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Question
If A = `[(1, 2, 3),(3, -2, 1),(4, 2, 1)]`, then show that A3 – 23A – 40I = 0.
Solution
A = `[(1, 2, 3),(3, -2, 1),(4, 2, 1)]`
A3 – 23A – 40I = 0
A2 = `[(1, 2, 3),(3, -2, 1),(4, 2, 1)][(1, 2, 3),(3, -2, 1),(4, 2, 1)]`
= `[(1 + 6 + 12, 2 - 4 + 6, 3 + 2 + 3),(3 - 6 + 4, 6 + 4 + 2, 9 - 2 + 1),(4 + 6 + 4, 8 - 4 + 2, 12 + 2 + 1)]`
= `[(19, 4, 8),(1, 12, 8),(14, 6, 15)]`
A3 = `[(19, 4, 8),(1, 12, 8),(14, 6, 15)][(1, 2, 3),(3, -2, 1),(4, 2, 1)]`
= `[(19 + 12 + 32, 38 - 8 + 16, 57 + 4 + 8),(1 + 36 + 32, 2 - 24 + 16, 3 + 12 + 8),(14 + 18 + 60, 28 - 12 + 30, 42 + 6 + 15)]`
= `[(63, 46, 69),(69, -6, 23),(92, 46, 63)]`
23A = `23[(1, 2, 3),(3, -2, 1),(4, 2, 1)] = [(23, 46, 69),(69, -46, 23),(92, 46, 23)]`
Now A3 – 23A – 40I
`[(63, 46, 69),(69, -6, 23),(92, 46, 63)] - [(23, 46, 69),(69, -46, 23),(92, 46, 23)] - [(40, 0, 0),(0, 40, 0),(0, 0, 40)]`
= `[(0, 0, 0),(0, 0, 0),(0, 0, 0)]`
= 0
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