COMEDK 2025 Afternoon Solved Paper

Concept:Subtract matrix $B$ from matrix $A$ entry‑wise and simplify the resulting elements using standard inverse trigonometric identities.Explanation:Both matrices have a common factor $\frac{1}{\pi}$.Subtract the entries inside the brackets and then divide by $\pi$.Top‑left entry:$\sin^{-1}\frac{1}{2} - \bigl(-\cos^{-1}\frac{1}{2}\bigr) = \sin^{-1}\frac{1}{2} + \cos^{-1}\frac{1}{2}$.We know $\sin^{-1}\frac{1}{2} = \frac{\pi}{6}$ and $\cos^{-1}\frac{1}{2} = \frac{\pi}{3}$.Their sum is $\frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$.Dividing by $\pi$ gives $\frac12$.Top‑right entry:$\tan^{-1}\frac{x}{\pi} - \tan^{-1}\frac{x}{\pi} = 0$.Bottom‑left entry:$\sin^{-1}\frac{x}{\pi} - \sin^{-1}\frac{x}{\pi} = 0$.Bottom‑right entry:$\cot^{-1}\sqrt{3} - \bigl(-\tan^{-1}\sqrt{3}\bigr) = \cot^{-1}\sqrt{3} + \tan^{-1}\sqrt{3}$.We know $\cot^{-1}\sqrt{3} = \frac{\pi}{6}$ and $\tan^{-1}\sqrt{3} = \frac{\pi}{3}$.Their sum is $\frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$.Dividing by $\pi$ gives $\frac12$.Thus $A - B = \begin{pmatrix} \frac12 & 0 \\[6pt] 0 & \frac12 \end{pmatrix} = \frac12 I$.Answer:Option B: $\frac12 I$.