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Question
Let `A =[(3,7),(2,5)] and B = [(6,8),(7,9)]`. Verify that `(AB)^(-1) = B^(-1)A^(-1).`
Solution
A = `[(3,7),(2,5)]`
`abs A = 15 - 14 = 1 ne 0 -> A^-1` exists.
`A^-1 = 1/abs A (adjA)`
`= 1/1 [(5,-7),(-2,3)]`
`= [(5,-7),(-1,3)]`
B = `[(6,8),(7,9)]`
`abs B = 54 - 56 = -2 ne 0 -> B^-1` exists.
`B^-1 = 1/abs B (adjB)`
`= 1/-2 [(9,-8),(-7,6)]`
`= [(-9/2,4),(7/2, -3)]`
`B^-1 A^-1 = [(-9/2,4),(7/2, -3)][(5,-7),(-2,3)]`
`= [(-45/2 - 8, 63/2 + 12),(35/2 + 6, -49/2 - 9)]`
`= [(-61/2, 87/2),(47/2, -67/2)]`
`AB = [(3,7),(2,5)][(6,8),(7,9)]`
`= [(18 + 49, 24 + 63),(12 + 35, 16 + 45)]`
`= [(67,87),(47,61)]`
`abs AB = 4087 - 4089 = -2 ne 0 -> (AB)^-1` exists.
`(AB)^-1 = 1/abs (AB) (adj AB)`
`= 1/-2 [(61,-87),(-47,67)]`
`= [(-61/2,87/2),(47/2,-67/2)]`
Hence, `(AB)^-1 = B^-1 A^-1`
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