Question

Let $J_{n,m}=\underset{0}{\overset{1/2}{\int }}\frac{x^n}{x^m-1}dx,\forall n>m\text{and n},m\in N.$
Consider a matrix $A={\left[a_{ij}\right]}_{3\times 3}$ where
$a_{ij}=\{\begin{array}{cc}{J}_{6+i,3}-J_{i+3,3}& i\le j\\ 0& i>j\end{array}.$ Then $|adj{A}^{-1}|$ is

• A. $(105{)}^2\times 2^{38}$
• B. $(15{)}^2\times 2^{42}$
• C. $(15{)}^2\times 2^{34}$
• D. $(105{)}^2\times 2^{36}$

Answer 1

The correct option is A $(105{)}^2\times 2^{38}$

$J_{6+i,3}-J_{i+3,3}=\underset{0}{\overset{1/2}{\int }}\frac{x^{6+i}}{x^3-1}dx-\underset{0}{\overset{1/2}{\int }}\frac{x^{i+3}}{x^3-1}dx$
$=\underset{0}{\overset{1/2}{\int }}\frac{x^{i+3}(x^3-1)}{(x^3-1)}dx$
$={\left[\frac{x^{i+4}}{i+4}\right]}_0^{1/2}=\frac{{\left(\frac{1}{2}\right)}^{i+4}}{i+4}$
$\therefore \text{det(A)}=a_{11}\cdot a_{22}\cdot a_{33}$
$=\frac{{\left(\frac{1}{2}\right)}^5}{5}\cdot \frac{{\left(\frac{1}{2}\right)}^6}{6}\cdot \frac{{\left(\frac{1}{2}\right)}^7}{7}=\frac{1}{105\times 2^{19}}$
Now, $|{\text{adj A}}^{-1}|=\frac{1}{{\text{det A}}^2}=(105{)}^2\times 2^{38}$