Question

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

`[(2, -3, 3),(2, 2, 3),(3, -2, 2)]`

Answer 1

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

∴ |A| = `|(2, -3, 3),(2, 2, 3),(3, -2, 2)|` 

= 2(4 + 6) + 3(4 – 9) + 3(–4 – 6)
= 20 – 15 – 30
= – 25 ≠ 0
∴ A^–1 exists.
Consider AA^–1 = I

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

Applying R₁ → 2R₁ – R₃, we get

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

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

`[(1, -4, 4),(0, 10, -5),(0, 10, -10)] "A"^-1 [(2, 0, -1),(-4, 1, 2),(-6, 0, 4)]`

Applying R₂ → `(1/10)` R₂ and R₃ → `(-1/10)`R3, we get

`[(1, -4, 4),(0, 1, -1/2),(0, -1, 1)] "A"^-1 = [(2, 0, -1),(-4/10, 1/10, 2/10),(6/10, 0, -4/10)]`

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

`[(1, 0, 2),(0, 1, -1/2),(0, 0, 1/2)] "A"^-1 = [(4/10, 4/10, -2/10),(-4/10, 1/10, 2/10),(2/10, 1/10, -2/10)]`

Applying R₃ → 2R₃ , we get

`[(1, 0, 2),(0, 1, -1/2),(0, 0, 1)] "A"^-1 = [(4/10, 4/10, -2/10),(-4/10, 1/10, 2/10),(4/10, 2/10, -4/10)]`

Applying R₁ → R₁ – 2R₃ and R₂ → R₂ + `(1/2)`R₃, we get

`[(1, 0, 0),(0, 1, 0),(0, 0,1)] "A"^-1 [(-4/10, 0, 6/10),(-2/10, 2/10, 0),(4/10, 2/10, -4/10)]`

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