Question

If `"A" = [(3 , 1),(2 , 1)] and "B" = [(1 , -2),(5 , 3)]`, then show that (A - B)² ≠ A2 - 2AB + B².

Answer 1

A - B = `[(3 , 1),(2 , 1)] - [(1 , -2),(5 , 3)]`

= `[(3 - 1, 1 + 2),(2 - 5 , 1 - 3)] = [(2 , 3),(-3 , -2)]`

(A - B)² = (A - B)(A - B)

⇒ (A - B)2 = `[(2 , 3),(-3 , -2)][(2 , 3),(-3 , -2)]`

= `[(4 - 9 , 6 - 6),(-6 + 6 , -9 + 4)]`

= `[(-5 , 0),(0 , -5)]`

and A² = `[(3 , 1),(2 , 1)][(3 , 1),(2 , 1)]`

= `[(9 + 2 , 3 + 1),(6 + 2 , 2 + 1)] = [(11 , 4),(8 , 3)]`

and B² = `[(1 , -2),(5 , 3)][(1 , -2),(5 , 3)]`

= `[(1 - 10 , -2 -6),(5 + 15 , -10 + 9)]`

= `[(-9 , -8),(20 , -1)]`

and AB = `[(3 , 1),(2 , 1)][(1 , -2),(5 , 3)]`

= `[(3 + 5 , -6 + 3),(2 + 5 , -4 + 3)] = [(8 , -3),(7 , -1)]`

Now A²  - 2AB + B²

= `[(11 , 4),(8 , 3)] -2 [(8 , -3),(7 , -1)] + [(-9 , -8),(20 , -1)]`

= `[(11 , 4),(8 , 3)] - [(16 , -6),(14 , -2)] + [(-9 , -8),(20 , -1)]`

= `[(11 - 6 - 9 , 4 + 6 - 8),(8 - 14 + 20 , 3 + 2 - 1)]`

= `[(-14 , 2),(14 , 4)]`
Hence, from above calculations, we get
(A - B)² ≠ A² - 2AB + B².