Let A = begin{bmatrix} 1 & -1 2 & alpha end{ | vedclass

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(A) Given $A = \begin{bmatrix} 1 & -1 \ 2 & \alpha \end{bmatrix}$ and $B = \begin{bmatrix} \beta & 1 \ 1 & 0 \end{bmatrix}$.$A + B = \begin{bmatrix}

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$2$

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(A) Given $A = \begin{bmatrix} 1 & -1 \ 2 & \alpha \end{bmatrix}$ and $B = \begin{bmatrix} \beta & 1 \ 1 & 0 \end{bmatrix}$.$A + B = \begin{bmatrix} \beta + 1 & 0 \ 3 & \alpha \end{bmatrix}$.$(A + B)^{2} = \begin{bmatrix} \beta + 1 & 0 \ 3 & \alpha \end{bmatrix} \begin{bmatrix} \beta + 1 & 0 \ 3 & \alpha \end{bmatrix} = \begin{bmatrix} (\beta + 1)^{2} & 0 \ 3(\beta + 1) + 3\alpha & \alpha^{2} \end{bmatrix}$.$A^{2} = \begin{bmatrix} 1 & -1 \ 2 & \alpha \end{bmatrix} \begin{bmatrix} 1 & -1 \ 2 & \alpha \end{bmatrix} = \begin{bmatrix} -1 & -1 - \alpha \ 2 + 2\alpha & \alpha^{2} - 2 \end{bmatrix}$.For $\alpha_{1}$,$(A + B)^{2} = A^{2} + \begin{bmatrix} 2 & 2 \ 2 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 1 - \alpha \ 4 + 2\alpha & \alpha^{2} \end{bmatrix}$.Comparing elements,$(\beta + 1)^{2} = 1 \implies \beta + 1 = \pm 1$. Also,$1 - \alpha = 0 \implies \alpha_{1} = 1$.For $\alpha_{2}$,$(A + B)^{2} = B^{2} = \begin{bmatrix} \beta & 1 \ 1 & 0 \end{bmatrix} \begin{bmatrix} \beta & 1 \ 1 & 0 \end{bmatrix} = \begin{bmatrix} \beta^{2} + 1 & \beta \ \beta & 1 \end{bmatrix}$.Comparing elements,the $(1,2)$ position gives $0 = \beta$,and the $(2,2)$ position gives $\alpha_{2}^{2} = 1$. From the $(2,1)$ position,$3(\beta + 1) + 3\alpha = \beta$. Substituting $\beta = 0$,we get $3(1) + 3\alpha = 0 \implies 3\alpha = -3 \implies \alpha_{2} = -1$.Thus,$|\alpha_{1} - \alpha_{2}| = |1 - (-1)| = |2| = 2$.

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