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Question
Find the inverse of A = `[(1,0,1),(0,2,3),(1,2,1)]` by using elementary column transformations.
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Solution
|A| = `|(1,0,1),(0,2,3),(1,2,1)|`
= 1(2 - 6) - 0 + 1(0 - 2)
= - 4 - 2
= - 6 ≠ 0
∴ A-1 exists.
Consider A-1A = I
∴ A-1 `[(1,0,1),(0,2,3),(1,2,1)] = [(1,0,0),(0,1,0),(0,0,1)]`
By `"C"_3 - "C"_1`, we get,
A-1 `[(1,0,0),(0,2,3),(1,2,0)] = [(1,0,-1),(0,1,0),(0,0,1)]`
By `"C"_2 ↔ "C"_3`we get
A-1 `[(1,0,0),(0,3,2),(1,0,2)] = [(1,-1,0),(0,0,1),(0,1,0)]`
By `"C"_2 - "C"_3`we get
A-1 `[(1,0,0),(0,1,2),(1,-2,2)] = [(1,-1,0),(0,-1,1),(0,1,0)]`
By `"C"_3 - 2"C"_2`we get
A-1 `[(1,0,0),(0,1,0),(1,-2,6)] = [(1,-1,2),(0,-1,3),(0,1,-2)]`
By `(1/6)"C"_3` we get
A-1 `[(1,0,0),(0,1,0),(1,-2,1)] = [(1,-1,1/3),(0,-1,1/2),(0,1,-1/3)]`
By `"C"_1 - "C"_3 "and" "C"_2 + 2"C"_3` we get
A-1 `[(1,0,0),(0,1,0),(0,0,1)] = [(2/3,-1/3,1/3),(-1/2,0,1/2),(1/3,1/3,-1/3)]`
∴ A-1 = `1/6[(4,-2,2),(-3,0,3),(2,2,-2)]`
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