Let X=begin{bmatrix} 0 & 1 & 0 0 & 0 & 1 0 | vedclass

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(A) Given $X = \begin{bmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ 0 & 0 & 0 \end{bmatrix}$,we find $X^{2} = \begin{bmatrix} 0 & 0 & 1 \ 0 & 0 & 0 \ 0 & 0 & 0

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$100$

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(A) Given $X = \begin{bmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ 0 & 0 & 0 \end{bmatrix}$,we find $X^{2} = \begin{bmatrix} 0 & 0 & 1 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{bmatrix}$ and $X^{3} = O$.Then $Y = \alpha I + \beta X + \gamma X^{2} = \begin{bmatrix} \alpha & \beta & \gamma \ 0 & \alpha & \beta \ 0 & 0 & \alpha \end{bmatrix}$.Since $Y \cdot Y^{-1} = I$,we have:$\begin{bmatrix} \alpha & \beta & \gamma \ 0 & \alpha & \beta \ 0 & 0 & \alpha \end{bmatrix} \begin{bmatrix} \frac{1}{5} & \frac{-2}{5} & \frac{1}{5} \ 0 & \frac{1}{5} & \frac{-2}{5} \ 0 & 0 & \frac{1}{5} \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$.Comparing the elements:$1$. $\frac{\alpha}{5} = 1 \Rightarrow \alpha = 5$.$2$. $-\frac{2\alpha}{5} + \frac{\beta}{5} = 0 \Rightarrow -2(5) + \beta = 0 \Rightarrow \beta = 10$.$3$. $\frac{\alpha}{5} - \frac{2\beta}{5} + \frac{\gamma}{5} = 0 \Rightarrow 5 - 2(10) + \gamma = 0 \Rightarrow 5 - 20 + \gamma = 0 \Rightarrow \gamma = 15$.Finally,calculate $(\alpha - \beta + \gamma)^{2} = (5 - 10 + 15)^{2} = (10)^{2} = 100$.

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