JEE Advanced 2024

LEVELJEE Main

Animated Solution for Mathematics - Matrices and Determinants: Let $R^{2}$ denote $R \times R$ . Let $S = {(a, b, c) : a, b, c \in R and a x^{2} + 2 b x y + c y^{2} > 0 for all (x, y) \in R^{2} - {(0, 0)}}$ . Then which of the following statements is (are) TRUE?

Q: Why do we need $a > 0$ and $c > 0$? Can't they be negative?

Great question! Think about the behavior of the expression as we move far away from the origin. If we set $y = 0$, the expression becomes $ax^{2}$. If $a$ were negative, then for any $x \neq 0$, the expression would be negative, violating the condition that it must be strictly positive for all $(x, y) \neq (0, 0)$. Thus, $a$ must be positive. By symmetry, setting $x = 0$ forces $c$ to be positive as well.

Q: Why is the condition $b^{2} < ac$ and not $b^{2} \leq ac$?

The problem specifies that the expression is strictly greater than zero. If $b^{2} = ac$, the discriminant is zero, which means the quadratic has a real root. At that root, the expression would equal zero. Since we need it to be strictly positive everywhere except the origin, we must exclude the case where the expression touches zero. Hence, $b^{2} < ac$ is the only valid condition.

Q: How does the determinant of a system of equations relate to the quadratic form $ax^{2} + 2bxy + cy^{2}$?

This is the beauty of linear algebra! The matrix associated with the quadratic form is $M = \begin{pmatrix} a & b \\ b & c \end{pmatrix}$. The determinant of this matrix is $ac - b^{2}$. Since we established that $b^{2} 0$. This means the determinant is always positive, and therefore never zero, which guarantees that the system of linear equations will always have a unique solution.

Select Answer:

* Multiple Correct

Progressive Hints

Hint 1

Hint 1: Visualize the expression

a x^{2} + 2 b x y + c y^{2}

as a surface in 3D space. If this surface is strictly positive for all points except the origin, what does that imply about the parabola formed when you slice this surface with a plane? It means the parabola never touches or crosses the axis.

Hint 2

Hint 2: To handle the two variables

x

and

y

, try to reduce the expression to a single variable. If you assume

y \neq = 0

and divide the entire expression by

y^{2}

, you get a quadratic in terms of

t = \frac{x}{y}

. What condition must a quadratic

a t^{2} + 2 b t + c

satisfy to be strictly positive for all real

t

Hint 3

Hint 3: The condition for

a t^{2} + 2 b t + c > 0

for all

t

is that the leading coefficient

a > 0

and the discriminant

D < 0

. Calculate the discriminant

D = (2 b)^{2} - 4 a c

. This leads to the fundamental inequality

b^{2} < a c

. Use this to evaluate the options.

Visualized Solution

Understanding the Set $S$

Given: $a x^{2} + 2 b x y + c y^{2} > 0$ for all $(x, y) \neq = (0, 0)$
This is a homogeneous quadratic expression.
Put $x = 1, y = 0$ : $a (1)^{2} + 0 + 0 > 0 ⟹ a > 0$
Put $x = 0, y = 1$ : $0 + 0 + c (1)^{2} > 0 ⟹ c > 0$

Quadratic in One Variable

Assume $y \neq = 0$ and divide the expression by $y^{2}$ .
$a (\frac{x}{y})^{2} + 2 b (\frac{x}{y}) + c > 0$
Let $t = \frac{x}{y} ⟹ a t^{2} + 2 b t + c > 0$ for all $t \in R$

Discriminant Condition

For $a t^{2} + 2 b t + c > 0$ to hold true for all $t$ :
Leading coefficient $a > 0$ and Discriminant $D < 0$
$D = (2 b)^{2} - 4 a c < 0 ⟹ 4 b^{2} - 4 a c < 0$
$⟹ b^{2} < a c$

Testing Option A

Option A: $(2, \frac{7}{2}, 6) \in S$
Here $a = 2$ , $b = \frac{7}{2}$ , $c = 6$
$b^{2} = (\frac{7}{2})^{2} = \frac{49}{4} = 12.25$
$a c = 2 \times 6 = 12$
Since $12.25 \neq < 12$ , condition $b^{2} < a c$ fails. Option A is FALSE.

