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Question
Find the inverse of `[(1,2,3),(1,1,5),(2,4,7)]` by the adjoint method.
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.
First 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 = |(1,5),(4,7)| = 7 - 20 = - 13`
`"A"_12 = (-1)^(1+2) "M"_12 = - |(1,5),(2,7)| = - (7 - 10) = 3`
`"A"_13 = (-1)^(1+3) "M"_13 = |(1,1),(2,4)| = 4 - 2 = 2`
`"A"_21 = (-1)^(2+1) "M"_21 = - |(2,3),(4,7)| = - (14 - 12) = - 2`
`"A"_22 = (-1)^(2+2) "M"_22 = |(1,3),(2,7)| = 7 - 6 = 1`
`"A"_23 = (-1)^(2+3) "M"_23 = - |(1,2),(2,4)| = - (4 - 4) = 0`
`"A"_31 = (-1)^(3+1) "M"_31 = |(2,3),(1,5)| = 10 - 3 = 7`
`"A"_32 = (-1)^(3+2) "M"_32 = - |(1,3),(1,5)| = - (5 - 3) = - 2`
`"A"_33 = (-1)^(3+3) "M"_33 = |(1,2),(1,1)| = 1 - 2 = - 1`
∴ the co-factor matrix =
`[("A"_11,"A"_12,"A"_13),("A"_21,"A"_22,"A"_23),("A"_31,"A"_32,"A"_33)] = [(-13,3,2),(-2,1,0),(7,-2,-1)]`
∴ adj A = `[(-13,-2,7),(3,1,-2),(2,0,-1)]`
∴ A-1 = `1/|"A"|`(adj A)
`= 1/-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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AB = [ ]
|AB| = `square`
M11 = –2 ∴ A11 = (–1)1+1 . (–2) = –2
M12 = –3 A12 = (–1)1+2 . (–3) = 3
M21 = 4 A21 = (–1)2+1 . (4) = –4
M22 = 3 A22 = (–1)2+2 . (3) = 3
Cofactor Matrix [Aij] = `[(-2, 3),(-4, 3)]`
adj (A) = [ ]
A–1 = `1/|A| . adj(A)`
A–1 = `square`
