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Question
If A = `[(-1,2,-2),(4,-3,4),(4,-4,5)]` then, show that the inverse of A is A itself.
Sum
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Solution
Given A = `[(-1,2,-2),(4,-3,4),(4,-4,5)]`
|A| = `-1|(-3,4),(-4,5)| - 2|(4,4),(4,5)| - 2|(4,-3),(4,-4)|`
= -1 [-15 + 16] - 2[20 - 16] - 2[-16 + 12]
= -1 [1] - 2[4] - 2[- 4]
= -1 - 8 + 8 ⇒ - 1 ≠ 0
[Aij] = `[(1,-4,-4),(-|(2,-2),(-4,5)|,|(-1,-2),(4,5)|,-|(-1,2),(4,-4)|),(|(2,-2),(-3,4)|,-|(-1,-2),(4,4)|,|(-1,2),(4,-3)|)]`
`= [(1,-4,-4),(-(10-8),(-5+8),-(4-8)),((8-6),-(-4+8),(3-8))]`
`= [(1,-4,-4),(-2,3,4),(2,-4,-5)]`
adj A = [Aij]T = `[(1,-2,2),(-4,3,-4),(-4,4,-5)]`
A-1 = `1/|"A"|` adj A
`= 1/(-1)[(1,-2,2),(-4,3,-4),(-4,4,-5)]`
`= -1[(1,-2,2),(-4,3,-4),(-4,4,-5)]`
`= [(-1,2,-2),(4,-3,4),(4,-4,5)]`
∴ A-1 = A
Hence proved.
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