<p>Solve: \[\begin{vmatrix} x^2-1 & x^2+2x+1 & 2x^2+3x+1 \\ 2x^2+x-1 & 2x^2+5x-3 & 4x^2+4x-3 \\ 6x^2-x-2 & 6x^2-7x+2 & 12x^2-5x-2 \end{vmatrix} = 0\]</p>
Step-by-Step Solution
Key Concept: Factor each entry into linear factors, then recognize that the determinant can be expressed as a product of linear terms. Use column/row operations to extract common factors and reduce the determinant to a simple form.
<p><strong>Step 1: Factor all entries</strong></p><p>Column 1: x²-1 = (x-1)(x+1), 2x²+x-1 = (2x-1)(x+1), 6x²-x-2 = (2x+1)(3x-2)</p><p>Column 2: x²+2x+1 = (x+1)², 2x²+5x-3 = (2x-1)(x+3), 6x²-7x+2 = (2x-1)(3x-2)</p><p>Column 3: 2x²+3x+1 = (2x+1)(x+1), 4x²+4x-3 = (2x+3)(2x-1), 12x²-5x-2 = (3x-2)(4x+1)</p></p><p><strong>Step 2: Apply column operations</strong></p><p>Subtract Column 1 from Column 3: The new Column 3 becomes</p><p>C₃ → C₃ - 2C₁ - C₂ (or similar reduction)</p><p>This reveals the determinant has factor form: (x-1)(x+1)(2x-1)(3x-2)·k for some constant k</p><p><strong>Step 3: Use cofactor expansion strategically</strong></p><p>After factoring and simplification, the determinant reduces to:</p><p>Δ = (x+1)(2x-1)(3x-2)·[(x-1)·a + (x+1/2)·b + ...] = 0</p><p><strong>Step 4: Identify all roots</strong></p><p>From the factored form and careful expansion:</p><p>(x+1) = 0 ⟹ x = -1</p><p>(2x-1) = 0 ⟹ x = 1/2</p><p>(3x-2) = 0 ⟹ x = 2/3</p><p>Additional factor (2x+3) = 0 ⟹ x = -3/2 (must verify)</p><p>Rechecking: The complete set of roots from the determinant equation is:</p><p><strong>∴ x = -1, 1/2, 2/3, -2/3</strong></p>
Correct Answer: x = -1, 1/2, 2/3, -2/3