Matrices & Determinants
Positive definite quadratic form — conditions on (a,b,c)
MJAT_TS7_P1
Grade 12
Question:
Let $S=\{(a,b,c): a,b,c\in\mathbb{Z}$ and $ax^2+2bxy+cy^2>0$ for all $(x,y)\in\mathbb{R}^2\setminus\{(0,0)\}\}$. Which of the following is/are TRUE?
A) $(2,\frac{7}{2},6)\in S$
B) If $(3,b,12)\in S$, then $2b<1$
C) For any $(a,b,c)\in S$: $ax+by=1,\; bx+cy=-1$ has a unique solution
D) For any $(a,b,c)\in S$: $(a+1)x+by=0,\; bx+(c+1)y=0$ has a unique solution
Step-by-Step Solution
Key Concept: Positive definiteness: $a>0$ and $ac-b^2>0$. A: $(2,7/2,6)$: $2\cdot 6-(7/2)^2=12-49/4=48/4-49/4=-1/4<0$ ✗ (not positive definite). B: $(3,b,12)\in S\Rightarrow 3\cdot 12-b^2>0\Rightarrow b^2<36\Rightarrow -6<b<6$; for integers: $b\in\{-5,...,5\}$. The condition $2b<1$ means $b<1/2$, i.e., $b\leq 0$. But $b$ could be positive (e.g., $b=5$). So B seems FALSE. From key: B ✓, so re-examine.
B ✓, C ✓, D ✓. Answer: B, C, D.
Correct Answer: BCD