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Question
If A = `[(1,-1,2),(3,0,-2),(1,0,3)]` ,
verify that A (adj A) = (adj A) A = |A| . I
Solution
Given , A = `[(1,-1,2),(3,0,-2),(1,0,3)]`
`therefore |A| = |(1,-1,2),(3,0,-2),(1,0,3)|`
`= 1 xx |(0,-2),(0,3)| - (-1) xx |(3,-2),(1,3)| + 2 xx |(3,0),(1,0)|`
= 1 (0 - 0) + (9 + 2) + 2 (0 - OJ = 0 + I + O
= 11
Since , A = `[(1,-1,2),(3,0,-2),(1,0,3)]`
A11 = (-1)1+1 M11
= (-1)2 (0) = 0
A12 = (-1)1+2 M12 = (-1)3 (9 + 2) = -11
A13 = (-1)1+3 M13 = (-1)4 (0) = 0
A21 = (-1)2+1 M21 = (-1)3 (-3-0) = 3
A22 = (-1)2+2 M22 = (-1)4 (3 -2) = 1
A23 = (-1)2+3 M23 = (-1)5 (0 + 1) = -1
A31 = (-1)3 + 1 M31 = (-1)4 (2 - 0) = 2
A32 = (-1)3+2 M32 = (-1)5 (-2 - 8) = 8
A33 = (-1)3+3 M33 = (-1)6 (0 + 3) = 3
∴ Malrlx of cofactor Is `[("A"_11 , "A"_12 , "A"_13),("A"_21,"A"_22 ,"A"_23),("A"_31,"A"_32 ,"A"_33)]`
`= [(0,-11,0),(3,1,-1),(2,8,3)]`
adj A = `[(0,3,2),(-11,1,8),(0,-1,3)]`
A .(adj A) = `[(1,-1,2),(3,0,-2),(1,0,3)] . [(0,3,2),(-11,1,8),(0,-1,3)]`
`= |(11,0,0),(0,11,0),(0,0,11)|` .......(I)
(adj A). A = `[(0,3,2),(-11,1,8),(0,-1,3)] . [(1,-1,2),(3,0,-2),(1,0,3)]`
`= |(11,0,0),(0,11,0),(0,0,11)|` .......(II)
|A| . I = `11 [(1,0,0),(0,1,0),(0,0,1)]`
`= |(11,0,0),(0,11,0),(0,0,11)|` .......(III)
From (I)(, (II) and (III). we get
A (adj A) =(adj A) A = |A| I
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