Question

All the elements of a square matrix A of order 'n' are non-negative. If $a_{ij} \forall i \neq j$, is the least non-negative value of the function, $f(x) = x^2 + 2x - 1$ and tr(A) = $\lambda$, then maximum value of |A| will be

• A. $\lambda^n$
• B. $(\frac{n}{\lambda})^n$
• C. $(\frac{\lambda}{n})^n$
• D. $(\lambda N)^n$

Answer 1

Answer: (C)

$(\frac{\lambda}{n})^n$

Answer 2

The given function is $f(x) = x^2 + 2x - 1$. We need to find its least non-negative value. $f(x) = (x^2 + 2x + 1) - 1 - 1 = (x+1)^2 - 2$. The minimum value of $f(x)$ is -2, which occurs at $x = -1$. We are looking for the least non-negative value, i.e., the smallest value of $f(x)$ such that $f(x) \ge 0$. We solve $f(x) = 0$: $x^2 + 2x - 1 = 0$. Using the quadratic formula, $x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-1)}}{2(1)} = \frac{-2 \pm \sqrt{8}}{2} = -1 \pm \sqrt{2}$. The roots are $x_1 = -1 - \sqrt{2}$ and $x_2 = -1 + \sqrt{2}$. Since the parabola opens upwards, $f(x) \ge 0$ when $x \le -1 - \sqrt{2}$ or $x \ge -1 + \sqrt{2}$. The values of $f(x)$ range from -2 upwards. The smallest non-negative value that $f(x)$ takes is 0, which occurs at $x = -1 + \sqrt{2}$ and $x = -1 - \sqrt{2}$.

The off-diagonal elements of the matrix A, $a_{ij}$ for $i \neq j$, are equal to this least non-negative value. So, $a_{ij} = 0$ for all $i \neq j$. This means the matrix A is a diagonal matrix: $A = \begin{pmatrix} a_{11} & 0 & \cdots & 0 \\ 0 & a_{22} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a_{nn} \end{pmatrix}$

All elements of A are non-negative, so $a_{ii} \ge 0$ for all $i$. The trace of A is given by tr(A) = $\sum_{i=1}^n a_{ii} = \lambda$. The determinant of a diagonal matrix is the product of its diagonal elements: $|A| = \prod_{i=1}^n a_{ii} = a_{11} a_{22} \cdots a_{nn}$.

We want to maximize the product $P = a_{11} a_{22} \cdots a_{nn}$ subject to the constraints $a_{ii} \ge 0$ for all $i$ and $\sum_{i=1}^n a_{ii} = \lambda$. By the Arithmetic Mean - Geometric Mean (AM-GM) inequality, for non-negative numbers $a_{11}, a_{22}, \dots, a_{nn}$: $\frac{a_{11} + a_{22} + \cdots + a_{nn}}{n} \ge \sqrt[n]{a_{11} a_{22} \cdots a_{nn}}$ Substituting the sum: $\frac{\lambda}{n} \ge (|A|)^{1/n}$ Raising both sides to the power of n (since n is a positive integer): $(\frac{\lambda}{n})^n \ge |A|$

The maximum value of $|A|$ is $(\frac{\lambda}{n})^n$. This maximum is achieved when the equality in the AM-GM inequality holds, which is when $a_{11} = a_{22} = \cdots = a_{nn}$. If $a_{ii} = c$ for all $i$, then $\sum_{i=1}^n c = nc = \lambda$, so $c = \frac{\lambda}{n}$. Since all elements of A are non-negative, $a_{ii} = \lambda/n$ must be non-negative. As $\lambda$ is the sum of non-negative numbers, $\lambda \ge 0$. Since n is the order of the matrix, $n \ge 1$. Thus, $\lambda/n \ge 0$ is valid.

The maximum value of $|A|$ is $(\frac{\lambda}{n})^n$.