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Question
If A = `[(2,-2,2),(2,3,0),(9,1,5)]` then, show that (adj A) A = O.
Solution
Given A = `[(2,-2,2),(2,3,0),(9,1,5)]`
[Aij] = `|(|(3,0),(1,5)|,-|(2,0),(9,5)|,|(2,3),(9,1)|),(-|(-2,2),(1,5)|,|(2,2),(9,5)|,-|(2,-2),(9,1)|),(|(-2,2),(3,0)|,-|(2,2),(2,0)|,|(2,-2),(2,3)|)|`
`= [(15-0,-(10-0),(2-27)),(-(-10-2),(10-18),-(2+18)),(0-6, -(0-4),(6+4))]`
`= [(15,-10,-25),(12,-8,-20),(-6,4,10)]`
adj A = [Aij]T = `[(15,12,-6),(-10,-8,4),(-25,-20,10)]`
Now (adj A) A = `[(15,12,-6),(-10,-8,4),(-25,-20,10)] [(2,-2,2),(2,3,0),(9,1,5)]`
`= [(30+24-54, -30+36-6, 30+0-30),(-20-16+36, 20-24+4, -20+0+20),(-50-40+90,50-60+10,-50+0+50)]`
`= [(0,0,0),(0,0,0),(0,0,0)]` = O
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If A = `[(4, 3, 2),(-1, 2, 0)]`, B = `[(1, 2),(-1, 0),(1, -2)]`
Find (AB)–1 by adjoint method.
Solution:
AB = `[(4, 3, 2),(-1, 2, 0)] [(1, 2),(-1, 0),(1, -2)]`
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`
