Question 6

The condition $ax^2 + 2bxy + cy^2 > 0$ for all $(x, y) \neq (0,0)$ implies that the quadratic form is positive definite. This occurs if and only if $a > 0$ and the determinant of the associated symmetric matrix $\begin{pmatrix} a & b \\ b & c \end{pmatrix}$ is positive, i.e., $ac - b^2 > 0$.

(A) For $(2, 7/2, 6)$, $a=2 > 0$ but $ac - b^2 = 2(6) - (7/2)^2 = 12 - 12.25 = -0.25 < 0$. Thus, it is not in $S$.

(B) For $(3, b, 1/12) \in S$, we must have $ac - b^2 > 0 \implies 3(1/12) - b^2 > 0 \implies 1/4 > b^2 \implies |b| < 1/2 \implies |2b| < 1$. This is correct.

(C) The system has a unique solution if the determinant $ac - b^2 \neq 0$. Since $(a, b, c) \in S \implies ac - b^2 > 0$, the determinant is non-zero. Correct.

(D) The determinant of this system is $(a+1)(c+1) - b^2 = ac + a + c + 1 - b^2 = (ac - b^2) + a + c + 1$. Since $ac - b^2 > 0$ and $a > 0$, it follows that $c > b^2/a \ge 0$, so $c > 0$. Thus, the determinant is the sum of positive terms plus 1, making it $> 1$. Therefore, it is never zero, and a unique solution exists. Correct.