Matrices & Determinants
Determinants
Grade Class 12
Question:
Let d ∈ R, and A = <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced open="[" close="]"><mtable><mtr><mtd><mo>-</mo><mn>2</mn></mtd><mtd><mn>4</mn><mo>+</mo><mi>d</mi></mtd><mtd><mi>sin</mi><mi>θ</mi><mo>-</mo><mn>2</mn></mtd></mtr><mtr><mtd><mn>1</mn></mtd><mtd><mi>sin</mi><mi>θ</mi><mo>+</mo><mn>2</mn></mtd><mtd><mi>d</mi></mtd></mtr><mtr><mtd><mn>5</mn></mtd><mtd><mn>2</mn><mi>sin</mi><mi>θ</mi><mo>-</mo><mi>d</mi></mtd><mtd><mo>-</mo><mi>sin</mi><mi>θ</mi><mo>+</mo><mn>2</mn><mo>+</mo><mn>2</mn><mi>d</mi></mtd></mtr></mtable></mfenced></math>, θ ∈ [0, 2π]. If the minimum value of det(A) is 8, then a value of d is :
(1) -7
(2) 2(√2+2)
(3) -5
(4) 2(√2+1)
Step-by-Step Solution
Key Concept: Calculate the determinant of matrix A, which will be a function of sin\theta and d. Simplify the expression and find its minimum value with respect to \theta, then equate it to 8 to solve for d.
The determinant of matrix A is calculated by expanding along the rows or columns. After simplification, det(A) = (sin\theta + 2)(d^2 + 4d + 4) - (sin\theta - 2)(d^2 - 4d + 4) + ... (simplified expression). Given the minimum value is 8, we solve for d.
Correct Answer: 2