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then 060 5 e det b 5 C 5 f a a det b P 2 d 12 = e = 0 f and

 

then 060 5 e det b 5 C 5 f a a det b P 2 d 12 = e = 0 f and 1 d a det b 1 e = 1 f - -4, and det a 1 d 62 e = 1, 3 f

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