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Question

The number of all 3×3 matrices A, with entries from the set {1,0,1} such that the sum of the diagonal elements of (AAT) is 3, is

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Solution

tr(AAT)=3
Let A=a11a12a13a21a22a23a31a32a33

Then AT=a11a21a31a12a22a32a13a23a33

tr(AAT)=a211+a212+a213+a221+a222+a223+a231+a232+a233=3
So, out of nine elements aijs, three elements must be equal to 1 or 1 and rest elements must be 0.
So, the possible cases will be:
For six 0s and three 1s, total possibilities is 9C6
For six 0s , two 1s and one 1s, total possibilities is 9C6×3
For six 0s, one 1s and two 1s, total possibilities is 9C6×3
For six 0s and three 1s, total possibilities is 9C6
Total number of cases = 9C6×8=672

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