Question

If A = `[(2, 3),(1, 2)]`, B = `[(1, 0),(3, 1)]`, find AB and (AB)⁻¹ 

Answer 1

AB = `[(2, 3),(1, 2)][(1, 0),(3, 1)]``

= `[(2 + 9, 0 + 3),(1 + 6, 0 + 2)]`

= `[(11, 3),(7, 2)]`

Now, |AB| = `|(11, 3),(7, 2)|`

= 22 − 21

= 1 ≠ 0

∴ (AB)⁻¹ exists.

Consider, (AB)(AB)⁻¹ = I

∴ `[(11, 3),(7, 2)]` (AB)⁻¹ = `[(1, 0),(0, 1)]`

Applying R₁ → 2R₁,

`[(22, 6),(7, 2)]` (AB)⁻¹ = `[(2, 0),(0, 1)]`

Applying R₁ → R₁ − 3R₂,

`[(1, 0),(7, 2)]` (AB)⁻¹ = `[(2, -3),(0, 1)]`

Applying R₂ → R₂ − 7R₁,

`[(1, 0),(0, 2)]` (AB)⁻¹ = `[(2, -3),(-14, 22)]`

Applying R₂ → `(1/2)` R₂

`[(1, 0),(0, 1)]` (AB)⁻¹ = `[(2, -3),(-7, 11)]`

∴ (AB)⁻¹ = `[(2, -3),(-7, 11)]`     .......(1)

|A| = `|(2,3),(1,2)| = 4 - 3 = 1 ne 0`

∴ A⁻¹ exists.

Consider, AA⁻¹ = I

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

Applying R₁ ↔ R₂, 

`[(1, 2),(2, 3)]` A⁻¹ = `[(0, 1),(1, 0)]`

Aplying 2R₂ → R₁,

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

Applying (−1)R₂ ↔ R₁,

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

Applying R₁ ↔ R₂,

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

∴ A⁻¹ = `[(2, -3),(-1, 2)]`

|B| = `|(1, 0),(3, 1)|`

= 1 − 0

= 1 ≠ 0

∴ B⁻¹ exists.

Consider, BB⁻¹ = I 

∴ `[(1, 0),(3, 1)]` B⁻¹ = `[(1, 0),(0, 1)]`

Applying R₂ ↔ 3R₁,

`[(1, 0),(0, 1)]` A⁻¹ = `[(1, 0),(-3, 1)]`

∴ B⁻¹ = `[(1, 0),(-3, 1)]`

∴ B⁻¹. A⁻¹ = `[(1, 0),(-3, 1)],[(2, -3),(-1, 2)]` 

= `[(2 - 0, -3+0),(-6-1, 9+2)]`

= `[(2, -3),(-7,11)]`      .......(2)

From (1) and (2), (AB)⁻¹ = B⁻¹. A⁻¹