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Question
If A = `[(1, 2),(-3, 4)], "B" = [(0, 1),(-2, 5)] and "C" = [(-2, 0),(-1, 1)]` find A(4B – 3C)
Solution
A = `[(1, 2),(-3, 4)], "B" = [(0, 1),(-2, 5)] and "C" = [(-2, 0),(-1, 1)]`
4B - 3C = `4[(0, 1),(-2, 5)] - 3[(-2, 0),(-1, 1)]`
= `[(0, 4),(-8, 20)] - [(-6, 0),(-3, 3)]`
= `[(0 - (-6), 4 - 0),(-8 - (-3), 20 - 3)]`
= `[(0 + 6, 4 - 0),(-8 + 3, 20 - 3)]`
= `[(6, 4),(-5, 17)]`
Now A(4B - 3C) = `[(1, 2),(-3, 4)][(6, 4),(-5, 17)]`
= `[(1 xx 6 + 2(-5), 1 xx 4 + 2 xx 17),(-3 xx + 6 + 4 x (-5), -3 xx 4 + 4 xx 17)]`
= `[(6 - 10, 4 + 34),(-18 - 20, -12 + 68)]`
= `[(-4, 38),(-38, 56)]`.
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