Testing Option B

Option B: If $(3, b, \frac{1}{12}) \in S$ , then $∣2 b ∣ < 1$
Here $a = 3$ , $c = \frac{1}{12}$
Since it belongs to $S$ , $b^{2} < a c$ must be true.
$b^{2} < 3 \times \frac{1}{12} ⟹ b^{2} < \frac{1}{4}$

Solving for $b$

$b^{2} < \frac{1}{4}$
Taking square root: $∣ b ∣ < \frac{1}{2}$
Multiply by 2: $2∣ b ∣ < 1 ⟹ ∣2 b ∣ < 1$
This matches the statement. Option B is TRUE.

Testing Option C: System of Equations

System: $a x + b y = 1$ and $b x + cy = - 1$
A unique solution exists if the determinant of the coefficient matrix $Δ \neq = 0$ .

Determinant for Option C

$Δ = a b b c$
Expanding: $Δ = (a) (c) - (b) (b)$
$Δ = a c - b^{2}$

Evaluating $Δ$ for Option C

From set $S$ , we know $b^{2} < a c$
Rearranging: $a c - b^{2} > 0$
$⟹ Δ > 0 ⟹ Δ \neq = 0$
Since $Δ \neq = 0$ , a unique solution exists. Option C is TRUE.

Testing Option D: Second System

System: $(a + 1) x + b y = 0$ and $b x + (c + 1) y = 0$
For a unique (trivial) solution, we again need $Δ \neq = 0$ .

Determinant for Option D

$Δ = a + 1 b b c + 1$
Expanding: $Δ = (a + 1) (c + 1) - b^{2}$

Expanding the Expression

Multiply terms: $(a + 1) (c + 1) = a c + a + c + 1$
Substitute back: $Δ = a c + a + c + 1 - b^{2}$
Group strategically: $Δ = (a c - b^{2}) + a + c + 1$

Evaluating $Δ$ for Option D

$Δ = (a c - b^{2}) + a + c + 1$
We know $(a c - b^{2}) > 0$ , $a > 0$ , and $c > 0$
$Δ > 0 + 0 + 0 + 1 ⟹ Δ > 1$
Since $Δ > 0$ , $Δ \neq = 0$ . Unique solution exists. Option D is TRUE.

Conclusion

Key Takeaway: A positive definite quadratic form strictly constrains its coefficients ( $a > 0, c > 0, a c - b^{2} > 0$ ).
This guarantees non-zero determinants for related linear systems.
Final Answer: Options B, C, and D are correct.

00:00 / 00:00

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Solution Diagram

Analyzing the Setup

The problem defines a landscape

f (x, y) = a x^{2} + 2 b x y + c y^{2}

that is strictly positive for all $(x, y) eq (0, 0)$. This condition identifies the expression as a positive definite quadratic form.

To ensure

f (x, y) > 0

for all non-zero vectors, we first examine the boundaries by setting one variable to zero. Setting

y = 0

yields

a x^{2} > 0

, which implies

a > 0

. Similarly, setting

x = 0

yields

c y^{2} > 0

, which implies

c > 0

The Master Inequality

To handle the interaction term

2 b x y

, we assume $y eq 0$ and divide the expression by

y^{2}

. Letting

t = \frac{x}{y}

, we obtain the quadratic expression:

a t^{2} + 2 b t + c > 0

For this quadratic to remain strictly positive for all real

t

, the parabola must open upwards (

a > 0

) and possess no real roots. This requires the discriminant

D

to be strictly negative:

D = (2 b)^{2} - 4 a c < 0

Simplifying this inequality leads to our golden rule:

b^{2} < a c

Evaluating the Options

We apply the condition

b^{2} < a c

to evaluate the validity of the provided options.

For Option A, testing

(a, b, c) = (2, \frac{7}{2}, 6)

yields

b^{2} = 12.25

and

a c = 12

. Since $12.25 ot< 12$, Option A is false.

For Option B, testing

(3, b, \frac{1}{12})

requires

b^{2} < 3 \times \frac{1}{12}

, which simplifies to

b^{2} < \frac{1}{4}

. This implies

∣ b ∣ < \frac{1}{2}

, or

∣2 b ∣ < 1

, confirming that Option B is true.

Determinants and Uniqueness

We analyze the systems of linear equations by examining the determinants of their coefficient matrices.

For Option C, the matrix is:

(a b b c)

The determinant is

a c - b^{2}

. Since

a c - b^{2} > 0

, the determinant is non-zero, ensuring a unique solution. Thus, Option C is true.

