Question

Let R² denote R × R. Let S = {(a, b, c) : a, b, c ∈ R and ax² + 2bxy + cy² > 0 for all (x, y) ∈ R² - {(0, 0}}.
Then which of the following statements is (are) TRUE ?

• A. \((2,\frac{7}{2},6)\in S\)
• B. If \((3,b,\frac{1}{12})\in S\), then |2b| < 1.
• C. For any given (a, b, c) ∈ S, the system of linear equations
  ax + by = 1
  bx + cy = -1
  has a unique solution.
• D. For any given (a, b, c) ∈ S, the system of linear equations
  (a + 1)x + by = 0
  bx + (c + 1)y = 0
  has a unique solution.

Answer 1

The Correct Option is B, C, D

To solve the problem, we analyze the set \( S \) and the properties of quadratic forms and linear systems associated with elements \((a,b,c) \in \mathbb{R}^3\).

Given:
\[ S = \{(a, b, c) : a,b,c \in \mathbb{R} \text{ and } a x^2 + 2 b x y + c y^2 > 0 \text{ for all } (x,y) \neq (0,0) \} \] So \( S \) is the set of coefficients defining a positive definite quadratic form.

Recall:
For the quadratic form \[ Q(x,y) = a x^2 + 2 b x y + c y^2, \] the form is positive definite if and only if: \[ a > 0, \quad \text{and} \quad \det \begin{pmatrix} a & b \\ b & c \end{pmatrix} = ac - b^2 > 0 \] and \[ c > 0 \] (since it must be positive for all \( (x,y) \neq (0,0) \)).

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1. Check if \((2, \frac{7}{2}, 6) \in S\):
\[ a = 2 > 0, \quad c = 6 > 0, \] Calculate the determinant: \[ ac - b^2 = 2 \times 6 - \left(\frac{7}{2}\right)^2 = 12 - \frac{49}{4} = \frac{48}{4} - \frac{49}{4} = -\frac{1}{4} < 0 \] Since the determinant is negative, the quadratic form is not positive definite.
Statement 1 is FALSE.

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2. If \((3, b, \frac{1}{12}) \in S\), then \(|2b| < 1\):
Conditions for positive definiteness:
\[ a = 3 > 0, \quad c = \frac{1}{12} > 0 \] and \[ ac - b^2 = 3 \times \frac{1}{12} - b^2 = \frac{1}{4} - b^2 > 0 \implies b^2 < \frac{1}{4} \] \[ |b| < \frac{1}{2} \] Multiply both sides by 2: \[ |2b| < 1 \] Statement 2 is TRUE.

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3. For any \((a, b, c) \in S\), the system:
\[ ax + by = 1 \\ bx + cy = -1 \] has a unique solution.

Analysis:
The coefficient matrix is: \[ M = \begin{pmatrix} a & b \\ b & c \end{pmatrix} \] Since \((a,b,c) \in S\), \(M\) is positive definite, so \( \det(M) = ac - b^2 > 0 \).
Therefore, \( M \) is invertible and the system has a unique solution.
Statement 3 is TRUE.

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4. For any \((a,b,c) \in S\), the system:
\[ (a+1)x + b y = 0 \\ b x + (c+1) y = 0 \] has a unique solution.

Analysis:
Coefficient matrix: \[ N = \begin{pmatrix} a+1 & b \\ b & c+1 \end{pmatrix} \] Determinant: \[ \det(N) = (a+1)(c+1) - b^2 \] Since \( a > 0, c > 0 \), clearly \( a+1 > 1, c+1 > 1 \).
Also, from positive definiteness of \( M \), \( ac - b^2 > 0 \).
We want to check if \( \det(N) > 0 \). Let's analyze: \[ \det(N) = ac + a + c + 1 - b^2 = (ac - b^2) + (a + c + 1) \] Since \( ac - b^2 > 0 \) and \( a, c, 1 > 0 \), sum is positive.
Therefore, \( \det(N) > 0 \) and \( N \) is invertible.
Hence, the system has a unique solution (only trivial \( x = y = 0 \) if right-hand side is zero).
Statement 4 is TRUE.

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Summary of Truth Values:
- Statement 1: FALSE
- Statement 2: TRUE
- Statement 3: TRUE
- Statement 4: TRUE

Final Answer:
\[ \boxed{ \text{Statements 2, 3, and 4 are TRUE; Statement 1 is FALSE} } \]

Answer 2

To determine which statements are true, we first examine each case:

We consider the constraint: \(ax^2 + 2bxy + cy^2 > 0\) for all \((x,y) \in \mathbb{R}^2 - \{(0,0)\)\). 
This implies that the quadratic form must be positive definite, and one of the conditions for a quadratic form \(ax^2 + 2bxy + cy^2\) to be positive definite is \(a > 0\), \(c > 0\), and the determinant \(ac - b^2 > 0\).

Option 1:

\((2,\frac{7}{2},6)\in S\)

For these values, \(a = 2\), \(b = \frac{7}{2}\), \(c = 6\). 
Check positive definiteness: \(ac - b^2 = (2)(6) - \left(\frac{7}{2}\right)^2 = 12 - \frac{49}{4} = \frac{48}{4} - \frac{49}{4} = -\frac{1}{4} < 0\). 
Thus, this form is not positive definite, so this option is false.

Option 2:

If \((3,b,\frac{1}{12})\in S\), then |2b|<1.

Given that \((a,b,c)\in S\) implies \(a > 0\), \(c > 0\), and \(ac - b^2 > 0\), check for positive definiteness: \(3 \cdot \frac{1}{12} - b^2 = \frac{1}{4} - b^2 > 0\Rightarrow b^2 < \frac{1}{4}\Rightarrow |b| < \frac{1}{2}\). 
Hence, \(|2b| < 1\) since \(|b| < \frac{1}{2}\). 
This option is true.

Option 3:

For any \((a, b, c) \in S\), the system ax + by = 1 bx + cy = -1 has a unique solution.

The determinant \(D\) of the system is \(a \cdot c - b^2\). 
Since \(ac - b^2 > 0\) for \((a,b,c)\in S\), \(D\neq0\), implying a unique solution by the non-zero determinant. 
Therefore, this option is true.

Option 4:

For any \((a, b, c) \in S\), the system (a+1)x + by = 0 bx + (c+1)y = 0 has a unique solution.

Check the determinant \(D = (a+1)(c+1) - b^2 = ac + a + c + 1 - b^2\). 
We already have \(ac - b^2 > 0\) which implies \(ac + a + c + 1 > b^2 + a + c + 1 > 0\). 
Therefore, the determinant is non-zero, and the system has a unique solution; thus, this option is true.

Hence, the true statements are: If \((3,b,\frac{1}{12}) \in S\), then |2b| < 1; for any given \((a, b, c) \in S\), the system \(ax+by=1\), \(bx+cy=-1\) has a unique solution; and for any given \((a, b, c) \in S\), the system \((a+1)x+by=0\), \(bx+(c+1)y=0\) has a unique solution.