^{\theta}A$ then $AB(AB)^{\theta}$ is equal to

$\begin{array}{1 1}(a)\;B^{\theta}A^{\theta}AB&(b)\;BA^{\theta}AB\\(c)\;BA^{\theta}A^{\theta}B&(d)\;None\;of\;these\end{array}$

Given :
$AB^{\theta}=B^{\theta}A$
So that
$(AB^{\theta})^{\theta}=(B^{\theta}A)^{\theta}$
$\Rightarrow (B^{\theta})^{\theta}A^{\theta}=A^{\theta}B$
Now
$AB(AB)^{\theta}=AB.B^{\theta}A^{\theta}$
$\Rightarrow AB^{\theta}BA^{\theta}$
$\Rightarrow B^{\theta}AA^{\theta}B$
$\Rightarrow B^{\theta}A^{\theta}AB=(AB)^{\theta}AB$