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Question
Answer the following question:
If A = `[(0, 1),(1, 0)]` and B = `[(0, -1),(1, 0)]` show that (A + B)(A – B) ≠ A2 – B2
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Solution
A + B = `[(0, 1),(1, 0)] + [(0, -1),(1, 0)]`
= `[(0, 0),(2, 0)]`
A – B = `[(0, 1),(1, 0)] - [(0, -1),(1, 0)]`
= `[(0, 2),(0, 0)]`
∴ (A + B)(A – B) = `[(0, 0),(2, 0)] [(0, 2),(0, 0)]`
= `[(0 + 0, 0 + 0),(0 + 0, 4 + 0)]`
= `[(0, 0),(0, 4)]` ...(1)
Also, A2 = A·A = `[(0, 1),(1, 0)] [(0, 1),(1, 0)]`
= `[(0 + 1, 0 + 0),(0 + 0, 1 + 0)]`
= `[(1, 0),(0, 1)]`
B2 = B·B = `[(0, -1),(1, 0)] [(0, -1),(1, 0)]`
= `[(0 - 1, 0 - 0),(0 + 0, -1 + 0)]`
= `[(-1, 0),(0, -1)]`
∴ A2 – B2 = `[(1, 0),(0, 1)] - [(-1, 0),(0, -1)]`
= `[(2, 0),(0, 2)]`
From (1) and (2),
(A+ B)(A – B) ≠ A2 – B2.
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