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Question
Let A = `[(1, 2),(-1, 3)]`, B = `[(4, 0),(1, 5)]`, C = `[(2, 0),(1, -2)]` and a = 4, b = –2. Show that: (A – B)C = AC – BC
Solution
We have,
A = `[(1, 2),(-1, 3)]`
B = `[(4, 0),(1, 5)]`
C = `[(2, 0),(1, -2)]` a
And a = 4, b = –2
(A – B) = `[(1, 2),(-1, 3)] - [(4, 0),(1, 5)]`
= `[(1 - 4, 2 - 0),(-1 - 1, 3 - 5)]`
= `[(-3, 2),(-2, -2)]`
∴ (A – B)C = `[(-3, 2),(-2, -2)] [(2, 0),(1, -2)]`
= `[(-4, -4),(-6, 4)]`
Now, AC = `[(1, 2),(-1, 3)] [(2, 0),(1, -2)]`
= `[(4, -4),(1, -6)]`
And BC = `[(4, 0),(1, 5)] [(2, 0),(1, -2)]`
= `[(8, 0),(7, -10)]`
∴ AC – BC = `[(4 - 8, -4 -0),(1 - 7, -6 + 10)]`
= `[(-4, -4),(-6, 4)]`
= (A – B)C
Hence proved.
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