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Question
Verify A(BC) = (AB)C of the following case:
A = `[(1, 0, 1),(2, 3, 0),(0, 4, 5)], "B" = [(2, -2),(-1, 1),(0, 3)] and "C" = [(3, 2, -1),(2, 0, -2)]`
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Solution
BC = `[(2, -2),(-1, 1),(0, 3)] [(3, 2, -1),(2, 0, -2)]`
= `[(6 - 4, 4 - 0, -2 + 4),(-3 + 2, -2 + 0, 1 - 2),(0 + 6, 0 + 0, 0 - 6)]`
= `[(2, 4, 2),(-1, -2, -1),(6, 0, -6)]`
∴ A(BC) = `[(1, 0, 1),(2, 3, 0),(0, 4, 5)][(2, 4, 2),(-1, -2, -1),(6, 0, -6)]`
= `[(2 - 0 + 6, 4 - 0 + 0, 2 - 0 - 6),(4 - 3 + 0, 8 - 6 + 0, 4 - 3 - 0),(0 - 4 + 30, 0 - 8 + 0, 0 - 4 - 30)]`
∴ A(BC) = `[(8, 4, -4),(1, 2, 1),(26, -8, -34)]` ...(i)
AB = `[(1, 0, 1),(2, 3, 0),(0, 4, 5)][(2, -2),(-1, 1),(0, 3)]`
= `[(2 + 0 + 0, -2 + 0 + 3),(4 - 3 + 0, -4 + 3 + 0),(0 - 4 + 0, 0 + 4 + 15)]`
= `[(2, 1),(1, -1),(-4, 19)]`
∴ (AB)C = `[(2, 1),(1, -1),(-4, 19)][(3, 2, -1),(2, 0, -2)]`
= `[(6 + 2, 4 + 0, -2 - 2),(3 - 2, 2 - 0, -1 + 2),(-12 + 38, -8 + 0, 4 - 38)]`
∴ (AB)C = `[(8, 4, -4),(1, 2, 1),(26, -8, -34)]` ...(ii)
From (i) and (ii), we get
A(BC) = (AB)C
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