Question

Find the inverse of  A = `[(1, 0, 1),(0, 2, 3),(1, 2, 1)]` by elementary column transformations.

Answer 1

|A| = `|(1, 0, 1),(0, 2, 3),(1, 2, 1)|`

= 1(2 – 6) – 0 + 1)0 – 2)

= – 4 – 2

= – 6 ≠ 0

∴ A^–1 exists.

Consider A^–1A = I

∴ A^–1 = `[(1, 0, 1),(0, 2, 3),(1, 2, 1)] = [(1, 0, 0),(0, 1, 0),(0, 0, 1)]`

Applying C₃ → C₃ – C₁, we get

A^–1 = `[(1, 0, 0),(0, 2, 3),(1, 2, 0)] = [(1, 0, -1),(0, 1, 0),(0, 0, 1)]`

Applying C₂ ↔ C₃, we get

A^–1 = `[(1, 0, 0),(0, 3, 2),(1, 0, 2)] = [(1, -1, 0),(0, 0, 1),(0, 1, 0)]`

Applying C₂ → C₂ – C₃, we get

A^–1 = `[(1, 0, 0),(0, 1, 2),(1, -2, 2)] = [(1, -1, 0),(0, -1, 1),(0, 1, 0)]`

Applying C₃ → C₃ – 2C₂, we get

A^–1 = `[(1, 0, 0),(0, 1, 0),(1, -2, 6)] = [(1, -1, 2),(0, -1, 3),(0, 1, -2)]`

Applying C₃ → `(1/6)` C₃, we get

A^–1 = `[(1, 0, 0),(0, 1, 0),(1, -2, 1)] = [(1, -1, 1/3),(0, -1, 1/2),(0, 1, -1/3)]`

Applying C₁ → C₁ – C₃ and C₂ → C₂ + 2C₃, we get

A^–1 = `[(1, 0, 0),(0, 1, 0),(0, 0, 1)] = [(2/3, -1/3, 1/3),(-1/2, 0, 1/2),(1/3, 1/3, -1/3)]`

∴ A^–1 = `(1)/(6)[(4, -2, 2),(-3, 0, 3),(2, 2, -2)]`.