Question

Solve the following equations by using the method of inversion:

x − y + z = 4, 2x + y − 3z = 2, x + y + z = 2 

Answer 1

The given equations can be written in matrix form as:

`[(1, -1, 1),(2, 1, -3),(1, 1, 1)] [(x),(y),(z)] = [(4),(0),(2)]`

This is of the form AX = B, where

`A = [(1, -1, 1),(2, 1, -3),(1, 1, 1)], X = [(x),(y),(z)] and B = [(4),(0),(2)]`

Let us find A⁻¹.

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

= 1(1 + 3) + 1(2 + 3) + 1(2 − 1)

= 4 + 5 + 1

= 10 ≠ 0

∴ A⁻¹ exists.

Consider AA⁻¹ = I

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

By R₂ − 2R₁ and R₃ − R₁, we get

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

By R₂ − R₃, we get

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

By R₁ + R₂ and R₃ − 2R₂, we get

`[(1, 0, -4),(0, 1, -5),(0, 0, 10)] A^-1 = [(0, 1, -1),(-1, 1, -1),(1, -2, 3)]`

By `(1/10)R_3`, we get

`[(1, 0, -4),(0, 1, -5),(0, 0, 1)] A^-1 = [(0, 1, -1),(-1, 1, -1),(1/10, -2/10, 3/10)]`

By R₁ + 4R₃ and R₂ + 5R₃, we get

`[(1, 0, 0),(0, 1, 0),(0, 0, 1)] A^-1 = [(4/10, 2/10, 2/10),(-5/10, 0, 5/10),(1/10, -2/10, 3/10)]`

∴ `A^-1 = 1/10 [(4, 2, 2),(-5, 0, 5),(1, -2, 3)]`

Now, premultiply AX = B by A⁻¹, we get

A⁻¹(AX) = A⁻¹B

∴ (A⁻¹A)X = A⁻¹B

∴ IX = A⁻¹B

∴ `X = 1/10[(4, 2, 2),(-5, 0, 5),(1, -2, 3)][(4),(2),(2)]`

∴ `[(x),(y),(z)] = 1/10 [(16 + 0 + 4),(-20 + 0 + 10),(4 - 0 + 6)] = [(2),(-1),(1)]`

∴ By equality of matrices, x = 2.4, y = −1, z = 0.6 is the required solution.