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Question
Find the adjoint of matrix A = `[(6, 5),(3, 4)]`
Solution
A = `[(6, 5),(3, 4)]`
A11 = (–1)1+1 M11 = M11 = 4
A12 = (–1)1+2 M12 = – M12 = – 3
A21 = (–1)2+1 M21 = – M21 = – 5
A22 = (–1)2+2 M22 = M22 = 6
adj A = `[("A"_11, "A"_12),("A"_21, "A"_22)]^"T"`
= `[(4, -3),(-5, 6)]^"T"`
= `[(4, -5),(-3, 6)]`
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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`
If A = `[(3, 1),(-1, 2)]`, show that A2 – 5A + 7I = 0
