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,$(I + A)^8 = I + 255A = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} + 255 \begin{bmatrix} 3 & -1 \ 6 & -2 \end{bmatrix} = \begin{bmatrix} 1 + 765 & 0 - 255 \ 0 + 1530 & 1 - 510 \end{bmatrix} = \begin{bmatrix} 766 & -255 \ 1530 & -509 \end{bmatrix}$.

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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