<p>Given \[\begin{vmatrix} x & -6 & -1 \\ 2 & -3x & x-3 \\ -3 & 2x & x+2 \end{vmatrix} = 0\] Find the sum of real roots.</p>
Step-by-Step Solution
Key Concept: Since the determinant equals zero, the matrix is singular. Expand the determinant to get a polynomial equation in x, then find all real roots and sum them using Vieta's formulas or direct calculation.
<p><strong>Step 1:</strong> Expand the determinant along the first row:</p><p>x·((-3x)(x+2) - (x-3)(2x)) + 6·(2(x+2) - (x-3)(-3)) - 1·(2(2x) - (-3x)(-3)) = 0</p><p><strong>Step 2:</strong> Simplify the first cofactor: (-3x)(x+2) - (x-3)(2x) = -3x² - 6x - 2x² + 6x = -5x²</p><p>So: x(-5x²) = -5x³</p><p><strong>Step 3:</strong> Simplify the second cofactor: 2(x+2) + 3(x-3) = 2x + 4 + 3x - 9 = 5x - 5</p><p>So: 6(5x - 5) = 30x - 30</p><p><strong>Step 4:</strong> Simplify the third cofactor: 4x - 9x = -5x</p><p>So: -1(-5x) = 5x</p><p><strong>Step 5:</strong> Combine: -5x³ + 30x - 30 + 5x = 0</p><p>-5x³ + 35x - 30 = 0</p><p>Divide by -5: x³ - 7x + 6 = 0</p><p><strong>Step 6:</strong> Factor: (x - 1)(x² + x - 6) = 0 → (x - 1)(x - 2)(x + 3) = 0</p><p><strong>Step 7:</strong> Real roots are x = 1, 2, -3</p><p>∴ Sum of real roots = 1 + 2 + (-3) = <strong>0</strong></p>
Correct Answer: 0