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Question
Find A–1 if A = `[(0, 1, 1),(1, 0, 1),(1, 1, 0)]` and show that A–1 = `("A"^2 - 3"I")/2`.
Solution
We have, A = `[(0, 1, 1),(1, 0, 1),(1, 1, 0)]`
Co-factors are:
A11 = –1,
A12 = 1
A13 = 1
A21 = 1
A22 = –1
A23 = 1
A31 = 1
A31 = 1
A32 = 1
A33 = –1
∴ adj A = `[(-1, 1, 1),(1, -1, 1),(1, 1, -1)]^"T"`
= `[(-1, 1, 1),(1, -1, 1),(1, 1, -1)]`
|A| = 0 – 1(–1) + 1.1 = 2
∴ A–1 = `("adj A")/|"A"|`
= `1/2 [(-1, 1, 1),(1, -1, 1),(1, 1, -1)]`
Now, A2 = `[(0, 1, 1),(1, 0, 1),(1, 1, 0)] * [(0, 1, 1),(1, 0, 1),(1, 1, 0)]`
= `[(2, 1, 1),(1, 2, 1),(1, 1, 2)]`
∴ `("a"^2 - 3"I")/2 = 1/2{[(2, 1, 1),(1, 2, 1),(1, 1, 2)] - [(3, 0, 0),(0, 3, 0),(0, 0, 3)]}`
= `1/2 [(-1, 1, 1),(1, -1, 1),(1, 1, -1)]`
= A–1
Hence proved.
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