If $A = \frac{1}{\pi} \begin{bmatrix} \sin^{-1}(x\pi) & \tan^{-1}\left(\frac{x}{\pi}\right) \\ \sin^{-1}\left(\frac{x}{\pi}\right) & \cot^{-1}(\pi x) \end{bmatrix}$ and $B = \frac{1}{\pi} \begin{bmatrix} -\cos^{-1}(x\pi) & \tan^{-1}\left(\frac{x}{\pi}\right) \\ \sin^{-1}\left(\frac{x}{\pi}\right) & -\tan^{-1}(\pi x) \end{bmatrix}$, then $A - B$ is equal to

$\frac{1}{2} I$

The correct answer is Option (4) → $\frac{1}{2} I$ ##

We have, $A = \begin{bmatrix} \frac{1}{\pi} \sin^{-1} x\pi & \frac{1}{\pi} \tan^{-1} \frac{x}{\pi} \\ \frac{1}{\pi} \sin^{-1} \frac{x}{\pi} & \frac{1}{\pi} \cot^{-1} \pi x \end{bmatrix}$

and $B = \begin{bmatrix} -\frac{1}{\pi} \cos^{-1} x\pi & \frac{1}{\pi} \tan^{-1} \frac{x}{\pi} \\ \frac{1}{\pi} \sin^{-1} \frac{x}{\pi} & -\frac{1}{\pi} \tan^{-1} \pi x \end{bmatrix}$

$∴A - B = \begin{bmatrix} \frac{1}{\pi}(\sin^{-1} x\pi + \cos^{-1} x\pi) & \frac{1}{\pi}(\tan^{-1} \frac{x}{\pi} - \tan^{-1} \frac{x}{\pi}) \\ \frac{1}{\pi}(\sin^{-1} \frac{x}{\pi} - \sin^{-1} \frac{x}{\pi}) & \frac{1}{\pi}(\cot^{-1} \pi x + \tan^{-1} \pi x) \end{bmatrix}$

$= \begin{bmatrix} \frac{1}{\pi} \cdot \frac{\pi}{2} & 0 \\ 0 & \frac{1}{\pi} \cdot \frac{\pi}{2} \end{bmatrix} \quad \left[ ∵\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \text{ and } \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2} \right]$

$= \begin{bmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \frac{1}{2} I$