Matrices & Determinants
Properties of determinants
Grade Class 12
Question:
Let a - 2b + c = 1. If f(x) = <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced open="|"><mtable><mtr><mtd><mi>x</mi><mo>+</mo><mi>a</mi></mtd><mtd><mi>x</mi><mo>+</mo><mn>2</mn></mtd><mtd><mi>x</mi><mo>+</mo><mn>1</mn></mtd></mtr><mtr><mtd><mi>x</mi><mo>+</mo><mi>b</mi></mtd><mtd><mi>x</mi><mo>+</mo><mn>3</mn></mtd><mtd><mi>x</mi><mo>+</mo><mn>2</mn></mtd></mtr><mtr><mtd><mi>x</mi><mo>+</mo><mi>c</mi></mtd><mtd><mi>x</mi><mo>+</mo><mn>4</mn></mtd><mtd><mi>x</mi><mo>+</mo><mn>3</mn></mtd></mtr></mtable></mfenced></math>, then :
(1) f(-50) = 501
(2) f(-50) = -1
(3) f(50) = 1
(4) f(50) = -501
Step-by-Step Solution
Key Concept: Apply row operations R2 -> R2 - R1 and R3 -> R3 - R2, then R3 -> R3 - R2 to simplify the determinant. Given a - 2b + c = 1, the determinant becomes independent of x.
Let the determinant be f(x). Applying R2 -> R2 - R1 and R3 -> R3 - R2: f(x) = |x+a x+2 x+1; b-a 1 1; c-b 1 1|. Since R2 and R3 are not identical, apply R3 -> R3 - R2: f(x) = |x+a x+2 x+1; b-a 1 1; c-2b+a 0 0|. Expanding along R3: f(x) = (c-2b+a) * (2-1) = c-2b+a. Given a-2b+c=1, f(x) = 1. Thus, f(50) = 1.
Correct Answer: 3