By Vieta's formulas, we have:
\[ \alpha + \beta + \gamma = \frac{3}{2} \] \[ \alpha\beta + \beta\gamma + \gamma\alpha = \frac{5}{2} \] \[ \alpha\beta\gamma = \frac{7}{2} \]
We want to find \( \Sigma \alpha^2 \beta^2 = \alpha^2 \beta^2 + \beta^2 \gamma^2 + \gamma^2 \alpha^2 \).
We know that:
\[ (\alpha\beta + \beta\gamma + \gamma\alpha)^2 = \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 + 2\alpha\beta\gamma(\alpha + \beta + \gamma) \]
So,
\[ \Sigma \alpha^2 \beta^2 = (\alpha\beta + \beta\gamma + \gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha + \beta + \gamma) \]
Substituting the values from Vieta's formulas:
\[ \Sigma \alpha^2 \beta^2 = \left(\frac{5}{2}\right)^2 - 2\left(\frac{7}{2}\right)\left(\frac{3}{2}\right) \] \[ = \frac{25}{4} - 2\left(\frac{21}{4}\right) \] \[ = \frac{25}{4} - \frac{42}{4} \] \[ = \frac{25 - 42}{4} \] \[ = -\frac{17}{4} \]
Therefore, \( \Sigma \alpha^2 \beta^2 = -\frac{17}{4} \).
Final Answer: The final answer is \( \boxed{(1)} \).
If \( A = \begin{pmatrix} x & y & y \\ y & x & y \\ y & y & x \end{pmatrix} \) and \( 5A^{-1} = \begin{pmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{pmatrix} \), then \( A^2 - 4A \) is: