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

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:
$27$