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Mathematics Question on Matrices

If A=1π[sin⁡−1(πx)tan⁡−1(xπ)\[0.3em]sin⁡−1xπcot⁡−1(πx)]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=1π[−cos⁡−1(πx)tan⁡−1(xπ)\[0.3em]sin⁡−1(xπ)−tan⁡−1(πx)]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−BA - B is equal to

A

II

B

00

C

2I2I

D

12I\frac {1}{2}I

Answer

12I\frac {1}{2}I

Explanation

Solution

We have, A=1π[sin⁡−1(πx)tan⁡−1(xπ) sin⁡−1(xπ)cot⁡−1(πx)]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=1π[−cos⁡−1(πx)tan⁡−1(xπ) sin⁡−1(xπ)−tan⁡−1(πx)]]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}
∴A−B\therefore A-B
=1π[sin⁡−1(πx)+cos⁡−1(πx)tan⁡−1(xπ)−tan⁡−1(xπ) sin⁡−1(xπ)−sin⁡−1(xπ)cot⁡−1(πx)+tan⁡−1(πx)]=\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}
=1π[π20 0π2][∵sin⁡−1x+cos⁡−1x=π/2  and cot⁡−1x+tan⁡−1x=π2]=\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}
=[120 012]=12[10 01]=12I=\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