Question

If A = `[(1, 0, 2),(0, 2, 1),(2, 0, 3)]`, prove that A² – 6A² + 7A + 2I = 0

Answer 1

Given: A = `[(1, 0, 2),(0, 2, 1),(2, 0, 3)]`

∴ A² = A × A = `[(1, 0, 2),(0, 2, 1),(2, 0, 3)] [(1, 0, 2),(0, 2, 1),(2, 0, 3)]`

= `[(1 + 0 + 4, 0 + 0 + 0, 2 + 0 + 6),(0 + 0 + 2, 0 + 4 + 0, 0 + 2 + 3),(2 + 0 + 6, 0 + 0 + 0, 4 + 0 + 9)]`

= `[(5, 0, 8),(2, 4, 5),(8, 0, 13)]`

A³ = A²·A = `[(5,0,8),(2,4,5),(8,0,13)] [(1,0,2),(0,2,1),(2,0,3)]`

= `[(5 + 0 + 16, 0 + 0 + 0, 10 + 0 + 24), (2 + 0 + 10, 0 + 8 + 0, 4 + 4 + 15), (8 + 0 + 26, 0 + 0 + 0, 16 + 0 + 39)]`

= `[(21, 0, 34),(12, 8, 23),(34, 0, 55)]`

∴ A³ – 6A² + 7A + 2I = `[(21, 0, 34),(12, 8, 23),(34, 0, 55)] - 6[(5, 0, 8),(2, 4, 5),(8, 0, 13)] + 7[(1, 0, 2),(0, 2, 1),(2, 0, 3)] + 2[(1, 0, 0),(0, 1, 0),(0, 0, 1)]`

= `[(21, 0, 34),(12, 8, 23),(34, 0, 55)] - [(30, 0, 48),(12, 24, 30),(48, 0, 78)] + [(7, 0, 14),(0, 14, 7),(14, 0, 21)] + [(2, 0, 0),(0, 2, 0),(0, 0, 2)]`

= `[(21 + 7 + 2, 0 + 0 + 0, 34 + 14 + 0),(12 + 0 + 0, 8 + 14 + 2, 23 + 7 + 0),(34 + 14 + 0, 0 + 0 + 0, 55 + 21 + 2)] - [(30, 0, 48),(12, 24, 30),(48, 0, 78)]`

= `[(30, 0, 48),(12, 24, 30),(48, 0, 78)] - [(30, 0, 48),(12, 24, 30),(48, 0, 78)]`

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

= 0

∴ A³ – 6A² + 7A + 2I = 0