2 $$\times$$ (i) $$-$$ (ii) $$-$$ (iii) gives :

$$-$$ (1 + $$\beta$$)z = 3 $$-$$ $$\alpha$$

For infinitely many solution

$$\beta$$ + 1 = 0 = 3 $$-$$ $$\alpha$$ $$\Rightarrow$$ ($$\alpha$$, $$\beta$$) = (3, $$-$$1)

Hence, $$\alpha$$ + $$\beta$$ $$-$$ $$\alpha$$$$\beta$$ = 5

adj (2A) = 2² adjA

$$\Rightarrow$$ adj(adj (2A)) = adj(4 adjA) = 16 adj (adj A)

= 16 | A | A

$$\Rightarrow$$ adj (32 | A | A) = (32 | A |)² adj A

12(32| A |)² |adj A | = 2³ (32 | A |)⁶ | adj A |

2³ . 2³⁰ | A |⁶ . | A |² = 2⁴¹

| A |⁸ = 2⁸ $$\Rightarrow$$ | A | = $$\pm$$2

| A |² = | A |² = 4

$$\left| {\matrix{ { - 2} & { - 2} & 0 \cr 2 & 0 & { - 1} \cr {{{\sin }^2}x} & {{{\cos }^2}x} & {1 + \cos 2x} \cr } } \right|\left( \matrix{ {R_1} \to {R_1} - {R_2} \hfill \cr \& \,{R_2} \to {R_2} - {R_3} \hfill \cr} \right)$$

= $$ - 2({\cos ^2}x) + 2(2 + 2\cos 2x + {\sin ^2}x)$$

= $$4 + 4\cos 2x - 2({\cos ^2}x - {\sin ^2}x)$$

$$ \therefore $$ $$f(x) = 4 + \underbrace {2\cos 2x}_{\max = 1}$$

$$ \Rightarrow $$ $$f{(x)_{\max }} = 4 + 2 = 6$$

For infinite solutions

$$\Delta$$ = $$\Delta$$₁ = $$\Delta$$₂ = $$\Delta$$₃ = 0

$$\Delta$$ = $$\left| {\matrix{ 1 & 1 & { - 1} \cr 1 & 2 & \alpha \cr 2 & { - 1} & 1 \cr } } \right| = 0$$

$$\Delta = \left| {\matrix{ 3 & 0 & 0 \cr 1 & 2 & \alpha \cr 2 & { - 1} & 1 \cr } } \right| = 0$$

$$\Delta$$ = 3(2 + $$\alpha$$) = 0

$$\Rightarrow$$ $$\alpha$$ = $$-$$2

$${\Delta _2} = \left| {\matrix{ 1 & 2 & { - 1} \cr 1 & 1 & { - 2} \cr 2 & \beta & 1 \cr } } \right| = 0$$

1(1 + 2$$\beta$$) $$-$$2(1 + 4) $$-$$ ($$\beta$$ $$-$$ 2) = 0

$$\beta$$ $$-$$ 7 = 0

$$\beta$$ = 7

$$\therefore$$ $$\alpha$$ + $$\beta$$ = 5