A = [a_{ij}] is a 3 times 3 matrix with posit | vedclass

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(D) Consider $A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \ a_{21} & a_{22} & a_{23} \ a_{31} & a_{32} & a_{33} \end{bmatrix}$.Each row sum is $6$:

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$12$

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(D) Consider $A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \ a_{21} & a_{22} & a_{23} \ a_{31} & a_{32} & a_{33} \end{bmatrix}$.Each row sum is $6$:$a_{11} + a_{12} + a_{13} = 6 \dots (i)$$a_{21} + a_{22} + a_{23} = 6 \dots (ii)$$a_{31} + a_{32} + a_{33} = 6 \dots (iii)$From the given condition on diagonal entries:$a_{11} = a_{12} + a_{21} \dots (iv)$$a_{22} = a_{23} + a_{32} = 2 \dots (v)$$a_{33} = a_{13} + a_{31} \dots (vi)$Since entries are positive integers, from $(v)$, $a_{23} = 1$ and $a_{32} = 1$. Substituting into $(ii)$, $a_{21} + 2 + 1 = 6 \Rightarrow a_{21} = 3$.From $(iv)$, $a_{11} = a_{12} + 3$. Substituting into $(i)$, $(a_{12} + 3) + a_{12} + a_{13} = 6 \Rightarrow 2a_{12} + a_{13} = 3$. Since $a_{ij} \ge 1$, we must have $a_{12} = 1$, $a_{13} = 1$, and $a_{11} = 4$.From $(iii)$ and $(vi)$, $a_{31} + a_{32} + a_{33} = 6 \Rightarrow a_{31} + 1 + a_{33} = 6 \Rightarrow a_{31} + a_{33} = 5$. Also $a_{33} = a_{13} + a_{31} = 1 + a_{31}$.Substituting $a_{33}$ in the sum, $a_{31} + (1 + a_{31}) = 5 \Rightarrow 2a_{31} = 4 \Rightarrow a_{31} = 2$, so $a_{33} = 3$.Thus, $A = \begin{bmatrix} 4 & 1 & 1 \ 3 & 2 & 1 \ 2 & 1 & 3 \end{bmatrix}$.$|A| = 4(6 - 1) - 1(9 - 2) + 1(3 - 4) = 4(5) - 1(7) + 1(-1) = 20 - 7 - 1 = 12$.

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