Question

For the matrix A = `[(1,1,1),(1,2,-3),(2,-1,3)]` show that A³ − 6A² + 5A + 11 I = 0. Hence, find A⁻¹.

Answer 1

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

A² = AA

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

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

= `[(4,2,1),(-3,8,-14),(7,-3,14)]`

A³ = A²A

= `[(4,2,1),(-3,8,-14),(7,-3,14)] [(1,1,1),(1,2,-3),(2,-1,3)]`

= `[(4+2+2,4+4-1,4-6+3),(-3+8-28,-3+16+14,-3-24-42),(7-3+28,7-6-14,7+9+42)]`

= `[(8,7,1),(-23,27,-69),(32,-13,58)]`

L.H.S. = A³ − 6A² + 5A + 11 I

= `[(8,7,1),(-23,27,-69),(32,-13,58)] - 6 [(4,2,1),(-3,8,-14),(7,-3,14)] + 5 [(1,1,1),(1,2,-3),(2,-1,3)] + 11 [(1,0,0),(0,1,0),(0,0,1)]`

= `[(8,7,1),(-23,27,-69),(32,-13,58)] - [(24,12,6),(-18,48,-84),(42,-18,84)] + [(5,5,5),(5,10,-15),(10,-5,15)] + [(11,0,0),(0,11,0),(0,0,11)]`

= `[(8 - 24 + 5 + 11, 7 - 12 + 5 + 0, 1 - 6 + 5 + 0),(-23 + 18 + 5 + 0, 27 - 48 + 10 + 11, -69 + 84 - 15 + 0),(32 - 42 + 10 + 0,-13 + 18 - 5 + 0, 58 - 84 + 15 + 11)]`

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

= 0

Hence, A³ − 6A² + 5A + 11 I = 0

Now, A³ − 6A² + 5A + 11 I = 0

A³ − 6A² + 5A = −11 I

A²AA⁻¹ = 6AAA⁻¹ + 5AA⁻¹ = 11 I A⁻¹

11A⁻¹ = −A² + 6A − 5I

= `[(-4,-2,-1),(3,-8,14),(-7,3,-14)] + 6 [(1,1,1),(1,2,-3),(2,-1,3)] - 5 [(1,0,0),(0,1,0),(0,0,1)]`

= `[(-4,-2,-1),(3,-8,14),(-7,3,-14)] + [(6,6,6),(6,12,-18),(12,-6,18)] - [(5,0,0),(0,5,0),(0,0,5)]`

= `[(-3,4,5),(9,-1,-4),(5,-3,-1)]`

A⁻¹ = `1/11 [(-3,4,5),(9,-1,-4),(5,-3,-1)]`