Question

Find the inverse of matrix A = `[(1, 0, 1),(0, 2, 3),(1, 2, 1)]` by using elementary row transformations 

Answer 1

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

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

= 1(2 – 6) – 0 + 1(0 – 2)

= 1(– 4) + 1(– 2)

= – 4 – 2

= – 6 ≠ 0

∴ A^–1 exists.

Consider AA^–1 = I

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

Applying R₃ → R₃ – R₁, we get

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

Applying R₂ → `(1/2)` R₂, we get

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

Applying R₃ → R₃ – 2R₂, we get

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

Applying R₃ → `(-1/3)` R₃, we get

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

Applying R₁ → R₁ – R₃, R₂ → R₂ – `(3/2)` R₃, we get

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

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