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Question
A–1 exists if |A| = 0.
Options
True
False
Solution
This statement is False.
Explanation:
A–1 exists if |A| ≠ 0.
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Find (AB)–1 by adjoint method.
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`