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