Question

If $A= \frac {1} { \pi}\begin{bmatrix} \sin^{-1} (\pi x) & \tan^{-1} (\frac {x}{\pi}) \\[0.3em] \sin^{-1}\frac {x}{\pi}& \cot^{-1}(\pi x) \end{bmatrix}$ , $B= \frac {1} { \pi}\begin{bmatrix} -\cos^{-1} (\pi x) & \tan^{-1} (\frac {x}{\pi}) \\[0.3em] \sin^{-1}(\frac {x}{\pi})& -\tan^{-1}(\pi x) \end{bmatrix}$ then $A - B$ is equal to

• A. $I$
• B. $0$
• C. $2I$
• D. $\frac {1}{2}I$

Answer 1

The Correct Option is D

We have, $A=\frac{1}{\pi} \begin{bmatrix}\sin ^{-1}(\pi x) & \tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1}\left(\frac{x}{\pi}\right) & \cot ^{-1}(\pi x)\end{bmatrix}$
and $B=\frac{1}{\pi}\begin{bmatrix}-\cos ^{-1}(\pi x) & \tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1}\left(\frac{x}{\pi}\right) & \left.-\tan ^{-1}(\pi x)\right]\end{bmatrix}$
$\therefore A-B$
$=\frac{1}{\pi}\begin{bmatrix} \sin ^{-1}(\pi x)+\cos ^{-1}(\pi x) & \tan ^{-1}\left(\frac{x}{\pi}\right)-\tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1}\left(\frac{x}{\pi}\right)-\sin ^{-1}\left(\frac{x}{\pi}\right) & \cot ^{-1}(\pi x)+\tan ^{-1}(\pi x)\end{bmatrix}$
$=\frac{1}{\pi}\begin{bmatrix}\frac{\pi}{2} & 0 \\ 0 & \frac{\pi}{2}\end{bmatrix} \begin{bmatrix}\because \sin ^{-1} x+\cos ^{-1} x=\pi / 2 \\ \text { and } \cot ^{-1} x+\tan ^{-1} x=\frac{\pi}{2}\end{bmatrix}$
$=\begin{bmatrix}\frac{1}{2} & 0 \\ 0 & \frac{1}{2}\end{bmatrix} = \frac{1}{2}\begin{bmatrix}1&0\\ 0&1\end{bmatrix} = \frac{1}{2}I$