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