JEE Main 2019 (10 January)

LEVELJEE Main

Animated Solution for Mathematics - Matrices and Determinants: Let $A = 2 b 1 b b^{2} + 1 b 1 b 2$ where $b > 0$ . Then the minimum value of $\frac{d e t ( A )}{b}$ is :

Select Answer:

Progressive Hints

Hint 1

Hint 1: Look at the matrix

A

. It is symmetric. Before you dive into the determinant expansion, take a deep breath and visualize the structure. The expansion will yield a polynomial in

b

. Don't panic if it looks messy at first; the terms are designed to simplify beautifully.

Hint 2

Hint 2: Once you have the expression for

det (A)

, you will be looking at a function of the form

f (b) = \frac{b ^{2} + k}{b}

. When you split this into

b + \frac{k}{b}

, you are staring at a classic structure. What inequality relates the sum of two positive numbers to their product?

Hint 3

Hint 3: You are now ready to apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality. Remember, for any two positive real numbers

x

and

y

\frac{x + y}{2} \geq x y

. Here, let

x = b

and

y = \frac{3}{b}

. The variables will cancel out, leaving you with a constant.

Visualized Solution

Matrix $A$ and Objective

Given matrix $A = 2 b 1 b b^{2} + 1 b 1 b 2$
Condition: $b > 0$
Objective: Find the minimum value of $\frac{d e t ( A )}{b}$

Expanding $det (A)$

To find $det (A)$ , we expand along the first row ( $R_{1}$ ).

First Term Calculation

$2 b^{2} + 1 b b 2$
$= 2 ((b^{2} + 1) (2) - (b) (b))$
$= 2 (2 b^{2} + 2 - b^{2})$
$= 2 (b^{2} + 2)$

Second Term Calculation

$- b b 1 b 2$
$= - b (2 b - b)$
$= - b^{2}$

Third Term Calculation

$+ 1 b 1 b^{2} + 1 b$
$= 1 (b^{2} - (b^{2} + 1))$
$= - 1$

Simplifying $det (A)$

$det (A) = 2 (b^{2} + 2) - b^{2} - 1$
$det (A) = 2 b^{2} + 4 - b^{2} - 1$
$det (A) = b^{2} + 3$

Target Expression $f (b)$

$f (b) = \frac{d e t ( A )}{b}$
$f (b) = \frac{b ^{2} + 3}{b}$
$f (b) = b + \frac{3}{b}$

AM-GM Inequality

For $x, y > 0$ , $\frac{x + y}{2} \geq x y$
Let $x = b$ and $y = \frac{3}{b}$

Finding the Minimum

$\frac{b + \frac{3}{b}}{2} \geq b \cdot \frac{3}{b}$
$\frac{b + \frac{3}{b}}{2} \geq 3$
$b + \frac{3}{b} \geq 23$

Final Answer

Minimum value of $\frac{d e t ( A )}{b} = 23$
Correct Option: (4) $23$

00:00 / 00:00

The Sigma Insight: Properties of Determinants

Solution Diagram

Analyzing the Setup

We are given the matrix

A

defined as:

A = 2 b 1 b b^{2} + 1 b 1 b 2

We are working under the constraint

b > 0

. Our objective is to determine the minimum value of the expression

\frac{d e t ( A )}{b}

The Determinant Expansion

To find the determinant, we expand along the first row. The expression is given by:

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

Let us compute these minors step-by-step:

The first term is

2 ((b^{2} + 1) (2) - b^{2}) = 2 (2 b^{2} + 2 - b^{2}) = 2 (b^{2} + 2) = 2 b^{2} + 4

The second term is

- b (2 b - b) = - b^{2}

The third term is

1 (b^{2} - (b^{2} + 1)) = - 1

The Simplification

Now, we combine these components to find the total determinant:

det (A) = (2 b^{2} + 4) - b^{2} - 1 = b^{2} + 3

The complexity of the matrix structure collapses into this elegant quadratic form.

The Optimization

We define the function to be minimized as

f (b) = \frac{b ^{2} + 3}{b}

, which simplifies to:

f (b) = b + \frac{3}{b}

Since

b > 0

, we apply the AM-GM inequality (Arithmetic Mean

\geq

Geometric Mean):

\frac{b + \frac{3}{b}}{2} \geq b \cdot \frac{3}{b} = 3

Multiplying by

2

, we obtain:

b + \frac{3}{b} \geq 23

The minimum value of the expression is $23$ .

Common Traps & FAQs

Wait, why can't I just use calculus to find the minimum? Can't I take the derivative and set it to zero?

