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Question
If A = `[(0, -x),(x, 0)]`, B = `[(0, 1),(1, 0)]` and x2 = –1, then show that (A + B)2 = A2 + B2.
Solution
We have, A = `[(0, -x),(x, 0)]`, B = `[(0, 1),(1, 0)]` and x2 = –1
∴ (A + B) = `[(0, -x + 1),(x + 1, 0)]`
∴ (A + B)2 = `[(0, -x + 1),(x + 1, 0)] [(0, -x + 1),(x + 1, 0)]`
= `[(1 - x^2, 0),(0, 1 - x^2)]` .....(i)
Also, A2 = A · A
= `[(0, -x),(x, 0)] [(0, -x),(x, 0)]`
= `[(-x^2, 0),(0, -x^2)]`
And B2 = B · B
= `[(0, 1),(1, 0)] [(0, 1),(1, 0)]`
= `[(1, 0),(0, 1)]`
∴ A2 + B2 = `[(-x^2, 0),(0, -x^2)] + [(1, 0),(0, 1)]`
= `[(1 - x^2, 0),(0, 1 - x^2)]` ......(ii)
From equations (i) and (ii), we have
(A + B)2 = A2 + B2
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