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(B) Given the system of linear equations:$1) \beta x + \alpha y - z = -1$$2) 3x - \beta y + \alpha z = 0$$3) \alpha x + \beta y + z = 1$Using Cramer's

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$(2, 1)$

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(B) Given the system of linear equations:$1) \beta x + \alpha y - z = -1$$2) 3x - \beta y + \alpha z = 0$$3) \alpha x + \beta y + z = 1$Using Cramer's rule, the solutions are $x = \frac{\Delta_1}{\Delta} = -1$, $y = \frac{\Delta_2}{\Delta} = 1$, and $z = \frac{\Delta_3}{\Delta} = 2$.Substitute these values into the equations:From $(1)$: $\beta(-1) + \alpha(1) - 2 = -1 \Rightarrow \alpha - \beta = 1$From $(2)$: $3(-1) - \beta(1) + \alpha(2) = 0 \Rightarrow 2\alpha - \beta = 3$From $(3)$: $\alpha(-1) + \beta(1) + 2 = 1 \Rightarrow -\alpha + \beta = -1 \Rightarrow \alpha - \beta = 1$Now solve the system of equations for $\alpha$ and $\beta$:$\alpha - \beta = 1$$2\alpha - \beta = 3$Subtract the first equation from the second:$(2\alpha - \beta) - (\alpha - \beta) = 3 - 1$$\alpha = 2$Substitute $\alpha = 2$ into $\alpha - \beta = 1$:$2 - \beta = 1 \Rightarrow \beta = 1$Thus, $(\alpha, \beta) = (2, 1)$.

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