Let A be a 3 × 3 matrix of non-negative real elements such that A [1, 1, 1]ᵀ = 3 [1, 1, 1]ᵀ. Then th

Q: Wait, why does the determinant represent volume? I thought it was just a calculation.

That is a common misconception! In linear algebra, the determinant is a scaling factor for volume. If you have a unit cube, the determinant tells you how much that cube is stretched or compressed after the linear transformation $A$. Maximizing the determinant is literally maximizing the volume of the transformed space. Visualizing this is the key to mastering JEE Advanced linear algebra.

Q: Why can't we just have some negative numbers to make the determinant larger?

The problem explicitly restricts us to non-negative real elements ($a_{ij} \ge 0$). This constraint is the 'cage' that limits our volume. If we could use negative numbers, we could potentially create vectors that are not just long, but oriented in ways that might seem to increase volume, but the non-negativity forces us to stay in the first octant of 3D space, which simplifies our optimization significantly.

Q: How do we know for sure that the diagonal matrix is the absolute maximum?

This is related to Hadamard's Inequality, which states that the determinant of a matrix is less than or equal to the product of the lengths of its row vectors. Since our row sums are fixed at $3$, the maximum length of a row vector with non-negative entries is $3$ (achieved when one element is $3$ and others are $0$). When these vectors are orthogonal, we achieve the equality case of Hadamard's inequality, giving us the absolute maximum.

Question

Let $A$ be a $3 	imes 3$ matrix of non-negative real elements such that $A \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} = 3 \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix}$. Then the maximum value of $\det(A)$ is ______

Sigma Percentile · Accepted Answer

### The Hidden Geometry of Matrices

Welcome, future engineer. Today, we are not just solving a matrix problem; we are embarking on a journey to understand the soul of linear algebra. Often, students look at a matrix like $A$ and see a grid of numbers.

But I want you to see something else. I want you to see a transformation, a geometric engine that stretches and rotates space. When we talk about the determinant, $\det(A)$, we are talking about the volume of the space that this matrix creates. Let's unlock this problem together.

### Decoding the Constraint

We are given the equation $A \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} = 3 \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix}$. At first glance, this looks like a simple eigenvalue problem where $3$ is an eigenvalue and the vector of ones is the eigenvector.

When you perform the multiplication of a $3 	imes 3$ matrix $A$ by a column vector of ones, each element of the resulting vector is the sum of the elements in the corresponding row of $A$.

If the result is $\begin{bmatrix} 3 \ 3 \ 3 \end{bmatrix}$, it tells us that the sum of the elements in each row is $3$. Mathematically, for each row $i$, we have:

$$\sum_{j=1}^{3} a_{ij} = 3$$

This is our boundary. This is the constraint that defines the limits of our world.

### The Geometry of Volume

Now, let's talk about the determinant. In the realm of 3D space, the determinant of a matrix is the volume of the parallelepiped formed by its row vectors $\vec{r}_1, \vec{r}_2, \vec{r}_3$.

Imagine these three vectors originating from the origin. They define the edges of a box, and the volume of this box is the determinant.

We want to maximize this volume. Think about a box with fixed edge lengths; the volume is largest when the edges are perpendicular to each other—when they are orthogonal. If the edges are slanted, the box 'collapses' and the volume decreases.

### The Optimization

We have a constraint: $a_{ij} \ge 0$ and the sum of elements in each row is $3$. Let's look at a single row vector $\vec{r}_i = (a_{i1}, a_{i2}, a_{i3})$.

We know $a_{i1} + a_{i2} + a_{i3} = 3$. To maximize the length of this vector, we want to concentrate all the 'weight' into one component. If we set one element to $3$ and the others to $0$, the length of the vector is maximized.

If we do this for all three rows, we get a matrix that looks like this:

$$A_{max} = \begin{bmatrix} 3 & 0 & 0 \ 0 & 3 & 0 \ 0 & 0 & 3 \end{bmatrix}$$

Look at this matrix. The row vectors are $(3, 0, 0)$, $(0, 3, 0)$, and $(0, 0, 3)$. These are the basis vectors of our 3D space, scaled by $3$. They are perfectly orthogonal and as long as they can possibly be under our constraints.

### Final Calculation

Finally, we calculate the determinant. For a diagonal matrix, the determinant is simply the product of the diagonal entries:

$$\det(A_{max}) = 3 	imes 3 	imes 3 = 27$$

This is the maximum volume. We have taken a seemingly abstract algebraic constraint and turned it into a geometric optimization problem.

We visualized the rows as vectors, understood that orthogonality maximizes volume, and arrived at the solution with elegance. Remember, in JEE Advanced, the math is just the language; the geometry is the story. **The maximum value is 27.**

Animated Solution for Mathematics - Matrices and Determinants: 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 ______

Enter Numerical Value:

Visualized Solution

Defining the Matrix $A$

Interpreting the Matrix Equation

The Non-negativity Constraint

Geometric Meaning of Determinant

Maximizing the Volume

The Optimal Matrix

Calculating Maximum Determinant

The Sigma Insight: Properties of Determinants

The Hidden Geometry of Matrices

Decoding the Constraint

The Geometry of Volume

The Optimization

Final Calculation

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