If the Boolean expression (p ∧ q) ○ (p ⊕ q) is a tautology, then ○ and ⊕ are respectively by:

  $A^2=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

$A^3=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^3\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]$        

$A^4\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^3\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^4\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$          

Similarly we get A¹⁹ =   $=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^{19}\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]\&A^{20}=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^{20}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

=   $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 4\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

$\Rightarrow 1+\alpha +\beta =1gives\alpha +\beta =0........\left(i\right)$        

  $2^{20}+\left(2^{19}-2\right)\alpha =4from\left(i\right)$

$\Rightarrow \alpha =\frac{4-2^{20}}{2^{19}-2}=\frac{4\left(1-2^{18}\right)}{-2\left(1-2^{18}\right)}=-2$          

So, b = 2

Hence b - a = 4

Given x + 2y – 3z = a

2x + 6y – 11z = b

x – 2y + 7z = c

Here    $\Delta =\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill -3\hfill \\ \hfill 2\hfill & \hfill 6\hfill & \hfill -11\hfill \\ \hfill 1\hfill & \hfill -2\hfill & \hfill 7\hfill \end{array}\right|\begin{array}{cc}\hfill \begin{array}{l}=\left(42-22\right)-2\left(14+11\right)-3\left(-4-6\right)\\ \end{array}\hfill & \hfill =20-50+30=0\hfill \end{array}$

For infinite solution 

20a – 8b – 4c = 0 Þ 5a = 2b + c

Sum of all elements of $A\cap B=2$   [Sum of natural number upto 100 which are neither divisible by 3 nor by 5]

$=2\left[\frac{100\times 101}{2}-3\left(\frac{33\times 34}{2}\right)-5\left(\frac{20\times 21}{2}\right)+15\left(\frac{6\times 7}{2}\right)\right]$

= 10100 – 3366 – 2100 + 630

              = 5264

Kindly go through the solution

B = (I – adjA)⁵

$N=M^2+M^4+.....+M^{98}$

$=\left(-{\alpha }^{2}I\right)+{\left(-{\alpha }^{2}I\right)}^2+....+{\left(-{\alpha }^{2}I\right)}^{49}$

$=I\left(-{\alpha }^2+{\alpha }^4-{\alpha }^6+....-{\alpha }^{98}\right)$

N =  $-I\left({\alpha }^2-{\alpha }^4+{\alpha }^6.......+{\alpha }^{98}\right)$

$=-I\frac{{\alpha }^2\left(1-{\left(-{\alpha }^2\right)}^{49}\right)}{1-\left(-{\alpha }^2\right)}$

N =  $\frac{-I{\alpha }^2\left(1+{\alpha }^{98}\right)}{1+{\alpha }^2}$  

Now  $\left(I-m^2\right)N=-2I$

$\left(I+{\alpha }^{2}I\right)\frac{\left(-I{\alpha }^2\cdot (1+{\alpha }^{98}\right)}{1+{\alpha }^2}=-2I$  

-> a¹⁰⁰ + a² = 2

->a = $\pm 1$