The number of real values of x satisfying <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced open="|" close="|"><mtable><mtr><mtd><mi>x</mi></mtd><mtd><mn>3</mn><mi>x</mi><mo>+</mo><mn>2</mn></mtd><mtd><mn>2</mn><mi>x</mi><mo>-</mo><mn>1</mn></mtd></mtr><mtr><mtd><mn>2</mn><mi>x</mi><mo>-</mo><mn>1</mn></mtd><mtd><mn>4</mn><mi>x</mi></mtd><mtd><mn>3</mn><mi>x</mi><mo>+</mo><mn>1</mn></mtd></mtr><mtr><mtd><mn>7</mn><mi>x</mi><mo>-</mo><mn>2</mn></mtd><mtd><mn>17</mn><mi>x</mi><mo>+</mo><mn>6</mn></mtd><mtd><mn>12</mn><mi>x</mi><mo>-</mo><mn>1</mn></mtd></mtr></mtable></mfenced></math> = 0 is -
Step-by-Step Solution
Key Concept: Perform row operations to simplify the determinant. Applying R3 -> R3 - (R1 + 3*R2) shows that the third row becomes a zero row, meaning the determinant is identically zero for all x.
Let the determinant be D. Applying R3 -> R3 - R1 - 3R2: The elements of the third row become: (7x-2) - (x) - 3(2x-1) = 7x-2-x-6x+3 = 1. Wait, let's recheck. R3 -> R3 - R1 - 3R2: (7x-2) - x - 3(2x-1) = 0. (17x+6) - (3x+2) - 3(4x) = 14x+4-12x = 2x+4. Actually, applying R3 -> R3 - 2R1 - 3R2: (7x-2) - 2(x) - 3(2x-1) = 7x-2-2x-6x+3 = -x+1. Let's try R3 -> R3 - 3R1 - 2R2: (7x-2) - 3(x) - 2(2x-1) = 7x-2-3x-4x+2 = 0. (17x+6) - 3(3x+2) - 2(4x) = 17x+6-9x-6-8x = 0. (12x-1) - 3(2x-1) - 2(3x+1) = 12x-1-6x+3-6x-2 = 0. Since R3 becomes a zero row, the determinant is 0 for all real x.
Correct Answer: D