#### Provide Solution for R. D. Sharma Maths Class 12 Chapter 4 Algebra of Matrices Exercise Multiple Choice Questions Question 41 Maths Textbook Solution.

Answer:$\frac{1}{2} I$

Given: $A=\frac{1}{\pi}\left[\begin{array}{ll} \sin ^{-1}(\pi x) & \tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1}\left(\frac{x}{\pi}\right) & \cot ^{-1}(\pi x) \end{array}\right] ; B=\frac{1}{\pi}\left[\begin{array}{ll} -\cos ^{-1}(\pi x) & \tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1}\left(\frac{x}{\pi}\right) & -\tan ^{-1}(\pi x) \end{array}\right]$

Hint:$\text { Use } \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2} \text { and } \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}$

Solution:

\begin{aligned} &A-B=\frac{1}{\pi}\left[\left[\begin{array}{cc} \sin (\pi x) & \tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1}\left(\frac{x}{\pi}\right) & \cot ^{-1}(\pi x) \end{array}\right]-\left[\begin{array}{cc} -\cos ^{-1}(\pi x) & \tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1}\left(\frac{x}{\pi}\right) & -\tan ^{-1}\left(\frac{\pi}{x}\right) \end{array}\right]\right]\\ &=\frac{1}{\pi}\left[\begin{array}{ll} \sin (\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}\left(\frac{\pi}{x}\right) \end{array}\right]\\ &=\frac{1}{\pi}\left[\begin{array}{cc} \frac{\pi}{2} & 0 \\ 0 & \frac{\pi}{2} \end{array}\right] \quad\left[\begin{array}{l} \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2} \\ \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2} \end{array}\right]\\ &=\frac{1}{\pi} \times \frac{\pi}{2} I=\frac{1}{2} I \end{aligned}