Question

Find the inverse `[(1, 2, 3 ),(1, 1, 5),(2, 4, 7)]` of the elementary row tranformation.

Answer 1

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)]`