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Question
If A = `[(1, -1, 2),(3, 0, -2),(1, 0, 3)]`, verify that A(adj A) = (adj A)A
Solution
A = `[(1, -1, 2),(3, 0, -2),(1, 0, 3)]`
A11 = (−1)1+1 M11 = `1|(0, -2),(0, 3)|` = 1(0 − 0) = 1 × 0 = 0
A12 = (−1)1+2 M12 = `1|(3, 0),(1, 0)|` = −1(9 + 2) = −11
A13 = (−1)1+3 M13 = `1|(3, 0)(1, 0)|` = 1(0 − 0) = 0
A21 = (−1)2+1 M21 = `-1|(-1, 2),(0, 3)|` = −1(−3 − 0) = 3
A22 = (−1)2+2 M22 = `1|(1, 2),(1, 3)|` = 1(3 − 2) = 1
A23 = (−1)2+3 M23 = `-1|(1, -1),(1, 0)|` = −1(0 + 1) = −1
A31 = (−1)3+1 M31 = `1|(1, -1),(1, 0)|` = 1(2 − 0) = 2
A32 = (−1)3+2 M32 = `-1|(1, 2),(3, -2)|` = −1(−2 − 6) = 8
A33 = (−1)3+3 M33 = `1|(1, -1),(3, 0)|` = 1(0 + 3) = 3
Hence, matrix of the co-factors 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)]`
= `["A"_"ij"]_(3 xx 3)`
Now, adj A = `["A"_"ij"]_(3 xx 3)^"T"`
= `[(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)]`
= `[(0 + 11 + 0, 3 - 1 - 2, 2 - 8 + 6),(0 + 0 + 0, 9 + 0 + 2, 6 + 0 - 6),(0 + 0 + 0, 3 + 0 - 3, 2 + 0 + 9)]`
= `[(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)]`
= `[(0 + 9 + 2, 0 + 0 + 0, 0 - 6 + 6),(-11 + 3 + 8, 11 + 0 + 0, -22 - 2 + 24),(0 - 3 + 3, 0 - 0 + 0, 0 + 2 + 9)]`
= `[(11, 0, 0,(0, 11, 0),(0 0 11)]` .......(ii)
From equations (i) and (ii), we get
A(adj A) = (adj A)A
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