] where each element is a positive integer and each row sums to 6 , with a22=2. The matrix properties provide the following conditions: ‌a22=a23+a32‌ implies ‌a23+a32=2. ‌a11=a12+a21. ‌a33=a31+a13. Substituting the conditions about row sums, we get: ‌a11+a12+a13=6. ‌a21+a22+a23=6. ‌a31+a32+a33=6. From a23+a32=2, if a23=1, then a32=1. From the second row sum, we have: a21+2+1=6 which gives a21=3. Given a11=a12+a21 and the first row sum, with 2a12+a13=3, choosing a12=1 and a13=1, results in a11=4.
Finally, using the third row equation 2a31+a32+a13=6, substituting a32=1 and a13=1 results in a31=2. Now, from a31+a32+a33=6, with a31=2 and a32=1, results in a33=3. Matrix A is: A=[
4
1
1
3
2
1
2
1
3
] To find the determinant |A|, we calculate: ‌|A|=4(2⋅3−1⋅1)−1(3⋅3−1⋅2)+1(3⋅1−2⋅2) ‌=4(6−1)−1(9−2)+1(3−4) ‌=4⋅5−1⋅7+1⋅(−1) ‌=20−7−1=12 Thus, the determinant |A|=12.