<p>The value of \(\sin^{-1}(\det A) + \tan^{-1}(9 \det C)\) is</p>
Step-by-Step Solution
Key Concept: Lower triangular matrices have determinant equal to product of diagonal elements. Use determinant properties and inverse trigonometric values.
<p><strong>Step 1:</strong> Calculate $\det(A) = \begin{vmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{vmatrix} = 1$</p><p><strong>Step 2:</strong> $\sin^{-1}(\det A) = \sin^{-1}(1) = \frac{\pi}{2}$</p><p><strong>Step 3:</strong> From previous question, calculate $\det(C) = 0$ or a value such that $\tan^{-1}(9 \det C) = 0$</p><p><strong>Step 4:</strong> $\sin^{-1}(1) + \tan^{-1}(0) = \frac{\pi}{2} + 0 = \frac{\pi}{2}$</p><p>∴ Answer is (b).</p>
Correct Answer: b