Let A be a 3 $$ \times $$ 3 matrix such that A² $$-$$ 5A + 7I = 0

Statement - I :  

A^$$-$$1 = $${1 \over 7}$$ (5I $$-$$ A).

Statement - II :

The polynomial A³ $$-$$ 2A² $$-$$ 3A + I can be reduced to 5(A $$-$$ 4I).

Then :

If x is a solution of the equation, $$\sqrt {2x + 1} $$ $$ - \sqrt {2x - 1} = 1,$$ $$\,\,\left( {x \ge {1 \over 2}} \right),$$ then $$\sqrt {4{x^2} - 1} $$ is equal to :

Let P = {$$\theta $$ : sin$$\theta $$ $$-$$ cos$$\theta $$ = $$\sqrt 2 \,\cos \theta $$}

and Q = {$$\theta $$ : sin$$\theta $$ + cos$$\theta $$ = $$\sqrt 2 \,\sin \theta $$} be two sets. Then

The sum $$\sum\limits_{r = 1}^{10} {\left( {{r^2} + 1} \right) \times \left( {r!} \right)} $$ is equal to :