Let $A$ be a $3 \times 3$ matrix such that $A^2 - 5A + 7I = 0$.
Statement-$I$: ${A^{-1}} = \frac{1}{7}(5I - A)$.
Statement-$II$: The polynomial $A^3 - 2A^2 - 3A + I$ can be reduced to $5(A - 4I)$.

List $I$	List $II$
$P.$ Let $y(x)=\cos \left(3 \cos ^{-1} x\right), x \in[-1,1], x \neq \pm \frac{\sqrt{3}}{2}$. Then $\frac{1}{y(x)}\left\{\left(x^2-1\right) \frac{d^2 y(x)}{d x^2}+x \frac{d y(x)}{d x}\right\}$ equals	$1. \ 1$
$Q.$ Let $A_1, A_2, \ldots, A_n(n>2)$ be the vertices of a regular polygon of $n$ sides with its centre at the origin. Let $\vec{a}_k$ be the position vector of the point $A_k, k=1,2, \ldots, n$. If $\left|\sum_{k=1}^{n-1}\left(\vec{a}_k \times \vec{a}_{k+1}\right)\right|=\left|\sum_{k=1}^{n-1}\left(\vec{a}_k \cdot \vec{a}_{k+1}\right)\right|$,then the minimum value of $n$ is	$2. \ 2$
$R.$ If the normal from the point $P(h, 1)$ on the ellipse $\frac{x^2}{6}+\frac{y^2}{3}=1$ is perpendicular to the line $x+y=8$,then the value of $h$ is	$3. \ 8$
$S.$ Number of positive solutions satisfying the equation $\tan ^{-1}\left(\frac{1}{2 x+1}\right)+\tan ^{-1}\left(\frac{1}{4 x+1}\right)=\tan ^{-1}\left(\frac{2}{x^2}\right)$ is	$4. \ 9$

Codes: $P \quad Q \quad R \quad S$

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