Let A = begin{bmatrix} alpha & -1 6 & beta e | vedclass

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(D) Given $\operatorname{det}(A) = 0$,we have $\alpha \beta - (-6) = 0$,which implies $\alpha \beta = -6$. Given $\alpha + \beta = 1$,we solve the qua

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$\begin{bmatrix} 766 & -255 \ 1530 & -509 \end{bmatrix}$

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(D) Given $\operatorname{det}(A) = 0$,we have $\alpha \beta - (-6) = 0$,which implies $\alpha \beta = -6$. Given $\alpha + \beta = 1$,we solve the quadratic equation $x^2 - (\alpha + \beta)x + \alpha \beta = 0$,which is $x^2 - x - 6 = 0$. Factoring gives $(x - 3)(x + 2) = 0$,so $x = 3$ or $x = -2$. Since $\alpha > 0$,we have $\alpha = 3$ and $\beta = -2$. Thus,$A = \begin{bmatrix} 3 & -1 \ 6 & -2 \end{bmatrix}$. Calculating $A^2$: $A^2 = \begin{bmatrix} 3 & -1 \ 6 & -2 \end{bmatrix} \begin{bmatrix} 3 & -1 \ 6 & -2 \end{bmatrix} = \begin{bmatrix} 9-6 & -3+2 \ 18-12 & -6+4 \end{bmatrix} = \begin{bmatrix} 3 & -1 \ 6 & -2 \end{bmatrix} = A$. Since $A^2 = A$,it follows that $A^n = A$ for all $n \geq 1$. Using the binomial expansion: $(I + A)^8 = I + \binom{8}{1}A + \binom{8}{2}A^2 + \dots + \binom{8}{8}A^8$. Since $A^k = A$ for $k \geq 1$,$(I + A)^8 = I + A(\binom{8}{1} + \binom{8}{2} + \dots + \binom{8}{8})$. The sum $\sum_{k=1}^{8} \binom{8}{k} = 2^8 - 1 = 256 - 1 = 255$. Therefore,$(I + A)^8 = I + 255A = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} + 255 \begin{bmatrix} 3 & -1 \ 6 & -2 \end{bmatrix} = \begin{bmatrix} 1 + 765 & 0 - 255 \ 0 + 1530 & 1 - 510 \end{bmatrix} = \begin{bmatrix} 766 & -255 \ 1530 & -509 \end{bmatrix}$.

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	$4. \ 9$

Let $A = \begin{bmatrix} \alpha & -1 \\ 6 & \beta \end{bmatrix}$,$\alpha > 0$,such that $\operatorname{det}(A) = 0$ and $\alpha + \beta = 1$. If $I$ denotes the $2 \times 2$ identity matrix,then the matrix $(I + A)^8$ is:

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