Question

If A = `[(1, 2, 3),(3, -2, 1),(4, 2, 1)]`, then show that A³ – 23A – 40I = 0.

Answer 1

A = `[(1, 2, 3),(3, -2, 1),(4, 2, 1)]`

A³ – 23A – 40I = 0

A² = `[(1, 2, 3),(3, -2, 1),(4, 2, 1)][(1, 2, 3),(3, -2, 1),(4, 2, 1)]`

= `[(1 + 6 + 12, 2 - 4 + 6, 3 + 2 + 3),(3 - 6 + 4, 6 + 4 + 2, 9 - 2 + 1),(4 + 6 + 4, 8 - 4 + 2, 12 + 2 + 1)]`

= `[(19, 4, 8),(1, 12, 8),(14, 6, 15)]`

A³ = `[(19, 4, 8),(1, 12, 8),(14, 6, 15)][(1, 2, 3),(3, -2, 1),(4, 2, 1)]`

= `[(19 + 12 + 32, 38 - 8 + 16, 57 + 4 + 8),(1 + 36 + 32, 2 - 24 + 16, 3 + 12 + 8),(14 + 18 + 60, 28 - 12 + 30, 42 + 6 + 15)]`

= `[(63, 46, 69),(69, -6, 23),(92, 46, 63)]`

23A = `23[(1, 2, 3),(3, -2, 1),(4, 2, 1)] = [(23, 46, 69),(69, -46, 23),(92, 46, 23)]`

Now A³ – 23A – 40I

`[(63, 46, 69),(69, -6, 23),(92, 46, 63)] - [(23, 46, 69),(69, -46, 23),(92, 46, 23)] - [(40, 0, 0),(0, 40, 0),(0, 0, 40)]`

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

= 0