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Question
Find the co-factor of the element of the following matrix.
`[(1,-1,2),(-2,3,5),(-2,0,-1)]`
Solution
Let A = `[(1,-1,2),(-2,3,5),(-2,0,-1)]`
The co-factor of aij is given by Aij = (−1)i+j Mij
Now, M11 = `|(3,5),(0,-1)|` = − 3 − 0 = − 3
∴ A11 = (− 1)1+1(− 3) = − 3
M12 = `|(-2,5),(-2,-1)|` = 2 + 10 = 12
∴ A12 = (− 1)1+2(12) = − 12
M13 = `|(-2,3),(-2,0)|` = 0 + 6 = 6
∴ A13 = (− 1)1+3(6) = 6
M21 = `|(-1,2),(0,-1)|` = 1 − 0 = 1
∴ A21 = (− 1)2+1(1) = − 1
M22 = `|(1,2),(-2,-1)|` = − 1 + 4 = 3
∴ A22 = (− 1)2+2(3) = 3
M23 = `|(1,-1),(-2,0)|` = 0 − 2 = − 2
∴ A23 = (−1)2+3(− 2) = 2
M31 = `|(-1,2),(3,5)|` = − 5 − 6 = − 11
∴ A31 = (− 1)3+1(− 11) = − 11
M32 = `|(1,2),(-2,5)|` = 5 + 4 = 9
∴ A32 = (− 1)3+2(9) = − 9
M33 = `|(1,-1),(-2,3)|` = 3 − 2 = 1
∴ A33 = (− 1)3+3(1) = 1
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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`
