Let J_(n,m) = ∫₀^(1/2) xⁿ/(xᵐ - 1)dx, ∀ n > m and n, m ∈ N. Consider a matrix A = [aᵢⱼ]_(3 × 3) where aᵢⱼ = { j_(6 + i,3) - j_(i + 3,3),, i ≤ j; 0,, i > j . Then | adjA⁻¹ | is :

Solution

Maths sir · 4 steps

Combine the two integrals first, and the awkward x³ - 1 cancels to leave a plain power of x.

1. Simplify each entry. Both integrals have the same denominator, so subtract the numerators: xⁱ⁺⁶ - xⁱ⁺³ = xⁱ⁺³(x³ - 1). The factor x³ - 1 cancels, leaving an easy integral.

   J_(6+i,3) - J_(i+3,3) = ∫₀^(1/2) xⁱ⁺³ dx = 1/((i+4) 2ⁱ⁺⁴)

2. Upper triangular matrix. Every entry below the diagonal is 0, so |A| is the product of the diagonal entries. Putting i = 1, 2, 3 gives |A| = 1/(5 × 2⁵) × 1/(6 × 2⁶) × 1/(7 × 2⁷) = 1/(210 × 2¹⁸) = 1/(105 × 2¹⁹).
3. Adjoint of the inverse. For any 3 × 3 invertible matrix B, |adj B| = |B|². Take B = A⁻¹, which has |A⁻¹| = 1/|A| = 105 × 2¹⁹.
4. So. |adj A⁻¹| = (105 × 2¹⁹)² = (105)² × 2³⁸, which is option (c).

The idea to keep. For a triangular matrix, the determinant is the product of the diagonal entries. Also remember |adj B| = |B|ⁿ⁻¹ for an n × n matrix.

Answer from the NTA final answer key. Solution by PrepWiser.