Question

Solve the following system of equations by using inversion method

x + y = 1, y + z = `5/3`, z + x = `4/3`

Answer 1

Matrix form of the given system of equations is

`[(1, 1, 0),(0, 1, 1),(1, 0, 1)] [(x),(y),(z)] [(1),(5/3),(4/3)]`

This is of the form AX = B,

where A = `[(1, 1, 0),(0, 1, 1),(1, 0, 1)]` X = `[(x),(y),(z)]` B = `[(1),(5/3),(4/3)]`

Pre-multiplying AX = B by A⁻¹, we get

A⁻¹(AX) = A−1B

∴ (A⁻¹A)X = A−1B

∴ IX = A⁻¹B

∴ X = A⁻¹B      .......(i)

To determine X, we have to find A⁻¹

|A| = `1|(1, 1),(0, 1)| - 1|(0, 1),(1, 1)| + 0`

= 1(1 – 0) –1(0 – 1)

= 1 + 1

= 2 ≠ 0

∴ A⁻¹ exists.

Consider, AA⁻¹ = I

∴ `[(1, 1, 0),(0, 1, 1),(1, 0, 1)]` A⁻¹ = `[(1, 0, 0),(0, 1, 0),(0, 0, 1)]`

Applying R₃ → R₃ – R₁, we get

`[(1, 1, 0),(0, 1, 1),(1, -1, 1)]` A⁻¹ = `[(1, 0, 0),(0, 1, 0),(-1, 0, 1)]`

Applying R₂ → R₂ − R₃, we get

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

Applying R₁ → `(1/2)` R₂, we get

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

Applying R₁ → R₁ − R₂, we get

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

Applying R₃ → R₃ + R₂, we get

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

∴ A⁻¹ = `[(1/2, 1/2, 1/2),(1/2, 1/2, -1/2),(-1/2, 1/2, 1/2)]`

∴ X = `[(1/2, 1/2, 1/2),(1/2, 1/2, -1/2),(-1/2, 1/2, 1/2)] [(1),(5/3),(4/3)]`       .......[From (i)]

∴ `[(x),(y),(z)] = [(1/3),(2/3),(3/1)]`

∴ By equality of matrices, we get

x = `1/3` y = `2/3` and z = 1