Question

If A = `[(3,1),(-1,2)]` show that A² – 5A + 7I = 0. Hence, find A^–1.

Answer 1

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

L.H.S. = A² – 5A + 7I

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

= `[(9 - 1,3 + 2),(-3 -2,-1 + 4)] - [(15,5),(-5,10)] + [(7,0),(0,7)]`

= `[(8 - 15 + 7,5 -5+0),(-5 +5+0,3 -10+7)]`

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

= 0

Hence proved.

Now multiplying by A⁻¹ both sides, we get

(A⁻¹A)A − 5AA⁻¹ + 7IA⁻¹ = 0

⇒ IA − 5I + 7A⁻¹ = 0

⇒ A − 5I + 7A⁻¹ = 0

⇒ 7A⁻¹ = 5I − AI

7A⁻¹ = `5[(1,0),(0,1)] - [(3,1),(-1,2)]`

7A⁻¹ = `[(5,0),(0,5)] - [(3,1),(-1,2)]`

7A⁻¹ = `[(2,-1),(1,3)]`

A⁻¹ = `1/7 [(2,-1),(1,3)]`