Positive Definite Quadratic Form and System of Linear Equations

To determine which of the statements are true, we first analyze the condition for a tuple $(a, b, c) \in \mathbb{R}^3$ to belong to the set $S$.

The set $S$ is defined as: $S = \left\{(a,b,c) \in \mathbb{R}^3 : ax^2 + 2bxy + cy^2 > 0 \text{ for all } (x,y) \in \mathbb{R}^2 \setminus \{(0,0)\}\right\}$

The expression $P(x,y) = ax^2 + 2bxy + cy^2$ is a binary quadratic form. For $P(x,y) > 0$ for all non-zero $(x,y)$, the quadratic form must be positive definite.

1. Setting $y = 0$ and $x = 1$, we get: $P(1,0) = a > 0$

2. Setting $x = 0$ and $y = 1$, we get: $P(0,1) = c > 0$

3. For $y \neq 0$, we can divide $P(x,y)$ by $y^2$: $\frac{P(x,y)}{y^2} = a\left(\frac{x}{y}\right)^2 + 2b\left(\frac{x}{y}\right) + c > 0$ Let $t = \frac{x}{y} \in \mathbb{R}$. The quadratic polynomial $Q(t) = at^2 + 2bt + c$ must be strictly positive for all $t \in \mathbb{R}$. Since $a > 0$, this requires the discriminant to be strictly negative: $\Delta = (2b)^2 - 4ac < 0 \implies 4b^2 - 4ac < 0 \implies ac - b^2 > 0$

Thus, $(a, b, c) \in S \iff a > 0, \, c > 0, \text{ and } ac - b^2 > 0$.

---

Analysis of Options:

Option (A): Check if $\left(2, \frac{7}{2}, 6\right) \in S$

Here, $a = 2$, $b = \frac{7}{2}$, and $c = 6$. We check the condition $ac - b^2 > 0$: $ac - b^2 = 2 \times 6 - \left(\frac{7}{2}\right)^2 = 12 - \frac{49}{4} = \frac{48 - 49}{4} = -\frac{1}{4} < 0$ Since $ac - b^2 < 0$, the quadratic form is not positive definite. Therefore, $\left(2, \frac{7}{2}, 6\right) \notin S$. Option (A) is FALSE.

---

Option (B): If $\left(3, b, \frac{1}{12}\right) \in S$, then $|2b| < 1$

Here, $a = 3$ and $c = \frac{1}{12}$. Since $\left(3, b, \frac{1}{12}\right) \in S$, we must have: $ac - b^2 > 0 \implies 3 \times \frac{1}{12} - b^2 > 0 \implies \frac{1}{4} - b^2 > 0$ $b^2 < \frac{1}{4} \implies |b| < \frac{1}{2} \implies |2b| < 1$ Option (B) is TRUE.

---

Option (C): System of linear equations for $(a,b,c) \in S$:

$\begin{array}{l} ax + by = 1 \\ bx + cy = -1 \end{array}$

The determinant of the coefficient matrix $M = \begin{pmatrix} a & b \\ b & c \end{pmatrix}$ is: $\det(M) = \begin{vmatrix} a & b \\ b & c \end{vmatrix} = ac - b^2$

Since $(a,b,c) \in S$, we know that $ac - b^2 > 0$, which implies $\det(M) \neq 0$. A non-zero determinant guarantees that the system of linear equations has a unique solution. Option (C) is TRUE.

---

Option (D): System of linear equations for $(a,b,c) \in S$:

$\begin{array}{l} (a+1)x + by = 0 \\ bx + (c+1)y = 0 \end{array}$

The determinant of the coefficient matrix $M' = \begin{pmatrix} a+1 & b \\ b & c+1 \end{pmatrix}$ is: $\det(M') = \begin{vmatrix} a+1 & b \\ b & c+1 \end{vmatrix} = (a+1)(c+1) - b^2 = ac + a + c + 1 - b^2 = (ac - b^2) + a + c + 1$

Since $(a,b,c) \in S$, we have:

• $ac - b^2 > 0$
• $a > 0$
• $c > 0$

Summing these positive terms: $\det(M') = (ac - b^2) + a + c + 1 > 0 + 0 + 0 + 1 = 1 > 0$

Since $\det(M') \neq 0$, the homogeneous system has only the trivial solution $(x, y) = (0, 0)$, which is a unique solution. Option (D) is TRUE.

---

Conclusion:

The correct statements are (B), (C), and (D).