Question

If A = `[(1, 3, 2),(2, 0, -1),(1, 2, 3)]`, then show that A³ – 4A² – 3A + 11I = 0, Hence find A^–1 .

Answer 1

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

A² = A.A

= `[(1, 3, 2),(2, 0, -1),(1, 2, 3)][(1, 3, 2),(2, 0, -1),(1, 2, 3)]`

= `[(9, 7, 5),(1, 4, 1),(8, 9, 9)]`

A³ = A².A

= `[(9, 7, 5),(1, 4, 1),(8, 9, 9)][(1, 3, 2),(2, 0, -1),(1, 2, 3)]`

= `[(28, 37, 26),(10, 5, 1),(35, 42, 34)]`

∴ A³ – 4A² – 3A + 11I = `[(28, 37, 26),(10, 5, 1),(35, 42, 34)] - 4[(9, 7, 5),(1, 4, 1),(8, 9, 9)] -3[(1, 3, 2),(2, 0, -1),(1, 2, 3)] + 11[(1, 0, 0),(0, 1, 0),(0, 0, 1)]`

= `[(28, 37, 26),(10, 5, 1),(35, 42, 34)] - [(36, 28, 20),(4, 16, 4),(32, 36, 36)] - [(3, 9, 6),(6, 0, -3),(3, 6, 9)] + [(11, 0, 0),(0, 11, 0),(0, 0, 11)]`

= `[(28 - 36 - 3 + 11, 37 - 28 - 9 + 0, 26 - 20 - 6 + 0),(10 - 4 - 6 + 0, 5 - 16 - 0 + 11, 1 - 4 + 3 + 0),(35 - 32 - 3 + 0, 42 - 36 - 6 + 0, 34 - 36 - 9 + 11)]`

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

= 0

A³ – 4A² – 3A + 11I = 0

A³·A^–1 – 4A²·A^–1 – 3A·A^–1 + 11IA^–1 = 0.A^–1

A²I – 4AI – 3I + 11A^–1 = 0

A² – 4A – 3I + 11A^–1 = 0

11A^–1 = 3I + 4A – A²

= `3[(1, 0, 0),(0, 1, 0),(0, 0, 1)] + 4[(1, 3, 2),(2, 0, -1),(1, 2, 3)] - [(9, 7, 5),(1, 4, 1),(8, 9, 9)]`

11A^–1 = `[(3 + 4 - 9, 0 + 12 - 7, 0 + 8 - 5),(0 + 8 - 1, 3 + 0 - 4, 0 - 4 - 1),(0 + 4 - 8, 0 + 8 - 9, 3 + 12 - 9)]`

11A^–1 = `[(-2, 5, 3),(7, -1, -5),(-4, -1, 6)]`

 A^–1 = `[(-2/11, 5/11, 3/11),(7/11, -1/11, -5/11),(-4/11, -1/11, 6/11)]`