Question

If A = `[(3, -5),(-4, 2)]`, then find A² – 5A – 14I. Hence, obtain A³.

Answer 1

Given that: A = `[(3, -5),(-4, 2)]`

A² = A . A

= `[(3, -5),(-4, 2)] [(3, -5),(-4, 2)]`

= `[(9 + 20, -15 - 10),(-12 - 8, 20 + 4)]`

= `[(29, -25),(-20, 24)]`

∴ A² – 5A – 14I = `[(29, -25),(-20, -24)] -5[(3, -5),(-4, 2)] -14[(1, 0),(0, 1)]`

= `[(29, -25),(-20, 24)] - [(15, -25),(-20, 10)] - [(14, 0),(0, 14)]`

= `[(29, -25),(-20, 24)] - [(29, -25),(-20, 24)]`

= `[(29 - 29, -25 + 25),(-20 + 20, 24 - 24)]`

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

Hence, A² – 5A – 14I = 0

Now, multiplying both sides by A, we get,

A² . A – 5A . A – 14IA = 0A

⇒ A³ – 5A² – 14A = 0

⇒ A³ = 5A² + 14A

⇒ A³ = `5[(29, -25),(-20, 24)] + 14[(3, -5),(-4, -2)]`

= `[(145, -125),(-100, 120)] + [(42, -70),(-56, 28)]`

= `[(145 + 42, -125 - 70),(-100 - 56, 120 + 28)]`

= `[(187, -195),(-156, 148)]`

Hence, A³ = `[(187, -195),(-156, 148)]`