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Question
If A = `[(2, 1),(0, 0)]`, B = `[(2, 3),(4, 1)]` and C = `[(1, 4),(0, 2)]`; then show that (B – A)C = BC – AC.
Solution
B – A = `[(2, 3),(4, 1)] - [(2, 1),(0, 0)] = [(0, 2),(4, 1)]`
(B – A)C = `[(0, 2),(4, 1)][(1, 4),(0, 2)]`
= `[(0, 0 + 4),(4 + 0, 16 + 2)]`
= `[(0, 4),(4, 18)]`
BC = `[(2, 3),(4, 1)][(1, 4),(0, 2)]`
= `[(2 + 0, 8 + 6),(4 + 0, 16 + 2)]`
= `[(2, 14),(4, 18)]`
AC = `[(2, 1),(0, 0)][(1, 4),(0, 2)]`
= `[(2 + 0, 8 + 2),(0, 0)]`
= `[(2, 10),(0, 0)]`
BC – AC = `[(2, 14),(4, 18)] - [(2, 10),(0,0)]`
= `[(0, 4),(4, 18)]`
Hence, (B – A)C = BC – AC
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