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Question
Let A = `[(1, 2),(1, 3)]`, B = `[(4, 0),(1, 5)]`, C = `[(2, 0),(1, 2)]` Show that A(BC) = (AB)C
Solution
A = `[(1, 2),(1, 3)]`, B = `[(4, 0),(1, 5)]`, C = `[(2, 0),(1, 2)]`
BC = `[(4, 0),(1, 5)] xx [(2, 0),(1, 2)]`
= `[(8 + 0, 0 + 0),(2 + 5, 0 + 10)]`
= `[(8, 0),(7, 10)]`
A(BC) = `[(1, 2),(1, 3)] xx [(8, 0),(7, 10)]`
= `[(8 + 14, 0 + 20),(8 + 21, 0 + 30)]`
= `[(22, 20),(29, 30)]` ...(1)
AB = `[(1, 2),(1, 3)] xx [(4, 0),(1, 5)]`
= `[(4 + 2, 0 + 10),(4 + 3, 0 +15)]`
= `[(6, 10),(7, 15)]`
(AB)C = `[(6, 10),(7, 15)] xx [(2, 0),(1, 2)]`
= `[(12 + 10, 0 + 20),(14 + 15, 0 + 30)]`
= `[(22, 20),(29, 30)]` ...(2)
From (1) and (2) we get
A(BC) = (AB)C
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