Question

Find inverse of the following matrices (if they exist) by elementary transformations:

`[(2, 0, -1),(5, 1, 0),(0, 1, 3)]`

Answer 1

Let A = `[(2, 0, -1),(5, 1, 0),(0, 1, 3)]`

∴ |A| = `|(2, 0, -1),(5, 1, 0),(0, 1, 3)|`

= 2(3 – 0) –0 –1(5 – 0)
= 6 – 0 – 5
= 1 ≠ 0
∴ A^–1 exists.
Consider AA^–1 = I

∴ `[(2, 0, 1),(5, 1, 0),(0, 1, 3)] "A"^-1 = [(1, 0, 0),(0, 1, 0),(0, 0, 1)]`

Applying R₁ ↔ R₂, we get

`[(5, 1, 0),(2, 0, -1),(0, 1, 3)] "A"^-1 = [(-2, 1, 0),(1, 0, 0),(0, 0, 1)]`

Applying R₁ → R₁ – 2R₂, we get

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

Applying R₂ → R₂ – 3R₃, we get

`[(1, 1, 2),(0, 1, 4),(0, 1, 3)] "A"^-1 = [(-2, 1, 0),(5, -2, 3),(0, 0, 1)]`

Applying R₁ → R₁ – R₂ and R₃ → R₃ – R₂, we get

`[(1, 0, -2),(0, 1, 4),(0, 0, -1)] "A"^-1 = [(-7, 3, -3),(5, -2, 3),(-5, 2, -2)]`

Applying R₃ → (– 1) R₃, we get

`[(1, 0, -2),(0, 1, 4),(0, 0, 1)] "A"^-1 = [(-7, 3, -3),(5, -2, 3),(5, -2, 2)]`

Applying R₁ → R₁ + 2R₃ and R₂ → R₂ – 4R₃, we get

`[(1, 0, 0),(0, 1, 0),(0, 0, 1)] "A"^-1 = [(3, -1, 1),(-15, 6, -5),(5, -2, 2)]`

∴ A^–1 = `[(3, -1, 1),(-15, 6, -5),(5, -2, 2)]`.