Question

If A = `[(0, -x),(x, 0)]`, B = `[(0, 1),(1, 0)]` and x² = –1, then show that (A + B)² = A² + B². 

Answer 1

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