Let P be a matrix of order 3 × 3 such that all the entries in P are from the set {–1, 0, 1}. Then, the maximum possible value of the determinant of P is _____ .
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(4)
Sol. det (P) = 
= a 1 (b 2 c 3 – b 3 c 2 ) – a 2 (b 1 c 3 – b 3 c 1 ) + a 3 (b 1 c 2 – b 2 c 1 ) ≤ 6
value can be 6 only if a 1 = 1, a 2 = –1, a 3 = 1, b 2 c 3 = b 1 c 3 = b 1 c 2 = 1, b 3 c 2 = b 3 c 1 = b 2 c 1 = – 1
⇒ (b 2 c 3 ) (b 3 c 1 ) (b 1 c 2 ) = – 1 & (b 1 c 3 )(b 3 c 2 ) (b 2 c 1 ) = 1
i.e. b 1 b 2 b 3 c 1 c 2 c 3 = 1 and – 1
hence not possible
Similar contradiction occurs when a 1 = 1, a 2 = 1, a 3 = 1, b 2 c 2 = b 3 c 1 = b 1 c 2 = 1 b 3 c 2 = b 1 c 3 = b 1 c 2 = – 1
Now for value to be 5 one the terms must be zero but that will make 2 terms zero which means answer cannot be 5
Now 
= 4 Hence max value = 4
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