Let A be a 3 times 3 matrix of non-negative r | vedclass

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(C) Let $A = \begin{bmatrix} a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \ c_1 & c_2 & c_3 \end{bmatrix}$.Given $A\begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} =

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$27$

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(C) Let $A = \begin{bmatrix} a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \ c_1 & c_2 & c_3 \end{bmatrix}$.Given $A\begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} = 3\begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix}$,we have:$a_1 + a_2 + a_3 = 3$$b_1 + b_2 + b_3 = 3$$c_1 + c_2 + c_3 = 3$Since all elements are non-negative,by the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality,the product of elements in each row is maximized when all elements in that row are equal.For row $1$: $a_1 a_2 a_3 \le (\frac{a_1+a_2+a_3}{3})^3 = (\frac{3}{3})^3 = 1$.Similarly,for row $2$ and row $3$,the product is at most $1$.The determinant $\operatorname{det}(A)$ is the sum of products of elements. By Hadamard's inequality or by considering the row sums,the maximum value of the determinant of a matrix with row sums equal to $S$ is $S^n$ for an $n 	imes n$ matrix if the matrix is diagonal.Here,$S=3$ and $n=3$,so $\operatorname{det}(A) \le 3^3 = 27$.The maximum value is attained when $A = \begin{bmatrix} 3 & 0 & 0 \ 0 & 3 & 0 \ 0 & 0 & 3 \end{bmatrix}$,which gives $\operatorname{det}(A) = 27$.

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 $\operatorname{det}(A)$ is:

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