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Question
If A = `[(0, 1),(1, 1)]` and B = `[(0, -1),(1, 0)]`, show that (A + B)(A – B) ≠ A2 – B2
Solution
Given that A = `[(0, 1),(1, 1)]` and B = `[(0, -1),(1, 0)]`
A + B = `[(0, 1),(1, 1)] + [(0, -1),(1, 0)]`
⇒ A + B = `[(0 + 0, 1 - 1),(1 + 1, 1 + 0)]`
⇒ A + B = `[(0, 0),(2, 1)]`
A – B = `[(0, 1),(1, 1)] - [(0, -1),(1, 0)]`
⇒ A – B = `[(0 - 0, 1 + 1),(1 - 1, 1 - 0)]`
⇒ A – B = `[(0, 2),(0, 1)]`
∴ `("A" + "B") * ("A" – "B") = [(0, 0),(2, 1)],[(0, 2),(0, 1)]`
= `[(0 + 0, 0 + 0),(0 + 0, 4 + 1)]`
= `[(0, 0),(0, 5)]`
Now, R.H.S. = A2 – B2
= `"A" * "A" – "B" * "B"`
= `[(0, 1),(1, 1)][(0, 1),(1, 1)] - [(0,-1),(1, 0)][(0, -1),(1, 0)]`
= `[(0 +1,0 +1),(0 + 1, 1 + 1)] - [(0 - 1, 0 + 0),(0 + 0, -1 + 0)]`
= `[(1, 1),(1, 2)] - [(-1, 0),(0, -1)]`
= `[(1 + 1, 1 -0),(1 -0, 2 + 1)]`
= `[(2, 1),(1, 3)]`
Hence, `[(0, 0),(0, 5)] ≠ [(2, 10),(1, 3)]`
Hence, (A + B) . (A – B) ≠ A2 – B2
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