If $A=\left[\begin{array}{cc}\hfill \sqrt[]{2}\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill \sqrt[]{2}\hfill \end{array}\right]$ , $B=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right]$ , C = ABA^T and X = AC²A^T then |X| is equal to

|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| = 2

$\underset{2024times}{\underset{?}{adj\left(adj\left(adj....\left(a\right)\right)\right)}}={\left|A\right|}^{{\left(n-1\right)}^{2024}}$            

$={\left|A\right|}^{{\left(n-1\right)}^{2024}}$                                             

$=2^{2^{2024}}$                       &nb

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

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

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

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

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

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

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