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(C) Given $A = \begin{bmatrix} 1 & 0 & 0 \ 0 & 2 & 0 \ 3 & 0 & -1 \end{bmatrix}$.Calculating powers of $A$:$A^2 = \begin{bmatrix} 1 & 0 & 0 \ 0 & 4

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$4$

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(C) Given $A = \begin{bmatrix} 1 & 0 & 0 \ 0 & 2 & 0 \ 3 & 0 & -1 \end{bmatrix}$.Calculating powers of $A$:$A^2 = \begin{bmatrix} 1 & 0 & 0 \ 0 & 4 & 0 \ 0 & 0 & 1 \end{bmatrix}$,$A^3 = \begin{bmatrix} 1 & 0 & 0 \ 0 & 8 & 0 \ 3 & 0 & -1 \end{bmatrix}$,$A^4 = \begin{bmatrix} 1 & 0 & 0 \ 0 & 16 & 0 \ 0 & 0 & 1 \end{bmatrix}$.By induction,for even $n$,$A^n = \begin{bmatrix} 1 & 0 & 0 \ 0 & 2^n & 0 \ 0 & 0 & 1 \end{bmatrix}$ and for odd $n$,$A^n = \begin{bmatrix} 1 & 0 & 0 \ 0 & 2^n & 0 \ 3 & 0 & -1 \end{bmatrix}$.Thus,$A^{20} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 2^{20} & 0 \ 0 & 0 & 1 \end{bmatrix}$ and $A^{19} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 2^{19} & 0 \ 3 & 0 & -1 \end{bmatrix}$.Substituting into the equation $A^{20} + \alpha A^{19} + \beta A = I_3$ (where $I_3$ is the identity matrix):$\begin{bmatrix} 1+\alpha+\beta & 0 & 0 \ 0 & 2^{20}+\alpha 2^{19}+2\beta & 0 \ 3\alpha+3\beta & 0 & 1-\alpha-\beta \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 4 & 0 \ 0 & 0 & 1 \end{bmatrix}$.Comparing elements: $1+\alpha+\beta = 1 \Rightarrow \alpha+\beta = 0 \Rightarrow \beta = -\alpha$.Also,$3\alpha+3\beta = 0$ (consistent with $\alpha+\beta=0$) and $1-\alpha-\beta = 1 \Rightarrow \alpha+\beta = 0$.Finally,$2^{20} + \alpha 2^{19} + 2\beta = 4$. Substituting $\beta = -\alpha$:$2^{20} + \alpha 2^{19} - 2\alpha = 4 \Rightarrow \alpha(2^{19}-2) = 4 - 2^{20}$.$\alpha = \frac{4 - 2^{20}}{2^{19}-2} = \frac{2(2 - 2^{19})}{2^{19}-2} = -2$.Since $\beta = -\alpha$,$\beta = 2$.Therefore,$\beta - \alpha = 2 - (-2) = 4$.

If the matrix $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 3 & 0 & -1 \end{bmatrix}$ satisfies the equation $A^{20} + \alpha A^{19} + \beta A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ for some real numbers $\alpha$ and $\beta$,then $\beta - \alpha$ is equal to ........ .

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