You absolutely can! Using calculus is a perfectly valid and robust method. You would find

f^{'} (b) = 1 - \frac{3}{b ^{2}}

, set it to zero, and solve for

b = 3

. However, in JEE Advanced, time is your most precious currency. The AM-GM inequality allows you to bypass the differentiation and critical point analysis entirely, saving you those crucial seconds for the tougher problems.

The problem says

b > 0

. Does that matter? What if

b

was negative?

That condition is the key that unlocks the door! The AM-GM inequality is only valid for positive real numbers. If

b

were negative, the function

b + \frac{3}{b}

would behave very differently (it would be negative and have a maximum value). The constraint

b > 0

ensures we are working in the domain where the inequality holds true.

I got stuck during the determinant expansion. How do I avoid making sign errors?

Sign errors are the silent killers of JEE scores. The best strategy is to write out the expansion explicitly:

+ a_{11} C_{11} - a_{12} C_{12} + a_{13} C_{13}

. Always write the alternating signs (

+ - +

) on your scratchpad before you start plugging in the values. It acts as a visual safety net.

Animated Solution for Mathematics - Matrices and Determinants: Let $A = 2 b 1 b b^{2} + 1 b 1 b 2$ where $b > 0$ . Then the minimum value of $\frac{d e t ( A )}{b}$ is :

Select Answer:

Visualized Solution

Matrix $A$ and Objective

Expanding $det (A)$

First Term Calculation

Second Term Calculation

Third Term Calculation

Simplifying $det (A)$

Target Expression $f (b)$

AM-GM Inequality

Finding the Minimum

Final Answer

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Determinant Expansion

The Simplification

The Optimization

Similar Questions

Let $d \in R$ , and $A = - 2 15 4 + d (sin θ) + 2 (2 sin θ) - d (sin θ) - 2 d (- sin θ) + 2 + 2 d$ , $θ \in [0, 2 π]$ . If the minimum value of $det (A)$ is 8, then a value of d is :

Let $A$ be a $3 \times 3$ matrix of non-negative real elements such that $A 111 = 3111$ . Then the maximum value of $det (A)$ is ______

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The value of the determinant $111 a b c a^{2} - b c b^{2} - c a c^{2} - ab$ is .........

Let $P = [a_{ij}]$ be a $3 \times 3$ matrix and let $Q = [b_{ij}]$ , where $b_{ij} = 2^{i + j} a_{ij}$ for $1 \leq i, j \leq 3$ . If the determinant of $P$ is 2, then the determinant of the matrix $Q$ is

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Animated Solution for Mathematics - Matrices and Determinants: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A=​2b1​bb2+1b​1b2​​ where b>0. Then the minimum value of bdet(A)​ is :

Select Answer:

Visualized Solution

Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and Objective

Expanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } det(A)

First Term Calculation

Second Term Calculation

Third Term Calculation

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } det(A)

Target Expression .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(b)

AM-GM Inequality

Finding the Minimum

Final Answer

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Determinant Expansion

The Simplification

The Optimization

Similar Questions

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Animated Solution for Mathematics - Matrices and Determinants: Let $A = 2 b 1 b b^{2} + 1 b 1 b 2$ where $b > 0$ . Then the minimum value of $\frac{d e t ( A )}{b}$ is :

Matrix $A$ and Objective

Expanding $det (A)$

Simplifying $det (A)$

Target Expression $f (b)$

Let $d \in R$ , and $A = - 2 15 4 + d (sin θ) + 2 (2 sin θ) - d (sin θ) - 2 d (- sin θ) + 2 + 2 d$ , $θ \in [0, 2 π]$ . If the minimum value of $det (A)$ is 8, then a value of d is :

Let $A$ be a $3 \times 3$ matrix of non-negative real elements such that $A 111 = 3111$ . Then the maximum value of $det (A)$ is ______

If $A = 1 - sin θ - 1 sin θ 1 - sin θ 1 sin θ 1$ ; then for all $θ \in (3 π /4, 5 π /4)$ , $det (A)$ lies in the interval :

The value of the determinant $111 a b c a^{2} - b c b^{2} - c a c^{2} - ab$ is .........

Let $P = [a_{ij}]$ be a $3 \times 3$ matrix and let $Q = [b_{ij}]$ , where $b_{ij} = 2^{i + j} a_{ij}$ for $1 \leq i, j \leq 3$ . If the determinant of $P$ is 2, then the determinant of the matrix $Q$ is

If $A = [2335]$ , then the determinant of the matrix $(A^{2025} - 3 A^{2024} + A^{2023})$ is

Let $A = 100 0 α β 0 β α$ and $∣2 A ∣^{3} = 2^{21}$ where $α, β \in Z$ , Then a value of $α$ is

Let $A$ be a matrix of order $3 \times 3$ and $det (A) = 2$ . Then $det (det (A) adj (5 adj (A^{3})))$ is equal to