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Matrices and Determinants question

2024 · Shift 1 · Q23
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Matrices and Determinants question

2024 · Shift 1 · Q23

JEE AdvancedMathematicsMatrices and DeterminantsMultiple correct+4 / −2
Let R2\mathbb{R}^2R2 denote R×R\mathbb{R} \times \mathbb{R}R×R. Let S={(a,b,c):a,b,c∈R and ax2+2bxy+cy2>0 for all (x,y)∈R2−{(0,0)}}.S=\left\{(a, b, c): a, b, c \in \mathbb{R} \text { and } a x^2+2 b x y+c y^2\gt 0 \text { for all }(x, y) \in \mathbb{R}^2-\{(0,0)\}\right\} .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?
  1. A
    (2,72,6)∈S\left(2, \frac{7}{2}, 6\right) \in S(2,27​,6)∈S
  2. B
    If (3,b,112)∈S\left(3, b, \frac{1}{12}\right) \in S(3,b,121​)∈S, then ∣2b∣<1|2 b|\lt 1∣2b∣<1.
  3. C
    For any given (a,b,c)∈S(a, b, c) \in S(a,b,c)∈S, the system of linear equations ax+by=1bx+cy=−1\begin{aligned} & a x+b y=1 \\ & b x+c y=-1 \end{aligned}​ax+by=1bx+cy=−1​ has a unique solution.
  4. D
    For any given (a,b,c)∈S(a, b, c) \in S(a,b,c)∈S, the system of linear equations (a+1)x+by=0bx+(c+1)y=0\begin{aligned} & (a+1) x+b y=0 \\ & b x+(c+1) y=0 \end{aligned}​(a+1)x+by=0bx+(c+1)y=0​ has a unique solution.
View written solutionFree

Correct answer: B, C, D

We are given S={(a,b,c):a,b,c∈R and ax2+2bxy+cy2>0 for all (x,y)≠(0,0)}.S=\left\{(a,b,c): a,b,c\in\mathbb R \text{ and } ax^2+2bxy+cy^2>0 \text{ for all }(x,y)\neq (0,0)\right\}.S={(a,b,c):a,b,c∈R and ax2+2bxy+cy2>0 for all (x,y)=(0,0)}.

This means the quadratic form Q(x,y)=ax2+2bxy+cy2Q(x,y)=ax^2+2bxy+cy^2Q(x,y)=ax2+2bxy+cy2 is positive definite.

For a quadratic form in two variables, the necessary and sufficient conditions for positive definiteness are:

a>0,c>0,ac−b2>0.a>0,\qquad c>0,\qquad ac-b^2>0.a>0,c>0,ac−b2>0.

Equivalently, the symmetric matrix A=(abbc)A=\begin{pmatrix}a&b\\ b&c\end{pmatrix}A=(ab​bc​) is positive definite iff its leading principal minors are positive: a>0,det⁡A=ac−b2>0.a>0,\qquad \det A=ac-b^2>0.a>0,detA=ac−b2>0.

We now check each option.

1. Option A

Given (a,b,c)=(2,72,6)(a,b,c)=\left(2,\frac72,6\right)(a,b,c)=(2,27​,6).

Check: a=2>0,a=2>0,a=2>0, ac−b2=2⋅6−(72)2=12−494=48−494=−14<0.ac-b^2=2\cdot 6-\left(\frac72\right)^2=12-\frac{49}{4}=\frac{48-49}{4}=-\frac14<0.ac−b2=2⋅6−(27​)2=12−449​=448−49​=−41​<0.

So the quadratic form is not positive definite. Hence, (2,72,6)∉S.\left(2,\frac72,6\right)\notin S.(2,27​,6)∈/S.

So A is false.

2. Option B

Suppose (3,b,112)∈S.\left(3,b,\frac1{12}\right)\in S.(3,b,121​)∈S.

Then positive definiteness gives ac−b2>0.ac-b^2>0.ac−b2>0. Substitute a=3a=3a=3, c=112c=\frac1{12}c=121​: 3⋅112−b2>03\cdot \frac1{12}-b^2>03⋅121​−b2>0 14−b2>0\frac14-b^2>041​−b2>0 b2<14b^2<\frac14b2<41​ ∣b∣<12.|b|<\frac12.∣b∣<21​. Therefore, ∣2b∣<1.|2b|<1.∣2b∣<1.

So B is true.

3. Option C

For any (a,b,c)∈S(a,b,c)\in S(a,b,c)∈S, consider the system

a x+b y&=1\\ b x+c y&=-1 \end{aligned}$$ Its coefficient matrix is $$A=\begin{pmatrix}a&b\\ b&c\end{pmatrix}.$$ A linear system has a unique solution iff the determinant of the coefficient matrix is nonzero. Since $(a,b,c)\in S$, we know $$ac-b^2>0.$$ Hence $$\det A=ac-b^2\neq 0.$$ Therefore the system has a unique solution. So **C is true**. --- ## 4. Option D Now consider the system $$\begin{aligned} (a+1)x+by&=0\\ bx+(c+1)y&=0 \end{aligned}$$ Its coefficient matrix is $$B=\begin{pmatrix}a+1&b\\ b&c+1\end{pmatrix}.$$ For uniqueness, we need $$\det B=(a+1)(c+1)-b^2\neq 0.$$ Expand: $$\det B=ac+a+c+1-b^2=(ac-b^2)+(a+c+1).$$ Since $(a,b,c)\in S$, we have $$ac-b^2>0,\qquad a>0,\qquad c>0.$$ Thus $$a+c+1>0.$$ So $$\det B=(ac-b^2)+(a+c+1)>0.$$ Hence $\det B\neq 0$, and the system has a unique solution. So **D is true**. --- ## Final conclusion True statements are: $$\boxed{B,\ C,\ D}$$
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