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Question
Find the inverse `[(1, 2, 3 ),(1, 1, 5),(2, 4, 7)]` of the elementary row tranformation.
Solution
Let A = `[(1, 2, 3 ),(1, 1, 5),(2, 4, 7)]`
∴ |A| = `|(1, 2, 3 ),(1, 1, 5),(2, 4, 7)|`
= 1(7 – 20) – 2(7 – 10) + 3(4 – 2)
= 1 × (–13) – 2 × (–3) + 3 × 2
= – 13 + 6 + 6
= – 1 ≠ 0
∴ A–1 exists.
Consider AA–1 = I
∴ `[(1, 2, 3 ),(1, 1, 5),(2, 4, 7)] "A"^-1 = [(1, 0, 0),(0, 1, 0),(0, 0, 1)]`
Applying R2 → R2 + R1
`[(1, 2, 3),(0, -1, 2),(2, 4, 7)]"A"^-1= [(1, 0, 0),(-1, 1, 0),(0, 0, 1)]`
Applying R3 → R3 – 2R1
`[(1, 2, 3),(0, -1, 2),(0, 0, 1)]"A"^-1= [(1, 0, 0),(-1, 1, 0),(-2, 0, 1)]`
Applying R2 → (– 1) R2, we get
`[(1, 2, 3),(0, 1, -2),(0, 0, 1)]"A"^-1 = [(1, 0, 0),(1, -1, 0),(-2, 0, 1)]`
Applying R2 → R2 + 2R3
`[(1, 2, 3),(0, 1, 0),(0, 0, 1)]"A"^-1 = [(1, 0, 0),(-3, -1, 2),(-2, 0, 1)]`
Applying R1 → R1 + 3R3
`[(1, 2, 0),(0, 1, 0),(0, 0, 1)]"A"^-1 = [(7, 0, -3),(-3, -1, 2),(-2, 0, 1)]`
Applying R1 → R1 + R2
`[(1, 0, 0),(0, 1, 0 ),(0, 0, 1)]"A"^-1 = [(13, 2, -7),(-3, -1, 2),(-2, 0, 1)]`
∴ A–1 = `[(13, 2, -7),(-3, -1, 2),(-2, 0, 1)]`
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