<p>If \(y = \sin(mx)\), then the value of the determinant</p><p>\[\Delta = \begin{vmatrix} \sin mx & m\cos mx & -m^2 \sin mx \\ -m^3 \cos mx & m^4 \sin mx & m^5 \cos mx \\ -m^6 \sin mx & -m^7 \cos mx & m^8 \sin mx \end{vmatrix}\]</p><p>is</p>
Step-by-Step Solution
Key Concept: Recognize that each row can be factored out with powers of m, revealing that one row is a scalar multiple of another row, making the determinant zero.
<p><strong>Step 1:</strong> Factor out powers of m from each row.</p><p>From Row 1: factor out m⁰ = 1</p><p>From Row 2: factor out m³, giving Row 2 = m³[−cos mx, m sin mx, m² cos mx]</p><p>From Row 3: factor out m⁶, giving Row 3 = m⁶[−sin mx, −m cos mx, m² sin mx]</p><p><strong>Step 2:</strong> Rewrite the determinant with factored form:</p><p>Δ = 1 · m³ · m⁶ × |sin mx, m cos mx, −m² sin mx | / [m³ × m⁶]</p><p>More directly, observe Row 2 = −m³ × Row 1 (after proper coefficient matching)</p><p><strong>Step 3:</strong> Notice that after factoring m⁶ from Row 3:</p><p>Row 3 has the form m⁶[−sin mx, −m cos mx, m² sin mx]</p><p>Comparing rows: Row 1 and Row 3 show a pattern where Row 3 = −m⁶ × (Row 1 structure)</p><p><strong>Step 4:</strong> The critical observation is that the three rows are linearly dependent. Specifically, the determinant has a repeated pattern in the arguments (sin mx and cos mx derivatives), making rows proportional.</p><p>Since rows are linearly dependent, the determinant = 0.</p><p>∴ Answer: <strong>0</strong></p>
Correct Answer: 0