Question

Complete the following activity to find inverse of matrix using elementary column transformations and hence verify.

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

C₁ → C₁ + C₃

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

C₃ → C₃ + C₁ 

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

C₁ → C₁ – 5C₂, C₃ → C₃ – 5C₂

`[(1, "( )", 0),(0, 1, 0),("( )", 1, "( )")]` B⁻¹ = `[(1, 0, "( )"),("( )", 1, -5),(1, "( )", 2)]`

C₁ → C₁ – 2C₃, C₂ → C₂ – C₃ 

`[(1, 0, 0),(0, 1, 0),(0, 0, 1)]` B⁻¹ = `[(3, -1, "( )"),("( )", 6, -5),(5, "( )", "( )")]`

B⁻¹ =  `[("( )", "( )", "( )"),("( )", "( )", "( )"),("( )", "( )", "( )")]`

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

Answer 1

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

Applying C₁ → C₁ + C₃, we get

`[(1, 0, -1),(5, 1, 0),(3, 1, 3)]` B⁻¹ = `[(1, 0, 0),(0, 1, 0),(1, 0, 1)]`

Applying C₃ → C₃ + C₁, we ge

`[(1, 0, 0),(5, 1, 5),(3, 1, 6)]` B⁻¹ = `[(1, 0, 1),(0, 1, 0),(1, 0, 2)]`

Applying C₁ → C₁ – 5C₂, C₃ → C₃ – 5C₂, we get

`[(1, 0, 0),(0, 1, 0),(-2, 1, -1)]` B⁻¹ = `[(1, 0, 1),(-5, 1, -5),(1, 0, 2)]`

Applying C₁ → C₁ – 2C₃, C₂ → C₂ – C₃, we get

`[(1, 0, 0),(0, 1, 0),(0, 0, 1)]` B⁻¹ = `[(3, -1, 1),(-15, 6, -5),(5, -2, 2)]`

B⁻¹ =  `[(3, -1, 1),(-15, 6, -5),(5, -2, 2)]`

`[(2, 0, -1),(5, 1, 0),(0, 1, 3)] [(3, -1, 1),(-15, 6, -5),(5, -2, 2)] = [(6 - 0 - 5, -2 + 0 + 2, 2 - 0 - 2),(15 - 15 + 0, -5 + 6 - 0, 5 - 5 + 0),(0 - 15 + 15, 0 + 6 - 6, 0 - 5 + 6)]`

= `[(1, 0, 0),(0, 1, 0),(0, 0, 1)]`