& a_{33} \end{bmatrix}$ where each element is a positive integer and each row sums to 6, with $a_{22} = 2$.

The matrix properties provide the following conditions:

$a_{22} = a_{23} + a_{32}$ implies $a_{23} + a_{32} = 2$.

$a_{11} = a_{12} + a_{21}$.

$a_{33} = a_{31} + a_{13}$.

Substituting the conditions about row sums, we get:

$a_{11} + a_{12} + a_{13} = 6$.

$a_{21} + a_{22} + a_{23} = 6$.

$a_{31} + a_{32} + a_{33} = 6$.

From $a_{23} + a_{32} = 2$, if $a_{23} = 1$, then $a_{32} = 1$.

From the second row sum, we have: $a_{21} + 2 + 1 = 6$ which gives $a_{21} = 3$.

Given $a_{11} = a_{12} + a_{21}$ and the first row sum,

with $2a_{12} + a_{13} = 3$,

choosing $a_{12} = 1$ and $a_{13} = 1$, results in $a_{11} = 4$.

Finally, using the third row equation $2a_{31} + a_{32} + a_{13} = 6$,

substituting $a_{32} = 1$ and $a_{13} = 1$ results in $a_{31} = 2$.

Now, from $a_{31} + a_{32} + a_{33} = 6$, with $a_{31} = 2$ and $a_{32} = 1$,

results in $a_{33} = 3$.

Matrix $A$ is:

$ A = \begin{bmatrix} 4 & 1 & 1 \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{bmatrix} $

To find the determinant $|A|$, we calculate:

$ |A| = 4(2 \cdot 3 - 1 \cdot 1) - 1(3 \cdot 3 - 1 \cdot 2) + 1(3 \cdot 1 - 2 \cdot 2) $

$ = 4(6 - 1) - 1(9 - 2) + 1(3 - 4) $

$ = 4 \cdot 5 - 1 \cdot 7 + 1 \cdot (-1) $

$ = 20 - 7 - 1 = 12 $

Thus, the determinant $|A| = 12$.

Matrices And Determinants

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