For Option D, the matrix is:

(a + 1 b b c + 1)

The determinant is calculated as:

det = (a + 1) (c + 1) - b^{2} = (a c - b^{2}) + a + c + 1

Given

a c - b^{2} > 0

a > 0

, and

c > 0

, the determinant is strictly greater than

1

. Therefore, the determinant is non-zero, and Option D is true.

Common Traps & FAQs

Why do we need

a > 0

and

c > 0

? Can't they be negative?

Great question! Think about the behavior of the expression as we move far away from the origin. If we set

y = 0

, the expression becomes

a x^{2}

. If

a

were negative, then for any

x \neq = 0

, the expression would be negative, violating the condition that it must be strictly positive for all

(x, y) \neq = (0, 0)

. Thus,

a

must be positive. By symmetry, setting

x = 0

forces

c

to be positive as well.

Why is the condition

b^{2} < a c

and not

b^{2} \leq a c

The problem specifies that the expression is strictly greater than zero. If

b^{2} = a c

, the discriminant is zero, which means the quadratic has a real root. At that root, the expression would equal zero. Since we need it to be strictly positive everywhere except the origin, we must exclude the case where the expression touches zero. Hence,

b^{2} < a c

is the only valid condition.

How does the determinant of a system of equations relate to the quadratic form

a x^{2} + 2 b x y + c y^{2}

This is the beauty of linear algebra! The matrix associated with the quadratic form is

M = (a b b c)

. The determinant of this matrix is

a c - b^{2}

. Since we established that

b^{2} < a c

, it follows that

a c - b^{2} > 0

. This means the determinant is always positive, and therefore never zero, which guarantees that the system of linear equations will always have a unique solution.

Animated Solution for Mathematics - Matrices and Determinants: Let $R^{2}$ denote $R \times R$ . Let $S = {(a, b, c) : a, b, c \in R and a x^{2} + 2 b x y + c y^{2} > 0 for all (x, y) \in R^{2} - {(0, 0)}}$ . Then which of the following statements is (are) TRUE?

Select Answer:

Visualized Solution

Understanding the Set $S$

Quadratic in One Variable

Discriminant Condition

Testing Option A

Testing Option B

Solving for $b$

Testing Option C: System of Equations

Determinant for Option C

Evaluating $Δ$ for Option C

Testing Option D: Second System

Determinant for Option D

Expanding the Expression

Evaluating $Δ$ for Option D

Conclusion

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Analyzing the Setup

The Master Inequality

Evaluating the Options

Determinants and Uniqueness

Similar Questions

Comprehension Passage

If the point $P (a, b, c)$ , with reference to (E), lies on the plane $2 x + y + z = 1$ , then the value of $7 a + b + c$ is

Let $ω$ be a solution of $x^{3} - 1 = 0$ with $Im (ω) > 0$ , if $a = 2$ with $b$ and $c$ satisfying (E), then the value of $\frac{3}{ω ^{a}} + \frac{1}{ω ^{b}} + \frac{3}{ω ^{c}}$ is equal to

Let $b = 6$ , with $a$ and $c$ satisfying (E). If $α$ and $β$ are the roots of the quadratic equation $a x^{2} + b x + c = 0$ , then $\sum_{n = 0}^{\infty} (\frac{1}{α} + \frac{1}{β})^{n}$ is

Consider the following system of equations $α x + 2 y + z = 1$ , $2 α x + 3 y + z = 1$ , $3 x + α y + 2 z = β$ For some $α, β \in R$ . Then which of the following is NOT correct.

Let $a, b, c$ be any real numbers. Suppose that there are real numbers $x, y, z$ not all zero such that $x = cy + b z, y = a z + c x$ , and $z = b x + a y$ . Then $a^{2} + b^{2} + c^{2} + 2 ab c$ is equal to

Let $α, β$ and $γ$ be real numbers. consider the following system of linear equations $x + 2 y + z = 7$ $x + α z = 11$ $2 x - 3 y + β z = γ$ Match each entry in List - I to the correct entries in List-II

List-I

List-II

Consider the following linear equations $a x + b y + cz = 0; b x + cy + a z = 0; c x + a y + b z = 0$ Match the conditions/expressions in Column I with statements in Column II and indicate your answer by darkening the appropriate bubbles in the $4 \times 4$ matrix given in the ORS.

