Let $$I_{n,m} = \int_0^{1/2} \frac{x^n}{x^m-1} dx$$, $$\forall n > m$$ and $$n, m \in N$$. Consider a matrix $$A = a_{ij_{3 \times 3}}$$ where $$a_{ij} = \begin{cases} I_{6+i,3} - I_{i+3,3}, & i \leq j \\ 0, & i > j \end{cases}$$. Then adj $$A^{-1}$$ is:

We have been given the integral

$$I_{n,m}= \int_{0}^{1/2}\frac{x^{\,n}}{x^{\,m}-1}\,dx,\qquad n>m,\;n,m\in\mathbb N,$$

and in the present problem the value of the parameter $$m$$ is fixed at $$m=3$$.

The entries of the $$3\times3$$ matrix $$A=[a_{ij}]$$ are defined as

$$a_{ij}= \begin{cases} I_{6+i,\,3}-I_{i+3,\,3}, & i\le j,\\[4pt] 0, & i>j. \end{cases}$$

Because each non-zero element depends only on the row-index $$i$$ and because every element below the main diagonal is zero, the matrix $$A$$ is upper-triangular. Consequently the determinant of $$A$$ equals the product of its diagonal elements. To find these diagonal elements we must first evaluate the difference $$I_{6+i,\,3}-I_{i+3,\,3}\;(i=1,2,3).$$

Let us put $$q=i+3\;(i=1,2,3)\;\Longrightarrow\;q=4,5,6$$. Then

$$I_{q+3,\,3}-I_{q,\,3}= \int_{0}^{1/2}\frac{x^{\,q+3}-x^{\,q}}{x^{\,3}-1}\,dx.$$

Observe the algebraic factorisation

$$x^{\,q+3}-x^{\,q}=x^{\,q}(x^{\,3}-1).$$

Substituting this in the integrand we get

$$\frac{x^{\,q+3}-x^{\,q}}{x^{\,3}-1}= \frac{x^{\,q}(x^{\,3}-1)}{x^{\,3}-1}=x^{\,q}.$$

Hence

$$I_{q+3,\,3}-I_{q,\,3}= \int_{0}^{1/2}x^{\,q}\,dx.$$

Using the standard power-rule integral

$$\int x^{\,n}\,dx=\frac{x^{\,n+1}}{n+1}+C,$$

we find

$$I_{q+3,\,3}-I_{q,\,3}= \left[\frac{x^{\,q+1}}{q+1}\right]_{0}^{1/2}= \frac{(1/2)^{\,q+1}}{q+1}\;-\;0= \frac{1}{q+1}\;2^{-(q+1)}.$$

Remembering that $$q=i+3$$, we finally obtain

$$I_{6+i,\,3}-I_{i+3,\,3}= \frac{1}{(i+3)+1}\;2^{-(i+3+1)}=\frac{1}{i+4}\;2^{-(i+4)}.$$

Denote this quantity by $$d_i$$ for ease of writing:

$$d_i=\frac{1}{i+4}\,2^{-(i+4)},\qquad i=1,2,3.$$

Let us list the three required values explicitly:

$$ \begin{aligned} d_1 &=\frac{1}{1+4}\,2^{-(1+4)}=\frac{1}{5}\,2^{-5}= \frac{1}{5\cdot2^{5}}=\frac{1}{160},\\[6pt] d_2 &=\frac{1}{2+4}\,2^{-(2+4)}=\frac{1}{6}\,2^{-6}= \frac{1}{6\cdot2^{6}}=\frac{1}{384},\\[6pt] d_3 &=\frac{1}{3+4}\,2^{-(3+4)}=\frac{1}{7}\,2^{-7}= \frac{1}{7\cdot2^{7}}=\frac{1}{896}. \end{aligned} $$

Since $$A$$ is upper-triangular, its determinant is simply

$$\det A=d_1\,d_2\,d_3.$$

Multiplying the three fractions we get

$$ \begin{aligned} \det A &=\left(\frac{1}{5\cdot2^{5}}\right) \left(\frac{1}{6\cdot2^{6}}\right) \left(\frac{1}{7\cdot2^{7}}\right)\\[6pt] &=\frac{1}{(5\cdot6\cdot7)\;2^{5+6+7}}\\[6pt] &=\frac{1}{210\;2^{18}}. \end{aligned} $$

Clearly $$A$$ is nonsingular, so $$A^{-1}$$ exists. Now recall two standard identities from matrix theory:

1. For any invertible matrix $$B$$, $$\det(B^{-1})=\dfrac{1}{\det B}.$$

2. For any invertible $$n\times n$$ matrix $$B$$, the adjugate satisfies $$\operatorname{adj}B=\det(B)\,B^{-1}$$  and consequently

$$\det(\operatorname{adj}B)=(\det B)^{\,n-1}.$$

We will apply these with $$B=A^{-1}$$ and note that our size is $$n=3$$.

First,

$$\det(A^{-1})=\frac{1}{\det A}=210\;2^{18}.$$

Next, the determinant of the adjugate of $$A^{-1}$$ is, by the second identity,

$$ \det(\operatorname{adj}A^{-1})=\bigl(\det(A^{-1})\bigr)^{\,3-1} =\bigl(210\;2^{18}\bigr)^{2}. $$

Let us expand this square step by step:

$$ \begin{aligned} \bigl(210\;2^{18}\bigr)^{2}&=210^{2}\;(2^{18})^{2}\\[6pt] &=210^{2}\;2^{36}. \end{aligned} $$

Because $$210=2\cdot105,$$ we can factor out a further power of two:

$$ 210^{2}=(2\cdot105)^{2}=2^{2}\cdot105^{2}=4\,105^{2}. $$

Therefore

$$ \det(\operatorname{adj}A^{-1})=4\;105^{2}\;2^{36}=105^{2}\;2^{38}. $$

The numerical expression that matches this result among the options is exactly $$ (105)^{2}\times2^{38}$$, which is listed as Option 4.

Hence, the correct answer is Option 4.