Question

Solve the following equations by the method of inversion.
2x – y + z = 1, x + 2y + 3z = 8 and 3x + y – 4z = 1

Answer 1

A = `[(2, -1, 1),(1, 2, 3),(3, 1, -4)], "B" = [(1),(8),(1)]`

x = `[(x),(y),(z)]`

|A| = 2(-8 - 3) + 1(-4 - 9) + 1(1 - 6)

= 2(-11) + 1(-13) + 1(-5)

= -22 - 13 - 5

|A| = -40 ≠ 0

∴ A^-1 exist.

M₁₁ = `|(2,3), (1,-4)| = -8 - 3 = -11`

M₁₂ = `|(1,3),(3,-4)| = -4 - 9 = -13`

M₁₃ = `|(1,2),(3,1)| = 1 - 6 = -5`

M₂₁ = `|(-1,1),(1,-4)| = 4 - 1 = 3`

M₂₂ = `|(2,1),(3,-4)| = -8 - 3 = -11`

M₂₃ = `|(2,-1),(3,1)| = 2 + 3 = 5`

M₃₁ = `|(-1,1),(2,3)| = -3 - 2 = -5`

M₃₂ = `|(2,1),(1,3)| = 6 - 1 = 5`

M₃₃ = `|(2,-1),(1,2)| = 4 + 1 = 5`

A_ij = (-1)^{i + j}.M_ij

A₁₁ = -11, A₁₂ = 13, A₁₃ = -5

A₂₁ = -3, A₂₂ = -11, A₂₃ = -5

A₃₁ = -5, A₃₂ = -5, A₃₃ = 5

cofactor matrix = `[(-11,13,-5),(-3,-11,-5),(-5,-5,5)]`

Adjoint Method = `[(-11,-3,-5),(13,-11,-5),(-5,-5,5)]`

`A^{-1} = 1/|A|`(Adjoint Method)

`A^{-1} = (-1)/40[(-11,-3,-5),(13,-11,-5),(-5,-5,5)]`

By method of Inversion,

X = A^-1.B

= `(-1)/40[(-11,-3,-5),(13,-11,-5),(-5,-5,5)][(1),(8),(1)]`

= `(-1)/40[(-11 - 24 - 5),(13 - 88 - 5),(-5 - 40 + 5)]`

= `(-1)/40[(-40),(-80),(-40)] = [(1),(2),(1)]`

∴ x = 1, y = 2, z = 1