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Question
Find the inverses of the following matrices by the adjoint method:
`[(1,2,3),(0,2,4),(0,0,5)]`
Solution
Let A = `[(1,2,3),(0,2,4),(0,0,5)]`
∴ |A| = `[(1,2,3),(0,2,4),(0,0,5)]`
= 1(10 - 0) - 0 + 0
= 1(10) - 0 + 0
= 10 ≠ 0
∴ A-1 exists.
First we have to find the co-factor matrix
`= ["A"_"ij"]_(3xx3)`, where `"A"_"ij" = (-1)^("i"+"j") "M"_"ij"`
Now, A11 = `(- 1)^(1 + 1) "M"_11 = |(2,4),(0,5)| = 10 - 0 = 10`
A12 = `(- 1)^(1 + 2) "M"_12 = - |(0,4),(0,5)| = - 0 - 0 = 0`
A13 = `(- 1)^(1 + 3) "M"_13 = |(0,2),(0,0)| = 0 - 0 = 0`
A21 = `(- 1)^(2 + 1) "M"_21 = - |(2,3),(0,5)| = - 10 - 0 = - 10`
A22 = `(- 1)^(2 + 2) "M"_22 = |(1,3),(0,5)| = 5 - 0 = 5`
A23 = `(- 1)^(2 + 3) "M"_23 = - |(1,2),(0,0)| = - 0 - 0 = 0`
A31 = `(- 1)^(3 + 1) "M"_31 = |(2,3),(2,4)| = 8 - 6 = 2`
A32 = `(- 1)^(3 + 2) "M"_32 = - |(1,3),(0,4)| = - 4 - 0 = - 4`
A33 = `(- 1)^(3 + 3) "M"_33 = |(1,2),(0,2)| = 2 - 0 = 2`
∴ the co-factor matrix
`= [("A"_11,"A"_12,"A"_13),("A"_21,"A"_22,"A"_23),("A"_31,"A"_32,"A"_33)] = [(10,0,0),(-10,5,0),(2,-4,2)]`
∴ adj A = `[(10,-10,2),(0,5,-4),(0,0,2)]`
∴ A-1 = `1/|"A"|`(adj A)
`= 1/10 [(10,-10,2),(0,5,-4),(0,0,2)]`
∴ A-1 = `1/10 [(10,-10,2),(0,5,-4),(0,0,2)]`
Notes
The answer in the textbook is incorrect.
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