Let A be a 3×3 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 ______.

We are given a $$3 \times 3$$ matrix $$A$$ with non-negative real entries such that:

$$A\begin{bmatrix}1\\1\\1\end{bmatrix} = 3\begin{bmatrix}1\\1\\1\end{bmatrix}$$

This means the sum of each row of $$A$$ equals 3. That is, if $$A = [a_{ij}]$$, then for each row $$i$$: $$a_{i1} + a_{i2} + a_{i3} = 3$$, and all $$a_{ij} \geq 0$$.

We want to maximize $$\det(A)$$.

By the AM-GM inequality applied to matrices, or by direct optimization, the maximum determinant of a doubly stochastic-like matrix with row sums equal to a constant occurs at a specific structure.

The Hadamard inequality states that for a matrix with rows $$\mathbf{r}_1, \mathbf{r}_2, \mathbf{r}_3$$:

$$|\det(A)| \leq |\mathbf{r}_1||\mathbf{r}_2||\mathbf{r}_3|$$

But this bound is achieved when rows are orthogonal.

Let us try the matrix $$A = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix}$$. Then $$\det(A) = 27$$, and each row sums to 3. All entries are non-negative.

Can we do better? The rows must satisfy $$a_{i1} + a_{i2} + a_{i3} = 3$$ with $$a_{ij} \geq 0$$. Each row vector lies in the simplex $$\{(x,y,z) : x+y+z=3, x,y,z \geq 0\}$$, which lies on the plane $$x+y+z=3$$.

The maximum norm of a row vector on this simplex is achieved at the vertices $$(3,0,0)$$, $$(0,3,0)$$, $$(0,0,3)$$, where $$|\mathbf{r}| = 3$$.

For orthogonal rows on this simplex, the best we can do is the identity matrix scaled by 3, giving $$\det = 27$$.

Let us verify this is indeed the maximum. Consider any matrix with row sums 3 and non-negative entries. Write $$A = 3I + B$$ where $$B$$ has row sums 0. Actually, let us use Lagrange multipliers or just check.

Consider the general structure with rows $$\mathbf{r}_i$$ on the simplex. Since the simplex is a subset of the plane $$x+y+z=3$$, the normal to this plane is $$\mathbf{n} = (1,1,1)/\sqrt{3}$$. All row vectors have the same projection onto $$\mathbf{n}$$, namely $$3/\sqrt{3} = \sqrt{3}$$.

Writing each row as $$\mathbf{r}_i = \sqrt{3}\hat{n} + \mathbf{w}_i$$ where $$\mathbf{w}_i \perp \mathbf{n}$$:

$$\det(A) = \det(\sqrt{3}\hat{n}\mathbf{1}^T + W)$$ where $$W$$ has rows $$\mathbf{w}_i$$.

Since $$\hat{n}\mathbf{1}^T$$ has rank 1, and using the matrix determinant lemma approach, the determinant can be expanded. For the $$3I$$ matrix, the rows are $$(3,0,0)$$, $$(0,3,0)$$, $$(0,0,3)$$ which are mutually orthogonal with norm 3 each.

By Hadamard's inequality: $$|\det(A)| \leq 3 \times 3 \times 3 = 27$$.

This bound is achieved by $$A = 3I$$, so the maximum determinant is $$27$$.

The answer is $$27$$.