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(B) For the system of linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero:$\begin{vmatrix} \alpha &

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$3$

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(B) For the system of linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero:$\begin{vmatrix} \alpha & \beta & \gamma \ \beta & \gamma & \alpha \ \gamma & \alpha & \beta \end{vmatrix} = 0$Expanding the determinant:$\alpha(\gamma\beta - \alpha^{2}) - \beta(\beta^{2} - \alpha\gamma) + \gamma(\beta\alpha - \gamma^{2}) = 0$$\alpha\beta\gamma - \alpha^{3} - \beta^{3} + \alpha\beta\gamma + \alpha\beta\gamma - \gamma^{3} = 0$$3\alpha\beta\gamma - (\alpha^{3} + \beta^{3} + \gamma^{3}) = 0$Using the identity $\alpha^{3} + \beta^{3} + \gamma^{3} - 3\alpha\beta\gamma = (\alpha + \beta + \gamma)(\alpha^{2} + \beta^{2} + \gamma^{2} - (\alpha\beta + \beta\gamma + \gamma\alpha))$,we have:$-(\alpha + \beta + \gamma)(\alpha^{2} + \beta^{2} + \gamma^{2} - (\alpha\beta + \beta\gamma + \gamma\alpha)) = 0$From Vieta's formulas,$\alpha + \beta + \gamma = -a$ and $\alpha\beta + \beta\gamma + \gamma\alpha = b$. Also,$\alpha^{2} + \beta^{2} + \gamma^{2} = (\alpha + \beta + \gamma)^{2} - 2(\alpha\beta + \beta\gamma + \gamma\alpha) = a^{2} - 2b$.Substituting these values:$-(-a)(a^{2} - 2b - b) = 0$$a(a^{2} - 3b) = 0$Since $a 
eq 0$,we must have $a^{2} - 3b = 0$,which implies $a^{2} = 3b$.Therefore,$\frac{a^{2}}{b} = 3$.

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