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Question
If A = `[(-4, -3, -3),(1, 0, 1),(4, 4, 3)]`, find adj (A).
Solution
A11 = (–1)1+1 M11 = `1|(0, 1),(4, 3)|` = 1(0 – 4) = – 4
A12 = (–1)1+2 M12 = `(-1)|(1, 1),(4, 3)|` = (–1)(3 – 4) = (–1)(–1) = 1
A13 = (–1)1+3 M13 = `1|(1, 0),(4, 4)|` = 1(4 – 0) = (1)(4) = 4
A21 = (–1)2+1 M21 = `(-1)|(-3, -3),(4, 3)|` = (–1)(– 9 + 12) = (–1)(3) = – 3
A22 = (–1)2+2 M22 = `1|(-4, -3),(4, 3)|` = 1(–12 + 12) = 1(0) = 0
A23 = (–1)2+3 M23 = `(-1)|(-4, -3),(4, 4)|` = (–1)(–16 + 12) = (–1)(–4) = 4
A31 = (–1)3+1 M31 = `1|(-3, -3),(0, 1)|` = 1(– 3 – 0) = – 3
A32 = (–1)3+2 M32 = `(-1)|(-4, -3),(1, 1)|` = (–1)(– 4 + 3) = (–1)(–1) = 1
A33 = (–1)3+3 M33 = `1|(-4, -3),(1, 0)|` = 1(0 + 3) = (1)(3) = 3
adj (A) = `[("A"_11, "A"_12, "A"_13),("A"_21, "A"_22, "A"_23),("A"_31, "A"_32, "A"_33)]``
= `[(-4, 1, 4),(-3, 0, 4),(-3, 1, 3)]`
= `[(-4, -3, -3),(1, 0, 1),(4, 4, 3)]`
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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`
