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Let A = [a_ij] be a 3x3 matrix where a_ij + a_ji = 0 and each element a_ij belongs to the set {0, ±1, ±2, ±3, ±4, ±5, ±6, ±7}. How many such matrices A are possible?
- 3370
- 3300
- 3375
- None of these
Correct answer: 3375
Solution
Condition a_ij + a_ji = 0: skew-symmetric matrix. Diagonal: a_ii + a_ii = 0 -> a_ii = 0 (3 elements fixed). Off-diagonal pairs (i<j): a_ij = -a_ji. For each pair, choose a_ij from the valid set. Valid values: a_ij in {0, ±1, ±2, ±3, ±4, ±5, ±6, ±7} and a_ji = -a_ij must also be in this set. Since the set is symmetric about 0, any choice of a_ij from {0, ±1,...,±7} automatically gives a valid a_ji. Set size = 15 (0, ±1, ±2, ±3, ±4, ±5, ±6, ±7 = 1 + 2*7 = 15). Number of off-diagonal pairs in 3x3 = 3 (pairs: (1,2),(1,3),(2,3)). Total matrices = 15³ = 3375.
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