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: 
\[ 27 \]