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Question
If A = `[(0, 1, 1),(1, 0, 1),(1, 1, 0)]`, show that `"A"^-1 = 1/2("A"^2 - 3"I")`
Solution
A = `[(0, 1, 1),(1, 0, 1),(1, 1, 0)]`
A2 = A × A
= `[(0, 1, 1),(1, 0, 1),(1, 1, 0)] [(0, 1, 1),(1, 0, 1),(1, 1, 0)]`
= `[(0 + 1 + 1, 0 + 0 + 1, 0 + 1 + 0),(0 + 0 + 1, 1 + 0 + 1, 1 + 0 + 0),(0 + 1 + 0, 1 + 0 + 0, 1 + 1 + 0)]`
= `[(2, 1, 1),(1, 2, 1),(1, 1, 2)]`
A2 – 3I = `[(2, 1, 1),(1, 2, 1),(1, 1, 2)] - 3[(1, 0, 0),(0, 1, 0),(0, 0, 1)]`
= `[(2, 1, 1),(1, 2, 1),(1, 1, 2)] + [(-3, 0, 0),(0, -3, 0),(0, 0, -3)]`
A2 – 3I = `[(-1, 1, 1),(1, -1, 1),(1, 1, -1)]` .........(1)
adj A = `[(+|(0, 1),(1, 0)|, -|(1, 1),(1, 0)|, +|(1, 0),(1, 1)|),(-|(1, 1),(1, 0)|, +|(0, 1),(1, 0)|, -|(0, 1),(1, 1)|),(+|(1, 1),(0, 1)|, -|(0, 1),(1, 1)|, +|(0, 1),(1, 0)|)]^"T"`
= `[(+(0 - 1) , -(0 - 1), +(1 - 0)),(-(0 - 1), +(0 - 1), -(0 - 1)),(+(1 - 0), -(0 - 1), +(0 - 1))]^"T"`
= `[(-1, 1, 1),(1, -1, 1),(1, 1, -1)]^"T"`
adj A = `[(-1, 1, 1),(1, -1, 1),(1, 1, -1)]`
A–1 = `1/|"A"|` adj A = `1/2 [(-1, 1, 1),(1, -1, 1),(1, 1, -1)]`
A–1 = `1/2 ("A"^2 - 3"I")` ......(Using (1))
Hence proved
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