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Question
Find A-1 by the adjoint method and by elementary transformations, if A = `[(1,2,3),(-1,1,2),(1,2,4)]`
Solution
|A| = `|(1,2,3),(-1,1,2),(1,2,4)|`
= 1(4 - 4) - 2(- 4 - 2) + 3(- 2 - 1)
= 0 + 12 - 9
= 3 ≠ 0
∴ A-1 exists.
A-1 by adjoint method:
We have to find the cofactor matrix
`= ["A"_"ij"]_(3xx3)`, where `"A"_"ij" = (-1)^("i" + "j") "M"_"ij"`
Now `"A"_11 = (-1)^(1+1) "M"_11 = |(1,2),(2,4)| = 4 - 4 = 0`
`"A"_12 = (-1)^(1+2) "M"_12 = - |(-1,2),(1,4)| = - (- 4 - 2) = 6`
`"A"_13 = (-1)^(1+3) "M"_13 = |(-1,1),(1,2)| = - 2 - 1 = - 3`
`"A"_21 = (-1)^(2+1) "M"_21 = - |(2,3),(2,4)| = - (8 - 6) = - 2`
`"A"_22 = (-1)^(2+2) "M"_22 = |(1,3),(1,4)| = 4 - 3 = 1`
`"A"_23 = (-1)^(2+3) "M"_23 = - |(1,2),(1,2)| = - (2 - 2) = 0`
`"A"_31 = (-1)^(3+1) "M"_31 = |(2,3),(1,2)| = 4 - 3 = 1`
`"A"_32 = (-1)^(3+2) "M"_32 = - |(1,3),(- 1,2)| = - (2 + 3) = - 5`
`"A"_33 = (-1)^(3+3) "M"_33 = |(1,2),(- 1,1)| = 1 + 2 = 3`
∴ the cofactor matrix =
`[("A"_11,"A"_12,"A"_13),("A"_21,"A"_22,"A"_23),("A"_31,"A"_32,"A"_33)] = [(0,6,-3),(-2,1,0),(1,-5,3)]`
∴ adj A = `[(0,-2,1),(6,1,-5),(-3,0,3)]`
∴ `"A"^-1 = 1/|"A"| ("adj A")`
`= 1/3 [(0,-2,1),(6,1,-5),(-3,0,3)]`
A-1 by elementary transformations:
Consider AA-1 = I
∴ `[(1,2,3),(-1,1,2),(1,2,4)] "A"^-1 = [(1,0,0),(0,1,0),(0,0,1)]`
By `"R"_2 + "R"_1` and `"R"_3 - "R"_1`we get,
∴ `[(1,2,3),(0,3,5),(0,0,1)] "A"^-1 = [(1,0,0),(1,1,0),(-1,0,1)]`
By `(1/3) "R"_2` we get,
`[(1,2,3),(0,1,5/3),(0,0,1)] "A"^-1 = [(1,0,0),(1/3,1/3,0),(-1,0,1)]`
By `"R"_1 - 2"R"_2` we get,
`[(1,0,-1/3),(0,1,5/3),(0,0,1)] "A"^-1 = [(1/3,-2/3,0),(1/3,1/3,0),(-1,0,1)]`
By `"R"_1 + 1/3"R"_3` and `"R"_2 - 5/3"R"_3`, we get,
∴ `[(1,0,0),(0,1,0),(0,0,1)] "A"^-1 = [(0,-2/3,1/3),(2,1/3,-5/3),(-1,0,1)]`
∴ `"A"^-1 = 1/3[(0,-2,1),(6,1,-5),(-3,0,3)]`
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