Question

Find the inverse of the following matrix by elementary row transformations if it exists.
`A = [(1, 2, -2), (0, -2, 1), (-1, 3, 0)]`

Answer 1

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

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

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

∴ | A | = 1( 0 - 3 ) -2( 0 + 1 ) - 2( 0 - 2 )
           = - 3 - 2 + 4
           = -1 ≠ 0
∴ A^-1 exists
    We have

                       AA^-1 = I

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

R₃ → R₃ + R₁

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

R₃ → R₃ + 2R₂

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

R₂ ↔ R₃

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

R₁ → R₁ - 2R₂, R₃ → R₃ + 2R₂

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

R₁ →  R₁ + 2R₃

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

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

Answer 2

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

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

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

∴ | A | = 1( 0 - 3 ) -2( 0 + 1 ) - 2( 0 - 2 )
           = - 3 - 2 + 4
           = -1 ≠ 0
∴ A^-1 exists
    We have

                       AA^-1 = I

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

R₃ → R₃ + R₁

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

R₃ → R₃ + 2R₂

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

R₂ ↔ R₃

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

R₁ → R₁ - 2R₂, R₃ → R₃ + 2R₂

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

R₁ →  R₁ + 2R₃

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

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

Answer 3

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

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

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

∴ | A | = 1( 0 - 3 ) -2( 0 + 1 ) - 2( 0 - 2 )
           = - 3 - 2 + 4
           = -1 ≠ 0
∴ A^-1 exists
    We have

                       AA^-1 = I

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

R₃ → R₃ + R₁

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

R₃ → R₃ + 2R₂

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

R₂ ↔ R₃

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

R₁ → R₁ - 2R₂, R₃ → R₃ + 2R₂

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

R₁ →  R₁ + 2R₃

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

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