Matrix A of order 3 * 3 is such that |A| = 2 if $n=\underset{2024times}{\underbrace{\left|adj\left(adj\left(adj....\left(a\right)\right)\right)\right|}}$  then remainder when n is divided by 9 is

|2A| = 2⁷

8|A| = 2⁷

Now |A| = α²–β² = 2⁴

α² = 16 + β²

α²– β² = 16

(α–β) (α+β) = 16

->α + β = 8 and

α – β = 2

->α = 5 and β = 3

|A| = 3

|B| = 1

->|C| = |ABA^T| = |A|B|A⁷| = |A|²|B|

= 9

->|X| = |A|C|²|A^T|

= 3 * 9² * 3 = 9 * 9² = 729

${\displaystyle \underset{0}{\overset{2}{\int }}\sqrt[]{2xdx}}-{\displaystyle \underset{0}{\overset{2}{\int }}\sqrt[]{2x-x^{2}dx}}={\displaystyle \underset{0}{\overset{1}{\int }}dy}-{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{1-y^2}dy-{\displaystyle \underset{0}{\overset{2}{\int }}\frac{y^2}{2}dy}+{\displaystyle \underset{1}{\overset{2}{\int }}2dy+I}}$

$\Rightarrow \frac{8}{3}-{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{1-t^2}dt}=1-\frac{8}{6}+2+I$

$I=1-{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{1-t^2}dt}$

${\displaystyle \underset{0}{\overset{2}{\int }}\sqrt[]{2xdx}}-{\displaystyle \underset{0}{\overset{2}{\int }}\sqrt[]{2x-x^{2}dx}}={\displaystyle \underset{0}{\overset{1}{\int }}dy}-{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{1-y^2}dy-{\displaystyle \underset{0}{\overset{2}{\int }}\frac{y^2}{2}dy}+{\displaystyle \underset{1}{\overset{2}{\int }}2dy+I}}$

$\Rightarrow \frac{8}{3}-{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{1-t^2}dt}=1-\frac{8}{6}+2+I$

$I=1-{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{1-t^2}dt}$

Given a > b

Area common to x² + y² $\le a^{2}and\frac{x^2}{a^2}+\frac{y^2}{b^2}\le 1$  

is  $\pi a^2-\pi ab=30\pi ..............\left(i\right)$  

Similarly  $\pi ab-\pi b^2=18\pi .................\left(ii\right)$  

Equation (i) and equation (ii)  $\frac{a}{b}=\frac{5}{3}$  

Equation (i) + equation (ii)  $a^2-b^2=48$  

a² = 75, b² = 27