Quadratic Forms and Linear Systems
DAILY_CHALLENGE
Grade None
Question:
Let $\mathbb{R}^2$ denote $\mathbb{R}\times\mathbb{R}$. Let
$$S=\{(a,b,c):a,b,c\in\mathbb{R}\text{ and }ax^2+2bxy+cy^2>0\text{ for all }(x,y)\in\mathbb{R}^2-\{(0,0)\}\}.$$
Then which of the following statements is (are) TRUE?
$\left(2,\dfrac{7}{2},6\right)\in S$
If $\left(3,b,\dfrac{1}{12}\right)\in S$, then $|2b|<1$.
For any given $(a,b,c)\in S$, the system of linear equations $ax+by=1$ and $bx+cy=-1$ has a unique solution.
For any given $(a,b,c)\in S$, the system of linear equations $(a+1)x+by=0$ and $bx+(c+1)y=0$ has a unique solution.
Step-by-Step Solution
Key Concept: Positive definiteness equivalent to a>0 and ac−b²>0; positive definite matrix has positive determinant
$(a,b,c)\in S$ iff the quadratic form is positive definite: $a>0$ and $ac-b^2>0$.
(A) $(2,7/2,6)$: $ac-b^2=12-49/4=-1/4<0$. NOT in $S$. FALSE.
(B) $(3,b,1/12)\in S$: $3\cdot\frac{1}{12}-b^2>0\Rightarrow\frac{1}{4}>b^2\Rightarrow|b|<\frac{1}{2}\Rightarrow|2b|<1$. TRUE.
(C) System matrix $\begin{pmatrix}a&b\\b&c\end{pmatrix}$ has determinant $ac-b^2>0$ (positive definite). Unique solution. TRUE.
(D) Matrix $\begin{pmatrix}a+1&b\\b&c+1\end{pmatrix}$: det $=(a+1)(c+1)-b^2=ac+a+c+1-b^2=(ac-b^2)+(a+c+1)>0$ since $ac-b^2>0$, $a>0$, $c>0$. Unique (trivial) solution. TRUE.
Correct Answer: B, C, D