# Solve the Following Questions

Question:

Let $J_{n, m}=\int_{0}^{\frac{1}{2}} \frac{x^{n}}{x^{m}-1} d x, \quad \forall n>m$ and $n, m \in N$

Consider a matrix $A=\left[a_{i j}\right]_{3 \times 3}$ where

$a_{i j}=\left\{\begin{array}{cl}J_{6+i, 3}-J_{i+3,3}, & i \leq j \\ 0, & i>j\end{array} .\right.$ Then $\left|\operatorname{adj} \mathrm{A}^{-1}\right|$ is :

1. $(15)^{2} \times 2^{42}$

2. $(15)^{2} \times 2^{34}$

3. $(105)^{2} \times 2^{38}$

4. $(105)^{2} \times 2^{36}$

Correct Option: , 3

Solution:

$\left[\begin{array}{ccc}a & v & v \\ a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right]$

$\mathrm{J}_{6+\mathrm{i}, 3}-\mathrm{J}_{\mathrm{i}+3,3} ; \mathrm{i} \leq \mathrm{j}$

$\Rightarrow \int_{0}^{\frac{1}{2}} \frac{x^{6+i}}{x^{3}-1}-\int_{0}^{\frac{1}{2}} \frac{x^{i+3}}{x^{3}-1}$

$\Rightarrow \int_{0}^{1 / 2} \frac{x^{i+3}\left(x^{3}-1\right)}{x^{3}-1}$

$\Rightarrow \frac{x^{3+i+1}}{3+i+1}=\left(\frac{x^{4+i}}{4+i}\right)_{0}^{1 / 2}$

$\mathrm{a}_{i j}=\mathrm{j}_{6+i, 3}-\mathrm{j}_{i+3,3}=\frac{\left(\frac{1}{2}\right)^{4+i}}{4+\mathrm{i}}$

$a_{11}=\frac{\left(\frac{1}{2}\right)^{5}}{5}=\frac{1}{5.2^{5}}$

$\mathrm{a}_{12}=\frac{1}{5.2^{5}}$

$\mathrm{a}_{13}=\frac{1}{5.2^{5}}$

$\mathrm{a}_{22}=\frac{1}{6.2^{6}}$

$\mathrm{a}_{23}=\frac{1}{6.2^{6}}$

$\mathrm{a}_{33}=\frac{1}{7.2^{7}}$

$A=\left[\begin{array}{ccc}\frac{1}{5.2^{5}} & \frac{1}{5.2^{5}} & \frac{1}{5.2^{5}} \\ 0 & \frac{1}{6.2^{6}} & \frac{1}{6.2^{6}} \\ 0 & 0 & \frac{1}{7.2^{7}}\end{array}\right]$

$|\mathrm{A}|=\frac{1}{5.2^{5}}\left[\frac{1}{6.2^{6}} \times \frac{1}{7.2^{7}}\right]$

$|\mathrm{A}|=\frac{1}{210.2^{18}}$

$\left|\operatorname{adj} \mathrm{A}^{-1}\right|=\left|\mathrm{A}^{-1}\right|^{n-1}=\left|\mathrm{A}^{-1}\right|^{2}=\frac{1}{(|\mathrm{~A}|)^{2}}$

$\Rightarrow\left(210.2^{18}\right)^{2}$

$(105)^{2} \times 2^{38}$