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Question
Find the inverse of `[(1,2,3),(1,1,5),(2,4,7)]` by using elementary row transformations.
Solution
Let A = `[(1,2,3),(1,1,5),(2,4,7)]`
|A| = `|(1,2,3),(1,1,5),(2,4,7)|`
= 1(7 - 20) - 2(7 - 10) + 3(4 - 2)
= - 13 + 6 + 6
= - 1 ≠ 0
∴ A-1 exists.
Consider AA-1 = I
∴ `[(1,2,3),(1,1,5),(2,4,7)] "A"^-1= [(1,0,0),(0,1,0),(0,0,1)]`
By `"R"_2 - "R"_1 "and" "R"_3 - 2"R"_1` , we get,
`[(1,2,3),(0,-1,2),(0,0,1)] "A"^-1= [(1,0,0),(-1,1,0),(-2,0,1)]`
By `(- 1)"R"_2`we get
`[(1,2,3),(0,1,-2),(0,0,1)] "A"^-1 = [(1,0,0),(1,-1,0),(-2,0,1)]`
By `"R"_1 - 2"R"_2`we get
`[(1,0,7),(0,1,-2),(0,0,1)] "A"^-1 = [(-1,2,0),(1,-1,0),(-2,0,1)]`
By `"R"_1 - 7"R"_3` and `"R"_2 + 2"R"_3` we get
`[(1,0,0),(0,1,0),(0,0,1)] "A"^-1 = [(13,2,-7),(-3,-1,2),(-2,0,1)]`
∴ A-1 = `[(13,2,-7),(-3,-1,2),(-2,0,1)]`
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