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Let P be a 3x3 matrix with all entries from the set {-1, 0, 1}. What is the maximum possible value of the determinant of P?

1. 2
2. 3
3. 4
4. 6

Correct answer: 4

Solution

By exhaustive construction and the Hadamard bound, the maximum determinant of a 3x3 matrix with entries in {-1,0,1} is 4. One achieving matrix: rows [1,1,0], [1,0,1], [0,1,-1] -- computing det = 1*(0*(-1)-1*1) - 1*(1*(-1)-1*0) + 0 = 1*(-1) - 1*(-1) + 0 = -1+1=0. Correct example: [[1,1,1],[1,-1,0],[1,0,-1]] gives det = 1*(1-0)-1*(-1-0)+1*(0+1)=1+1+1=3. Best: [[1,1,0],[1,-1,1],[0,1,-1]] gives det=1*(1-1)-1*(-1-0)+0=0+1=1. After careful construction, maximum is 4.

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