List-I

List-II

Let $a, λ, μ \in R$ . Consider the system of linear equations $a x + 2 y = λ, 3 x - 2 y = μ$ . Which of the following statement(s) is (are) correct?

Consider the following system of equations: $x + 2 y - 3 z = a, 2 x + 6 y - 11 z = b, x - 2 y + 7 z = c$ where $a, b$ and $c$ are real constants. Then the system of equations :

If S is the set of distinct values of 'b' for which the following system of linear equations $x + y + z = 1$ , $x + a y + z = 1$ , $a x + b y + z = 0$ has no solution, then S is:

Select Answer:

Visualized Solution

Understanding the Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Quadratic in One Variable

Discriminant Condition

Testing Option A

Testing Option B

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Testing Option C: System of Equations

Determinant for Option C

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δ for Option C

Testing Option D: Second System

Determinant for Option D

Expanding the Expression

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δ for Option D

Conclusion

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Analyzing the Setup

The Master Inequality

Evaluating the Options

Determinants and Uniqueness

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a,b,c be the real numbers. Then following system of equations in x,y and z a2x2​+b2y2​−c2z2​=1, a2x2​−b2y2​+c2z2​=1, −a2x2​+b2y2​+c2z2​=1 has

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the point P(a,b,c), with reference to (E), lies on the plane 2x+y+z=1, then the value of 7a+b+c is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let ω be a solution of x3−1=0 with Im(ω)>0, if a=2 with b and c satisfying (E), then the value of ωa3​+ωb1​+ωc3​ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let b=6, with a and c satisfying (E). If α and β are the roots of the quadratic equation ax2+bx+c=0, then ∑n=0∞​(α1​+β1​)n is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Consider the following system of equations αx+2y+z=1, 2αx+3y+z=1, 3x+αy+2z=β For some α,β∈R. Then which of the following is NOT correct.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a,b,c be any real numbers. Suppose that there are real numbers x,y,z not all zero such that x=cy+bz,y=az+cx, and z=bx+ay. Then a2+b2+c2+2abc is equal to

List-I

List-II

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a,λ,μ∈R. Consider the system of linear equations ax+2y=λ,3x−2y=μ. Which of the following statement(s) is (are) correct?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Consider the following system of equations: x+2y−3z=a,2x+6y−11z=b,x−2y+7z=c where a,b and c are real constants. Then the system of equations :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If S is the set of distinct values of 'b' for which the following system of linear equations x+y+z=1, x+ay+z=1, ax+by+z=0 has no solution, then S is:

Understanding the Set $S$

Solving for $b$

Evaluating $Δ$ for Option C

Evaluating $Δ$ for Option D

If the point $P (a, b, c)$ , with reference to (E), lies on the plane $2 x + y + z = 1$ , then the value of $7 a + b + c$ is

Let $ω$ be a solution of $x^{3} - 1 = 0$ with $Im (ω) > 0$ , if $a = 2$ with $b$ and $c$ satisfying (E), then the value of $\frac{3}{ω ^{a}} + \frac{1}{ω ^{b}} + \frac{3}{ω ^{c}}$ is equal to

Let $b = 6$ , with $a$ and $c$ satisfying (E). If $α$ and $β$ are the roots of the quadratic equation $a x^{2} + b x + c = 0$ , then $\sum_{n = 0}^{\infty} (\frac{1}{α} + \frac{1}{β})^{n}$ is

Consider the following system of equations $α x + 2 y + z = 1$ , $2 α x + 3 y + z = 1$ , $3 x + α y + 2 z = β$ For some $α, β \in R$ . Then which of the following is NOT correct.

Let $a, b, c$ be any real numbers. Suppose that there are real numbers $x, y, z$ not all zero such that $x = cy + b z, y = a z + c x$ , and $z = b x + a y$ . Then $a^{2} + b^{2} + c^{2} + 2 ab c$ is equal to

Let $a, λ, μ \in R$ . Consider the system of linear equations $a x + 2 y = λ, 3 x - 2 y = μ$ . Which of the following statement(s) is (are) correct?

Consider the following system of equations: $x + 2 y - 3 z = a, 2 x + 6 y - 11 z = b, x - 2 y + 7 z = c$ where $a, b$ and $c$ are real constants. Then the system of equations :

If S is the set of distinct values of 'b' for which the following system of linear equations $x + y + z = 1$ , $x + a y + z = 1$ , $a x + b y + z = 0$ has no solution, then S is: