DeterminantHard
Question
A is a (3 × 3) diagonal matrix having integral entries such that det (A) = 120, number of such matrices is
Options
A.360
B.390
C.240
D.270
Solution
given A = 
abc = 120 = 23 × 31 × 51
Case - I All are Positive = 5C2 × 3C2 × 3C2 = 90
Case - I one Positive and two negative = 3 × (5C2 × 3C2 × 3C2) = 270
So number of possible matrices = 90 + 270 = 360
abc = 120 = 23 × 31 × 51
Case - I All are Positive = 5C2 × 3C2 × 3C2 = 90
Case - I one Positive and two negative = 3 × (5C2 × 3C2 × 3C2) = 270
So number of possible matrices = 90 + 270 = 360
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