Matrices & Determinants
Arithmetic Progression
Grade Class 12
Question:
If <i>a, b, c</i> are in AP, then <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced open="|" close="|"><mtable><mtr><mtd><mi>x</mi><mo>+</mo><mn>1</mn></mtd><mtd><mi>x</mi><mo>+</mo><mn>2</mn></mtd><mtd><mi>x</mi><mo>+</mo><mi>a</mi></mtd></mtr><mtr><mtd><mi>x</mi><mo>+</mo><mn>2</mn></mtd><mtd><mi>x</mi><mo>+</mo><mn>3</mn></mtd><mtd><mi>x</mi><mo>+</mo><mi>b</mi></mtd></mtr><mtr><mtd><mi>x</mi><mo>+</mo><mn>3</mn></mtd><mtd><mi>x</mi><mo>+</mo><mn>4</mn></mtd><mtd><mi>x</mi><mo>+</mo><mi>c</mi></mtd></mtr></mtable></mfenced></math> equals -
(A) <i>a</i> + <i>b</i> + <i>c</i>
(B) <i>x</i> + <i>a</i> + <i>b</i> + <i>c</i>
(C) 0
(D) none of these
Step-by-Step Solution
Key Concept: Since a, b, c are in AP, 2b = a + c. Applying row operations R1 -> R1 + R3 - 2R2 will make the third column zero.
Let the determinant be \Delta. Since a, b, c are in AP, we have 2b = a + c. Applying R1 -> R1 + R3, the third column becomes (2x + a + c) = (2x + 2b). Now, applying R1 -> R1 - 2R2, the third column becomes (2x + 2b) - 2(x + b) = 0. Since one column is zero, the determinant is 0.
Correct Answer: C