Problem 6 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(\mathrm{A}=\left[\begin{array}{ccc}2 & \mathrm{~b} & 1 \\ \mathrm{~b} & \mathrm{~b}^{2}+1 & \mathrm{~b} \\ 1 & \mathrm{~b} & 2\end{array}\right]\) where \(\mathrm{b}>0\). Then the minimum value of \(\frac{\operatorname{det}(\mathrm{A})}{\mathrm{b}}\) is: \(\quad\) [Jan. 10,2019 (II)] (a) \(2 \sqrt{3}\) (n) \(-2 \sqrt{3}\) (c) \(-\sqrt{3}\) (d) \(\sqrt{3}\)

Step-by-Step Solution

Answer

The minimum value is \( 2\sqrt{3} \).

Step 1: Compute the Determinant

To find the determinant of the matrix \( A = \begin{bmatrix} 2 & b & 1 \ b & b^2 + 1 & b \ 1 & b & 2 \end{bmatrix} \), use the formula for the determinant of a 3x3 matrix: \[ \operatorname{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]where the elements from matrix \( A \) are given by \( a = 2, b = b, c = 1, d = b, e = b^2 + 1, f = b, g = 1, h = b, i = 2 \). First, calculate each term: - Submatrix 1: \( ei - fh = (b^2+1) \cdot 2 - b \cdot b = 2b^2 + 2 - b^2 = b^2 + 2 \)- Submatrix 2: \( di - fg = b \cdot 2 - b \cdot 1 = 2b - b = b \)- Submatrix 3: \( dh - eg = b \cdot b - (b^2+1) \cdot 1 = b^2 - b^2 - 1 = -1 \)Plug the results into the formula:\[ \operatorname{det}(A) = 2(b^2 + 2) - b(b) + 1(-1) \]\[ \operatorname{det}(A) = 2b^2 + 4 - b^2 - 1 = b^2 + 3 \].

Step 2: Express the Function to be Minimized

Now, express the function to be minimized. We need to find the minimum value of \( \frac{\operatorname{det}(A)}{b} \). Substitute the expression for the determinant found in Step 1:\[ \frac{\operatorname{det}(A)}{b} = \frac{b^2 + 3}{b} = b + \frac{3}{b} \].

Step 3: Differentiate the Function

Differentiate the function \( f(b) = b + \frac{3}{b} \) with respect to \( b \) to find the minimum value. The derivative is:\[ f'(b) = 1 - \frac{3}{b^2} \].Set the derivative equal to zero to find critical points:\[ 1 - \frac{3}{b^2} = 0 \]Solving gives:\[ b^2 = 3 \]\[ b = \sqrt{3} \] (since \( b > 0 \)).

Step 4: Verify Minimization Condition

Verify the second derivative condition:Find \( f''(b) \):\[ f''(b) = \frac{6}{b^3} \].Since \( f''(b) > 0 \) for \( b > 0 \), \( f(b) \) has a local minimum at \( b = \sqrt{3} \).

Step 5: Calculate the Minimum Value of the Function

Substitute \( b = \sqrt{3} \) back into \( f(b) \) to find the minimum value:\[ f(\sqrt{3}) = \sqrt{3} + \frac{3}{\sqrt{3}} = \sqrt{3} + \sqrt{3} = 2\sqrt{3} \].Thus, the minimum value is \( 2\sqrt{3} \).

Key Concepts

3x3 matrix determinant

Understanding the determinant of a 3x3 matrix is a fundamental concept in linear algebra. A 3x3 matrix has three rows and three columns, formatted as follows:

• a, b, c
• d, e, f
• g, h, i

To find the determinant, denoted as \( \operatorname{det}(A) \), a specific formula is used:\[ \operatorname{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]This formula involves calculated minors derived from the matrix, where each minor is computed by excluding the row and column of a particular element and finding the determinant of the resulting 2x2 matrix.
For the matrix in our exercise "A", applying these concepts allows us to calculate the determinant efficiently. Knowing how to compute the determinant helps in various areas like solving systems of equations, understanding matrix inverses, and evaluating eigenvalues.
Determinants have practical applications in physics, engineering, and computer graphics, making them a critical tool in analyzing linear transformations.

minimization problem

Minimization problems are a type of optimization problem where the objective is to find the lowest point of a function. In this exercise, the function we are trying to minimize is \( f(b) = b + \frac{3}{b} \), which arises from dividing the determinant by a variable \( b \).
To find the minimum point, we first calculate the derivative of \( f(b) \), which is \( f'(b) = 1 - \frac{3}{b^2} \). Setting the derivative equal to zero identifies the critical points, showing where the function's rate of change is zero, which can indicate a minimum value point. Solving \( 1 - \frac{3}{b^2} = 0 \) gives us \( b^2 = 3 \), leading to \( b = \sqrt{3} \), since \( b > 0 \).
Calculating the second derivative \( f''(b) = \frac{6}{b^3} \) helps confirm that the critical point is a minimum. Since \( f''(b) > 0 \) at \( b = \sqrt{3} \), the function indeed has a local minimum there. The minimum value of \( f(b) \) at this point is \( 2\sqrt{3} \), showcasing how calculus tools apply to solving optimization problems.

matrix calculus

Matrix calculus extends regular calculus concepts to functions involving matrices. Understanding matrix calculus is vital in fields like engineering, physics, and computer science, especially in optimization and control systems.
In our context, matrix calculus played a role in simplifying expressions involving determinants and optimization. When dealing with functions of matrices, differentiation helps in analyzing how small changes in matrix elements affect the function's output. This can be extended to multi-variable functions by using partial differentiation and applying matrix operations to evaluate gradients and Hessians.
The exercise demonstrates a fundamental application: optimizing the value of an expression derived from a matrix determinant. By differentiating the function \( f(b) = b + \frac{3}{b} \), a process akin to matrix calculus principles, we pinpoint where minimization occurs. Tools from matrix calculus are invaluable in achieving precision in optimization problems, making them an essential component of modern computational techniques.