Let $J_{n,m}=\int_{0}^{1/2} \frac{x^{n}}{x^{m}-1} dx, \forall n>m$ and $n, m \in N$. Consider a matrix $A=[a_{ij}]_{3 \times 3}$ where $a_{ij}=\begin{cases}J_{6+i,3}-J_{i+3,3}, & i \leq j \\ 0, & i>j\end{cases}$. Then $|\text{adj } A^{-1}|$ is:

JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Determinants

Let

J_{n, m} = \int_{0}^{1/2} \frac{x ^{n}}{x ^{m} - 1} d x, \forall n > m

and

n, m \in N

. Consider a matrix

A = [a_{ij}]_{3 \times 3}

where

a_{ij} = {J_{6 + i, 3} - J_{i + 3, 3}, 0, i \leq j i > j

. Then

∣ adj A^{- 1} ∣

is:

Answer: (C)

For an upper-triangular matrix the determinant is the product of the diagonal entries, and

a_{ii} = J_{i + 6, 3} - J_{i + 3, 3} = \int_{0}^{1/2} \frac{x ^{i + 6} - x ^{i + 3}}{x ^{3} - 1} d x = \int_{0}^{1/2} x^{i + 3} d x = \frac{( 1/2 ) ^{i + 4}}{i + 4} .

Thus

a_{11} = \frac{1}{160}

a_{22} = \frac{1}{384}

a_{33} = \frac{1}{896}

, giving

∣ A ∣ = \frac{1}{160 \cdot 384 \cdot 896} = \frac{1}{2 ^{19} \cdot 105} .

For a

3 \times 3

matrix

adj A^{- 1} = A^{- 1}^{2} = \frac{1}{∣ A ∣ ^{2}} = (105)^{2} \times 2^{38}

JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Single correct option