Question

Let \( A \) be a \( 3 \times 3 \) 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 _____

Answer 1

Correct Answer: 27

Let \(A\) be a \(3 \times 3\) matrix: 
\[ A = \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{bmatrix} \] 

Given that: 
\[ A \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = 3 \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \] 
This implies: \[ a_1 + a_2 + a_3 = 3 \quad \dots (1) \]
\[ b_1 + b_2 + b_3 = 3 \quad \dots (2) \]
\[ c_1 + c_2 + c_3 = 3 \quad \dots (3) \] 

Now, we want to maximize \(\det(A)\): 
\[ \det(A) = a_1b_2c_3 + a_2b_3c_1 + a_3b_1c_2 - (a_3b_2c_1 + a_1b_3c_2 + a_2b_1c_3). \] 

To achieve the maximum value, we set \(a_1 = b_2 = c_3 = 3\) and all other elements to zero:
\[ \det(A) = 3 \times 3 \times 3 = 27. \] 
Thus, the maximum value of \(\det(A)\) is: 
\[ 27 \]

Answer 2

Step 1: Given information.
We are given a \( 3 \times 3 \) matrix \( A \) with non-negative real elements such that:
\[ A \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = 3 \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}. \] This means that the vector \( \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \) is an eigenvector of \( A \) corresponding to the eigenvalue \( 3 \).

Step 2: Understanding what this implies.
If \( A \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = 3 \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \), then each row of \( A \) has elements that sum up to 3.
That is, if \( A = [a_{ij}] \), then for each \( i = 1, 2, 3 \):
\[ a_{i1} + a_{i2} + a_{i3} = 3, \] and each \( a_{ij} \ge 0. \)

Step 3: Aim.
We need to find the maximum possible value of \( \det(A) \) under the above constraints.

Step 4: Using properties of non-negative matrices with row sums equal to a constant.
If every row of \( A \) sums to 3, then \( 3 \) is one eigenvalue of \( A \).
The sum of all eigenvalues (trace of \( A \)) is the sum of diagonal elements, which can vary depending on the distribution of the entries.

To maximize \( \det(A) \), we want to make all eigenvalues as large and equal as possible, because for a fixed sum of eigenvalues, the product (which is the determinant) is maximized when all eigenvalues are equal.

If all rows of \( A \) are identical, say \( [a, b, c] \), with \( a + b + c = 3 \), then \( A \) will have rank 1, giving \( \det(A) = 0 \). Hence, identical rows are not optimal.

However, if \( A = 3I \), the identity matrix multiplied by 3, we get \( \det(A) = 3^3 = 27 \).
This matrix clearly satisfies the given condition because: \[ A \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = 3 \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}, \] and all elements of \( A \) are non-negative.

Step 5: Verify maximality.
For any stochastic-type non-negative matrix with row sum 3, by Hadamard’s inequality or determinant bounds for positive matrices, the determinant cannot exceed the product of the diagonal entries when they are all equal to 3. Hence, \( \det(A) \le 27 \).

Final Answer:
\[ \boxed{27} \]