1. The problem is to analyze the quadratic form $ (a-1)x^2 + a^2 xy + (a+1)y^2 $. 2. We can write it as a quadratic form matrix: $$ Q = \begin{bmatrix} a-1 & \frac{a^2}{2} \\ \frac{a^2}{2} & a+1 \end{bmatrix} $$ 3. To understand the nature of this quadratic form, we can compute its discriminant $\Delta$: $$ \Delta = (a^2)^2 - 4(a-1)(a+1) = a^4 - 4(a^2 - 1) = a^4 - 4a^2 + 4 $$ 4. Simplify the discriminant: $$ \Delta = (a^2 - 2)^2 $$ 5. Since $\Delta = (a^2 - 2)^2 \geq 0$ for all real $a$, the quadratic form is always either positive definite, negative definite, or indefinite depending on $a$. 6. To classify definiteness, check the leading principal minor: $$ D_1 = a - 1 $$ 7. Check the determinant of $Q$: $$ D_2 = \det(Q) = (a-1)(a+1) - \left(\frac{a^2}{2}\right)^2 = a^2 - 1 - \frac{a^4}{4} $$ 8. Simplify $D_2$: $$ D_2 = -\frac{a^4}{4} + a^2 - 1 $$ 9. The quadratic form is positive definite if $D_1 > 0$ and $D_2 > 0$. 10. So, $a - 1 > 0 \Rightarrow a > 1$ and $D_2 > 0$. 11. Analyze $D_2 > 0$: $$ -\frac{a^4}{4} + a^2 - 1 > 0 \Rightarrow a^4 - 4a^2 + 4 < 0 $$ 12. But from step 4, $a^4 - 4a^2 + 4 = (a^2 - 2)^2 \geq 0$, so $D_2 \leq 0$ always. 13. Therefore, the quadratic form is never positive definite. 14. Similarly, check for negative definiteness or indefiniteness by testing values of $a$. 15. For example, at $a=0$, the form is $-1 x^2 + 0 + 1 y^2 = -x^2 + y^2$, which is indefinite. 16. At $a=2$, $D_1 = 1 > 0$, $D_2 = -\frac{16}{4} + 4 - 1 = -4 + 4 - 1 = -1 < 0$, so indefinite. 17. Conclusion: The quadratic form is indefinite for all real $a$. Final answer: The quadratic form $ (a-1)x^2 + a^2 xy + (a+1)y^2 $ is indefinite for all real values of $a$.